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# Test: Analysis of Financial Statements- Case Based Type Questions
Test Description
## 12 Questions MCQ Test Accountancy Class 12 | Test: Analysis of Financial Statements- Case Based Type Questions
Test: Analysis of Financial Statements- Case Based Type Questions for Commerce 2023 is part of Accountancy Class 12 preparation. The Test: Analysis of Financial Statements- Case Based Type Questions questions and answers have been prepared according to the Commerce exam syllabus.The Test: Analysis of Financial Statements- Case Based Type Questions MCQs are made for Commerce 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Analysis of Financial Statements- Case Based Type Questions below.
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Test: Analysis of Financial Statements- Case Based Type Questions - Question 1
### Read the following information and answer the given questions:The following data is available from Pitambar Ltd.Q. What is the Net Profit after Tax in March 2020?
Detailed Solution for Test: Analysis of Financial Statements- Case Based Type Questions - Question 1 Net Profit before Tax = ₹4,00,000 + ₹2,00,000 - ₹2,40,000 - ₹24,000 = ₹3,36,000
Tax = ₹3,36,000 x 30 / 100 = ₹1,00,800
Net Profit after Tax = ₹3,36,000 - ₹1,00,800 = ₹2,35,200
Test: Analysis of Financial Statements- Case Based Type Questions - Question 2
### Read the following information and answer the given questions:The following data is available from Pitambar Ltd.Q. What is the percentage of the Total Revenue from Revenue from operations in 2019?
Detailed Solution for Test: Analysis of Financial Statements- Case Based Type Questions - Question 2 Revenue from operations = 200% of ₹1,50,000 = ₹3 ,00,000
Total Revenue = ₹3,00,000 + ₹1,50,000 = ₹4 ,50,000
Required Percentage = 4,50,000 / 3,00,000 × 100 = 150%
Test: Analysis of Financial Statements- Case Based Type Questions - Question 3
### Read the following information and answer the given questions:The following data is available from Pitambar Ltd.Q. What is the revenue from operations on 31st March, 2020?
Detailed Solution for Test: Analysis of Financial Statements- Case Based Type Questions - Question 3 200% of ₹2,00,000 = ₹4,00,000
Test: Analysis of Financial Statements- Case Based Type Questions - Question 4
The following data is available from Pitambar Ltd.
Q. What is the value of the tax paid by the firm in 2019?
Detailed Solution for Test: Analysis of Financial Statements- Case Based Type Questions - Question 4 Tax Paid = (₹3,00,000 + ₹1,50,000 - ₹1,50,000 - ₹30,000) × 30 / 100
= ₹2,70,000 × 30 / 100
= ₹81,000
Test: Analysis of Financial Statements- Case Based Type Questions - Question 5
Following information are taken from the books of Agarwal Pvt Ltd.:
Q. What is the percentage change in the tax paid?
Detailed Solution for Test: Analysis of Financial Statements- Case Based Type Questions - Question 5 Net Profit before Tax for 2020 = ₹12,60,000
2019 = ₹12,00,000
Tax paid for 2020 = ₹6,30,000
2019 = ₹6 ,00,000
Percentage Change = ₹30,000 / ₹6 ,00,000 x 100 = 5%
Test: Analysis of Financial Statements- Case Based Type Questions - Question 6
Following information are taken from the books of Agarwal Pvt Ltd.:
Q. What is the percentage change in the profit earned after tax?
Detailed Solution for Test: Analysis of Financial Statements- Case Based Type Questions - Question 6 Net Profit after Tax for 2020 = ₹6,30,000
2019 = ₹6,00,000
Percentage Change = ₹30,000 / ₹6 ,00,000 x 100 = 5%
Test: Analysis of Financial Statements- Case Based Type Questions - Question 7
Following information are taken from the books of Agarwal Pvt Ltd.:
Q. What is the value of absolute change in the Total Revenue?
Detailed Solution for Test: Analysis of Financial Statements- Case Based Type Questions - Question 7 Total Revenue = Revenue from operations + Other Income
For 2019 = ₹24,00,000
For 2020 = ₹33,60,000
Absolute Change = ₹24,00,000 - ₹33,60,000 = ₹9,60,000
Test: Analysis of Financial Statements- Case Based Type Questions - Question 8
Following information are taken from the books of Agarwal Pvt Ltd.:
Q. The absolute change in the Expenses is:
Detailed Solution for Test: Analysis of Financial Statements- Case Based Type Questions - Question 8 ₹21,00,000 - ₹12,00,000 = ₹9,00,000
Test: Analysis of Financial Statements- Case Based Type Questions - Question 9
Following is the data given of Lalit Ltd.
Q. What is the percentage change in the Total Expenses?
Detailed Solution for Test: Analysis of Financial Statements- Case Based Type Questions - Question 9 1,50,000 / 5,50,000 = 27.27%
Test: Analysis of Financial Statements- Case Based Type Questions - Question 10
Following is the data given of Lalit Ltd.
Q. What is the percentage change in the Total Revenue earned?
Detailed Solution for Test: Analysis of Financial Statements- Case Based Type Questions - Question 10 3,00,000 + 10,000 / 12,00,000+20,000 x 100 = 24.81%
Test: Analysis of Financial Statements- Case Based Type Questions - Question 11
Following is the data given of Lalit Ltd.
Q. What is the percentage change in the Revenue from operations?
Detailed Solution for Test: Analysis of Financial Statements- Case Based Type Questions - Question 11 3,00,000 / 12,00,000 x 100 = 25%
Test: Analysis of Financial Statements- Case Based Type Questions - Question 12
Following is the data given of Lalit Ltd.
Q. What is the change in the profit before tax?
Detailed Solution for Test: Analysis of Financial Statements- Case Based Type Questions - Question 12 Total Revenue in 2013 = ₹12,00,000 + ₹20,000 = ₹12,20,000
Total Expenses = ₹5 ,50,000
Profit = ₹1 2,20,000 - ₹5,50,000 = ₹6,70,000
In 2014, Total Revenue = ₹15,00,000 + ₹30,000 = ₹15,30,000
Total Expenses = ₹7,00,000
Profit = ₹1 5,30,000 - ₹7,00,000 = ₹8,30,000
Change in profit = ₹8,30,000 - ₹6,70,000 = ₹1,60,000
## Accountancy Class 12
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## Accountancy Class 12
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# Online Help
###### All Products Maple MapleSim
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Qrew Member
2 years ago
# Reminder Notification for Upcoming Birthdays
We want to know when an employee's birthday is upcoming. We capture their DOB in a date field, so how do I get that month/day information in a format that I can use in a Reminder notification?
------------------------------
Keith Hendrick
------------------------------
### 4 Replies
• Here is an example of a formula Date Field which will calculate an employees Next anniversary of their hire date. You could just adapt it for your purposes
var date HireDate=[Hire Date];
var date Datetest = Date(Year(Today()),Month(\$HireDate),Day(\$HireDate));
------------------------------
mark.shnier@gmail.com
------------------------------
• Worked perfectly. Thanks!
------------------------------
Keith Hendrick
------------------------------
• I actually have this very thing in an app I built for another entity:
First, I built a report that displays "Birthdays (Next Month)", then I created a Quickbase Notification that sends the report to a specified user on the 15th day of each month.
The report is built off of the Birthday field (Formula Date) and an "Age (Now)" field (formula-numeric). "Birthdate" field below is a standard Date field.
Birthday Formula:
`AdjustYear([Birthdate], [Age (Now)]+1)`
Age Now Formula:
`//AGE and decimal months`
`// replace the [DOB] field with your date of birth field`
`var date DOB = [Birthdate];`
`var number Years = `
`Year(Today())-Year(\$DOB)`
`-`
`If(`
`Month(Today())<Month(\$DOB)`
`or`
`(Month(Today())=Month(\$DOB) and Day(Today())<Day(\$DOB)),1,0);`
`var number MonthsDifference = Month(Today()) - Month(\$DOB) ;`
`var number Months = If(\$MonthsDifference >= 0, \$MonthsDifference, 12 + \$MonthsDifference);`
`Round(\$Years + \$Months/12,0.1)`
Hope this helps.
------------------------------
Ryan Buschmeyer
------------------------------
• Thanks, Ryan.
------------------------------
Keith Hendrick
------------------------------ | 463 | 1,983 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-22 | longest | en | 0.704809 |
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# FOUR SQUARE QUESTIONS! 四方块问题 - PowerPoint PPT Presentation
FOUR SQUARE QUESTIONS! 四方块问题. 这是一个富有哲理的智力游戏。特此翻译为中文，并推荐给大家。 [email protected] 4 Square Questions. B. A. Look at the diagram carefully. Now, I will ask you FOUR questions about this square. Are you ready? 仔细瞧瞧左边这个图 现在让我来问您四个问题 准备好了吗？. D. C.
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
## PowerPoint Slideshow about ' FOUR SQUARE QUESTIONS! 四方块问题' - sissy
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Presentation Transcript
4 SquareQuestions
B
A
Look at the diagram carefully.
D
C
4 SquareQuestions
B
A
Q
Q 1 问题1
Divide the white area in square A into two equal pieces.
Easy! Isn't it?
D
C
4 SquareQuestions
B
A
Q 1
Q 1
Divide the white area in square A into two equal pieces.
D
C
4 SquareQuestions
B
A
Q 2
Q 2
Divide the white area in square B into three equal pieces.
Not so difficult!
D
C
4 SquareQuestions
B
A
Q 2
Q 2
Divide the white area in squareBinto three equal pieces.
D
C
4 SquareQuestions
B
A
Q 3
Q 3
Divide the white area in squareCinto four equal pieces.
Very difficult? That's right!
D
C
4 SquareQuestions
B
A
Q 3
Q 3
Divide the white area in squareCinto four equal pieces.
D
C
4 SquareQuestions
Here comes the last question!
B
A
D
C
4 SquareQuestions
Q 4
B
A
Q 4
Divide the white area in squareDinto seven equal pieces.
World Record is SEVEN seconds!
D
C
4 SquareQuestions
B
A
Q 4
Q 4
Divide the white area in squareDinto seven equal pieces.
Any ideas?
D
C
4 SquareQuestions
B
A
Q 4
Q 4
Divide the white area in squareDinto seven equal pieces.
Are you still puzzled?
D
C
4 SquareQuestions
B
A
Q 4
Q 4
Divide the white area in squareDinto seven equal pieces.
D
C
4 SquareQuestions
B
A
Q 4
Q 4
Divide the white area in squareDinto seven equal pieces.
Was it really that tough?
D
C
4 SquareQuestions
Q 4
B
A
Q 4
If you didn't get the answer, it was just that your mind was conditioned to seek a complex solution!
D
C
4 SquareQuestions
• Lessons learnt: 从这个智力游戏中获得的教训
• Most of the times, our mind gets conditioned so much by the circumstances that we can not see the obvious.
• 在很多情况下，我们的心灵被周围的环境或氛围引导到“想得太复杂”的路子上去，以至于对显而易见的解决办法反而视而不见。
BE SIMPLE IN LIFE them complicated. !
THE GREATNESS OF THIS MAN WAS HIS SIMPLICITY!
HAVE A NICE DAY! them complicated. | 887 | 2,831 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2017-34 | longest | en | 0.763079 |
https://mattermodeling.stackexchange.com/questions/4090/rotation-of-crystal-structure-to-match-another-structure-of-the-same-compound-po/4131 | 1,719,234,354,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865383.8/warc/CC-MAIN-20240624115542-20240624145542-00507.warc.gz | 326,843,458 | 29,770 | # Rotation of crystal structure to match another structure of the same compound/polymorph
Say, I have two crystal structures of a particular organic molecule, the crystal structures are basically identical, apart from a rotation and redefinition of the lattice vectors/angles, a simple result of different choices during the structure refinement process after the xray or neutron experiment for structure determination. As a practical example consider triclinic (P-1) malonic acid for which two structures have been published:
one (MALNAC02) with: a,b,c, alpha,beta,gamma=5.156,5.341,8.407,71.48,76.12,85.09
the other (MALNAC) with a,b,c, alpha,beta,gamma=5.33,5.14,11.25,102.7,135.17,85.17
If I wanted to convert one structure to the other so that they perfectly overlap (apart from a trivial translation step), by rotating it successively through one or more of the cartesian (x,y,z) axis through appropriate angles, how would I find these angles?
I understand there are different ways to do this, other than the successive rotation around the cartesian axes (use a rotation matrix or Eulerian angles) but here I need to know these angles (around the cartesian axes), also I know there is not only one unique version of these angles - but I am fine with having just one of them.
If you know which atoms correspond to each other in the two structures, you can use a structural superposition method. Least-squares superposition methods find the rotation matrix and translation that minimizes the RMSD between given points.
There are a few well-established methods. Recently, I had to use one and I picked QCP (because it comes with BSD-licensed C code), but for a small set of points like in your problem any method should work.
• thanks, QCP looks like a rather useful tool, however, my goal is not to align two structures but to figure out the angles mentioned in the original post, it might be trivial for an expert, but I don;t know how to get these angles from the rotation matrix that a tool like QCP would supply. Commented Jan 9, 2021 at 13:11
• Commented Jan 12, 2021 at 10:30
If you're familiar with Python, this can be done using pymatgen fairly easily.
To illustrate, let's first define our problem:
from pymatgen.core.lattice import Lattice
from pymatgen.core.structure import Structure
malnac_lattice = Lattice.from_parameters(5.33, 5.14, 11.25, 102.7, 135.17, 85.17)
malnac02_lattice = Lattice.from_parameters(5.156, 5.341, 8.407, 71.48, 76.12, 85.09)
# here, because we're only interested in the lattices,
# we can define a "dummy structure" with a single H atom at the origin
malnac_struct = Structure(malnac_lattice, ["H"], [[0.0, 0.0, 0.0]])
malnac02_struct = Structure(malnac02_lattice, ["H"], [[0.0, 0.0, 0.0]])
Now we have the structures defined, we can use StructureMatcher to compare them and find the transformation to change one into the other.
from pymatgen.analysis.structure_matcher import StructureMatcher
# StructureMatcher can accept different tolerances for judging equivalence
matcher = StructureMatcher(primitive_cell=False)
# first, we can verify these lattices are equivalent
matcher.fit(malnac_struct, malnac02_struct) # returns True
# and we can get the transformation matrix from one to the other
# this returns the supercell matrix (e.g. change of basis),
# as well as any relevant translation, and mapping of atoms from one
# crystal to the other
matcher.get_transformation(malnac_struct, malnac02_struct)
# returns (array([[ 0, -1, 0], [-1, 0, 0], [ 0, 1, 1]]), array([0., 0., 0.]), [0])
$$\begin{pmatrix} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 1 & 1 \end{pmatrix}$$
This matrix defines the transformation from the one set of lattice vectors (a, b, c) into the other, and the angles can be obtained from here.
More information can be found in the pymatgen docs, or you can ask the developers (myself included) at matsci.org/pymatgen.
Hope this helps!
• I still have to try that, but it looks exactly like what i was looking for, thanks! Commented Jan 15, 2021 at 13:37
It seems to me that there is a discrepancy about the c-axis in the 2 unit cells, producing a significant difference in the unit cell volume (213.1 and 210.8, respectively).
If you apply the transformation matrix [-b,-a,c] to the first unit cell 5.156 5.341 8.407 71.48 76.12 85.09 and compare with the second one you obtain:
5.341 5.156 8.407 103.88 108.52 85.09 Transformed Unit Cell 5.33 5.14 11.25 102.7 135.17 85.17 Second Unit Cell
You can see that a and b axes, as well as gamma (the angle between a and b) are in good agreement in both cases (within experimental differences), whereas the parameters where the c-axis plays a role diverge (c-axis and alfa and beta angles). I would suggest to draw both structures using the corresponding structural data (if available) as well as to compare the occupation of the special positions (if any) for the different atoms.
• the lattice parameters are indeed different, but the crystal structures are still identical (within exptl error bars) the crystallographers simply made different decisions about how to represent the structure in both cases (think about P21/n vs P21/c) so that is not the issue really - the question is how to rotate one cell so that the content of both cells are aligned/superimposed (again within exptl error bars)... Commented Jan 9, 2021 at 14:52
• no, this is not a simple problem of rotation. here you are in the triclinic system (space group P-1); you have no symmetry axis (in this case the unique symmetry operation is the inversion). So there are no different settings in this group as in the P21/c SG where different main axes can be selected. In any case, the unit cell volume must be the same, whichever the cell choice (if both structures are really the same). at first sight I would say that the c-axis of the 2nd cell is actually a combination of the c-axis in the first unit cell with other axes. Commented Jan 9, 2021 at 16:53
• In fact if you apply the transformation [-b,-a,b+c] to the first unit cell you obtain 5.341 5.156 11.3017 102.64 135.14 85.09 that are by about the same paramters of the 2nd cell. The problem is that this 2nd cell has a higher unit cell volume, whereas you should choose the lower one. For this reason the structural model of the 2nd cell could be correct in principle, but should be rearranged according to the 1st one model. Commented Jan 9, 2021 at 17:02
• [-b,-a,b+c] ... sorry i'm not familiar with this type notation, could you explain? also, both structures are xray and according to the CCDC (content of cif files) taken at room temperature, but one (MALNAC) is from 1957 and the other (MALNAC02) from 1994, so a difference in cell volume of about 1 percent should be acceptable/tolerable i believe... Commented Jan 9, 2021 at 18:12
• you have to consider them as vectors. for example, -b means that in the 2nd cell the a-axis has the length of the b-axis in the first one, but the vector is directed in an opposite direction. the c-axis of the 2-nd cell is given b the vector summation of the b and axis in th first one. in matrix form corresponds to a 3x3 matrix: [ 0 -1 0] [ -1 0 1] [ 0 0 1] for details look at International Tables for Crystallography (2006). Vol. A, Chapter 5.1, pp. 78. Commented Jan 10, 2021 at 10:17 | 1,951 | 7,293 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-26 | latest | en | 0.929914 |
http://precisioncalc.com/xlpISLESS.html | 1,623,586,644,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487608702.10/warc/CC-MAIN-20210613100830-20210613130830-00436.warc.gz | 39,166,536 | 2,013 | PrecisionCalc
xl
Precision
xlpISLESS
Determines whether one number is less than another, with up to 32,767 significant digits of precision. Returns TRUE if the first number is less than the second; returns FALSE if the first number is larger than the second number, or if they are equal. Returns #NUM! if either input cannot be resolved to a valid number.
Syntax
xlpISLESS(number1,number2)
number1 Required. The first number to be compared. number2 Required. The second number to be compared.
Remarks
• number1 and number2 can accept both numbers and text.
• number1 and number2 can accept text formatted with the local currency symbol and thousands separators, and negatives can be formatted with either a leading hyphen or parentheses.
• number1 and number2 can accept text up to 32,767 characters long.
• If any xlPrecision edition other than the 32,767 SD edition is being used, number1 and number2 are rounded down to the maximum number of significant digits allowed by the edition in use.
Examples
Formula Description Result =xlpISLESS(2,2) 2 < 2 FALSE =xlpISLESS(2.1,-3.1) 2.1 < -3.1 FALSE =xlpISLESS("2abc2",3) 2abc2 < 3 #NUM! =xlpISLESS(A1,A2) A1 = -0.3 A2 = "-0.3" A1 < A2 -0.3 < -0.3 (FALSE) | 330 | 1,216 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2021-25 | latest | en | 0.777526 |
https://www.cp2k.org/exercises:2015_ethz_mmm:monte_carlo_ice?rev=1426692231 | 1,558,277,577,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232254889.43/warc/CC-MAIN-20190519141556-20190519163556-00073.warc.gz | 749,133,064 | 9,542 | # Open SourceMolecular Dynamics
### Sidebar
#### For Developers
exercises:2015_ethz_mmm:monte_carlo_ice
# Properties of Ice from Monte Carlo Simulations
In this exercise we will use Monte Carlo sampling to calculate the dielectric constant of the disordered hexagonal phase (Ih) of water ice.
In order to speed up the calculation, we are going to use the SPC/E water model. The model is a non polarizable forcefield model, with parameters for:
• Charge of the atomic species
• Harmonic O-H bond elongation $U_{bond}= k_{bond}⋅(r-r_{eq})^2$
• Harmonic H-O-H angle bending $U_{angle}= k_{angle}⋅(θ-θ_{eq})^2$
• Lennard Jones interaction between non bonded species (interaction between atoms belonging to different water molecules)
(A DFT-version of the calculation can be found here: 10.1021/jp4103355).
## Introduction
Water molecules are arranged according to the so called “ice rules”:
1. Water molecules are present as neutral H2O
2. Each molecule makes four hydrogen bonds with its four nearest neighbors, two as a hydrogen bond donor and two as an acceptor.
All the possible proton arrangements in the ice strucrure are not energetically equivalent and we sample and weight them with a Monte Carlo approach. For this purpose, a specific algorithm was created (see: 10.1063/1.1568337), that allows to employ special proton reordering moves, which lead to an effective sampling of the ice dipole (N.B: different proton arrangement = different dipole).
The dielectric constant of a system describes its response to an external electric field. If the dipole moment is properly sampled, one can compute the dielectric constant of ice, by applying the Kubo Formula. This is valid in the approximation that the response of the system to the time-dependent perturbation (the field) is linear.
The dielectric constant can then be calculated from the dipole moments via: $$\epsilon = 1 + \left(\frac{4 \pi}{3 \epsilon_0 V k_B T } \right ) \operatorname{Var}(M) \ ,$$
where $M$ denotes the dipole moment of the entire simulation cell and $\operatorname{Var}(M)$ denotes the variance of the dipole moment of the sampling: $$\operatorname{Var}(M) = (\langle M \cdot M\rangle - \langle M\rangle\langle M\rangle ) \ .$$
## Task 1: Calculate the dielectric constant from one simulation
• As usual, download the 4 required files (at the end of this page: input, auxiliary input, ice coordinates and topology) in the same directory.
• Run the input-file mc_exercise.inp
⇒ It will create a file tmc_trajectory_T200.00.dip_cl, which contains the dipole moment for each accepted step.
You should run these calculations on 3 cores with :
bsub -W 01:00 -n 3 mpirun cp2k.popt -i mc_exercise.inp -o mc_exercise.out
• Plot the histogram of the z-component of the dipole moment.
⇒ The distribution should be symmetric, because the simulation cell is also symmetric in the z-direction.
• Compute the dielectric constant of ice. To simplify this task you can use the following python-script:
calc_dielectric_constant.py
#!/usr/bin/env python
import sys
import numpy as np
if(len(sys.argv) < 3):
print "Usage: calc_dielectric_constant.py <temperature> <traj_1.dip_cl> ... <traj_N.dip_cl>"
sys.exit()
T = float(sys.argv[1])
print "Temperature: %.2f K"%T
weights = []; M_vec = []; cell_vol = []
for fn in sys.argv[2:]:
assert(fn.endswith(".dip_cl"))
weights.append(raw_data[1:,0]-raw_data[:-1,0])
M_vec.append(raw_data[:-1,1:4])
weights = np.concatenate(weights)
M_vec = np.concatenate(M_vec)
cell_vol = np.concatenate(cell_vol)
V = np.mean(cell_vol) # [Angstrom^3]
print "Total number of samples: %d"%np.sum(weights)
print "Mean cell volume: %.2f Angstrom^3"%V
M = np.sqrt(np.sum(np.square(M_vec), axis=1)) # take norm of dipol moment
var = np.average(np.square(M), weights=weights) - np.square(np.average(M, weights=weights))
epsilon_0 = 8.8541878176e-12 #[F/m] vacuum permittivity
e = 1.602176565e-19 # [C] elementary charge
kb = 1.3806488e-23 #[J/K] Boltzmann constant
angstrom2meter = 1e-10
s = (e*e*4*np.pi)/(3*V*kb*T*angstrom2meter*epsilon_0)
epsilon = 1 + s*var
print "Dielectric constant: %.2f"%epsilon
Before you can run the python-script you have to load a newer python-module and set the executable-bit:
you@eulerX ~$module load python/2.7.6 you@eulerX ~$ chmod +x calc_dielectric_constant.py
you@eulerX ~$./calc_dielectric_constant.py 200 tmc_trajectory_T200.00.dip_cl ## Task 2: Gather more samplings You can gather more samples by launching multiple independent runs in parallel with different random number seeds. The seed is given by the RND_DETERMINISTIC keyword. The gathered trajectories can then be analyzed collectively with the python-script: you@eulersX ~$ ./calc_dielectric_constant.py 200 trj_1_T200.00.dip_cl trj_1_T200.00.dip_cl ...
You can also share trajectories with your colleges.
## Required Files
### Main Input-File (run this)
This file specifies the type of algorithm do adopt and the simulation parameters.
mc_exercise.inp
&GLOBAL
PROJECT H2O_MC
PROGRAM TMC ! Tree Monte Carlo algorithm
RUN_TYPE TMC
PRINT_LEVEL LOW ! Low amount of information written in the output
WALLTIME 1:00:00 ! The simulation will last one hour and then will stop
&END GLOBAL
&MOTION
&TMC
RND_DETERMINISTIC 42 !<=== Change this number to obtain different samplings
PRINT_COORDS .FALSE. !this avoids the printing of all coordinates and file-size problems
GROUP_CC_SIZE 0
NUM_MC_ELEM 100000
ENERGY_FILE_NAME H2O_ice.inp ! refers to the auxiliary input for the force specification
TEMPERATURE 200
&MOVE_TYPE MOL_TRANS ! specifies "proton moves" for better sampling
SIZE 0.05
PROB 3
&END
&MOVE_TYPE ATOM_TRANS
SIZE 0.01
PROB 3
&END
&MOVE_TYPE MOL_ROT
SIZE 5
PROB 3
&END
&MOVE_TYPE PROT_REORDER
PROB 5
&END
&MOVE_TYPE VOL_MOVE
SIZE 0.1
PROB 1
&END
PRESSURE 0.01
ESIMATE_ACC_PROB .TRUE. !accuracy parameters, do not change
RESTART_OUT 0
NUM_MV_ELEM_IN_CELL 5
PRINT_ONLY_ACC .FALSE.
&TMC_ANALYSIS
CLASSICAL_DIPOLE_MOMENTS
RESTART .FALSE.
&CHARGE
ATOM O
CHARGE -0.8476
&END CHARGE
&CHARGE
ATOM H
CHARGE 0.4238
&END CHARGE
&END TMC_ANALYSIS
&END TMC
&END MOTION
### Auxiliary Input-File (do NOT run this directly)
This auxiliary input specifies the system properties
H2O_ice.inp
&FORCE_EVAL
METHOD FIST
&MM
&NEIGHBOR_LISTS
GEO_CHECK OFF
VERLET_SKIN 4.0
&END NEIGHBOR_LISTS
&FORCEFIELD
&SPLINE
EMAX_SPLINE 10.0
&END
&CHARGE
ATOM O
CHARGE -0.8476
&END CHARGE
&CHARGE
ATOM H
CHARGE 0.4238
&END CHARGE
&BOND
ATOMS O H
K [nm^-2kjmol] 502080.0
R0 [nm] 0.09572
KIND G87
&END BOND
&BEND
ATOMS H O H
THETA0 [deg] 104.500
KIND G87
&END BEND
&NONBONDED
&LENNARD-JONES
ATOMS O O
EPSILON [kjmol] 0.650
SIGMA [angstrom] 3.166
RCUT 8.0
&END LENNARD-JONES
&LENNARD-JONES
ATOMS O H
EPSILON 0.0
SIGMA 3.1655
RCUT 8.0
&END LENNARD-JONES
&LENNARD-JONES
ATOMS H H
EPSILON 0.0
SIGMA 3.1655
RCUT 8.0
&END LENNARD-JONES
&END NONBONDED
&END FORCEFIELD
&POISSON
&EWALD
EWALD_TYPE ewald
ALPHA .45
GMAX 19
&END EWALD
&END POISSON
&END MM
&SUBSYS
&CELL
ABC 13.5084309945 15.6001709945 14.7072509945
&END CELL
&TOPOLOGY
COORD_FILE_NAME ./ice_ih_96.xyz
COORD_FILE_FORMAT xyz
CONNECTIVITY MOL_SET
&MOL_SET
&MOLECULE
NMOL 96
CONN_FILE_NAME topology_H2O.psf
&END
&END
&END TOPOLOGY
&END SUBSYS
&END FORCE_EVAL
&GLOBAL
&END GLOBAL
### Topology File
This file specifies the topology of the system (which atoms, bonds, angles in the water molecule). It allows the program to distinguish which atoms belong to which water molecule, and therefore discriminate between inter and intramolecular interactions.
topology_H2O.psf
PSF
1 !NTITLE
Topology file for water
3 !NATOM
1 WAT 1 WAT O O 0.000000 15.999400 0
2 WAT 1 WAT H H 0.000000 1.008000 0
3 WAT 1 WAT H H 0.000000 1.008000 0
2 !NBOND
1 2 1 3
1 !NTHETA
2 1 3
0 !NPHI
0 !NIMPHI
0 !NDON
0 !NACC
0 !NNB | 2,450 | 8,033 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2019-22 | longest | en | 0.84595 |
www.arrowacademy.co.za | 1,709,636,278,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948234904.99/warc/CC-MAIN-20240305092259-20240305122259-00760.warc.gz | 622,043,933 | 31,635 | # One solution for most maths problems
### Article By: Hilda Erasmus: Foundation Phase Specialist | 31 May 2019
Mathematics is one of the school subjects that require logic and insight. It often happens that learners struggle to master the subject because it simply becomes too complicated. There are many books available on how to make Maths easier to enable children to understand it better.
However, I want to focus on a very simple mathematical symbol – the = (equals) sign. This symbol is neglected in the Foundation Phase. Learners are taught to read sums from left to right and to make groups. The = is often replaced by arrows and sums are presented as follows: 2 + 3 → gives you 5. Learners memorise these methods and symbols but show limited comprehension once they have to apply it.
Let’s look at an example: 3 + 4 = __ + 2. Learners who don’t fully understand the = will try to answer the sum from left to right: 3 + 4 = 7 + 2 = 9. They will add an extra = to answer the question based on the methods they were taught.
Another big mistake occurs when the sum is presented differently, for example: 4 + __ = 7. The learner was taught to add or subtract groups. In this case, the learner sees a 4, a 7 and a +. The answer to the problem is then written as follows: 4 + 11 = 7. This is because the learner adds the 4 and the 7. The same happens with subtraction, for example: __ – 4 = 6 becomes 2 – 4 = 6.
By the time learners are faced with more challenging sums, the = is merely the symbol next to the answer. The secret to any Maths problem is to balance the sum. The centre of the sum is the = and the golden rule is to balance both sides of the =. Before learners are taught about + and -, they must learn about =. This can be done using simple exercises where two sides are matched:
Balance the circles so that there are an equal number of dots on either side of the =.
Which number must be in the blank circle so that the circles on either side of the = are the same?
When learners understand that sums must be balanced and that the = is the starting point of every sum, they will master problems such as 6 = 4 + __ and __ – 2 = 3 in no time.
Learners must thus get out of the habit of doing sums from left to right. Instead, they must start at the = and balance the sums. | 550 | 2,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2024-10 | latest | en | 0.964925 |
https://www.jiskha.com/display.cgi?id=1330494685 | 1,503,278,023,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886107065.72/warc/CC-MAIN-20170821003037-20170821023037-00186.warc.gz | 930,010,523 | 3,973 | # Math
posted by .
A farmer needs to enclose three sides of a pasture with a fence (the fourth side is a river). The farmer has 42 meters of fence and wants the pasture to have an area of 220 sq-meters. What should the dimensions of the pasture be? (For the purpose of this problem, the width will be the smaller dimension (needing two sides); the length with be the longer dimension (needing one side). Additionally, the length should be as long as possible.)
• Math -
If the width is w, then the length is 42-2w. So, the area is
y = w(42-2w)
y = -2w^2 + 42w
this is just a parabola, and y reaches its maximum when w = -42/-4 = 10.5
so, the pasture is 10.5 x 21. Area = 220.5
If we want just 220 m^2, then the field is 11x20
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More Similar Questions | 758 | 3,057 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-34 | latest | en | 0.948994 |
www.whitemarmotte.com | 1,718,252,607,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.11/warc/CC-MAIN-20240613025523-20240613055523-00183.warc.gz | 998,373,179 | 59,816 | # The Avalanche Risk Scale – Why 3 is a Dangerous Number
The International Avalanche Risk Scale is an important tool in off-piste decision making, but it is frequently misunderstood. If you have ever said, “It’s only three out of five, it’ll be fine,” you need to read this right now.
I am not going to say that you should never venture off-piste on a cat. 3 day, but you need to be aware of the risks when you do. Category 3 does not mean ‘moderately safe’ as a lot of skiers seem to think.
Most good skiers and snowboarders I meet think they understand the avalanche scale because they think they understand a scale of 1 to 5. Category 1 is obviously as safe as it gets and category 5 must be really dangerous. So far so good. The next assumption is that category 2 is fairly safe and category 4 is pretty dangerous. Not exactly, but close enough. And the last assumption is that category 3 is middling, moderate or average. Not too dangerous anyway. And this is where people get into trouble.
English speaking countries are promoting the wordy version of the scale, rather than the numbers. In this version, the five categories are low, moderate, considerable, high and very high. Would you say, “It’s only considerable risk today,” with the same nonchalance as you might say “it’s only three out of five”? The other takeaway from the wordy version is that category 2 is ‘moderate’, not ‘fairly safe’. Unfortunately, the need to cater for speakers of different languages means that European countries are unlikely to abandon the numerical scale in the near future.
## Some Statistics
Almost every time I read about an avalanche accident there will be a sentence somewhere saying the avalanche risk was ‘considerable’. My own impression from reading newspaper or online reports is that the vast majority of all fatal avalanche accidents happen when the risk is category three. Let’s have a look at whether the statistics agree.
### France
From 2001 to 2011, l’ANENA compiled statistics on 238 fatal avalanches in which 317 people died. Looking at the 207 avalanches which occurred in the winter months (when the avalanche bulletin was published), gives the following graph. Backcountry refers to ski tourers, walkers and snowshoers. Off-piste refers to skiers and snowboarders in lift-accessed terrain.
### Switzerland
SLF have compiled statistics of the number of ‘avalanche victims per danger level for persons caught while engaging in backcountry or off-piste activities, or in buildings or on transportation routes (accidents between 1997/98 and 2015/16’.
### North America
Bruce Tremper’s excellent book, Staying Alive in Avalanche Terrain gives some statistics for avalanche accidents in the United States and Canada.
Source: Staying Alive in Avalanche Terrain, 2nd Edition, Bruce Tremper, Pub: The Mountaineers Books 2008
While the graphs above differ considerably at hazard levels of two and four, they all show clearly that category 3, or ‘considerable’, is the most dangerous level in all four countries in terms of the number of fatalities.
## What do the numbers really mean?
There are many different wordings of the International Avalanche Hazard Scale in use, but all of them give a similar interpretation. The table below is from Wikipedia.
### Avalanche Risk
1 – Low Snow is generally very stable. Avalanches are unlikely except when heavy loads are applied on a very few extreme steep slopes. Any spontaneous avalanches will be minor sloughs. In general, safe conditions.
2 – Moderate On some steep slopes the snow is only moderately stable. Elsewhere it is very stable. Avalanches may be triggered when heavy loads are applied, especially on a few generally identified steep slopes. Large spontaneous avalanches are not expected.
3 – Considerable On many steep slopes the snow is only moderately or weakly stable. Avalanches may be triggered on many slopes even if only light loads are applied. On some slopes, medium or even fairly large spontaneous avalanches may occur.
4 – High On most steep slopes the snow is not very stable. Avalanches are likely to be triggered on many slopes even if only light loads are applied. In some places, many medium or sometimes large spontaneous avalanches are likely.
5 – Very High The snow is generally unstable. Even on gentle slopes, many large spontaneous avalanches are likely to occur.
If we look at the second column we can see that the stability ranges from very stable snow to generally unstable snow, as we would expect. Now let’s re-read the description for 3 – Considerable. ‘On many steep slopes the snow is only moderate or weakly stable.’ In other words, there are many slopes where the snow is weakly or moderately bonded. Reading along we see that ‘Avalanches may be triggered on many slopes even if only light loads are applied.’ A light load means a single skier, snowboarder or walker. To put it another way, there are many slopes where the weight of a single skier can trigger an avalanche.
In order to ski safely off-piste when the risk is considerable, you need the knowledge to avoid the ‘many’ steep slopes which are dangerous. Or you need to avoid altogether those slopes which are steep enough to avalanche.
## Why do so many people get caught in avalanches when the risk is ‘considerable’?
With both the statistics and the wording of the avalanche forecast telling us that category 3 is dangerous, why do so many people get caught in avalanches on those days? Perhaps more tellingly, why do so many expert skiers, snowboarders and mountaineers get caught on those days?
There are many factors to consider here. Moderate risk could be considered as green light conditions, where most routes are practicable. High risk indicates red light conditions where most routes are dangerous. Everybody knows they have to be careful when the risk is high. Considerable risk means amber light conditions – dangerous but not so dangerous that all off-piste routes should be avoided. In any situation, the borderline conditions are the hardest to judge.
## Human factors
Many users of the mountain environment tend to overestimate their avalanche skills. This can be especially true of those who consider themselves experts at their sport. All too often their avalanche skills do not match their skiing or snowboarding ability. People either misinterpret the avalanche scale as discussed above, or they understand what considerable risk means but overestimate their ability to travel safely in avalanche conditions. A bad decision in avalanche terrain will often have no adverse consequences. People will get away with it most of the time. Bad decision-making is reinforced until the day an accident happens.
Additionally, the set of human factors known as ‘heuristic traps’ tend to lead people to make bad decisions. A proper discussion of these will need a post of its own, but here are some examples of how they might influence people’s behaviour on a considerable risk day. The ‘scarcity’ trap leads people to ignore danger signs in the rush for fresh tracks on a powder day. The ‘acceptance’ trap can make them avoid voicing concerns in front of the group. The ‘expert halo’ trap abdicates the decision-making to somebody else. Finally, the ‘commitment’ trap ignores evidence that might interfere with completing the day’s objective.
Just knowing that category 3 is statistically the most dangerous level will not necessarily keep you safe. But at least you will not be ingorant of the risks. Read the avalanche bulletins, choose appropriate terrain, and above all, stay safe out there.
### 2 Comments on “The Avalanche Risk Scale – Why 3 is a Dangerous Number”
1. This is the best and most simple explanation I’ve read. We lost my son and his teammate in a level 3 “Considerable” danger avalanche in Austria.
Both boys were in the middle of the resort in between two groomed runs when they were swept away and buried 12 feet deep after hitting a terrain trap. None of Bryce’s teammates nor he had any idea that they were in danger, while skiing off of the groomed run into the powder.
I learned from Bryce and Ronnie’s tragedy that more than 50% of avalanche deaths occur in a Level 3 “Considerable danger rating. The rating scale goes from 1-5.
Considerable is a 3, so it sounds pretty safe, doesn’t it? Thanks for your great explanation! You also must check the avalanche report in the middle of the resort if you are planning on skiing off the groomed run in Europe and I feel that this is not explained enough over there.
Thank you for putting this piece together.
It is really great.
Laura Astle
Sandy, Utah
P.S. We put a movie out and it also does not explain the precautions that you must take while skiing in Europe.
“Off Piste” Tragedy in the Alps. Vimeo
2. Hi Laura, I’m so sorry to hear about your loss. Thank you for sharing your story, I think that the more people know about these accidents the more we can raise awareness of the risks. Off-piste skiing is a wonderful sport but it can be more dangerous than people realise. I hope your excellent video can help to prevent a future tragedy, and this article was written with the same intention.
You raise a really important point about the differences between skiing in Europe and North America. We don’t have the same concept of in-bounds and out-of-bounds skiing over here. Many skiers, including Europeans, don’t appreciate that stepping off the groomed runs means you are on uncontrolled off-piste slopes. This is an important message to get out, especially to people visiting from different parts of the world where the safety norms can be quite different.
Thank you again for your comment,
Ian
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Posted by2 months ago
I’ve only taken calc 1 (and obv the courses required before it) and I was wondering if i could start learning number theory. Would you suggest a course i should learn before number theory or is it okay if i learn number theory now?
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i just finished a course in discreet mathematics and would like to continue down the rabbit hole. Do you have any suggestions for some fun summer reading?
Thanks
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Posted by1 year ago
Archived
Firstly, since this is the most appropriate sub but sadly also very inactive, consider this a practice run for me asking elsewhere.
This has bugged me since childhood, it still does, but it's hard to put into words. Please bear with me.
For the most part, it seems that the most interesting mathematical things happen somewhere in the region of 0. Predictably, things that happen at 0 tend to be nothing of interest except when nothing is interesting, but there's a lot of action at ranges from zero that easily fit on graphs and models - that is to say, as "small" numbers. Before you point out that anything that can be plotted can be scaled/translated and so easily be made to fit, let me point out that I already know that, but also that it doesn't count because it just goes toward further illustrating my point.
Would it somehow be true to say that "numberspace" is "flat" and that the "flatness" is a direct consequence of the "flatness" of spacetime?
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Posted by1 year ago
Archived
Hey everyone! I have been trying to remember how, the shape formed when the vertices of a polygon are joined by straight lines and the edges of the polygon are deleted, is called. I was hoping one of you knows. Thank you!
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Posted by2 years ago
Archived
I've been plagued with the following problem that I cannot figure out. Let a,b,c,d belong to the integers. Given a/b + c/d, under what specific conditions in choosing the values of a,b,c,d will the gcd( ad+bc , bd ) <>1 .
I'm looking for a catch all formula/algorithm (if there is one) as oppose to trial and error.
I have a proof, but it ends up leading to some obvious statement ie chasing my tail.
I suspect the answer has to do with modular arithmetic or, more elementary,
dividend / divisor = quotient + remainder/ divisor
1
Posted by2 years ago
Archived
Prove (1-a)(1-b)(1-c) = abc only if a>1/2. Could someone help me?
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Moderators | 610 | 2,566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2018-39 | latest | en | 0.970467 |
https://forum.math.toronto.edu/index.php?topic=803.0 | 1,709,101,378,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474697.2/warc/CC-MAIN-20240228044414-20240228074414-00649.warc.gz | 252,610,053 | 5,797 | ### Author Topic: Difference between what we talk about in Chapter 3 & 4 (Read 2760 times)
#### ziyao hu
• Newbie
• Posts: 4
• Karma: 0
##### Difference between what we talk about in Chapter 3 & 4
« on: October 19, 2016, 10:54:31 AM »
I'm wondering why in Chapter 4, we need 2 boundary conditions, if we can solve the problem with only 1 b.c. just as the problem 5 in Assignment 5?
Pictures shown as below,
#### Victor Ivrii
• Elder Member
• Posts: 2607
• Karma: 0
##### Re: Difference between what we talk about in Chapter 3 & 4
« Reply #1 on: October 19, 2016, 12:10:28 PM »
You probably is confusing boundary conditions and initial conditions. For both wave equation and heat equation we need just 1 boundary condition at each end, however for other types of equations we may need more
Now, we need two initial conditions (as $t=0$) for wave equation, and one for heat equation because the former has an order 2 with respect to $t$, and the latter has order $1$. This was explained in details during the lecture.
Please do not post big chunks of screenshots but only exact equations you are talking about, and better type them. Then you questions would not require guesswork to answer | 316 | 1,195 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-10 | latest | en | 0.952731 |
excelautomationsolutions.com | 1,716,747,471,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058972.57/warc/CC-MAIN-20240526170211-20240526200211-00716.warc.gz | 202,057,991 | 42,464 | Current is a flow of electrical charge carriers, usually electrons or electron-deficient atoms. The common symbol for current is the uppercase letter I. The standard unit is the ampere, symbolized by A. One ampere of current represents one coulomb of electrical charge (6.24 x 1018 charge carriers) moving past a specific point in one second. Physicists consider current to flow from relatively positive points to relatively negative points; this is called conventional current or Franklin current. Electrons, the most common charge carriers, are negatively charged. They flow from relatively negative points to relatively positive points.
Electric current can be either direct or alternating. Direct current (DC) flows in the same direction at all points in time, although the instantaneous magnitude of the current might vary. In an alternating current (AC), the flow of charge carriers reverses direction periodically. The number of complete AC cycles per second is the frequency, which is measured in hertz. An example of pure DC is the current produced by an electrochemical cell. The output of a power-supply rectifier, prior to filtering, is an example of pulsating DC. The output of common utility outlets is AC.
Current per unit cross-sectional area is known as current density. It is expressed in amperes per square meter, amperes per square centimetre, or amperes per square millimetre. Current density can also be expressed in amperes per circular mil. In general, the greater the current in a conductor, the higher the current density. However, in some situations, current density varies in different parts of an electrical conductor. A classic example is the so-called skin effect, in which current density is high near the outer surface of a conductor, and low near the centre. This effect occurs with alternating currents at high frequencies. Another example is the current inside an active electronic component such as a field-effect transistor (FET).
An electric current always produces a magnetic field. The stronger the current, the more intense the magnetic field. A pulsating DC, or an AC, characteristically produces an electromagnetic field. This is the principle by which wireless signal propagation occurs.
EAS is a group of highly experienced professionals working in oil and gas industry whose goal is to be the best provider of Control and Automation based services from conception to Engineering, Design and detailing, installing, commissioning and hand over to customer. We achieve this goal by specializing exclusively in the project management of EAS plans.
Specialization means providing quality services to industries, professionals who are working in Industrial automation, students. In our pursuit of absolute quality, we will maintain a quantitative means of measuring our progress towards this objective.
Industrial Automation is the use of Control Systems, such as computers or robots, and information technologies for handling different processes and machineries in an industry to replace a human being. It is the second step beyond mechanization in the scope of industrialization
Ohm’s law
Main article: Ohm’s law
Ohm’s law states that the current through a conductor between two points is directly proportional to the potential difference across the two points. Introducing the constant of proportionality, the resistance, one arrives at the usual mathematical equation that describes this relationship
I = V \ R
where I is the current through the conductor in units of amperes, V is the potential difference measured across the conductor in units of volts, and R is the resistance of the conductor in units of ohms. More specifically, Ohm’s law states that the R in this relation is constant, independent of the current.
Alternating and direct current
In alternating current (AC) systems, the movement of electric charge periodically reverses direction. AC is the form of electric power most delivered to businesses and residences. The usual waveform of an AC power circuit is a sine wave, though certain applications use alternative waveforms, such as triangular or square waves. Audio and radio signals carried on electrical wires are also examples of alternating current. An important goal in these applications is recovery of information encoded (or modulated) onto the AC signal.
## Current measurement
Current can be measured using an ammeter.
Electric current can be directly measured with a galvanometer, but this method involves breaking the electrical circuit, which is sometimes inconvenient.
Current can also be measured without breaking the circuit by detecting the magnetic field associated with the current. Devices, at the circuit level, use various techniques to measure current:
## Current is a flow of electrical charge carriers, usually electrons or electron-deficient atoms. The common symbol for current is the uppercase letter I. The standard unit is the ampere, symbolized by A. One ampere of current represents one coulomb of electrical charge (6.24 x 1018 charge carriers) moving past a specific point in one second. Physicists consider current to flow from relatively positive points to relatively negative points; this is called conventional current or Franklin current. Electrons, the most common charge carriers, are negatively charged. They flow from relatively negative points to relatively positive points.
Electric current can be either direct or alternating. Direct current (DC) flows in the same direction at all points in time, although the instantaneous magnitude of the current might vary. In an alternating current (AC), the flow of charge carriers reverses direction periodically. The number of complete AC cycles per second is the frequency, which is measured in hertz. An example of pure DC is the current produced by an electrochemical cell. The output of a power-supply rectifier, prior to filtering, is an example of pulsating DC. The output of common utility outlets is AC.
Current per unit cross-sectional area is known as current density. It is expressed in amperes per square meter, amperes per square centimetre, or amperes per square millimetre. Current density can also be expressed in amperes per circular mil. In general, the greater the current in a conductor, the higher the current density. However, in some situations, current density varies in different parts of an electrical conductor. A classic example is the so-called skin effect, in which current density is high near the outer surface of a conductor, and low near the centre. This effect occurs with alternating currents at high frequencies. Another example is the current inside an active electronic component such as a field-effect transistor (FET).
An electric current always produces a magnetic field. The stronger the current, the more intense the magnetic field. A pulsating DC, or an AC, characteristically produces an electromagnetic field. This is the principle by which wireless signal propagation occurs.
EAS is a group of highly experienced professionals working in oil and gas industry whose goal is to be the best provider of Control and Automation based services from conception to Engineering, Design and detailing, installing, commissioning and hand over to customer.
We achieve this goal by specializing exclusively in the project management of EAS plans.
Specialization means providing quality services to industries, professionals who are working in Industrial automation, students. In our pursuit of absolute quality, we will maintain a quantitative means of measuring our progress towards this objective.
Industrial Automation is the use of Control Systems, such as computers or robots, and information technologies for handling different processes and machineries in an industry to replace a human being. It is the second step beyond mechanization in the scope of industrialization
Ohm’s law
Main article: Ohm’s law
Ohm’s law states that the current through a conductor between two points is directly proportional to the potential difference across the two points. Introducing the constant of proportionality, the resistance, one arrives at the usual mathematical equation that describes this relationship
I = V \ R
where I is the current through the conductor in units of amperes, V is the potential difference measured across the conductor in units of volts, and R is the resistance of the conductor in units of ohms. More specifically, Ohm’s law states that the R in this relation is constant, independent of the current.
Alternating and direct current
In alternating current (AC) systems, the movement of electric charge periodically reverses direction. AC is the form of electric power most delivered to businesses and residences. The usual waveform of an AC power circuit is a sine wave, though certain applications use alternative waveforms, such as triangular or square waves. Audio and radio signals carried on electrical wires are also examples of alternating current. An important goal in these applications is recovery of information encoded (or modulated) onto the AC signal.
## Current measurement
Current can be measured using an ammeter.
Electric current can be directly measured with a galvanometer, but this method involves breaking the electrical circuit, which is sometimes inconvenient.
Current can also be measured without breaking the circuit by detecting the magnetic field associated with the current. Devices, at the circuit level, use various techniques to measure current:
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http://www.talkstats.com/threads/non-parametric-test-alternative-to-2-way-anova.72081/ | 1,537,614,223,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158320.19/warc/CC-MAIN-20180922103644-20180922124044-00232.warc.gz | 412,215,742 | 11,778 | # Non-parametric test alternative to 2-way ANOVA?
#### ab8675309
##### New Member
Hello everyone. I recently attempted to run a 2-way ANOVA test (using SPSS) but have run into the difficulty of meeting the ANOVA test assumptions of normality (both Levene's test and the Shapiro-Wilk test were both significant). It's also apparent that the data does not appear to be normally distributed (just by looks of the data via a histogram alone). I'm trying to figure out what would be a good alternative to testing this data since the normality and equality of variances appear to be violated.
Here's a summary of my data:
• The hypothesis is I would like to see if there is an interaction between a person's BMI and their procedure type on the number of events that occurs during their procedure
• 2 independent variables (BMI, procedure type; both categorical) with 1 continuous (number of events; interval) dependent variable
• Sample sizes for each group are unequal
I've looked online for a lot of resources but I'm not sure if they apply to what I'm trying to do. I've heard of using other alternatives like Friedman, Welch's, and Brown-Forsythe tests, but I don't know if they're applicable to this scenario. I've also seen recommendations to use the log of my values or to do a robust analysis, but again...very confused as to how to approach this.
So right now I'm pretty lost. Any help would be greatly appreciated. Thank you for your time.
#### Attachments
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#### Karabiner
##### TS Contributor
Data need not be normally distributed in ANOVA. In small samples, it is assumed that the residuals are from normally distributed populations, but that is not an important assumpton if samples are not small (n > 30 or so). Some real problems seem to be
a) the categorization of the BMI, which sacrifices many degrees of freedom - why don't you use the BMI as interval scaled variable?
b) your dependent variable is a count variable with only a very small range, therefore ANOVA seems inappropriate; you should consider Poisson regression or negative binomial regression instead.
With kind regards
Karabiner
BTW, there are no "nonparametric" alternatives for 2-factorial problems.
#### ab8675309
##### New Member
Hello Karabiner,
Thank you for your helpful reply. I appreciate the information regarding ANOVA. I was under the impression that to run an ANOVA test you needed to meet these assumptions:
1. Independence of observations – this is an assumption of the model that simplifies the statistical analysis.
2. Normality – the distributions of the residuals are normal.
3. Equality (or "homogeneity") of variances, called homoscedasticity — the variance of data in groups should be the same.
By the way that's from the ANOVA Wiki page. I did test the residuals and plot them on a histogram, and they appeared to be very skewed. I also used a Q-Q plot and saw that they were not lining up either, which is why I started to become confused as to what would be the next step in this study if it did not meet normality and homoscedasticity. However, as you mentioned, this is count data and therefore ANOVA would not be appropriate as it is not a test for counts.
When you say that is not an important assumption do you mean when the total sample size from all groups is n > 30? Or when each factor's group (say for my example BMI: <18.5 * Procedure: Endoscopy) has to be n > 30? Because I have some groups (let's just say for example BMI: > 50 * Procedure: Colonscopy) that have fewer than 30 cases in it. Is that an issue even if I decide to consider using Poisson regression or negative binomial regression?
To answer your other question, unfortunately the people who collected the data did not have access to the actual BMI number: it was instead grouped into a range according to the forms they used. Out of curiosity if I did get BMI to the interval range (say for example I did happen to have this info), then would it change what type of test I would do? If so then what test?
Lastly -- and again I appreciate your assistance with everything -- how would I know whether Poisson or negative binomial is the better test for this study? I ask only because you give me an option for either or, but most flow charts/reference material I've looked at online don't even cover those studies. How would I know going forward which test to use?
Last edited:
#### Karabiner
##### TS Contributor
When you say that is not an important assumption do you mean when the total sample size from all groups is n > 30?
Yes.
Out of curiosity if I did get BMI to the interval range (say for example I did happen to have this info), then would it change what type of test I would do? If so then what test?
In that case, you would include BMI as an interval scaled variable in your analyses instead of BMI as a categorical variable with 8 levels. E.g. in a regression, you would only need 1 variable "BMI" instead of 7 dummy-variables which jointly represent "BMI".
how would I know whether Poisson or negative binomial is the better test for this study?
https://stats.stackexchange.com/que...mial-negative-binomial-and-poisson-regression
http://www.math.usu.edu/jrstevens/biostat/PoissonNB.pdf
https://www.theanalysisfactor.com/p...ng-count-model-diagnostics-to-select-a-model/
HTH
Karabiner
#### ab8675309
##### New Member
Hello Karabiner,
Thank you for your insight and help. | 1,244 | 5,421 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-39 | longest | en | 0.950267 |
https://www.physicsforums.com/threads/find-acceleration-of-blocks-on-table-solved.215362/ | 1,723,551,135,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641076695.81/warc/CC-MAIN-20240813110333-20240813140333-00684.warc.gz | 716,119,438 | 18,969 | # Find Acceleration of Blocks on Table [SOLVED]
• jesuslovesu
In summary, the conversation is about finding the acceleration of blocks A and B on a table, with weights of 5 lbs and 6 lbs respectively and a friction coefficient of 0.3. The equations for sum of forces in the y-axis are discussed, with the problem being the relationship between aa and ab. The issue is resolved by changing aa = 2ab to aa = -2ab to account for the direction of motion.
jesuslovesu
[SOLVED] Block on a table
## Homework Statement
http://img91.imageshack.us/img91/4285/fudgepq5.th.png
I am having some trouble finding the acceleration of the blocks, can anyone help me?
The weight of B is 6 lbs.
The weight of A is 5 lbs.
The friction coef is .3
## The Attempt at a Solution
1) The positive y-axis goes from left to right
sum of forces in y = (6/32)*ab = 2T - 1.8
2) positive y-axis goes from top to bottom
sum of forces in y = (5/32)*aa = 5-T
I know that aa = -2ab
The way my book derives the relationship between aa and ab is from 'dependent motion' they say that length = sa + 2sb and taking the time derivative twice is 0 = aa + 2ab, the problem is with the way I have my equations setup, when i use aa = -2ab i get the wrong answer, does anyone know how I can fix it so I am using aa = -2ab and getting the correct answer?
I don't see anything wrong with my equations, but i am getting the wrong accelerations. Can anyone help?
Last edited by a moderator:
sign problems
jesuslovesu said:
I know that aa = -2ab
Not exactly. You've defined both aa and ab to be positive when B moves right and A moves down, so it should be: aa = 2ab.
The way my book derives the relationship between aa and ab is from 'dependent motion' they say that length = sa + 2sb and taking the time derivative twice is 0 = aa + 2ab,
That's fine, but realize that you are defining the position of B to increase to the right, but sb actually decreases when mass B moves to the right. That's why you need to change the sign.
First, it is important to note that the question has been marked as [SOLVED], so it is likely that the original poster has already found the answer to their problem. However, for the sake of providing a response, I will give some suggestions for finding the acceleration of the blocks.
To start, it is important to identify the forces acting on each block. For block A, we have its weight (5lbs) acting downwards and the tension force (T) acting upwards. For block B, we have its weight (6lbs) acting downwards, the friction force (0.3*6lbs) acting in the opposite direction of motion, and the tension force (T) acting upwards.
Next, we can use Newton's second law (F=ma) to set up equations for each block. For block A, we have 5-T=5a (a is the acceleration of block A). For block B, we have 6-0.3*6-T=6a (a is the acceleration of block B). We also know that the blocks are connected by a string, so they must have the same acceleration (a).
Now, we can solve for the tension force (T) by setting the two equations equal to each other: 5-T=5a and 6-0.3*6-T=6a. This gives us T=5-5a and T=6-6a. Setting these equal to each other, we get 5-5a=6-6a, which gives us a=1. This means that the blocks are accelerating at a rate of 1 m/s^2.
I am not sure what equations you were using in your attempt at a solution, but it is possible that there was a mistake in your setup or calculations. It is always important to double check your work and make sure that all equations and values are correct. I hope this helps!
## 1. How do you find the acceleration of blocks on a table?
The acceleration of blocks on a table can be found by using the equation a = F/m, where a is the acceleration, F is the net force acting on the blocks, and m is the mass of the blocks. This equation is based on Newton's Second Law of Motion.
## 2. What is the difference between static and kinetic friction?
Static friction is the force that prevents an object from moving when a force is applied to it, while kinetic friction is the force that acts on an object that is already in motion. The magnitude of static friction is always equal to or greater than the magnitude of kinetic friction.
## 3. How do you calculate the net force on the blocks?
The net force on the blocks can be calculated by finding the vector sum of all the individual forces acting on the blocks. This can be done by using the formula Fnet = ΣF, where Fnet is the net force and ΣF is the sum of all the forces.
## 4. What factors affect the acceleration of blocks on a table?
The acceleration of blocks on a table can be affected by the net force acting on the blocks, the mass of the blocks, and the coefficient of friction between the blocks and the table. Other factors such as air resistance and the shape of the blocks can also have an impact on the acceleration.
## 5. Can the acceleration of blocks on a table be negative?
Yes, the acceleration of blocks on a table can be negative. This would indicate that the blocks are slowing down or moving in the opposite direction of the applied force. It is important to consider the direction of the net force and the initial velocity of the blocks when determining the sign of the acceleration.
• Introductory Physics Homework Help
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3K | 1,447 | 5,761 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-33 | latest | en | 0.947395 |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-7th-edition/chapter-p-prerequisites-section-p-3-integer-exponents-and-scientific-notation-p-3-exercises-page-24/44 | 1,537,396,578,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156311.20/warc/CC-MAIN-20180919220117-20180920000117-00476.warc.gz | 754,147,523 | 13,269 | ## College Algebra 7th Edition
$6.3*10^{38}$
$\frac{(73.1)(1.6341*10^{28})}{0.0000000019}=\frac{(7.31*10)(1.6341*10^{28})}{1.9*10^{-9}}=\frac{7.31*1.6341}{1.9}*10^{1+28--9}=6.286985*10^{38}\approx 6.3*10^{38}$ | 113 | 210 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2018-39 | latest | en | 0.23719 |
http://forums.wolfram.com/mathgroup/archive/1997/May/msg00008.html | 1,653,510,543,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662593428.63/warc/CC-MAIN-20220525182604-20220525212604-00623.warc.gz | 23,382,815 | 7,667 | nonlinear schrodinger PDF
• To: mathgroup at smc.vnet.net
• Subject: [mg7095] nonlinear schrodinger PDF
• From: "Weigang(Victor) Feng" <wfeng at cs.umb.edu>
• Date: Wed, 7 May 1997 01:58:18 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com
``` Hi, there, when I try to solve the below eq using NDSolve, I met the below
situation, I am novice to numerical PDF, is anybody can help to make it
work?
NDSolve[{1/2 D[y[x, t], t, t]+Abs[y[x,t]]^2 y[x,t] ==I D[y[x, t], x, x],
y[x, 0] == Exp[-x^2], Derivative[0,1][y][x, 0] == 0,
y[-5, t] == y[5, t]}, y, {x, -5, 5}, {t, 0, 5}]
NDSolve::"mxst":
"Maximum number of \!\(1000\) steps reached at the point \!\(t\) == \
\!\(3.93328131362742325`\)."
NDSolve::"eerr":
"Warning: Estimated spatial error at \!\(t\) =
\!\(3.93328131362742325`\) \
is much greater than prescribed error tolerance. Grid spacing \
\!\(\*StyleBox[\"0.175438596491228064`\", Rule[PrintPrecision, 16], \
Rule[StyleBoxAutoDelete, True]]\) which was based on the initial
conditions \
may be too large. A singularity may have formed or you may want to
specify a \
smaller grid spacing using the StartingStepSize option."
Out of memory. Exiting.
```
• Prev by Date: Re: a simple question
• Next by Date: Re: Is Mathematica on a PC Too Slow? | 446 | 1,261 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-21 | latest | en | 0.753292 |
https://www.avatto.com/computer-science/test/mcqs/questions-answers/c/24/1.html | 1,721,499,623,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517515.18/warc/CC-MAIN-20240720174732-20240720204732-00230.warc.gz | 589,239,469 | 18,355 | C Programming MCQ - Structure & Union
1:
Consider the declaration:
static struct
{ unsigned a:5;
unsigned b:5;
unsigned c:5;
unsigned d:5%:
}v=(1,2,3,4);
v occupies
A. 4 bytes B. 2 bytes C. 1 byte D. None of these Answer Report Discuss Option: A Explanation : Click on Discuss to view users comments. Write your comments here:
2:
A short integer occupies 2 bytes an, ordinary integer 4 bytes and a long integer occupies 8 bytes of memory
If a structure is defined as
struct TAB{
short a;
int b;
long c;
}TABLE[10];
the the total memory requirement for TABLE is
A. 14 B. 140 C. 40 D. 32 Answer Report Discuss Option: B Explanation : Click on Discuss to view users comments. sonali manohar shelke said: (8:56am on Monday 18th December 2017) a is short which occupies 2 bytes per element so there are 10 elements which equal to 20 bytes for ab is int which contains 4 bytes per element so again they alson contain 10 elements which is equal to 40 bytesand c is long which occupies 80 bytes for 10 elements' 80 total memory space occupied is 20 40 80=140so option B is correct Write your comments here:
3:
The expression a<<6 shifts all bits of a six places to the left. If a 0x6db7, then what is the value of a<<6?
A. 0xa72b B. 0xa3b C. 0x6dc0 D. 0x1111 Answer Report Discuss Option: C Explanation : Click on Discuss to view users comments. Write your comments here:
4:
The declaration
union id{
char color[12];
int size;}shirt,Jeans;
denotes shirt and Jeans are variable of type id and
A. Each can have a value of color and size B. Each can represent either a 12-character color or an integer size at a time C. Shirt and Jeans are same as struct variables D. Variable shirt and Jeans cannot be used simultaneously in a statement Answer Report Discuss Option: B Explanation : Click on Discuss to view users comments. Write your comments here:
5: Most appropriate sentence to describe unions is
A. Union are like structures B. Union contain members of different data types which share the same storage area in memory C. Union are less frequently used in program D. Union are used for set operations Answer Report Discuss Option: B Explanation : Click on Discuss to view users comments. Write your comments here:
Syllabus Covered in this section is
• Elements of C-Tokens
• Identifiers
• data types in C
• Control structures in, C
• Sequence, selection and iteration(s).
• Structured data types in C
• Arrays, Structures, union, String, and pointers
• Functions, Recursion
• Parameter passing, Scope
• Binding, Abstract data types
This Section covers Multiple Choice Questions Answers in C Programming . Here questions answers are given with explanation and references. These questions can be used for the preparation of various competitive and academic exams.
Who can benefit -
• Any undergraduate or postgraduate student who is seeking C Programming objective type questions answers can use this section.
• C Programming Objective type Questions Answers are also used at SSC and HSc level for Exam preparation.
• C Programming MCQ Questions Answers are also used by engineering students in the preparation of their Exams.
• Any candidate who has to appear for DOEACC A, B or C level Exam can also learn C Programming Questions Answers.
• All the teachers who are appearing for Kendriya Vidyalya Sangathan Entrance Exam can also use C Programming Multiple Choice Questions Answers.
• C Programming MCQs can also be used by the students who are pursuing B.Sc or Msc Computer Science.
• C Programming Questions Answers can also be used by BCA students for the preparation of their exams.
• Any student who is pursuing B.Sc. in Information Technology can also use this C Programming mcq Questions Answers.
• MCA students can also prepare for their exams using C Programming Objective Type Questions Answers.
Various Search Terms used for this section are | 904 | 3,879 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-30 | latest | en | 0.840755 |
https://www.teacherspayteachers.com/Product/Dinosaur-Doubles-Plus-One-Math-Game-Worksheets-Center-Sort-2218284 | 1,488,059,772,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171834.68/warc/CC-MAIN-20170219104611-00128-ip-10-171-10-108.ec2.internal.warc.gz | 889,944,008 | 24,281 | Total:
\$0.00
# Dinosaur Doubles Plus One- Math Game, Worksheets, Center Sort
Subjects
Resource Types
Common Core Standards
Product Rating
4.0
File Type
PDF (Acrobat) Document File
0.81 MB | 11 pages
### PRODUCT DESCRIPTION
Fun way for students to practice their doubles and doubles plus one!
Dinosaur Doubles Sort:
Cut out the dinosaur cards. Students match the “doubles plus one” cards to the “doubles” cards that will help them solve the problem. Then students record their answers on the recording sheet.
Dinosaur Doubles Game:
Cut out and laminate the spinner. Use a paper clip and a pencil to spin the spinner. Students can play the game individually, in partners, or groups of 3. They spin the spinner, double the number they land on and add one. They then cover that piece on their dinosaur board. The first one to fill up their board wins the game.
Dinosaur Worksheets:
-Students cut out the "doubles plus one" card and paste them under the double that will help them solve the problem. Then they write the answers in the blanks.
-Students color the dinosaur next to the domino a color depending on if it's a double, a double plus one, or neither.
Total Pages
11
N/A
Teaching Duration
N/A
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2 ratings
\$1.50
User Rating: 4.0/4.0
(32 Followers)
\$1.50 | 368 | 1,401 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-09 | longest | en | 0.865367 |
https://physics.stackexchange.com/questions/616448/total-forces-is-0-equivalent-that-the-torque-is-0 | 1,719,251,477,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865401.1/warc/CC-MAIN-20240624151022-20240624181022-00205.warc.gz | 393,322,762 | 41,310 | # Total forces is 0 equivalent that the torque is 0?
I came across a theorem that states that :
for a solid body (made of a collection of points $$M_i$$ of masses $$m_i$$ and speed $$\overrightarrow{v_i}$$ ) in rotation around a point $$O$$ under the action of a set of forces $$\overrightarrow{F_i}$$ each with its point $$M_i$$ of application; then we have the following equivalence : $$\overrightarrow{\mathcal{M}}(\sum_{i} \overrightarrow{F_i})=\frac d{dt}\overrightarrow{L}=\overrightarrow{0} \Longleftrightarrow\sum_{i} \overrightarrow{F_i}=\overrightarrow{0}$$ where $$\overrightarrow{L}=\sum_{i} \overrightarrow{L_i}=\sum_{i} \overrightarrow{OM_i}\wedge m_i \overrightarrow{v_i}$$
is the total moment. and $$\overrightarrow{\mathcal{M}}(\sum_{i} \overrightarrow{F_i})$$ is the total torque.
I tried to prove this theorem but I could not
The above is obviously not true when considering a force couple. Two separated equal and opposite forces have $$\sum F = 0$$, but a net torque, and thus $$\frac{\rm d}{{\rm d}t} L \neq 0$$.
This is because in general
$$\sum \vec{F}_i = 0$$ is not the same as $$\sum \vec{r}_i \times \vec{F}_i =0$$
I might be missing something here, but here is the development of momentum from particles to move together. Assume the body is rotating about the center of mass, and the center of mass is translating.
From the origin, the location of each particle $$\vec{r}_i$$ is split into the location of the center of mass $$\vec{r}_C$$ plus the location from the center of mass $$\vec{d}_i$$ such that
$$\sum_i m_i \vec{d}_i = \vec{0} \tag{1}$$
## Kinematics
Each particle moves with a different velocity based on the following vector field
$$\vec{v}_i = \vec{v}_C + \vec{\omega}\times \vec{d}_i \tag{2}$$
## Translational Momentum
Add up the momentum of each particle
\require{cancel} \begin{aligned} \vec{p} & = \sum_i m_i \vec{v}_i = \sum_i m_i ( \vec{v}_C + \vec{\omega}\times \vec{d}_i ) \\ & = \left( \sum_i m_i \right) \vec{v}_C + \vec{\omega}\times \cancel{ \left( \sum_i m_i \vec{d}_i \right)} = m \, \vec{v}_C \end{aligned} \tag{3}
where $$m = \sum_i m_i$$
### Rotational Momentum
\begin{aligned} \vec{L}_C & = \sum_i \vec{d}_i \times (m_i \vec{v}_i) = \sum_i \vec{d}_i \times m_i ( \vec{v}_C + \vec{\omega}\times \vec{d}_i ) \\ & = \cancel{\left(\sum_i m_i \vec{d}_i\right)} \times \vec{v}_C + \sum_i m_i \vec{d}_i \times ( \vec{\omega} \times \vec{d}_i ) \\ & = \mathbf{I}_C\, \vec{\omega} \end{aligned} \tag{4}
### Law of Motion
Newtons law of motion is
\begin{aligned} \sum_i \vec{F}_i & = \tfrac{\rm d}{{\rm d}t} \vec{p} \\ \sum_i \vec{d}_i \times \vec{F}_i & = \tfrac{\rm d}{{\rm d}t} \vec{L}_C \end{aligned} \tag{5}
Now without any specific definition of the force vector field $$\vec{F}_i$$ unlike the velocity field the above sum of torques is not equal to zero in general even if the sum of forces is zero.
• in the equation (4) you miss the $m_i$ ; and i dont know how you defined $\mathbf{I}_C$ Commented Feb 26, 2021 at 15:03
• you have to calculate vector product and you will how $I_c$ is defined. Commented Feb 28, 2021 at 14:40
• @El-Mo $$\mathbf{I}_C = \sum_i m_i \pmatrix{ y_i^2+z_i^2 & -x_i y_i & -x_i z_i \\ -x_i y_i & x_i^2 + z_i^2 & -y_i z_i \\ -x_i z_i & -y_i z_i & x_i^2+y_i^2 }$$ where $d_I = \pmatrix{x_i \\ y_i \\ z_i}$ for each particle. Commented Mar 1, 2021 at 14:38
The theorem claims that torque vanishing implies forces vanishing. Your question asks for the converse. Forces sum to zero implies that torque vanishes. The first statement is true provided the torque about every point vanishes, the second is clearly false.
the theorem is just saying the opposite of what I understood | 1,232 | 3,684 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 23, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-26 | latest | en | 0.742959 |
http://physicsinventions.com/net-force-on-a-charge-in-a-square/ | 1,542,794,211,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039747665.82/warc/CC-MAIN-20181121092625-20181121113915-00032.warc.gz | 246,316,705 | 9,053 | 1. The problem statement, all variables and given/known data
There are 4 charges in the corner of a square with the lengths of the sides measuring 18.2cm
q1= 2.34μC (top left) q2=-1.58μC (top right) q3= 2.03μC (bottom left) q4= -3.00μC (bottom right)
→Calculate net force on q4
→Calculate net force at the center of the square
2. Relevant equations
Fnet=|KQ|/r^2
K constant = 8.99×10^9
3. The attempt at a solution
q1= [(8.99×10^9)(2.34×10^-6)]/(0.182)^2 = 635086.3422 @ 315°
q2= [(8.99×10^9)(1.58×10^-6)]/(0.182)^2 = 428818.9832 @270°
q3= [(8.99×10^9)(2.03×10^-6)]/(0.182)^2 = 550950.9721 @ 0°
for my next step I was going to calculate x and y components and then plug it into the
d=*sqrt [(x^2)+(y^2)]
to calculate net force on q4… but I’m very uncertain about how to continue or if my previous calculations are even right.
http://ift.tt/1gn0MQc | 326 | 854 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-47 | latest | en | 0.77794 |
https://blogs.mathworks.com/steve/2015/01/20/divergent-colormaps/?from=cn | 1,714,029,443,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297290384.96/warc/CC-MAIN-20240425063334-20240425093334-00736.warc.gz | 119,420,555 | 28,412 | ## Steve on Image Processing with MATLABImage processing concepts, algorithms, and MATLAB
Note
Steve on Image Processing with MATLAB has been archived and will not be updated.
# Divergent colormaps
I promised earlier to post about divergent colormaps. A divergent colormap is usually constructed by concatenating two colormaps together that have different color schemes.
Here is an example of a divergent colormap from www.ColorBrewer.org by Cynthia A. Brewer, Geography, Pennsylvania State University.
rgb = [ ...
94 79 162
50 136 189
102 194 165
171 221 164
230 245 152
255 255 191
254 224 139
253 174 97
244 109 67
213 62 79
158 1 66 ] / 255;
b = repmat(linspace(0,1,200),20,1);
imshow(b,[],'InitialMagnification','fit')
colormap(rgb)
This colormap has the brightest color in this middle, which is typical for divergent colormaps.
A divergent colormap is used to compare data values to a reference value in a way that visually highlights whether values are above or below the reference.
Let me show you an example using sea surface temperatures. I learned today that 28 degrees Celsius has a specific meaning in hurricane modeling and prediction. It is considered to be the minimum temperature for hurricane formation.
Using the NASA EOSDIS Reverb tool, I found and downloaded sea surface temperature from the Aqua satellite for August 1, 2011. The data arrived as a netCDF file.
amsre = ncread('20110801-AMSRE-REMSS-L2P_GRIDDED_25-amsre_20110801rt-v01.nc',...
'sea_surface_temperature');
size(amsre)
ans =
1440 720 2
The two planes contain data from two different orbits.
orbit1 = flipud(amsre(:,:,1)');
orbit2 = flipud(amsre(:,:,2)');
imshow(orbit1,[],'InitialMagnification','fit')
title('Orbit 1')
imshow(orbit2,[],'InitialMagnification','fit')
title('Orbit 2')
The black regions are NaNs representing missing data. Combine the two orbits by taking the maximum.
sst = max(orbit1,orbit2);
imshow(sst,[],'InitialMagnification','fit')
Now grab a subset of the data that includes a portion of the Atlantic Ocean.
ssta = sst(180:400,330:700);
imshow(ssta,[],'InitialMagnification','fit')
Apply the divergent colormap.
colormap(rgb)
colorbar
OK, I see one problem right off the bat. It looks like our data is in Kelvin. Let's fix that.
ssta = ssta - 273.15;
imshow(ssta,[],'InitialMagnification','fit')
colormap(rgb)
colorbar
Now I want to make use of our divergent colormap to highlight the temperature of interest, 28 degrees. To do that, I'll use the CLim property of the axes object. The CLim property is a two-element vector. The first element tells us the data value that maps to the lowest colormap color, and the second element tells us the data value that maps to the highest colormap color. MATLAB computes these values automatically based on the range of the data.
get(gca,'CLim')
ans =
-1.9500 33.6000
To get the middle colormap color to represent 28 degrees, we need to pick our two CLim values so that they are equally far away from 28 degrees. After some experimentation, I settled on a range from 22 degrees to 34 degrees.
set(gca,'CLim',[22 34])
That's looking better, but it isn't helpful to have the missing data displayed using the bottom color of the colormap. I will fix that problem using this procedure.
1. Construct a gray mask image (in truecolor format).
2. Display it on top of our sea surface temperature image.
3. Set the AlphaData of the gray mask image so that the sea surface temperatures show through.
Here are just the first two steps.
mask = isnan(ssta);
hold on
hold off
The missing values are displayed as gray, but we can't see the sea surface temperatures. Fix that by using the NaN-mask as the overlay image's AlphaData.
h.AlphaData = mask; % This syntax requires MATLAB R2014b.
Temperatures close to the threshold are displayed using yellow, the brightest color in the colormap and the one right in the middle. Temperatures above the threshold are displayed using oranges and reds, while temperatures below are displayed using greens and blues.
Before I go, I want to give a shout to the new Developer Zone blog that just started today. If you are into advanced software developer topics, I encourage you to head over there and take a look.
Published with MATLAB® R2014b
| | 1,059 | 4,344 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-18 | latest | en | 0.747981 |
https://www.reddit.com/r/answers/comments/1c002n/if_a_woman_has_two_kids_with_different_fathers/ | 1,480,713,523,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540563.83/warc/CC-MAIN-20161202170900-00102-ip-10-31-129-80.ec2.internal.warc.gz | 1,019,247,848 | 20,249 | This is an archived post. You won't be able to vote or comment.
[–] 11 points12 points (10 children)
half siblings. full siblings are from the same two parents (father, mother). half siblings only need one parent in common.
[–] 31 points32 points (5 children)
They are actually called 3/4 siblings or sibling cousins. Their relatedness is 37.5%, more than your average half siblings' 25% but less than full siblings' 50%.
[–] 17 points18 points (2 children)
Interestingly, if a pair of identical twins marry and have kids, the children are siblings according to their genetics.
[–] 1 point2 points (1 child)
Aren't the kids of any two parents siblings? Or did that just go over my head.
[–] 5 points6 points (0 children)
When Sally and Amy marry Hank and Joe, their kids are genetically siblings.
[–] 0 points1 point (1 child)
I think all of those values are actually half that. Siblings share an average 25% of their alleles.
[–] 1 point2 points (0 children)
I'm pretty sure it's 50%. 25% is having the exact same genotype (ie. having the same maternal AND the same paternal allele) on a given locus.
If mom has alleles M1, M2 and dad F1, F2 of a gene, their kids may have M1-F1, M1-F2, M2-F1, M2-F2 genotype with 25% probability each.
So whatever genotype the first kid has on this locus, the second will have the same genotype with 25% chance, a genotype with one shared allele with 50% chance and a complementary genotype with 25% chance. So the expected number of shared alleles on this locus is 2*25% + 1*50% + 0*25% = 1.
So on average you share one allele with your sibling out of the two on any given locus which is 50%.
EDIT: I had to escape asterisk signs to show up
[–][S] 2 points3 points (3 children)
Plain and simple, I guess. I wasn't sure as the kids were kind of cousins, but also siblings.. bit mind boggling hah but I guess it's bound to happen.
[–] 4 points5 points (1 child)
but if 2 twin brothers married 2 twin sisters, their offsprings would genetically be brothers.
[–] 4 points5 points (0 children)
They would have to be identical twins.
[–] 1 point2 points (0 children)
The children's mother would also be their Aunt and their brother's dad would be their Uncle.
[–] 6 points7 points (1 child)
Three-quarter siblings*. Closer genetically than half-siblings, but not as close as full siblings.
*I made that term up just now.
[–] 7 points8 points (0 children)
It appears to be the correct term!
So even though you were right, I'm only awarding you 3/4 rightness, as it was a guess.
[–] 1 point2 points (0 children)
My dad has a child with a woman who also has a child by his brother. So my sister's half sister is also her cousin.
[–] 1 point2 points (0 children)
I would go with three-quarters. Seriously.
[–] 2 points3 points (1 child)
Interesting but unrelated fact, if two sets twins get together and each pair has kids, genetically the children would be siblings
[–] 0 points1 point (0 children)
That would be true if they were identical twins. | 823 | 3,031 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2016-50 | latest | en | 0.94251 |
https://idswater.com/2020/11/14/how-many-ml-is-equal-to-500mg/ | 1,653,023,759,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662531352.50/warc/CC-MAIN-20220520030533-20220520060533-00310.warc.gz | 385,610,388 | 39,343 | # How many ml is equal to 500mg?
November 14, 2020 Off By idswater
## How many ml is equal to 500mg?
Convert 500 Milligram to Milliliters
500 Milligram (mg) 0.500000 Milliliters (ml)
1 mg = 0.001000 ml 1 ml = 1,000 mg
## What is 300grams in grams?
Convert 300 Milligrams to Grams
mg g
300.00 0.3
300.05 0.30005
300.10 0.3001
300.15 0.30015
## What quantity is 300 ml?
Quick Conversions
U.S. Standard Metric
1 cup 200 ml and 2-15 ml spoons
1 1/4 cup 300 ml
1 1/3 cup 300 ml and 1-15 ml spoon
1 1/2 cup 350 ml
## Is 500mg the same as 5 ml?
How Many ML Is 500 MG? After converting 500 mg in ML, the answer will be 0.5 ml.
## Is 300ml 300 grams?
Cooking Ingredient: A density is required for converting between mL and grams….mL to grams conversions (water)
mL to Grams mL to Grams
5 mL = 5 grams 250 mL = 250 grams
6 mL = 6 grams 300 mL = 300 grams
7 mL = 7 grams 350 mL = 350 grams
8 mL = 8 grams 400 mL = 400 grams
## How many glasses is 300 ml of water?
Convert 300 Milliliters to Cups
mL cups
300.00 1.2680
300.05 1.2682
300.10 1.2684
300.15 1.2687
## How many glasses of water is 800 ml?
Convert 800 Milliliters to Glasses
mL glasses
800.00 5.4102
800.05 5.4106
800.10 5.4109
800.15 5.4113
## How many milligrams are in 300 milliliters of water?
300 Milliliters (ml) = 300,000 Milligram (mg) 1 ml = 1,000 mg. 1 mg = 0.001000 ml
## How to convert a milligram to a milliliter?
Since 1 milligram is equal to 0.001 milliliters, which can be written as 1 mg = 1/1000 mL. Deriving from this equation, 1/1000 mL = 1 mg, hence 1 mL = 1000 mg. Hence to convert mL to mg, multiply the entered milliliter with 1000 to get the result. For example, when the given number of milliliters is 25, then the conversion …
## How many milligrams are in 0.6 milliliters?
0 Milligrams 0.6 Milliliters 10000 Milligrams 10 Milliliters. How 25 ML to MG ? 25 ML = 25000 MG. How many milliliters are in a milligram? 1 milliliters ML = 1000 milligram MG. How much is 5 ml in MG? 5 ML to MG = 5000 MG. How many milligrams are in 2 milliliters? Convert 2 Milliliters to Milligram
## How many milligrams are in.500 ml?
0.5 ml =. 500 mg. 2.5 ml =. 2500 mg. 4.5 ml =. 4500 mg. 11 ml =. 11000 mg. 0.6 ml =.
## What does 300 mL equal?
300 Milliliters (mL) =. 10.14421 Ounces (fl oz) Milliliters : A milliliter (also written “milliliter”; SI symbol ml) is a non-SI metric system unit of volume which is commonly used as liquid unit.
## What is 300 milliliters in imperial fluid ounces?
300 Milliliters is equivalent to 10.558523918356 Imperial Fluid Ounces. The conversion factor from Milliliters to Imperial Fluid Ounces is 0.035195079727854. To find out how many Milliliters in Imperial Fluid Ounces, multiply by the conversion factor or use the Volume converter above.
## How many milliliters are in one ounce?
Quick Answer. There are approximately 29.574 milliliters (mL) per US liquid ounce (fl. oz). | 998 | 2,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2022-21 | latest | en | 0.762635 |
http://nasawavelength.org/resource-search?facetSort=1&educationalLevel=High+school&topicsSubjects=Mathematics&learningTime=4+to+6+hours | 1,534,375,288,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210362.19/warc/CC-MAIN-20180815220136-20180816000136-00473.warc.gz | 277,879,557 | 14,800 | ## Narrow Search
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# Identifying the Key Changing Conditions of the Earth System (Grades 7-9)
This unit consists of four activities. Students begin by examining temperature cycles (current, recent and historical) then add in factors such as carbon dioxide, precipitation and cloud cover to discover regional and global differences in the... (View More)
# MY NASA DATA: Nutrients in Chesapeake Bay
In this problem-based data analysis activity, students assume roles as members of an International Team of Marine Biologists, tasked with predicting and monitoring possible harmful algae blooms. Students use data maps and guiding questions, to... (View More)
# Earth Exploration Toolbook: Is Greenland Melting?
Users explore data, using My World GIS, that characterize the dynamic Greenland Ice Sheet. By examining photographs, map views, and tabular data, users gain an understanding of how and why scientists are monitoring the ice sheet and what they are... (View More)
# MY NASA DATA: Seasonal Cloud Cover Variations
Students will examine seasonal cloud coverage data in North Carolina (or their local area), and analyze data to determine whether there is a correlation between season, cloud cover and type of cloud most prevalent during each season. Step-by-step... (View More)
Audience: High school
Materials Cost: Free
# Using Mathematical Models to Investigate Planetary Habitability: Activity A Finding a Mathematical Description of a Physical Relationship
In this activity, student teams learn about research design and design a controlled experiment exploring the relationship between a hypothetical planet, an energy source, and distance. They analyze the data and derive an equation to describe the... (View More)
Audience: High school
Materials Cost: Free per student
# Sand or Rock: Finding Out from 1000 km
Learners will measure temperature of two different surfaces; sand and stone; on a sunny day, make a series of temperature measurements, and plot the results. Extensions include experimenting with different materials, using temperature sensors and... (View More)
# Who Will Feed the World?
Students confront the challenge of improving global agricultural production in order to feed increasingly larger populations. Students evaluate tables, maps, graphs, photos and satellite images detailing global population growth patterns and... (View More)
# Why Is Snow Important in the Southwestern United States?
Water supply in the Southwestern United States depends on snow. Students discover its importance through analysis and evaluation of data, satellite images, space shuttle photos, and ground-based observations, and then apply their findings in a... (View More)
Keywords: Snowpack
1 | 608 | 3,073 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-34 | longest | en | 0.874086 |
https://discuss.codechef.com/t/approach-for-problem-gamcount/14187 | 1,600,718,982,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400202007.15/warc/CC-MAIN-20200921175057-20200921205057-00267.warc.gz | 372,122,261 | 3,690 | # Approach for problem GAMCOUNT
The problem link is: https://www.codechef.com/problems/GAMCOUNT
Approach:
As we know Henry gonna win every game in which possible number of all valid moves is odd.
So we calculate total number of valid moves of each of Xi(see NOTE) and sum them.
More generally Henry wins if the sum of all these moves is odd ,also problem can be simplified if considering total moves of each number Xi as either even or odd, then just look if there are odd number of odds in initial given Xi set and count these games in answer.
• Is this approach is correct?
NOTE: we can simply get valid number of moves with Xi by just knowing how many times we can subtract 2^V(Xi) so that resulting number be >=0. | 164 | 720 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2020-40 | latest | en | 0.912401 |
https://math.answers.com/Q/What_is_8_over_9_divided_by_2_over_3 | 1,675,885,001,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500904.44/warc/CC-MAIN-20230208191211-20230208221211-00372.warc.gz | 414,436,287 | 40,507 | 0
# What is 8 over 9 divided by 2 over 3?
Wiki User
2011-09-27 20:27:04
4 over 3.
Wiki User
2011-09-27 20:27:04
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## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
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Q: What is 8 over 9 divided by 2 over 3? | 128 | 343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-06 | latest | en | 0.863732 |
http://www.finchrobot.com/teaching/finch-tells-time | 1,545,193,483,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376830479.82/warc/CC-MAIN-20181219025453-20181219051453-00061.warc.gz | 373,769,184 | 7,870 | # Finch Tells Time
Teacher Note: Start by creating a large clock face on posterboard, as shown below. This activity uses Snap! Level 3 but could be modified to use another level.
In this activity, the Finch will be the minute hand of a clock!
Place your Finch in the center of the clock face with its beak pointing at the 12.
Where should the Finch be pointing at 3:15? Write a program to make the Finch turn from the start position (pointing at 12) to the position of the minute hand at 3:15. You will need to adjust your speed and wait time carefully to make the Finch move exactly this amount. Remember, the Finch is a clock, so it should turn clockwise!
Where should the Finch be pointing at 7:40? Write a program to make the Finch turn from the start position (pointing at 12) to the position of the minute hand at 7:40.
How far does the minute hand of a clock move between 4:50 and 6:10? Write a program to show this with the Finch.
• You know that the start time is 4:50. Where will the Finch be pointing?
• How far should the Finch turn to show that one hour has passed? Where will the Finch be pointing now?
• How much farther does the Finch need to turn to end at 6:10? Where will the Finch be pointing at 6:10?
• What happens when you put it all together? Remember that you are trying to show more than one hour passing!
How far does the minute hand move between 11:30 and 1:00? Write a program to show this with the Finch.
• You know that the start time is 11:30. Where will the Finch be pointing?
• How far should the Finch turn to show that one hour has passed? Where will the Finch be pointing now?
• How much farther does the Finch need to turn to end at 1:00? Where will the Finch be pointing at 1:00?
• What happens when you put it all together? Remember that you are trying to show more than one hour passing!
Standards:
This activity is aligned with Common Core math standards that focus on telling time (1.MD.B and 2.MD.C). | 478 | 1,953 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2018-51 | latest | en | 0.938067 |
https://www.clutchprep.com/chemistry/practice-problems/136365/a-0-70-g-sample-of-a-weak-monoprotic-acid-was-dissolved-in-150-ml-of-water-and-t-1 | 1,638,743,337,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363226.68/warc/CC-MAIN-20211205221915-20211206011915-00167.warc.gz | 780,861,490 | 34,773 | # Problem: A 0.70 g sample of a weak monoprotic acid was dissolved in 150 mL of water and titrated with a 0.055 M solution of KOH at 25°C. At the half-equivalence point, after addition of 25.5 mL of the KOH solution, the pH was 3.8.Determine the molecular weight (in g/mol) of the acid.A. 84.8 g/molB. 249.6 g/molC. 499.1 g/molD. 269.4 g/molE. 49.9 g/mol
###### FREE Expert Solution
We’re being asked to calculate the molar mass of a weak monoprotic acid
Recall that at the equivalence point of a titration:
Also, recall that moles = molarity × volume
This means:
$\overline{){\left(\mathbf{MV}\right)}_{{\mathbf{acid}}}{\mathbf{=}}{\left(\mathbf{MV}\right)}_{{\mathbf{base}}}}$
We're going to calculate for the molar mass of the weak monoprotic acid using the following steps:
82% (39 ratings)
###### Problem Details
A 0.70 g sample of a weak monoprotic acid was dissolved in 150 mL of water and titrated with a 0.055 M solution of KOH at 25°C. At the half-equivalence point, after addition of 25.5 mL of the KOH solution, the pH was 3.8.
Determine the molecular weight (in g/mol) of the acid.
A. 84.8 g/mol
B. 249.6 g/mol
C. 499.1 g/mol
D. 269.4 g/mol
E. 49.9 g/mol | 396 | 1,182 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2021-49 | latest | en | 0.824411 |
https://cs.stackexchange.com/questions/166827/deciding-if-a-language-is-cfl-or-in-p | 1,716,623,783,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058789.0/warc/CC-MAIN-20240525065824-20240525095824-00014.warc.gz | 154,775,387 | 40,519 | # Deciding if a language is CFL or in $P$
I'm trying to decide whether $$L_c=$${$$w=uxu, | \ u,x\in \Sigma ^* \ and \ |u|=c$$} for some constant $$c\in \mathbb{N}$$ is context free or not.
initialliy, I've thought about choosing $$x=\epsilon$$ which will lead to $$w=uu, \ u\in \Sigma ^*$$ but I've been told that since $$c$$ is a constant, it wont be the same language as $$L=$${$$uu | u\in \Sigma ^*$$} which put me in a standstill.
I've also thought about some examples that will make the pumping lemma not work on $$L_c$$, yet they all seem to fail (I couldent find a pumping constant that throws me out of $$L_c$$)
any ideas on how to progress from here?
• cs.stackexchange.com/q/18524/755
– D.W.
Feb 27 at 17:56
• @D.W. This is truely an eye opening post, however I couldent seem to find a 'technique' that fits the type of language I'm facing with, could you direct me to the part in cs.stackexchange.com/q/18524/755 that discusses these kind of languages? Feb 27 at 18:06
## 1 Answer
Observe that for some $$c \in \mathbb{N}$$ and finite alphabet $$\Sigma$$, $$\Sigma^c$$ with $$|\Sigma^c| = |\Sigma|^c$$ is finite. Therefore
$$L_c = \bigcup_{u \in \Sigma^c} u\Sigma^*u$$
is the union of a finite number of regular languages. So $$L_c$$ must be regular/context-free.
Assume that $$\Sigma$$ and $$c$$ are chosen as above and that $$\sim_c$$ is the indistinguishability relation for $$L_c$$. By the Myhill-Nerode theorem, $$L_c$$ is regular if and only if $$\Sigma^*/\sim_c$$ is finite. We now prove $$L_c$$ regular by showing that each string in $$\Sigma^*$$ is indistinguishable to one of length $$\leq 2c$$, so the number of classes in $$\Sigma^*/\sim_c$$ must be bounded by $$|\Sigma|^{2c}$$.
Choose $$w \in \Sigma^*$$, if $$|w| \leq c$$ then $$w \sim_c w$$. If $$|w| > c$$ then there must be $$u \in \Sigma^c$$, $$y \in \Sigma^+$$ such that $$w = uy$$. Now take the greatest prefix $$u_1$$ of $$u$$ such that $$y = xu_1$$ for some $$x \in \Sigma^*$$. If $$z \in \Sigma^*$$ with $$|z| \geq c$$, then
$$wz = uyz \in L_c \iff u \text{ is a suffix of } z \iff uu_1z \in L_c.$$
Because $$u_1$$ is the greatest prefix of $$u$$ with $$w = uxu_1$$, it follows that if $$|z| < c$$ then also
$$wz = uyz \in L_c \iff u_2z = u \text{ for a suffix } u_2 \text{ of } u_1 \iff uu_1z \in L.$$
So $$w \sim_c uu_1$$ and $$|uu_1| \leq |uu| = 2c$$, done.
• this is interesting, I've proven that $L_c$ is not regular using the Myhill-Nerode theorm, I choose {$0^cx0^i$}$_{i=0}^{\infty}$ for some $c\in \mathbb{N}$ and $x\in \Sigma ^*$. and by choosing $z=0^{c-i}$ shown there are infinite equivalence classes which is a contradiction to the theorm, however I am unsure on how to show that it is context free, could you elaborate more on your solution? Feb 28 at 12:37
• Sorry, I don't really understand how you get an infinite number of equivalence classes here 😅 I've also added a proof of the regularity of $L_c$ using the the Myhill-Nerode theorem to my answer. Feb 28 at 14:40 | 986 | 2,986 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 48, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-22 | latest | en | 0.873722 |
http://stackoverflow.com/questions/17643867/collision-response-and-elastic-impulse-in-xna-4-0 | 1,455,430,844,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701169261.41/warc/CC-MAIN-20160205193929-00291-ip-10-236-182-209.ec2.internal.warc.gz | 205,230,116 | 19,058 | # Collision response and elastic impulse in XNA 4.0
I know there are physic plugins for C# or XNA, but I want to create my own, so I can learn about the topic.
My problems are the following:
I try to apply an elastic impulse to my character with the right angle and velocity. The velocity is calculated the right way, the angle is not and distorts the results! The next problem is, that my character gets into a shaking mode, though it should stand still. I know where the problem comes from, but I don't know how to fix it (edit: do I have to consider the penetration depth for that?)
The `IPhysicsObject` inherits the most important informations, the `Vector2[]` has the collisionPoint at index 0 and the penetration depth at index 1.
I have tried to work with this but yeah.. I don't know
`````` public void ElasticImpulse(IPhysicsObject Object, Vector2[] _colPos)
{
//this function is down below
if (checkCollidingObjects(m_cCharacter, Object))
return;
//this List is like this declined:
//public static List<IPhysicsObject[]> CollidingObjects = new List<IPhysicsObject[]>();
//this list contains every pair of objects, that collided this frame, it is cleared after all physics and game logic is done.
CollidingObjects.Add(new IPhysicsObject[] { m_cCharacter, Object });
//deltavelocity is the velocity between two frames
Vector2 v1 = Velocity - DeltaVelocity;
float lv1 = (float)Math.Sqrt(v1.X * v1.X + v1.Y * v1.Y);
float m1 = Mass;
float k1 = Damping;
Vector2 v2 = Object.Physik.Velocity - Object.Physik.DeltaVelocity;
float lv2 = (float)Math.Sqrt(v2.X * v2.X + v2.Y * v2.Y);
float m2 = Object.Mass;
float k2 = Object.Physik.Damping;
Vector2 colDir1 = _colPos[0] - m_cCharacter.Position;
Vector2 colDir2 = _colPos[0] - Object.Position;
colDir1.Normalize();
colDir2.Normalize();
Vector2 colNorm1 = new Vector2(colDir1.Y, -colDir1.X);
Vector2 colNorm2 = new Vector2(colDir2.Y, -colDir2.X);
float ldir1 = (float)Math.Sqrt(colNorm1.X * colNorm1.X + colNorm1.Y * colNorm1.Y);
float ldir2 = (float)Math.Sqrt(colNorm2.X * colNorm2.X + colNorm2.Y * colNorm2.Y);
float pi = MathHelper.Pi;
//float angle1 = pi - ((v1.X * colNorm1.X + v2.Y * colNorm1.Y) / (lv1 * ldir1)) / v1.Length();
float angle1 = pi - (float)Math.Acos(((v1.X * colNorm1.X + v2.Y * colNorm1.Y) / (lv1 * ldir1)) / v1.Length());
angle1 = (float.IsNaN(angle1)) ? 0 : angle1;
//float angle2 = pi - ((v2.X * colNorm2.X + v2.Y * colNorm2.Y) / (lv2 * ldir1)) / v2.Length();
float angle2 = pi - (float)Math.Acos(((v2.X * colNorm2.X + v2.Y * colNorm2.Y) / (lv2 * ldir1)) / v2.Length());
angle2 = (float.IsNaN(angle2)) ? 0 : angle2;
//calculating the new velocities u 1/2. Got this formula out of the wiki link i posted above (took the german wiki version)
Vector2 u1 = (m1 * v1 + m2 * v2 - (m2 * (v1 - v2) * k2)) / (m1 + m2) - v1;
Vector2 u2 = (m1 * v1 + m2 * v2 - (m1 * (v2 - v1) * k1)) / (m1 + m2) - v2;
//transform the new velocities by the correct angle
Vector2 newV1 = new Vector2(
u1.X * (float)Math.Cos(angle1) - u1.Y * (float)Math.Sin(angle1),
u1.X * (float)Math.Sin(angle1) + u1.Y * (float)Math.Cos(angle1));
Vector2 newV2 = new Vector2(
u2.X * (float)Math.Cos(angle2) - u2.Y * (float)Math.Sin(angle2),
u2.X * (float)Math.Sin(angle2) + u2.Y * (float)Math.Cos(angle2));
newV1 = new Vector2(
(float.IsNaN(newV1.X)) ? 0 : newV1.X,
(float.IsNaN(newV1.Y)) ? 0 : newV1.Y);
newV2 = new Vector2(
(float.IsNaN(newV2.X)) ? 0 : newV2.X,
(float.IsNaN(newV2.Y)) ? 0 : newV2.Y);
}
bool checkCollidingObjects(IPhysicsObject obj1, IPhysicsObject obj2)
{
if (CollidingObjects.Count > 0)
{
int a = CollidingObjects.FindIndex(x => (x[0] == obj1 && x[1] == obj2) ||
(x[1] == obj1 && x[0] == obj2));
return a != -1;
}
return false;
}
``````
-
The shaking I think you are talking about is when an object is fighting itself for its direction - such as when your object is being pulled by gravity downwards, but your collision handling is making it move back upwards – Sayse Jul 14 '13 at 21:37
yeah i know that. Normally i just canceled to add the gravity constant to stop it. But i wanted to do it a bit more realistic now. I add an anti gravity force now how should negate the gravity so the character stand still. I think the problem lays in the impulse function too. My prediction is it doesn't stop to bounce. i wanted to add a simple version of the real physics. you can say a miniminiminiminiminimized version. ;) – Napp Jul 14 '13 at 21:45
Generally, you should not mess with angles if you can avoid it. And in this case you can. Could you add a figure of the collision? It is not clear what the variables stand for (e.g. `DeltaVelocity`, `colNorm1/2`, `u` ...). How can you calculate the collision normal from the velocities? That depends on the body's shape. Or do you have standard objects (e.g. spheres)? Is one of the bodies fixed or are they both moveable? The shaking is probably caused by multiple collisions where there should be only one (and therefore multiple reflections of the velocity vector). – Nico Schertler Jul 15 '13 at 8:42
ok, sorry that i forgot to mention it! This is only a 2D Game and i'm just messing around with Rectangles. In my tests now i got only one moveable (my character) and the ground he should bounce of but i want to keep it as universal as possible (for enemy or item bouncing, idk). i managed to fix the angle Problem by taking ´float angle1 = pi - (float)....´ this and transformed it into ´float angle = Vector2.Dot(v1, colNorm1) / lv1;´ this I already did this in previous versions but always forgot to divide by the length (the velocity is no unit vector, i forgot that i need one... stupid me) – Napp Jul 15 '13 at 10:17
i added a piece of information, do you need more, or is this enough? – Napp Jul 15 '13 at 10:30 | 1,694 | 5,731 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2016-07 | latest | en | 0.763308 |
https://berkeleygw.org/tutorial-sigma/ | 1,597,296,449,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738960.69/warc/CC-MAIN-20200813043927-20200813073927-00583.warc.gz | 228,978,907 | 17,713 | # Introduction to GW approximation
## Theory
GW approximation is a many-body perturbation theory that describes the electron self-energy, $\Sigma$. In a compact form, $\Sigma = iGW$, which is a convolution of Green’s function ($G$) and screened Coulomb interaction ($W = \epsilon^{-1}V$, where $V$is the bare Coulomb interaction). The binaries sigma_real.x (for systems with inversion symmetry) and sigma_cplx.x (for systems without inversion symmetry) are aimed to calculate the self-energy $\Sigma$. With the inclusion of self-energy, we are able to (easily) calculate the quasiparticle (interacting electrons) band structure of materials.
The electron self-energy is in general non-local, and frequency-dependent. One makes several approximations (need references) and can express the electron self-energy in the following way, i.e. the so-called GW approximation,
$\Sigma(\bold{r}, \bold{r}'; \omega ) = i\int \frac{d\omega'}{2\pi} e^{-i\delta \omega'} G(\bold{r}, \bold{r}'; \omega-\omega')W(\bold{r}, \bold{r}'; \omega')$,
where $\delta=0^+$. Generally we use non-interacting Green’s function constructed with Kohn-Sham wavefunctions, and we use the RPA dielectric matrix calculated in the previous step to construct the screened Coulomb potential.
$G(\bold{r}, \bold{r}'; \omega) = \sum_{n,\bold{k}} \frac{\psi_{n\bold{k}}(\bold{r}) \psi^*_{n\bold{k}}(\bold{r}')}{\omega - \varepsilon_{n\bold{k}} - i\delta_{n\bold{k}}}$,
and the screened Coulomb potential is
$W(\bold{r}, \bold{r}'; \omega) = \int \epsilon^{-1}(\bold{r}, \bold{r}''; \omega)V(|\bold{r}' - \bold{r}''|) d\bold{r}'' = \int \epsilon^{-1}(\bold{r}, \bold{r}'; \omega)\frac{e^2}{|\bold{r}' - \bold{r}''|} d\bold{r}''$.
In DFT, people often use LDA or GGA (note: BGW also supports hybrid functional as starting point) to describe the exchange correlation in a local sense, whereas in the GW approximation, the exchange correlation is embedded in the electron self-energy. After we evaluate the $\Sigma$ matrix elements sandwiched by the Kohn-Sham states, we subtract the contribution from the local exchange-correlation potential $V_{xc}(\bold{r})$, and add the self-energy to the Kohn-Sham eigenvalues,
$E^{QP}_{n\bold{k}} = \varepsilon^{LDA}_{n\bold{k}} - \langle n\bold{k}|V^{LDA}_{xc}|n\bold{k}\rangle + \langle n\bold{k}|\Sigma(\omega)|n\bold{k}\rangle$.
In this way, we get the quasibarticle band structures.
In BerkeleyGW, there are several different options to account for the frequency-dependence and its integral, and this is introduced in the previous section Epsilon. One can use plasmon-pole model to describe the frequency-dependence, and most often we use generalized plasmon-pole model (GPP, this is the default of BerkeleyGW); or directly sampling at different frequencies (either real or imaginary axis) and integrate, and we call this full-frequency calculation.
In GPP, one uses the dielectric matrix at zero frequency (i.e. static limit) $\epsilon^{-1}(\omega = 0)$ and $f$-sum rule model the dielectric behavior at finite frequencies. For a GPP calculation, in BerkeleyGW, we separate the self-energy into two parts: the screened exchange (SEX) term which accounts for the contribution from the holes in the Green’s function, and the Coulomb-hole (COH) term, which accounts for the contribution from the holes in the screened Coulomb potential. The two terms read as
$\Sigma_{\text{SEX}} (\textbf{r},\textbf{r}';E) = -\sum_{n\textbf{k}}^{\text{occ}} \phi_{n\textbf{k}}(\textbf{r}) \phi_{n\textbf{k}}^*(\textbf{r}') W(\textbf{r},\textbf{r}';E - \epsilon_{n\textbf{k}})$,
$\Sigma_{\text{COH}} (\textbf{r},\textbf{r}';E) = \sum_{n\textbf{k}} \phi_{n\textbf{k}}(\textbf{r}) \phi_{n\textbf{k}}^*(\textbf{r}') \int_0^\infty dE' \frac{B(\textbf{r},\textbf{r}';E')}{E - \epsilon_{n\textbf{k}} - E'}$,
where $B$ is an anti-symmetric spectral function [need reference]. We observe that the first term requires a summation over all occupied states, and it appears to be the bare exchange interaction but instead of bare Coulomb potential, the screened Coulomb potential shows up, therefore the name screened exchange. However, the COH term requires a summation over all bands (occupied plus unoccupied), therefore a careful convergence test is needed w.r.t. number of empty states.
If we neglect the frequency-dependence, i.e. take the static limit, we are doing static COHSEX calculation. Furthermore, (as just touched before) if we use bare Coulomb potential, and consider only the bare exchange term (no COH, as we don’t even have $W$), we are at the Hartree-Fock level.
In full-frequency calculation, we separate the self-energy into bare-exchange $X$ part, and the correlation part $Cor$. One can directly sample the real axis, or sample the imaginary axis (contour deformatoin).
In BerkeleyGW, we calculate the physical quantities in reciprocal space with plane wave basis set.
## Running tips
BerkeleyGW is designed to efficiently perform massive parallelization, so is sigma. The sigma code is parallelized over bands, and calculate the matrix elements (diagonal, or off-diagonal) one $k$ point each time. For a moderate system of few to 10 atoms, the usage of around one hundred to few thousands CPUs is quite efficient. Please read the output carefully to investigate the parallelization report to get a better idea what is the optimal number of CPUs for your particular system. Since the $k$ points are calculated one-by-one, you can manually separate the $k$ points into several jobs to save human waiting time. | 1,493 | 5,526 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 24, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2020-34 | latest | en | 0.762839 |
https://www.physicsforums.com/threads/probability-and-byte.156221/ | 1,653,568,406,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604794.68/warc/CC-MAIN-20220526100301-20220526130301-00392.warc.gz | 1,088,550,244 | 14,202 | # Probability and byte
## Homework Statement
In a binary communication system all bit bytes are possible, and equally likely. Any bit is equallly likely to be a 1 or 0.
1- what is the probability of oberving a byte with exactly three 1's?
2-if a byte is observed with exactly three i's, what is the probability the first bit is a 1?
## Homework Equations
combination formula nCp
## The Attempt at a Solution
1- I did:
P(exactly three 1's)= C(3,8)/ 2^8
2-
Here I am confused:
I did P( three 1's and 1 is first)=C(2,7)/(2^8)
OR
P(three 1's and 1 is first)=C(3,7)/(2^8)
I cannot see which one is right.
Can someone tell me what I did right and wrong?
Thank you
B | 203 | 667 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2022-21 | latest | en | 0.948128 |
https://learn.careers360.com/engineering/question-the-shortest-distance-between-the-line-and-the-curve-is-option-1-option-2-option-3-option-4-0-120110/ | 1,621,159,748,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992516.56/warc/CC-MAIN-20210516075201-20210516105201-00006.warc.gz | 378,440,235 | 110,980 | # The shortest distance between the line $x - y = 1$ and the curve $x^2 = 2y$ is : Option: 1 $\frac{1}{2}$ Option: 2 $\frac{1}{2\sqrt{2}}$ Option: 3 $\frac{1}{\sqrt{2}}$ Option: 4 0
Answers (1)
The shortest distance between curves is always along common normal.
the slope of the line
$\frac{dy}{dx}=1=\text{slope of the line}$
P is any point on the parabola, and also tangent pass through point P
slope of the tangent to the parabola
$\\2x=2\frac{dy}{dx}\\\frac{dy}{dx}=x=1\\\Rightarrow y=\frac{1}{2}$
$\text{Point P}=\left ( 1,\frac{1}{2} \right )$
$\therefore \text { Shortest distance }=\left|\frac{1-\frac{1}{2}-1}{\sqrt{1^{2}+1^{2}}}\right|=\frac{1}{2 \sqrt{2}}$
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Questions | 592 | 1,993 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-21 | latest | en | 0.704025 |
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# Help ASAP
0
60
3
Triangle $$ABC$$ is drawn so that the angle bisector of $$\angle{BAC}$$ meets $$\overline{BC}$$ at $$D$$ and so that triangle $$ABD$$ is an isosceles triangle with $$AB=AD$$. Line segment $$\overline{AD}$$ is extended past $$D$$ to $$E$$ so that triangle $$CDE$$ is isosceles with $$CD = CE$$ and $$\angle DBE = \angle DAB$$. Show that triangle $$AEC$$ is isosceles.
Jan 29, 2020
#1
+370
+1
There are 4 isosceles triangles: ABD, CDE, BCE, and AEC.
Δ ABD => 45° + 67.5° + 67.5° = 180°
Δ CDE => 45° + 67.5° + 67.5° = 180°
Δ BCE => 45° + 90° + 45° = 180°
Δ AEC => 45° + 67.5° + 67.5° = 180°
Triangle AEC is an isosceles triangle, because 2 of its 3 angles are equal, and as a result of that, 2 of its sides are equal as well.
Jan 29, 2020
edited by Dragan Jan 30, 2020
edited by Dragan Jan 30, 2020
#2
0
Sorry, but this doesn't explain why
Guest Jan 29, 2020
#3
+107524
0
Why doesn't it?
Melody Feb 5, 2020 | 429 | 968 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2020-10 | longest | en | 0.591398 |
http://www.physicsforums.com/showthread.php?t=211578 | 1,369,351,533,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704075359/warc/CC-MAIN-20130516113435-00016-ip-10-60-113-184.ec2.internal.warc.gz | 650,001,274 | 7,556 | ## From 1D to cylindrical co-ordinates in PDEs
Hi all,
I have another post on here relating to Fick's law of diffusion, but before I asked that I really should have started with this question:
How do you go from a one dimensional version of the diffusion equation to a cylindrical co-ordinate system of the same equation?
I have found both equations in many books but they just seem to skip from the one or three dimensional case in Cartesian co-ordinates to the cylindrical version without any intermediate steps.
I have attached the diffusion equation in pdf entitled "ficks2" and also the other part which is an excerpt from a paper "diffusion equation". The paper is solving the PDE for the case of two dimensions in cylindrical co-ordinates but I fail to see how they get equation (1) in the paper from the Cartesian equivalent.
The next part of my question would then be how do they solve this version of the diffusion equation. (i.e. how do they get from (1) to (3))
I have very limited experience in this field so the more simple the replies are the better!
Gareth
PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor
Quote by gareth How do you go from a one dimensional version of the diffusion equation to a cylindrical co-ordinate system of the same equation? I have found both equations in many books but they just seem to skip from the one or three dimensional case in Cartesian co-ordinates to the cylindrical version without any intermediate steps.
Hi,
The following book
http://www.amazon.com/First-Course-D...1594486&sr=1-2
has a very good explanations on basic cylindrical problems, it shows some of the algebraical steps on how to go from Cartesian to cylindrical coordinates in 3D. The book also may give you some tips and could help you solving the diffusion equation.
Quote by gareth The next part of my question would then be how do they solve this version of the diffusion equation.
My experience in this field is also very limited if you don't come further let me know I may have some notes on the diffusion equation
Hope it helps
Thank you and best regards
phioder
Thanks for the reply Phioder, I'll let you know how I get on. | 503 | 2,339 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2013-20 | latest | en | 0.953472 |
http://www.answers.com/Q/How_many_miles_is_it_from_earth_to_the_moon | 1,508,621,470,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824899.43/warc/CC-MAIN-20171021205648-20171021225648-00785.warc.gz | 390,729,315 | 45,988 | What would you like to do?
# How many miles is it from earth to the moon?
Would you like to merge this question into it?
#### already exists as an alternate of this question.
Would you like to make it the primary and merge this question into it?
#### exists and is an alternate of .
The average distance from the Earth to the Moon is 238,857 miles, about thirty times the diameter of the Earth.
Earth's diameter is about 8,000 miles.
Earth's circumference is about 25,000 miles.
Earth's distance from the moon is 1 second (in light years).
85 people found this useful
Thanks for the feedback!
# How many miles away is the moon from the earth?
238857 miles but it is moving away from earth at around 12cm a year
# How many miles from the earth to the moon at apogee and perigee?
the correct aproximate answer is 250,000 miles
# How many miles away is the moon from Earth?
its about 238,857 miles away so be prepared for a long journy if your planning on going!
# How many miles is the moon away from Earth?
approximetly 390,019, 410 miles away
# How many miles away from Earth is the moon?
the moon is about 230,000 miles away from the surface of the earth
# How many miles is there from the surface of the Earth all the way to the moon?
That varies considerably between apogee (furthest) and perigee (closest approach). The average is roughly 234 thousand miles. (I subtracted 4000 miles--earth's radius--from th
# How many miles are between the earth and the moon?
Our moons average distance from the Earth is 238,854 miles (384,399 km). However the Moon's orbit is elliptical and the distance therefore varies during its orbit though, fr
# How many miles distant is the moon from earth?
250,000 miles apart.
# How many million miles away is the moon to earth?
The Moon averages about 250,000 miles away. So about one-quarter of a million miles.
# How many miles from the surface of the earth to the surface of the moon?
the distance form the earth to the moon changes however At its closest point, the Moon gets to 363,104 km (225,622 miles), and at its furthest point, it's 405,696 km (252,088
# How many miles each moon circle around the earth?
If you are asking how far the Moon is from the Earth, the easy answer is about 250,000 miles or 400,000 kilometers. If you want to know how long is the orbit of the Moon (the
# How many million miles away is the moon from earth?
The moon is 225,622.37 miles (363,104 km)at its closest approach (perihelion) and 252,087.81 (405,696 km) miles when it is furthest from earth (aphelion)
# How many miles does the moon Orbit the Earth?
The Moon does not orbit in a perfect circle around the Earth, so an average distance of 381,550 km is usually given, with the extremes being 356,400 km (closest) and 406,700 k
# How many million miles from the earth to the moon?
It is a bit less than a quarter of a million miles from the Earth to the moon.
# How many miles from the moon two earth?
200,000,500,000 Miles
238857 miles | 750 | 3,010 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-43 | longest | en | 0.955715 |
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11-14-2012, 13:39 #1 Dinky Numpty Join Date: May 2003 Posts: 16,414 Laws 1Law of Mechanical Repair - After your hands become coated with grease, your nose will begin to itch and you'll have to pee. 2.Law of Gravity - Any tool, nut, bolt, screw, when dropped, will roll to the least accessible place in the universe. 3.Law of Probability - The probability of being watched is directly proportional to the stupidity of your act. 4.Law of Random Numbers - If you dial a wrong number, you never get a busy signal - and someone always answers. 6.Variation Law - If you change lines (or traffic lanes), the one you were in will always move faster than the one you are in now (works every time). 7.Law of the Bath - When the body is fully immersed in water, the telephone rings. 8.Law of Close Encounters - The probability of meeting someone you know INCREASES dramatically when you are with someone you don't want to be seen with. 9.Law of the Result - When you try to prove to someone that a machine won't work, IT WILL!!! 10.Law of Biomechanics - The severity of the itch is inversely proportional to the reach. 11..Law of the Theater & Hockey Arena - At any event, the people whose seats are furthest from the aisle, always arrive last. They are the ones who will leave their seats several times to go for food, beer, or the toilet and who leave early before the end of the performance or the game is over. The folks in the aisle seats come early, never move once, have long gangly legs or big bellies and stay to the bitter end of the performance. The aisle people also are very surly folk. 12.The Coffee Law - As soon as you sit down to a cup of hot coffee, your boss will ask you to do something which will last until the coffee is cold. 13.Murphy's Law of Lockers - If there are only 2 people in a locker room, they will have adjacent lockers. 14.Law of Physical Surfaces - The chances of an open-faced jelly sandwich landing face down on a floor, are directly correlated to the newness and cost of the carpet or rug. 15.Law of Logical Argument - Anything is possible IF you don't know what you are talking about. 16.Brown's Law of Physical Appearance - If the clothes fit, they're ugly. 17.Oliver's Law of Public Speaking--A CLOSED MOUTH GATHERS NO FEET!!! 18.Wilson's Law of Commercial Marketing Strategy - As soon as you find a product that you really like, they will stop making it. 19.Doctors' Law - If you don't feel well, make an appointment to go to the doctor, by the time you get there you'll feel better.. But don't make an appointment, and you'll stay sick.
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Contact Us - Glock Pistols - Glock Pistol | Glock Talk - Archive - Top | 840 | 3,410 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2015-18 | longest | en | 0.951539 |
https://hsc.co.in/mathematics-important-question-bank-for-kerala-board-2-hse-hsc-board-exam-2019/ | 1,725,981,149,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651255.81/warc/CC-MAIN-20240910125411-20240910155411-00181.warc.gz | 285,498,787 | 31,008 | # Mathematics – Important Question Bank for Kerala Board (+ 2) HSE (HSC) Board Exam 2019
HSC Board Exams are fast approaching and students are getting anxious about how to prepare for their HSC Board Exams. So we had mentioned some
After the tremendous success of our last year Important Questions Bank for Kerala Board (+2) Board Exam 2016 and 2017 and 2018 we have also created a list of Most Important Question Bank for Kerala Board (+2) HSE (HSC) Board Exam 2019 which are likely to appear in HSC Board Exams this year.
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##### Maths Commerce
1. If a line has the directions ratios -18,12 , -4, then what are its direction cosines?
2. Show that the points(2,3,4),(-1,-2,1),(5,8,7) are collinear
3. If , then find the value of x+y
4. Assume that each born child is equally like to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl. (ii) at least one is a girl?
5. Find the area of the region bounded by y^2=9x, x=2, x=4 and the y-axis in the first quadrant.
6. solve by product rule
7. Find the vector and the Cartesian equations of the lines that pass through the origin and (5,-2,3).
8. There are two bags , one of which contains 3 black and 4 white balls while the other contains 4 black and 3 white balls. A die is thrown at random. If it shows up 1 or 3, a ball is taken from the 1st bag but it shows up any other number, a ball is chosen from the second bag. Find the probability of choosing a black ball.
9. Determine the direction cosines of the normal to the plane and the distance from the origin when z=2.
10. Is the function f defined by f(x)= continuous at x=0 or x=1 or x=2 ?
Hi, we’re trying to collate and gather the data and would be updating it here a few days before the exam. Please keep on visiting our website for updates.
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Don’t forget to read : MUST REMEMBER THINGS on the day of Exam for HSC Students | 638 | 2,484 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-38 | latest | en | 0.903865 |
https://excellup.com/six_science/6scChapters/6-science-light-shadow-reflection.aspx | 1,561,587,068,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000575.75/warc/CC-MAIN-20190626214837-20190627000837-00494.warc.gz | 428,363,836 | 4,040 | Class 6 Science
Light: Light is a form of energy which helps us in seeing objects. When light falls on an object, some of the light gets reflected. The reflected light comes to our eyes and we are able to see an object.
Luminous Object: An object which produces light is called a luminous object, e.g. sun, bulb, etc.
Non-luminous object: An object which does not produce light is called non-luminous object, e.g. moon.
## Propagation of Light
###### Fig: Propagation of light
Light always travels in a straight line. This can be proved by a simple experiment. For this, take a cylindrical tube; which can be easily bent. Try to see a source of light; like a bulb or a candle; through the straight tube. Once the tube is bent at some angle, it is not possible to see the source of light through it. This happens because light travels in a straight line.
Transparent Object: An object which allows complete passage to light is called a transparent object. We can clearly see through a transparent object.
Translucent Object: An object which allows partial passage to light is called a translucent object. We can see through a translucent object but the vision would be faint.
Opaque Object: An object which does not allows passage to light is called an opaque object. We cannot see through an opaque object.
###### Fig: Transparent translucent
When light falls on an opaque object, a dark patch is formed on the other side of the object; if a screen is present on the other side. Three things are required for formation of shadow, viz. a source of light, an object and a screen.
The size of shadow depends on the distance of source of light and on the angle at which the light rays fall on the object.
If the source of light is closer to the object, a larger shadow is formed than when the source of light is far from the object.
If the angle of incident light is smaller, the shadow is longer. On the other hand, if the angle of incident light is bigger, the shadow is smaller. This explains, why our shadows are longer in the morning and evening and smaller in the noon.
## Structure of a pinhole camera:
###### Fig: Pinhole camera
A pinhole camera is simple device. One can get an inverted and real image in a pinhole camera.
A pinhole camera is composed of two boxes. One of the boxes snugly fits inside another box and can be slid to and fro. There is a pin-sized hole at the front of the camera and a translucent screen at the back of the camera. The distance between the pin-hole and screen can be adjusted to obtain a clear image of an object.
##### Mirror and Reflection:
###### Fig: Reflection of Light
When light falls on a shiny surface, the light rays bounce back. This phenomenon is called reflection of light. Mirror is a surface from which most of the incident light gets reflected.
Light: NCERT Solution
Light: MCQ Quiz | 616 | 2,857 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-26 | longest | en | 0.916366 |
https://scholaron.com/homework-answers/xa0-21-the-end-of-the-2102153 | 1,624,157,344,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487655418.58/warc/CC-MAIN-20210620024206-20210620054206-00578.warc.gz | 452,131,794 | 19,423 | /
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21. The end of the cutting cord a gas-powered weed
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Question : 21. The end of the cutting cord a gas-powered weed : 2102153
21. The end of the cutting cord on a gas-powered weed cutter is 0.15 m in length. If the motor rotates at the rate of 20 rev/s, what is the tangential speed of the end of the cord?
A. 628 m/s
B. 25 m/s
C. 19 m/s
D. 63 m/s
22. A bucket in an old well is hoisted upward by a rope which winds up on a cylinder having a radius of 0.050 m. How many rev/s must the cylinder turn if the bucket is raised at a speed of 0.15 m/s?
A. 3.0 rev/s
B. 1.5 rev/s
C. 0.48 rev/s
D. 0.24 rev/s
23. Consider a point on a bicycle wheel as the wheel makes exactly four complete revolutions about a fixed axis. Compare the linear and angular displacement of the point.
A. Both are zero.
B. Only the angular displacement is zero.
C. Only the linear displacement is zero.
D. Neither is zero.
24. Consider a point on a bicycle wheel as the wheel turns about a fixed axis, neither speeding up nor slowing down. Compare the linear and angular velocities of the point.
A. Both are constant.
B. Only the angular velocity is constant.
C. Only the linear velocity is constant.
D. Neither is constant.
25. Consider a point on a bicycle wheel as the wheel turns about a fixed axis, neither speeding up nor slowing down. Compare the linear and angular accelerations of the point.
A. Both are zero.
B. Only the angular acceleration is zero.
C. Only the linear acceleration is zero.
D. Neither is zero.
Solution 5 (1 Ratings )
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Physics 1 Year Ago 286 Views
This Question has Been Answered! | 433 | 1,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2021-25 | longest | en | 0.861356 |
http://www.beatthegmat.com/mba/2009/10/18/all-about-exponents | 1,526,838,724,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863662.15/warc/CC-MAIN-20180520170536-20180520190536-00459.warc.gz | 349,119,566 | 14,411 | by on October 18th, 2009
An exponent is a fancy way of writing the value of a number multiplied several times (up to infinity) by itself. 3 x 3 x 3 x 3 can also be written as and can be expressed as “three raised to the fourth power” or “three to the fourth”. The large number (3) is called the base and the little number (in this case 4) is called the exponent. Where the exponent is an integer, exponentiation corresponds to repeated multiplication, whereas multiplication corresponds to repeated addition.
There are several properties and rules to keep in mind with exponents:
### Multiplying numbers with the same base (Product Rule for exponents)
When multiplying numbers with the same base, all you do is add their exponents.
### Dividing numbers with the same base (Quotient Rule for exponents)
When dividing numbers that have the same base, subtract the bottom number’s exponent from the top number’s exponent:
### Raising a power to a power
When you raise an exponent to another power, or raise a power to a power, you need to multiply the exponents:
### Distributing exponents
When you see several numbers within parentheses and an exponent outside of the parentheses, you need to make sure you distribute that exponent to all of the numbers within those parentheses:
### Adding/subtracting exponents with same base and exponent
You can add and subtract numbers with the same base and exponent, but they must have the same base and exponent:
### Other important exponent rules
Any number raised to the power of 0 is 1:
Any number raised to the power of 1 is itself:
Zero raised to any nonzero exponent equals zero:
The difference of squares (really get to know this, as the GMAT loves to test you on this). Whenever you see an equation that can be simplified into the difference of squares, definitely factor it, it will probably help you out tremendously:
If you raise a positive fraction that is less than 1 to a power, the fraction gets smaller:
If you raise a negative number to an odd power, the number gets smaller:
If you raise a negative number to an even power, the number becomes positive:
Any number to the negative power x is equal to the reciprocal of the same number to the positive power x:
### Example Question
A list contains 11 consecutive integers. What is the greatest integer on the list?
1) If x is the smallest integer on the list, then
2) If x is the smallest integer on the list, then
A) Statement 1 alone is sufficient but statement 2 alone is not sufficient to answer the question asked.
B) Statement 2 alone is sufficient but statement 1 alone is not sufficient to answer the question asked.
C) Both statements 1 and 2 together are sufficient to answer the question but neither statement is sufficient alone.
D) Each statement alone is sufficient to answer the question.
Solution
If we can determine the smallest integer on the list or a specific integer on the list when the list is written in increasing order, we can determine the greatest integer on the list.
1) Sufficient: We’re given one variable and one equation for the smallest integer on the list. That means we could solve for smallest integer and add 10 to find the greatest integer. If you don’t see this, consider:
Cubing both sides, x + 72 = 64. Then x + 72 = 64 and x = -8. Adding 10 to -8, the greatest integer is 2. Eliminate choices B, C and E.
2) Insufficient: If then and , so x could be -8 or 8.
There are two different possibilities for the smallest integer on the list, so there must be two different possibilities for the greatest integer on the list. Statement 2) is insufficient, leaving the correct answer choice as A.
Have a question on exponents? Please post them in the comment box below.
• Thanks for this.
I have seen a problem that came up with 0^0 and I have seen two different answers?
• David,
As the first link says (as well as the article), it is generally the best rule to follow that any number, including 0, raised to the zero, is 1. Don't get this confused with zero raised to any nonzero exponent is zero.
hope this helps.
• Can also perform operations (multiplication and division) on different bases that have same exponent. Perform the operation and keep the exponent.
4^2 * 2^2 = 8^2.
4^2 = 2 x 2 x 2 x 2
2^2 = 2 x 2
2x2x2x2x2x2 = 8 x 8 = 8^2
• what are applications of laws? i mean tell all about applications of laws.
• hi this doesnt help at all
• That was rude it may not help you, but it has helped some others like me that did not know certain ways like this
• Thanks this has helped me. | 1,053 | 4,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2018-22 | latest | en | 0.910102 |
https://www.bestpfe.com/mathematical-preliminaries-for-isogeny-based-cryptography-2/ | 1,653,386,114,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662570051.62/warc/CC-MAIN-20220524075341-20220524105341-00219.warc.gz | 740,798,744 | 18,554 | # Mathematical preliminaries for isogeny-based cryptography
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## Mathematical preliminaries for isogeny-based cryptography
The building block for isogeny-based cryptography is elliptic curves. An elliptic curve is a smooth curve of genus one with a distinguished rational point. To un-derstand this definition we start by recalling notions about quadratic fields and ideal class groups. We then study affine and projective plane curves, along with the notions of non-singularity, dimension, function fields, divisors and genus. Having gathered the tools to properly define elliptic curves, we turn to their properties: an elliptic curve is an algebraic variety but also an algebraic group. We define the addition in this group using divisors, then scalar multiplication and torsion subgroups. Finally, we introduce the invariant differential.
After studying properties of elliptic curves we focus on morphisms between such curves, namely isogenies. We define the notions of separable isogeny, dual and degree. We give V´elu’s formulae, which are used to compute isogenies in practice, as well as a concrete example. We conclude the section with an intro-duction to modular curves an their link to isogenies. We then focus on isogenies from one curve to itself, i.e. endomorphisms, and recall how the endomorphism ring of ordinary and supersingular curves differs. We describe the Deuring corre-spondence which links the world of isogenies with the world of fractional ideals, and introduce the action of the ideal class group on different subset of elliptic curves. Finally we introduce isogeny graphs to describe the structure of isoge-nies linking curves from a given set. With graphical example we highlight the differences that arise depending on the endomorphism ring properties.
Quadratic imaginary order and class groups
We start by recalling notions of quadratic fields, orders, (integral) ideals, invert-ible and principal fractional ideals, gathering the tools to formally define the ideal class group itself.
A quadratic field is Q(√d) where d �= 0, 1 is a squarefree integer. If d < 0 then the field is called an imaginary quadratic field. The discriminant of Q(√d) is D = d if d ≡ 1 (mod 4) or D = 4d otherwise.
An order in a field k containing Q is a subring R of k that is finitely generated as a Z-module and is such that R ⊗Z Q = k. An order O is maximal if every order O� such that O ⊂ O� ⊂ k is such that O� = O. Any order of a quadratic field is contained in a unique maximal order (see [BV07] Theorem 8.1.4).
Proposition 1. Let O be an order of an imaginary quadratic field k = Q(√d).
The maximal order√is Ok = Z + ωZ, where ω = 12 (1 + d) if d is congruent to 1 modulo 4 or ω = d otherwise. If O is an order of a quadratic field k, then O is a submodule of the maximal order, and can be written as O = Z + fωZ, where f = [O : Ok] is called the conductor.
Proof. See [BV07] Proposition 7.2.6 and [Gal12] Section A.12..
Proposition 2. Let I be an integral ideal of an order O = Z + fωZ in Q(√d), √ √ where ω = d if d �≡ 1 (mod 4) and ω =12 (1 + d) otherwise. We have I = c(aZ + (b + fω)Z), where a, b and c are integers such that c > 0, a > b � 0, and
• a divides b2 − d if d �≡ 1 (mod 4) ;
• a divides b(b + 1) − d−41 if d ≡ 1 (mod 4) .
Proof. See [BV07], Equation (8.8) and Proposition 8.4.5, with q = 1, and √ Δ =√ f −d.
Recall that the norm of an ideal I is defined as N(I) = |O/I| . The norm is multiplicative, i.e. N(IJ) = N(I)N(J) . An ideal I strictly included in a ring R is said to be a prime ideal if for every element a and b in R such that ab belongs to I, then a or b belongs to I. Using the link between integral ideals and integral binary quadratic forms, it is possible to show that every ideal whose norm is coprime to the conductor has unique factorization into a product of invertible prime ideals (see [BV07] Theorem 8.6.8).
Example 1. The order Ok = Z+ (1+ 2 −3) Z is a maximal order of the quadratic field k = Q(√ −3). The suborder O = Z + 5√ Z has conductor 10. Set −3 √ ω = (1+ 2 −3) . The ideal I = 5(21Z + (5 − ω)Z) is an ideal of Ok, with norm 525. It is the product of three prime ideals, namely 3Z + (1 − 2ω)Z with norm 3, 7Z + (4 + 2ω)Z with norm 7, and 5Z with norm 25.
Fractional ideals
From integral ideals we move on to fractional ideals of an order O. Since frac-tional ideals are not ideals, we will reserve the term ideal for integral ideals in order to avoid confusion. Fractional ideals will always be named as such.
A fractional ideal of an order O in a field k is a subset a of k such that aa is an (integral) ideal O of k for some positive integer a ∈ Z. A fractional ideal of an order O in a quadratic field k is said to be principal if it can be written as a = αO for some α in k. It is said to be invertible if there exist a fractional ideal b such that ab = O .
In the maximal order of a quadratic field, all nonzero fractional ideals are invertible. Moreover every principal fractional ideal is invertible (see [BV07] Corollary 8.4.15, with the second point following from the definition).
The set of fractional ideal forms an abelian semigroup under multiplication (with usual multiplication of ideals). The set of invertible fractional ideals I(O) is a subgroup, in which the set of principle ideals P r(O) is a normal subgroup (see [BV07] Proposition 8.4.10 and Theorem 8.4.13).
Example 2. Consider the quadratic imaginary field k = Q(√ −3) and its max-imal order Ok = Z + √−3Z. Consider the ideal I = 6Z + (3 + √−3)Z of Ok.
Taking α = 2 + 34 −3 ∈ Q( −3)∗, αOk and αI are fractional ideals (and not an integral ideal) of Ok.
### Ideal class group
Definition 4. Let O be an order of an imaginary quadratic field. The ideal class group is Cl(O) = I(O)/P r(O) .
Intuitively, this means that we will consider equivalence classes of invertible fractional ideals up to multiplication by an non-zero element of Q(√d). For example, the fractional ideals I and αI belong to the same class.
Proposition 3. The order of the ideal class group of Ok asymptotically satisfies log(# Cl(O)) ∼ log |d| .
Proof. The is a special case of the Brauer–Siegel theorem (see [Lan94]). See also [BV07] Theorem 9.3.10.
### Algebraic plane curves
This section aims to gather all the elements needed to define elliptic curves, namely the notion of affine and projective spaces, plane curves, dimensions, smoothness and genus. See [Gal12] for references of the results in this section.
Affine plane curves
Let K be a perfect field. The affine 2-space over K is the plane A2(K) = {(x, y) : x, y ∈ K}. The set of K-rational points of A2 is the set A2(K) = {(x, y) ∈ A2 : x, y ∈ K}.
An affine plane curve is defined by a single polynomial equation f(x, y). An algebraic plane curve C is defined over K if its defining polynomial is defined over K . We denote this by C/K. If C/K is a curve defined by f(x, y) = 0 with K K �/ K is an extension, then C( K �) = { (x, y) ∈ f a polynomial in ¯[x, y], and A2(K�) : f(x, y) = 0}.
Projective plane curves
Let K be a perfect field. The projective 2-space (over K ), denoted by P 2 or 2 3 ¯ such that at least one parameter is P (K) is the set of all triplets (X, Y, Z) ∈ A nonzero, modulo the equivalence relation (X, Y, Z) � (X�, Y �, Z�) if there exists a λ ¯∗ such that X = λX�, Y = λY �, Z = λZ�. ∈ K ¯∗ is denoted by (X : Y : Z), and An equivalence class (λX, λY, λZ) : λ ∈ K the individual X, Y, Z are called homogeneous coordinates for the corresponding point in P2 . The set of K-rational points in P2 is the set P2(K) = {(X : Y : Z) ∈ P2 : X,Y,Z ∈ K}.
A projective plane curve is defined by a single homogeneous polynomial equation f(X, Y, Z) = 0.1 It is defined over K if its generating polynomial is defined over K . We denote this by C/K. If C/K is a projective plane curve defined by f(X, Y, Z) = 0 with f a homogeneous polynomial in ¯[X, Y, Z], and
K K�/K is an extension, then C(K�) = {(X : Y : Z) ∈ P2(K�) : f(X, Y, Z) = 0}.
Example 3. We start with an example of an affine plane curve. Let f = x3 +7x+21−y 2 ∈ K[x, y]. Then f defines an affine plane curve C/K = {(x, y) ∈ A2 : x3 + 7x + 21 − y2 = 0}. The polynomial F = X3 + 7XZ 2 + 21Z3 − Y 2Z = Z3f(X/Z, Y /Z) ∈ K[X, Y, Z] is homogeneous, and defines a projective closure of C in P2.
Function field
Affine case
Definition 5. Let C be a affine algebraic plane curve defined over K generated by a polynomial f. The coordinate ring of C over K is K[C] = K[x, y]/(f). The function field is K(C) = {f1/f2 : f1, f2 ∈ K[C], f2 ∈/ (f)}S with the equivalence relation f1/f2 ≡ f3/f4 if and only if f1f4 − f2f3 ∈ (f), the ideal of K[C] generated by f.
In other words, K(C) is the field of fractions of the affine coordinate ring K[C] over K. Elements of K(C) are called rational functions. For a ∈ K the rational function f : V → k given by f(P ) = a is called a constant function.1 ¯ d f(X, Y, Z)
A polynomial f ∈ K[X, Y, Z] is homogeneous of degree d if f(λX, λY, λZ) = λ for all λ ¯. ∈ K
READ Generalities on first order ordinary differential equations
Projective case
Definition 6. Let C be a projective algebraic set defined over K. The homoge-nous coordinate ring of C over K is K[C] = K[X, Y, Z]/f. The function field is K(C) = {f1/f2 : f1, f2 ∈ K[C] homogeneous of the same degree , f2 ∈/ (f)} with the equivalence relation f1/f2 ≡ f3/f4 if and only if f1f4 − f2f3 ∈ (f).
In other words, K(C) is the field of fractions of the projective coordinate ring K[C] over K. Elements of K(C) are called rational functions.
#### Smooth algebraic plane curves
We study the regularity, or smoothness, of a curve. As an introduction to this notion, we provide two examples in figures 2.1 and 2.2. The first former is regular whilst the latter presents a singularity at the origin. We then formally define these two notions.
Projective case
Let C be a projective plane curve, let P ∈ C , and choose A2 ⊂ P2 with P ∈ A2. Then C is nonsingular (or smooth) at P if C ∩ A 2 is nonsingular at P . If C is nonsingular at every point, then we say that C is nonsingular.
Example 5. The curve C = {(X : Y : Z) ∈ P2 : X3 + 7XZ2 + 21Z3 −Y 2Z = 0 is a smooth projective plane curve. Indeed let P ∈ C different from the point at the infinity. Then taking an affine plane S with Z �= 0, C ∩ S = {(x, y) ∈ A2 : x3 + 7x + 21 − y 2 = 0), which is smooth. For the point at the infinity, we choose another affine plane S with Y �= 0 and proceed similarly.
Morphisms of plane curves
Let C be a plane curve and let f ∈ K(C). Then f is defined or regular at P if it can be written as f1/f2 with f2(P ) �= 0 with1f, f2 ∈ K[C].
Definition 7. Let C and C� be two plane curves defined over K. A rational map ϕ : C → C� over K which is regular at every point P ∈ C(K) is called a morphism. If there exists a morphism ψ : C� → C over K such that ϕ ◦ ψ and ψ ◦ ϕ are the identity on C� and C, respectively, then ϕ is a plane curve isomorphism.
#### Divisor of a function
The divisor of a function is a formal sum representing its zeros and poles counted with multiplicities. The formal definition of a divisor requires the notion of the valuation of a function at a point. The valuation carries the information of the behaviour of the function at this point: if it is a zero (resp. a pole), the valuation is equal to its multiplicity (resp. minus its multiplicity); otherwise, the valuation is simply zero. To formally define divisors of functions, we first introduce the local ring of a variety and its maximal ideal.
Definition 8. Let C be a plane curve over K. The local ring over K of C at a point P ∈ C(K) is OP,K(x,y) = {f ∈ K(x, y) : f is regular at P }. We write mP,K(x,y) = {f ∈ OP,K(x,y) : f(P ) = 0} ⊆ OP,K(x,y) for the maximal ideal of OP,K(x,y). A uniformizer for C at P is any generator of mP,K(x,y).
Definition 9. Let K be a field. A discrete valuation on K is a function v : K∗ → Z such that:
1. for all f, g ∈ K∗, v(f g) = v(f) + v(g);
2. for all f, g ∈ K∗ such that f + g �= 0, v(f + g) ≥ min(v(f), v(g));
3. there is some f ∈ K∗ such that v(f) = 1 (equivalently, v is surjective to Z).
Let C be a plane curve over K and P ∈ C(K). Let mP = mP,K(C) be as in Definition 8 and define m0P = OP,K(X). Let f ∈ OP,K(X) be such that f �= 0. Then the function f → Pv(f) = max{m ∈ N : f ∈ mmP} defines a discrete valuation. We say that vP (f) is the order of f at P . If vP (f) = 1 then f has a simple zero at P .
For each point P on the curve, let gP and hP be functions in OP,K(X)� such that gP /hP = f. The divisor of f is Div(f) = P ∈C(K) vP (gP )(P ) − � P ∈C(K) vP (hP )(P ). The divisor of a function is also called a principal divisor.
We write Prin(C) = {Div(f) : f ∈ K(C)∗}.
Divisor class group
The divisor group of a curve C defined over K, denoted by Div(C), is the free abelian group generated by the points of C. Thus a divisor D ∈ Div(C) is a formal sum D = nP (P ) , P ∈C(K) where nP ∈ Z and nP = 0 for all but finitely many P ∈ C(K). The degree of D is defined by � deg D = nP . P ∈C(K)
We write Div0(C) = {D ∈ Div(C) : deg(D) = 0} .
A divisor D = P ∈C(K) nP (P ) is effective, denoted by D ≥ 0, if nP ≥ 0 for every P ∈ C. Similarly, for any two divisors D1, D2 ∈ Div(C), we write D1 ≥ D2 to indicate that D1 − D2 is effective.
Definition 10. Let C be a curve defined over K and let D = P∈C( ) nP (P ) K ∈ K K ∈ K be a divisor on C. For σ Gal( / ) define σ(D) = C( ) nP (σ(P )).
Then D is defined over K if σ(D) = D for all σ ∈ Gal( /K)�.We write DivK(C) K for the set of divisors on C that are defined over K.
Lemma 4. Prin(C) is a subgroup of DivK0(C).
Proof. See [Gal12] Chapter 7 Section 7 Lemma 7.7.6. K 0 class group 0 is Pic0(C)
The degree zero divisor of a curve C over Div (C)/ Prin(C). We call two divisors D1, D2 ∈ Div (C) linearly equivalent and write D1 ≡ D2 if D1 −D2 ∈ Prin(C). The equivalence class (called a divisor class) of a divisor D ∈ Div0(C) under linear equivalence is denoted [D].
1 Introduction
1.1 Landscape of cryptology
1.2 Computationally hard problems
1.3 Original Diffie–Hellman
1.4 Quantum revolution
1.5 Isogeny history
1.6 Problematic
1.7 Overview
I Preliminaries
2 Mathematical preliminaries for isogeny-based cryptography
2.1 Quadratic imaginary order and class groups
2.1.1 Quadratic fields, orders and ideals
2.1.2 Fractional ideals
2.1.3 Ideal class group
2.2 Algebraic plane curves
2.2.1 Affine plane curves
2.2.2 Projective plane curves
2.2.3 Function field
2.2.4 Smooth algebraic plane curves
2.2.5 Morphisms of plane curves
2.2.6 Divisor of a function
2.2.7 Divisor class group
2.2.8 Genus
2.3 Elliptic curves
2.3.1 Representation of elliptic curves
2.3.2 Algebraic group
2.3.3 Torsion
2.3.4 Invariant differential
2.4 Isogenies
2.4.1 Definitions
2.4.2 V´elu’s formulae
2.4.3 Example
2.4.4 Modular curves
2.5 Endomorphisms and curve classification
2.5.1 The endomorphism ring
2.5.2 Supersingular and ordinary cases
2.6 Deuring correspondence and the action of the ideal class group
2.6.1 Action of the ideal class group on elliptic curves
2.6.2 Deuring correspondence
2.7 Isogeny graphs
2.7.1 Ordinary case
2.7.2 Supersingular case over Fp
2.7.3 Supersingular over Fp2
3 Isogeny-based key exchange protocols
3.1 Ordinary case (CRS)
3.1.1 Security of the scheme and parameter sizes
3.1.2 Couveignes key exchange protocol
3.1.3 Rostovstev–Stolbunov key exchange protocol
3.1.4 Computation
3.2 Supersingular case over Fp2 (SIDH and SIKE)
3.2.1 Commutative diagram
3.2.2 SIDH key exchange protocol
3.2.3 Underlying security problems
3.2.4 From SIDH to SIKE
3.3 Supersingular case over Fp (CSIDH)
3.3.1 The ideal class group action
3.3.2 CSIDH key exchange protocol
3.3.3 Security of the scheme
3.3.4 Computation
3.4 Key validation
3.5 Comparison of CRS, SIDH, SIKE and CSIDH
II CSIDH implementation
4 Protecting CSIDH against side-channel attacks
4.1 Preliminaries: side-channel attacks
4.1.1 Timing attacks
4.1.2 Power consumption analysis
4.1.3 Fault injection
4.1.4 Constant-time and dummy-free algorithms
4.2 Previous constant-time implementations
4.2.1 Meyer–Campos–Reith
4.2.2 Onuki–Aikawa–Yamazaki–Takagi
4.3 Contribution: Fault-attack resistance
4.4 Contribution: Derandomized CSIDH
4.4.1 Flawed pseudorandom number generators
4.4.2 Derandomized CSIDH with dummies
4.4.3 Derandomized dummy-free CSIDH
4.5 Following constant-time implementations
III CSIDH generalization: higher-degree supersingular group actions
5 (d, �)-structures
5.1 Curves with a d-isogeny to their conjugate
5.1.1 Galois conjugates
5.1.2 (d, �)-structures
5.1.3 Isogenies of (d, �)-structures
5.1.4 Twisting
5.1.5 Involutions
5.1.6 Supersingular (d, �)-structures
5.1.7 Curves with non-integer d2-endomorphisms
5.2 Action on supersingular (d, �)-structures
5.2.1 Preliminaries on orientations
5.2.2 Action on primitive O-oriented curves
5.2.3 Natural orientation for supersingular (d, �)-structures
5.2.4 Link between natural and induced orientation
5.2.5 Free and transitive class group action
5.3 The (d, �)-supersingular isogeny graph
5.3.1 General structure
5.3.2 Examples
5.3.3 Involutions
5.3.4 Automorphism of order 3
5.4 Crossroads: curves with multiple (d, �)-structures
5.5 Map from (d, �)-structures to modular curves
5.6 Parametrization
5.6.1 Representing (2, �)-structures
5.6.2 Representing (3, �)-structures
5.6.3 Representing (5, �)-structures
5.6.4 Representing (7, �)-structures
6 HD CSIDH: Higher degree commutative supersingular Diffie–nHellman
6.1 HD CSIDH: Higher degree CSIDH
6.1.1 Hard problems
6.1.2 HD CSIDH
6.2 Practical computation
6.2.1 V´elu approach
6.2.2 Modular approach
6.3 Example
6.4 Public key compression
6.4.1 Key compression with modular curves
6.4.2 Key compression with parametrization
6.5 Public key validation
6.5.1 CSIDH versus HD CSIDH
6.5.2 Checking (d, �)-structures
6.5.3 Checking supersingularity: Sutherland’s algorithm
6.5.5 Determining the level
6.5.6 Validation algorithm for HD CSIDH
6.5.7 CSIDH and HD CSIDH validation comparison
IV Cryptanalysis
7 Cryptanalysis for SIDH
7.1 The Delfs–Galbraith algorithm
7.1.1 The general supersingular isogeny problem
7.2 Generalization
7.2.1 Generalized Delfs–Galbraith algorithm
7.2.2 Choosing the set D
7.2.3 Comparisons
7.3 Application to SIDH/SIKE cryptanalysis
7.3.1 Specific case: weak public keys in SIKEp434
7.3.2 General case: SIDH, shortcut
Perspectives
Bibliography
GET THE COMPLETE PROJECT | 5,732 | 18,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-21 | latest | en | 0.906106 |
http://en.wikipedia.org/wiki/Large_deviation_principle | 1,432,325,259,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207926620.50/warc/CC-MAIN-20150521113206-00239-ip-10-180-206-219.ec2.internal.warc.gz | 77,889,136 | 13,480 | # Rate function
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In mathematics — specifically, in large deviations theory — a rate function is a function used to quantify the probabilities of rare events. It is required to have several properties which assist in the formulation of the large deviation principle.[clarification needed] In some sense, the large deviation principle is an analogue of weak convergence of probability measures, but one which takes account of how well the rare events behave.
A rate function is also called[who?] a Cramér function, after the Swedish probabilist Harald Cramér.
## Definitions
An extended real-valued function I : X → [0, +∞] defined on a Hausdorff topological space X is said to be a rate function if it is not identically +∞ and is lower semi-continuous, i.e. all the sub-level sets
$\{ x \in X | I(x) \leq c \} \mbox{ for } c \geq 0$
are closed in X. If, furthermore, they are compact, then I is said to be a good rate function.
A family of probability measures (μδ)δ>0 on X is said to satisfy the large deviation principle with rate function I : X → [0, +∞) (and rate 1 ⁄ δ) if, for every closed set F ⊆ X and every open set G ⊆ X,
$\limsup_{\delta \downarrow 0} \delta \log \mu_{\delta} (F) \leq - \inf_{x \in F} I(x), \quad \mbox{(U)}$
$\liminf_{\delta \downarrow 0} \delta \log \mu_{\delta} (G) \geq - \inf_{x \in G} I(x). \quad \mbox{(L)}$
If the upper bound (U) holds only for compact (instead of closed) sets F, then (μδ)δ>0 is said to satisfy the weak large deviation principle (with rate 1 ⁄ δ and weak rate function I).
### Remarks
The role of the open and closed sets in the large deviation principle is similar to their role in the weak convergence of probability measures: recall that (μδ)δ>0 is said to converge weakly to μ if, for every closed set F ⊆ X and every open set G ⊆ X,
$\limsup_{\delta \downarrow 0} \mu_{\delta} (F) \leq \mu(F),$
$\liminf_{\delta \downarrow 0} \mu_{\delta} (G) \geq \mu(G).$
There is some variation in the nomenclature used in the literature: for example, den Hollander (2000) uses simply "rate function" where this article — following Dembo & Zeitouni (1998) — uses "good rate function", and "weak rate function". Fortunately, regardless of the nomenclature used for rate functions, examination of whether the upper bound inequality (U) is supposed to hold for closed or compact sets tells one whether the large deviation principle in use is strong or weak.
## Properties
### Uniqueness
A natural question to ask, given the somewhat abstract setting of the general framework above, is whether the rate function is unique. This turns out to be the case: given a sequence of probability measures (μδ)δ>0 on X satisfying the large deviation principle for two rate functions I and J, it follows that I(x) = J(x) for all x ∈ X.
### Exponential tightness
It is possible to convert a weak large deviation principle into a strong one if the measures converge sufficiently quickly. If the upper bound holds for compact sets F and the sequence of measures (μδ)δ>0 is exponentially tight, then the upper bound also holds for closed sets F. In other words, exponential tightness enables one to convert a weak large deviation principle into a strong one.
### Continuity
Naïvely, one might try to replace the two inequalities (U) and (L) by the single requirement that, for all Borel sets S ⊆ X,
$\lim_{\delta \downarrow 0} \delta \log \mu_{\delta} (S) = - \inf_{x \in S} I(x). \quad \mbox{(E)}$
Unfortunately, the equality (E) is far too restrictive, since many interesting examples satisfy (U) and (L) but not (E). For example, the measure μδ might be non-atomic for all δ, so the equality (E) could hold for S = {x} only if I were identically +∞, which is not permitted in the definition. However, the inequalities (U) and (L) do imply the equality (E) for so-called I-continuous sets S ⊆ X, those for which
$I \big( \stackrel{\circ}{S} \big) = I \big( \bar{S} \big),$
where $\stackrel{\circ}{S}$ and $\bar{S}$ denote the interior and closure of S in X respectively. In many examples, many sets/events of interest are I-continuous. For example, if I is a continuous function, then all sets S such that
$S \subseteq \bar{\stackrel{\circ}{S}}$
are I-continuous; all open sets, for example, satisfy this containment.
### Transformation of large deviation principles
Given a large deviation principle on one space, it is often of interest to be able to construct a large deviation principle on another space. There are several results in this area:
## History and basic development
The notion of a rate function began with the Swedish mathematician Harald Cramér's study of a sequence of i.i.d. random variables (Zi)i∈ℕ at the time of the Great Depression. Namely, among some considerations of scaling, Cramér studied the behavior of the distribution of $X_n=\frac 1 n \sum_{i=1}^n Z_i$ as n→∞.[1] He found that the tails of the distribution of Xn decay exponentially as e(x) where the factor λ(x) in the exponent is the Legendre transform (a.k.a. the convex conjugate) of the cumulant-generating function $\Psi_Z(t)=\log \mathbb E e^{tZ}.$ For this reason this particular function λ(x) is sometimes called the Cramér function. The rate function defined above in this article is a broad generalization of this notion of Cramér's, defined more abstractly on a probability space, rather than the state space of a random variable.
## References
1. ^ Cramér, Harald (1938). "Sur un nouveau théorème-limite de la théorie des probabilités". Colloque consacré à la théorie des probabilités, Part 3, Actualités scientifiques et industrielles (in French) 731: 5–23. | 1,533 | 5,689 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 12, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2015-22 | latest | en | 0.872981 |
https://daitips.com/how-do-you-fail-a-grade/ | 1,656,142,880,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103034877.9/warc/CC-MAIN-20220625065404-20220625095404-00296.warc.gz | 241,096,775 | 15,173 | # How Do You Fail A Grade?
## How Do You Fail A Grade?
Normally a Fail grade means that you have attempted some or all of the assessment items, but not received a passing grade overall.
## What does it take to fail a grade?
To fail a grade a student usually must fail two or more core classes or fail the standardized test in their state. In some cases, the school may make social promotion or summer school available options. … To avoid this, they will often have to attend summer school to make up the class.
## Can you pass a grade with one F?
Originally Answered: Can you pass the 8th grade with one F? Varies by school, but junior high is “pass the grade” and one F will usually not hold you back. Starting in 9th, though, It’s usually “pass the class,” meaning you have to retake any class you fail.
## Is it possible to fail a grade?
Failing a grade is a possibility at an online school. However, the consequences are a little less dramatic than it would be the case of failing a grade at a traditional school. A student is not obliged to repeat the grade from the beginning, but they instead continue from the point they failed.
## Is 50% a fail grade?
Because in most cases, a grade is 50 is defined as non-passing performance. Because in most cases, a grade is 50 is defined as non-passing performance. A popular grading scale used in many school districts in the United States is a 10-point absolute scale, 90-100 = A, 80-89 = B, 70-79 = C, 60-69 = D, and 0-59 = F.
## Is 60 a passing grade?
In primary and secondary schools, a D is usually the lowest passing grade. However, there are some schools that consider a C the lowest passing grade, so the general standard is that anything below a 60% or 70% is failing, depending on the grading scale.
## Can I pass 9th grade with 2 F’s?
Can you pass 9th grade with 2 F’s? Typically, 9th & up you pass/ fail courses, not grades. You’ll have to retake those 3, plus whatever else you can fit in. It’s your school policy as to whether they will classify you as 9 or 10.
## Can I pass with an F in math?
C is anywhere between 70% and 79% D – this is still a passing grade, and it’s between 59% and 69% F – this is a failing grade.
## What grades can hold you back?
In the United States, grade retention can be used in kindergarten through to twelfth grade; however, students in grades seven through twelve are usually only retained in the specific failed subject due to each subject having its own specific classroom rather than staying in one classroom with all subjects taught for …
## Can I graduate high school with an F?
A student can score Ds in all subjects and still graduate high school in some states. … Some high schools won’t allow students with grades of F to graduate. Students who have failed a class must take it again during the summer. High school students in Arizona must pass 22 credits to graduate high school.
## Is 70 a passing grade?
C – this is a grade that rests right in the middle. C is anywhere between 70% and 79% D – this is still a passing grade, and it’s between 59% and 69% F – this is a failing grade.
## Can students get held back?
Yes, a school can retain or promote a student without parent or guardian approval. However, the district PPR policy approved by the district’s school board must provide an appeal process for parents who disagree with a principal’s promotion or retention decision for their student.
## Will my child have to repeat a grade?
Ideally, no. Repeating a grade―also known as “grade retention” ―has not been shown to help children learn. Children won’t outgrow learning and attention issues by repeating a grade. In fact, repeating a grade may contribute to long-term issues with low self-esteem, as well as emotional or social difficulties.
## Is 50% a pass?
It is a pass (P) and an average score in Australia.
P Pass 50 to 64
N Fail 49 and below
## What is 50 in a grade?
D+ 55% to 59% 57.5%
D 50% to 54% 52.5%
E 40% to 49% 45%
F 0% to 39% 20%
## Is 50 a pass in Ontario?
The following is the levels on the Ontario rubric, its meaning, and its corresponding letter/percentage grades: Level 4, beyond government standards (A; 80 percent and above) … Level 2, approaching government standards (C; 60–69 percent) Level 1, well below government standards (D; 50–59 percent)
## Is 60 A good grade on a test?
A B is 80% to 89% A C is 70% to 79% A D is 60% to 69% and finally an F is 59% and below – and it’s not a passing grade.
How to calculate test score.
D 63-66
D- 60-62
F Below 60
## What is the passing grade of 60 items?
60% to 69% earns a Merit. 50% to 59% is Pass.
67 – 69 D+
63 – 66 D
60 – 62 D-
< 60 F
## Can u repeat 9th grade?
Unsuccessful 9th grade transitions are correlated with repeating the grade, disconnecting from school, and dropping out. Approximately 22% of students repeat ninth grade (more than any other grade).
## How can I pass 9th grade?
1. Avoid Absences. As stated earlier, the material you’ll be learning in the 9th grade will play a big role in your advancement to graduation, so you need to be present in class. …
2. Set Goals. …
3. Take Notes. …
4. Use a Planner.
## What happens if you dont pass freshman year?
Usually, colleges will allow you to return after doing poorly for one year. … So, if you fail freshman year, you’re college will probably put you on academic probation as a first step. Then if you fail your third semester you’ll likely be suspended.
## How much does an F bring down your grade in a class?
How much does an F bring down your grade in a class?
D+ 1.3
D 1.0
D- 0.7
F 0.0
## What will an F do to my GPA?
The failing grade will NOT calculate in your GPA, but it will still show on your transcript. On your transcript, an “E” will show to the right of your failing grade to mark the course as “Excluded”. On your transcript, an “I” will show to the right of the second time you took the class, marking it as “Included”.
## What happens if you fail math in school?
A failing grade will likely hurt your GPA (unless you took the course pass/fail), which could jeopardize your financial aid. The failure will end up on your college transcripts and could hurt your chances of getting into graduate school or graduating when you originally planned to.
## Can a teacher hold you back a grade?
Legislation signed into law June 30 allows parents and students over 18 to decide for themselves whether they or their kids should repeat their 2020-21 grade. In other years, the decision to hold students back is made by school officials and teachers.
## Should I hold my 3rd grader back?
Current thinking on retention
Recent research shows that, for the most part, holding kids back a grade isn’t the best practice. The National Association of School Psychologists (NASP) reports that some kids do better in school the first year or two after being held back. But it also says that this effect doesn’t last.
## Can a child repeat Grade 1?
The earlier the repeat year is done, the better i.e. grade 1 or 2. Grade R repeats are generally only done if a child is delayed in language, has fine and gross motor delays or is emotionally very immature. Every child is unique in their needs and therefore each case should be treated carefully.
## What is the lowest GPA you can graduate high school with?
A 2.0 GPA meets high school graduation requirements but most colleges—and some trade schools—expect a high school GPA of at least 3.0 (B) or better.
## What is the minimum GPA to graduate?
2.0
At minimum, experts say, students must generally meet a GPA standard of 2.0, on a 4.0 scale, to graduate and remain eligible for federal financial aid. Institutional scholarships and program enrollment at many colleges often hinge on academic achievement above a C average.
## Can you graduate if you fail a class senior year?
If you’re failing a class before graduation, you won’t graduate. If you can’t adjust course and get a passing grade, or it’s too late, you’ll have to retake the class the following semester in order to get your degree (both for High School or College.)
## Is a 70 good?
When 70% is good enough
70-79% First Class Methodical, relevant and sustained critical engagement with the available literature/ sources
60-69% Upper 2nd Class (2.1) Engaging critically and consistently with the available literature/ sources, organised in a coherent and persuasive manner
## Is a 70 bad in college?
Is a 70 a bad grade in college? C is anywhere between 70% and 79% D – this is still a passing grade, and it’s between 59% and 69% F – this is a failing grade.
## Is 70 passing in college?
A passing percentage is 70% or higher. Each college has policies with respect to grades for transferred coursework. You should check with the college you plan to attend to determine its specific grade-posting policies. * All grades above a 69.5% are considered passing.
## Is it bad to be held back in school?
Holding a student back may create significantly more issues than solutions. Even though research says that student retention is ineffective and could even be harmful, some schools still believe that making children repeat grades is a good idea.
See more articles in category: Uncategorized | 2,224 | 9,219 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-27 | latest | en | 0.963575 |
https://math.stackexchange.com/questions/2445167/calculating-the-limit-of-an-integral/2445220 | 1,579,692,845,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250606975.49/warc/CC-MAIN-20200122101729-20200122130729-00309.warc.gz | 557,147,039 | 30,892 | # Calculating the limit of an integral.
I'm tasked with finding the following limit:
$$\lim_{n\to\infty} \int_0^1 \frac{(nx)^2}{(1+x^2)^n}dx$$
I think the right way to do this is to swap the integral and the limit to get the answer $0$, but I'm having trouble justifying this operation using e.g. monotone convergence or dominated convergence. Alternatively, I could find an upper bound on the integral which converges to zero.
• Are you sure it's actually $0$? From a computation, it looks like it actually increases as a function of $n$, and grows at least more quickly than $\log n$. – anomaly Sep 25 '17 at 22:27
Make the change of variables $x=y/\sqrt n$ to see the expression equals
$$\tag 1 \sqrt n \int_0^{\sqrt n} \frac{y^2}{(1+y^2/n)^n}\,dy.$$
Now $(1+u/n)^n \le e^u$ for $u\ge 0.$ Therefore $(1)$ is at least
$$\sqrt n \int_0^{\sqrt n} y^2e^{-y^2}\,dy.$$
The limit of this is $\infty,$ hence the limit in question is $\infty.$ | 298 | 942 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2020-05 | latest | en | 0.829749 |
https://es.mathworks.com/matlabcentral/cody/problems/564-how-to-subtract/solutions/1750844 | 1,610,733,515,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495936.3/warc/CC-MAIN-20210115164417-20210115194417-00412.warc.gz | 357,608,213 | 16,920 | Cody
# Problem 564. How to subtract?
Solution 1750844
Submitted on 14 Mar 2019 by Mark Sanderson
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
X='+68768686834554'; Y='+76574535435398'; Z_correct='-7805848600844'; assert(isequal(mysub(X,Y),Z_correct))
-7805848600844
2 Pass
X='+1'; Y='+2'; Z_correct ='-1'; assert(isequal(mysub(X,Y),Z_correct))
-1
3 Pass
X='+100'; Y='+20'; Z_correct ='+80'; assert(isequal(mysub(X,Y),Z_correct))
+80
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To add the vectors (x₁,y₁) and (x₂,y₂), we add the corresponding components from each vector: (x₁+x₂,y₁+y₂). Here's a concrete example: the sum of (2,4) and (1,5) is (2+1,4+5), which is (3,9). There's also a nice graphical way to add vectors, and the two ways will always result in the same vector.
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Algebra
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Salman Khan
02/07/2018
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This arithmetic lesson demonstrates more long division problems with remainders.[Arithmetic playlist: Lesson 26 of 38]
Subject:
Mathematics
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Lecture
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Author:
Salman Khan
12/15/2011
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You get the general idea of decimal is and what the digits in different places represent (place value). Now you're ready to do something with the decimals. Adding and subtracting is a good place to start. This will allow you to add your family's expenses to figure out if your little brother is laundering money (perhaps literally). Have fun! Common Core Standard: 5.NBT.B.7
Subject:
Numbers and Operations
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Lecture
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Author:
Khan
Salman
06/25/2014
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Remix and Share
Rating
You get the general idea of decimal is and what the digits in different places represent (place value). Now you're ready to do something with the decimals. Adding and subtracting is a good place to start. This will allow you to add your family's expenses to figure out if your little brother is laundering money (perhaps literally). Have fun! Common Core Standard: 5.NBT.B.7
Subject:
Numbers and Operations
Material Type:
Activity/Lab
Homework/Assignment
Lecture
Provider:
Provider Set:
Author:
Khan
Salman
06/25/2014
Conditions of Use:
Remix and Share
Rating
You get the general idea of decimal is and what the digits in different places represent (place value). Now you're ready to do something with the decimals. Adding and subtracting is a good place to start. This will allow you to add your family's expenses to figure out if your little brother is laundering money (perhaps literally). Have fun! Common Core Standard: 5.NBT.B.7
Subject:
Numbers and Operations
Material Type:
Activity/Lab
Homework/Assignment
Lecture
Provider:
Provider Set:
Author:
Khan
Salman
06/25/2014
Conditions of Use:
Remix and Share
Rating
The real world is seldom about whole numbers. If you precisely measure anything, you're likely to get a decimal. If you don't know how to multiply these decimals, then you won't be able to do all the powerful things that multiplication can do in the real world (figure out your commission as a robot possum salesperson, determining how much shag carpet you need for your secret lair, etc.). Common Core Standards: 5.NBT.B.5, 5.NBT.B.7
Subject:
Numbers and Operations
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Activity/Lab
Homework/Assignment
Lecture
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Provider Set:
Author:
Khan
Salman
06/25/2014
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Remix and Share
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The real world is seldom about whole numbers. If you precisely measure anything, you're likely to get a decimal. If you don't know how to multiply these decimals, then you won't be able to do all the powerful things that multiplication can do in the real world (figure out your commission as a robot possum salesperson, determining how much shag carpet you need for your secret lair, etc.). Common Core Standards: 5.NBT.B.5, 5.NBT.B.7
Subject:
Numbers and Operations
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Activity/Lab
Homework/Assignment
Lecture
Provider:
Provider Set:
Author:
Khan
Salman
06/25/2014
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Remix and Share
Rating
The real world is seldom about whole numbers. If you precisely measure anything, you're likely to get a decimal. If you don't know how to multiply these decimals, then you won't be able to do all the powerful things that multiplication can do in the real world (figure out your commission as a robot possum salesperson, determining how much shag carpet you need for your secret lair, etc.). Common Core Standards: 5.NBT.B.5, 5.NBT.B.7
Subject:
Numbers and Operations
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Activity/Lab
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Lecture
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Provider Set:
Author:
Khan
Salman
06/25/2014
Conditions of Use:
Remix and Share
Rating
The real world is seldom about whole numbers. If you precisely measure anything, you're likely to get a decimal. If you don't know how to multiply these decimals, then you won't be able to do all the powerful things that multiplication can do in the real world (figure out your commission as a robot possum salesperson, determining how much shag carpet you need for your secret lair, etc.). Common Core Standards: 5.NBT.B.5, 5.NBT.B.7
Subject:
Numbers and Operations
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Provider Set:
Author:
Khan
Salman
06/25/2014
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Remix and Share
Rating
The real world is seldom about whole numbers. If you precisely measure anything, you're likely to get a decimal. If you don't know how to multiply these decimals, then you won't be able to do all the powerful things that multiplication can do in the real world (figure out your commission as a robot possum salesperson, determining how much shag carpet you need for your secret lair, etc.). Common Core Standards: 5.NBT.B.5, 5.NBT.B.7
Subject:
Numbers and Operations
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Lecture
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Provider Set:
Author:
Khan
Salman
06/25/2014
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Remix and Share
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Anything you can do with whole numbers, you can do with decimals. Subtraction is no exception. In this tutorial, you'll get some good practice subtracting decimals! Common Core Standard: 5.NBT.B.7
Subject:
Numbers and Operations
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Activity/Lab
Homework/Assignment
Lecture
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Provider Set:
Author:
Khan
Salman
06/25/2014
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Remix and Share
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Anything you can do with whole numbers, you can do with decimals. Subtraction is no exception. In this tutorial, you'll get some good practice subtracting decimals! Common Core Standard: 5.NBT.B.7
Subject:
Numbers and Operations
Material Type:
Activity/Lab
Homework/Assignment
Lecture
Provider:
Provider Set:
Author:
Khan
Salman
06/25/2014
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Remix and Share
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This lesson explains leverage and insolvency and why it is good or bad. [Banking, Money, Finance playlist: Lesson 10 of 24]
Subject: | 1,619 | 6,457 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2020-24 | latest | en | 0.873153 |
https://www.powershow.com/view2a/4904d0-ZDkzN/If_playing_SOL-opoly_on_the_gameboard_students_select_a_playing_piece_If_playing_on_an_overhead_each_team_selects_a_color_or_transparency_game_piece_for_play_Roll_the_die_to_determine_who_goes_first_First_player_rolls_the_die_and_moves_that_many_powerpoint_ppt_presentation | 1,550,264,322,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247479159.2/warc/CC-MAIN-20190215204316-20190215230316-00630.warc.gz | 942,328,784 | 20,914 | # If playing SOL-opoly on the gameboard, students select a playing piece. If playing on an overhead each team selects a color or transparency game piece for play. Roll the die to determine who goes first. First player rolls the die and moves that many - PowerPoint PPT Presentation
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## If playing SOL-opoly on the gameboard, students select a playing piece. If playing on an overhead each team selects a color or transparency game piece for play. Roll the die to determine who goes first. First player rolls the die and moves that many
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### If playing SOL-opoly on the gameboard, students select a playing piece. If playing on an overhead each team selects a color or transparency game piece for play. – PowerPoint PPT presentation
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Title: If playing SOL-opoly on the gameboard, students select a playing piece. If playing on an overhead each team selects a color or transparency game piece for play. Roll the die to determine who goes first. First player rolls the die and moves that many
1
United States History to 1877 SOL-opoly
If playing SOL-opoly on the gameboard, students
select a playing piece. If playing on an overhead
each team selects a color or transparency game
piece for play. Roll the die to determine who
goes first. First player rolls the die and moves
that many spaces. The card for that property is
worth a particular number of points. The teacher
asks the student a question and if the response
is correct the student receives that many points
and the card for that property. If they are
they have received the card for that property,
each time someone now lands on that space they
are asked a question and if they are correct they
receive the amount of points that property is
worth but if they miss that question the property
owner has the opportunity to respond and if
correct receives the points. When a player lands
on Community Chest or Chance space they draw the
corresponding card and respond accordingly. Each
time a player passes United States History to
1877 SOLopoly they receive 50 points. Play
continues until all properties are owned or until
someone has more than 350 points.
2
(No Transcript)
3
LEWIS CLARK HIGHWAY
REDCOAT ALLEY
PATRIOT PLACE
CIVICS
4
(No Transcript)
5
RECONSTRUCTION RD
GREAT PLAINS PLACE
GEOGRAPHY
6
(No Transcript)
7
HISTORY
CONSTITUTION CUL DE SAC
FEDERAL FREEWAY
BATTLE BOULEVARD
8
(No Transcript)
9
DOUGLAS EXPRESSWAY
10
(No Transcript)
11
(No Transcript)
12
(No Transcript)
13
(No Transcript)
14
(No Transcript)
15
COMMUNITY CHEST COMMUNITY CHEST COMMUNITY CHEST
COMMUNITY CHEST COMMUNITY CHEST COMMUNITY CHEST
COMMUNITY CHEST COMMUNITY CHEST COMMUNITY CHEST
COMMUNITY CHEST COMMUNITY CHEST COMMUNITY CHEST
Maps are wrong move back one space
Rough seas lose one turn
spaces
Colonies have no representation in Parliament
lose a turn
British surrender at Yorktown move ahead 2
Ghana, Mali, Songhai become powerful through
British increase tax move back 2
Declaration of Independence approved move ahead 4
spaces
Thomas Paine authors Common Sense take an extra
turn
Boston Massacre occurs go back 4 spaces
16
COMMUNITY CHEST COMMUNITY CHEST COMMUNITY CHEST
COMMUNITY CHEST COMMUNITY CHEST COMMUNITY CHEST
COMMUNITY CHEST COMMUNITY CHEST COMMUNITY CHEST
COMMUNITY CHEST COMMUNITY CHEST COMMUNITY CHEST
Articles of Confederation provide a weak national
government go back 2 spaces
James Madison authors the Bill of Rights move
First Continental Congress meets go ahead 2 spaces
Disagree with England with how to govern lose a
turn
Eli Whitney invents the cotton gin take an extra
turn
Women denied right to vote go back 3 spaces
The North and South are divided over many issues
lose a turn
The steamboat is improved by Robert Fulton go
George Washington provides leadership take an
extra turn
Carpetbaggers take advantage of the South move
back one space
Frederick Douglas escapes to the North move ahead
one space
The South secedes from the Union go back 4 spaces
17
CHANCE CARD CHANCE CARD CHANCE CARD
CHANCE CARD CHANCE CARD CHANCE CARD
CHANCE CARD CHANCE CARD CHANCE CARD
CHANCE CARD CHANCE CARD CHANCE CARD
14th Amendment grants citizenship to all persons
born in the US move ahead one space
African America soldiers discriminated against
lose a turn
Disease a major killer during Civil War move back
one space
Jefferson arranges for Louisiana Purchase take an
extra turn
Families and friends pitted against each other
during Civil War move back 3 spaces
Stamp Act imposed on colonists lose a turn
Phillis Wheatley writes poems and plays
supporting American Independence go ahead 4
spaces
Proclamation of 1763 hampers western movement of
settlers go back 4 spaces
13th Amendment bas slavery in the US move ahead 3
spaces
Exploration creates an exchange of goods and
ideas take an extra turn
Kwakiutl inhabit the Pacific Northwest Coast take
an extra turn
Explorers have a lack of adequate supplies lose a
turn
18
CHANCE CARD CHANCE CARD CHANCE CARD
CHANCE CARD CHANCE CARD CHANCE CARD
CHANCE CARD CHANCE CARD CHANCE CARD
CHANCE CARD CHANCE CARD CHANCE CARD
George Washington becomes Commander of the
Continental Army move ahead one space
England taxes colonies to help maintain troops
lose a turn
Colonial governors appointed by the King move
back one space
Iroquois inhabit northeast North America, Eastern
Lees surrender to Grant at Appomattox ends the
Civil War take an extra turn
Slaves captured in Africa and sold to slave
The Spanish bring European diseases to American
Indians move back 3 spaces
Indentured servants have no money for passage to
the colonies lose a turn
Pennsylvania Colony settled by Quakers take an
extra turn
Robert Smalls honored as a hero move ahead four
spaces
Clara Barton creates the American Red Cross move
Slaves owned as property for life with no rights
go back four spaces
19
COMMUNITY CHEST
WHEATLEY WAY
JEFFERSON PLACE
WASHINGTON WALK
FEDERAL FREEWAY
CONSTITUTION CUL DE SAC
HISTORY
DOUGLAS EXPRESSWAY
BATTLE BLVD
US History to 1877
COLONY COURT
LINCOLN LANE
CHANCE CARDS
CHANCE CARDS
Monopoly
CIVIL WAR CIRCLE
LOUISIANA PURCHASE | 1,600 | 6,577 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-09 | latest | en | 0.890125 |
https://dml.cz/handle/10338.dmlcz/137479 | 1,675,818,037,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500664.85/warc/CC-MAIN-20230207233330-20230208023330-00005.warc.gz | 219,641,767 | 5,819 | # Article
Full entry | PDF (0.2 MB)
Keywords:
Prime numbers; truncatable primes; integer expansions; square-free numbers
Summary:
We are interested whether there is a nonnegative integer $u_0$ and an infinite sequence of digits $u_1, u_2, u_3, \dots$ in base $b$ such that the numbers $u_0 b^n+u_1 b^{n-1}+\dots + u_{n-1} b +u_n,$ where $n=0,1,2, \dots ,$ are all prime or at least do not have prime divisors in a finite set of prime numbers $S.$ If any such sequence contains infinitely many elements divisible by at least one prime number $p \in S,$ then we call the set $S$ unavoidable with respect to $b$. It was proved earlier that unavoidable sets in base $b$ exist if $b \in \lbrace 2,3,4,6\rbrace ,$ and that no unavoidable set exists in base $b=5.$ Now, we prove that there are no unavoidable sets in base $b \geqslant 3$ if $b-1$ is not square-free. In particular, for $b=10,$ this implies that, for any finite set of prime numbers $\lbrace p_1, \dots , p_k\rbrace ,$ there is a nonnegative integer $u_0$ and $u_1, u_2, \dots \in \lbrace 0,1,\dots ,9\rbrace$ such that the number $u_0 10^n + u_1 10^{n-1}+\dots +u_{n}$ is not divisible by $p_1, \dots , p_k$ for each integer $n \geqslant 0.$
References:
[1] Angell I.O., Godwin H.J.: On truncatable primes. Math. Comp., 31 (1977), 265–267. DOI 10.1090/S0025-5718-1977-0427213-2 | MR 0427213 | Zbl 0347.10009
[2] Bugeaud Y., Dubickas A.: Fractional parts of powers and Sturmian words. C. R. Acad. Sci. Paris, Ser. I, 341 (2005), 69–74. MR 2153958 | Zbl 1140.11318
[3] Dubickas A.: Integer parts of powers of Pisot and Salem numbers. Archiv der Math., 79 (2002), 252–257. DOI 10.1007/s00013-002-8311-4 | MR 1944949 | Zbl 1004.11059
[4] Dubickas A.: On the distance from a rational power to the nearest integer. J. Number Theory, 117 (2006), 222–239. DOI 10.1016/j.jnt.2005.07.004 | MR 2204744 | Zbl 1097.11035
[5] Dubickas A. A. Novikas : Integer parts of powers of rational numbers. Math. Zeitschr., 251 (2005), 635–648. DOI 10.1007/s00209-005-0827-4 | MR 2190349
[6] Forman W. H.N. Shapiro: An arithmetic property of certain rational powers. Comm. Pure Appl. Math., 20 (1967), 561–573. DOI 10.1002/cpa.3160200305 | MR 0211977
[7] Guy R.K.: Unsolved problems in number theory. Springer–Verlag, New York, 1994. MR 1299330 | Zbl 0805.11001
[8] Kahan S. S. Weintraub: Left truncatable primes. J. Recreational Math., 29 (1998), 254–264.
[9] Koksma J.F.: Ein mengen-theoretischer Satz über Gleichverteilung modulo eins. Compositio Math., 2 (1935), 250–258. MR 1556918
[10] Weisstein E.W.: Truncatable prime. From MathWorld - A Wolfram Web Resourse, http://mathworld.wolfram.com/TruncatablePrime.html
[11] Zaimi T.: On integer and fractional parts powers of Salem numbers. Archiv der Math., 87 (2006), 124–128. DOI 10.1007/s00013-006-1560-x | MR 2250906
Partner of | 1,006 | 2,821 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-06 | latest | en | 0.654594 |
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## sgu112
maksyuki 发表于 oj 分类，标签:
ab-ba
You are given natural numbers a and b. Find ab-ba.
Input
Input contains numbers a and b (1≤a,b≤100).
## sgu105
maksyuki 发表于 oj 分类，标签:
Div 3
There is sequence 1, 12, 123, 1234, ..., 12345678910, ... . Given first N elements of that sequence. You must determine amount of numbers in it that are divisible by 3.
Input
Input contains N (1<=N<=231 - 1).
Output
Write >>继续阅读
## sgu104
maksyuki 发表于 oj 分类，标签:
Little shop of flowers
PROBLEM
You want to arrange the window of your flower shop in a most pleasant way. You have F bunches of flowers, each being of a different kind, and at least as many vases ordered in a row. The vases are glued onto >>继续阅读
## sgu102
maksyuki 发表于 oj 分类，标签:
Coprimes
For given integer N (1<=N<=104) find amount of positive numbers not greater than N that coprime with N. Let us call two positive integers (say, A and B, for example) coprime if (and only if) their greatest common divisor is 1. >>继续阅读
## sgu100
maksyuki 发表于 oj 分类，标签:
A+B
Read integers A and B from input file and write their sum in output file.
Input
Input file contains A and B (0<A,B<10001). | 397 | 1,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-18 | longest | en | 0.770433 |
http://math.stackexchange.com/questions/130400/multivariable-calculus-maxima-minima?answertab=active | 1,462,335,358,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860122501.26/warc/CC-MAIN-20160428161522-00070-ip-10-239-7-51.ec2.internal.warc.gz | 186,750,684 | 17,736 | # Multivariable calculus maxima minima
I have got a multivariable function $f$ defined on an open set $V$. Suppose $f$ attains maximum at some point $(x,y)$ inside of $V$. At this point we also have $f_{xx}=f_{yy}=0$. And finally the Laplacian of $f$ on $V$ is greater than or equal to $0$. What can we say? such a point can really exist?
-
We have $V \subseteq \mathbb R^2$, I suppose? Or are $x$ and $y$ multi-dimensional? In either case: Such a point can exist, we can take a constant $f$. – martini Apr 11 '12 at 11:37
For $P=(x_0,y_0)\in\mathbb{R}^2$, $f(x,y)=a[(x-x_0)^2+(y-y_0)^2]+b$ has an extremum (max for $a<0$) at $P_0$ but the Laplacian is $\nabla^2f=4a<0$. So I'd be tempted to say that it's not possible if $f$ is $C^2$ on $V$. – bgins Apr 11 '12 at 11:41
What kind of smoothness assumptions are we making about $f$? – bgins Apr 11 '12 at 12:12
A $C^2$-function $f$ with $\Delta f(x,y)\geq0$ for all $(x,y)$ is subharmonic, see here. A nonconstant subharmonic function cannot have a maximum at an interior point.
On $\mathbb{R}^n$ with sufficient smoothness assumptions, the second derivative test implies such a scenario is impossible. If we define $M=\det H$ where $H_{ij}=\frac{\partial^2 f}{\partial x_i~\partial x_i}$ is the Hessian matrix, then the test is conclusive as long as $M\ne0$. If $M$ is positive/negative definite (at an interior point of the domain), then $f$ has a local extremum (minimum/maximum). If there are eigenvalues of $H$ of each sign, then the point is a saddle point (the restriction of $f$ along eigenlines of opposite-sign eigenvalues has a minimum on one line but a maximum on the other). Otherwise ($M=0$ or there are complex eigenvalues), the test is inconclusive. In two dimensions, this means that $M=f_{xx}f_{yy}-f_{xy}^2 > 0$ is a necessary condition for a point to be a local extremum. But if that holds, then $f_{xx}f_{yy} > f_{xy}^2 \ge 0$, so that $f_{xx}$ & $f_{yy}$ share the same sign and so the sign of the Laplacian determines whether the point is a minimum or maximum (if $\nabla^2f$ is positive or negative respectively). But the Laplacian alone does not determine whether a point is a local extremum. You will find a brief discussion of the above here, with more under the article on Morse theory. | 703 | 2,266 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2016-18 | latest | en | 0.821212 |
https://www.physicsforums.com/threads/coefficient-of-friction.535043/ | 1,542,553,182,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744381.73/warc/CC-MAIN-20181118135147-20181118161147-00321.warc.gz | 966,811,676 | 12,289 | # Homework Help: Coefficient of friction
1. Sep 29, 2011
### nickthrop101
1. The problem statement, all variables and given/known data
An object is moved with a speed of 2m/s on one side of a 1m table to the other, it stops right at the other and of the table without falling off. Does it have wheels or not?
2. Relevant equations
The teacher gave this just to me and said we havnt learnt about it yet but wanted to challenge me. I think it has something to do with the coefficient of friction and kinetic friction
3. The attempt at a solution
i dont want to be given an answer but just the relevant equations and how to use them thanks :)
2. Sep 29, 2011
### BobG
You should be able to find the equation for coefficient of friction on the internet, plus approximate values for coefficient of friction for different surfaces/situations.
The equation is simple enough:
$$F_{fric} = \mu * F_n$$
mu is your coefficient of friction
F_n is your normal force (in this case, the force of gravity)
If you can figure out your acceleration, then you have another way to figure out your force of friction:
F=ma
Between those two equations, you should be able to solve for your coefficient of friction and check the internet to see if that's a reasonable value for wheels or no wheels. | 302 | 1,287 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2018-47 | latest | en | 0.924184 |
https://www.convert-measurement-units.com/convert+Kilowattyear+to+Megawattyear.php | 1,726,704,329,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651944.55/warc/CC-MAIN-20240918233405-20240919023405-00799.warc.gz | 670,312,786 | 14,130 | Convert kWyr to MWyr (Kilowattyear to Megawattyear)
## Kilowattyear into Megawattyear
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Kilowattyear+to+Megawattyear.php
# Convert kWyr to MWyr (Kilowattyear to Megawattyear)
1. Choose the right category from the selection list, in this case 'Energy'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Kilowattyear [kWyr]'.
4. Finally choose the unit you want the value to be converted to, in this case 'Megawattyear [MWyr]'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
### Utilize the full range of performance for this units calculator
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '43 Kilowattyear'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Kilowattyear' or 'kWyr'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Energy'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '34 kWyr to MWyr' or '67 kWyr into MWyr' or '1 Kilowattyear -> Megawattyear' or '67 kWyr = MWyr' or '34 Kilowattyear to MWyr' or '1 kWyr to Megawattyear' or '34 Kilowattyear into Megawattyear'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(67 * 34) kWyr'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '34 Kilowattyear + 1 Megawattyear' or '1mm x 67cm x 34dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 1.884 271 587 791 4×1020. For this form of presentation, the number will be segmented into an exponent, here 20, and the actual number, here 1.884 271 587 791 4. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 1.884 271 587 791 4E+20. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 188 427 158 779 140 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 939 | 3,855 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-38 | latest | en | 0.83841 |
http://openstudy.com/updates/5585fbdbe4b091b59af161d2 | 1,477,637,332,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721558.87/warc/CC-MAIN-20161020183841-00250-ip-10-171-6-4.ec2.internal.warc.gz | 198,267,880 | 26,324 | anonymous one year ago Using the completing-the-square method, find the vertex of the function f(x) = –2x2 + 12x + 5 and indicate whether it is a minimum or a maximum and at what point.
1. anonymous
it is a maximum because the leading coefficient|dw:1434844163162:dw| is negative so the parabola opens down
2. campbell_st
|dw:1434843957618:dw|
3. anonymous
Ok:)
4. anonymous
$f(x) = –2x^2 + 12x + 5$to find the vertex, the first coordinate of the vertex is always $-\frac{b}{2a}$ which in your case is $-\frac{12}{2\times (-2)}$
5. anonymous
the second coordinate of the vertex is what you get when you replace $$x$$ by the first coordinate
6. anonymous
3
7. dmndlife24
Graph opens downward so it would be a maximum point Vertex formula|dw:1434844392512:dw|
8. anonymous
yeah that is the first coordinate the second coordinate if $$f(3)$$
9. anonymous
Now what do I do?
10. anonymous
have a beer you are done
11. campbell_st
well completing the square $f(x) = (-2x^2 + 12x) + 5$ factor out -2 $f(x) = -2(x^2 - 6x) + 5$ now complete the square in x
12. anonymous
So is it maximum or minimum??
13. anonymous
, find the vertex of the function $(3,23)$ it is a max have a nice day
14. anonymous
Solve x2 + 12x + 6 = 0 using the completing-the-square method.
15. anonymous
Thanks friend. I also need help with that one
16. anonymous
|dw:1434844578090:dw|
17. campbell_st
so inside the brackets you need to add (-6/2)^2 so the function is $f(x) = -2(x^2 - 6x + 9) + 5 -18$ which becomes f(x) = -2(x -3)^2 - 13 now you have completed the square have the equation in vertex form and can answer the question using the required method
18. campbell_st
|dw:1434844393236:dw|
19. anonymous
$x^2 + 12x + 6 = 0 \\ x^2+12x=-6$ is a start
20. anonymous
hahahah you're funny @satellite73 !!!
21. anonymous
I see, I see
22. anonymous
|dw:1434844713212:dw|
23. anonymous
24. campbell_st
|dw:1434844498447:dw|
25. anonymous
$x^2+12x=-6$takes two steps to complete the square what is half of 12?
26. anonymous
6
27. anonymous
right, and what is $$6^2\_? 28. anonymous oops what is \(6^2$$?
29. anonymous
36
30. anonymous
so go right from $x^2+12x=-6$ to $(x+6)^2=-6+36$ or $(x+6)^2=30$
31. anonymous
then take the square root of both sides, don't forget the $$\pm$$ and get $x+6=\pm\sqrt{30}$
32. anonymous
subtract 6 and you are done
33. anonymous
so is it 6 plus or negative radical 30?
34. anonymous
no
35. anonymous
SUBTRACT 6 to solve for $$x$$
36. anonymous
|dw:1434844993983:dw|
37. anonymous
nope
38. anonymous
|dw:1434845032667:dw|
39. anonymous
|dw:1434845026665:dw|
40. anonymous
yeah that one | 867 | 2,659 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2016-44 | longest | en | 0.754967 |
http://chestofbooks.com/architecture/Building-Construction-V4/Proportiontng-A-Hot-Water-System.html | 1,519,183,126,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813322.19/warc/CC-MAIN-20180221024420-20180221044420-00195.warc.gz | 73,093,281 | 6,109 | 209. The principal object to be sought in designing a system of hot-water piping is to adjust and equalize the resistance in each circuit and branch, so that the hot water will flow with equal readiness to each radiator. This is accomplished by making the diameter of each pipe just sufficient to pass the desired amount of water under the head, or driving force, which is available in that particular part of the system. Artificial resistances are also introduced at some points by putting in extra elbows or bends, and valves are sometimes used for the same purpose.
The fall of temperature of the hot water, as it passes through a radiator, is usually estimated at about 20° for good practice, and 35° is regarded as the limit in any case.
## 210. Height Of Circuit
Height Of Circuit. The horizontal pipes on the upper floors of a building, and also the risers leading thereto, may be made smaller in diameter than those upon the lower floors, because the driving force which impels the water increases with the height of the circuits.
The proper size of a pipe having been determined for a given service on the first floor, the diameter for equal service on higher floors, the temperatures remaining the same, may be found by multiplying by the following factors:
Story 2d 3d 4th 5th Factors........ .87 .80 .76 .73
No factors are given for heights above the fifth floor, or about 50 feet, because the decrease for the succeeding stories is so small that it is of little practical account.
211. Conversely, the area of heating surface that may be properly supplied by a pipe of given diameter will increase as the circuit is made higher. If the area which is known to be right for a given size of pipe on the first floor be taken as 1, the areas on the upper floors will increase in the following order:
Story 2d 3d 4th 5th Proper area heating surface. . 1.40 1.70 1.98 2.20
## 212. Resistance Of Circuit
Resistance Of Circuit. The resistance to the flow, caused by elbows, tees, and other fittings, is considerable. The resistance in a common elbow, the ends of the pipes being left square, is about equal to the frictional resistance of a piece of straight pipe having a length equal to 100 times its diameter. If the ends of the pipes are beveled to an edge, the resistance may be reduced to 70 diameters, or even to 60 in small sizes. With a long bend having a radius of 5 diameters, the resistance falls to 10 diameters, or less.
A plain T offers about the same resistance as an elbow, and a return bend from 1 1/2 to 2 times as much. The gain made by reaming the ends of the pipe is much less in the large diameters than in the small sizes.
The actual length of a circuit is always understood to be the actual distance traveled by the water in going from, and returning to, the boiler. | 633 | 2,809 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2018-09 | latest | en | 0.945238 |
https://www.coursehero.com/file/5798577/32p-Nested-loops-printable-/ | 1,498,717,026,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323870.46/warc/CC-MAIN-20170629051817-20170629071817-00495.warc.gz | 860,435,166 | 33,449 | 32p. Nested loops _printable_
# 32p. Nested loops _printable_ - Nested Loops For an...
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1 ©2009 by L. Lagerstrom Nested Loops • For an introduction to the basic concepts, see the associated video clip • Nested loop structure • Nested loop results • Nested loop execution • Nested loop example • Finding the location of the max value in an array • The indmax function ©2009 by L. Lagerstrom The Nested Loop Structure Often it's useful to put one loop inside another loop. We call this a double loop, or a nested loop. The basic structure is: for m = 1:3 for n = 1:4 fprintf('Value of m is %.0f and value of n is %.0f \n', m, n) end end See if you can figure out what will be displayed when this code is run. (We give the results on the next slide.) ©2009 by L. Lagerstrom Nested Loop Results Our nested loop example: for m = 1:3 for n = 1:4 fprintf('Value of m is %.0f and value of n is %.0f \n', m, n) end end The displayed results will be: Value of m is 1 and value of n is 1 Value of m is 1 and value of n is 2 Value of m is 1 and value of n is 3 Value of m is 1 and value of n is 4 Value of m is 2 and value of n is 1 Value of m is 2 and value of n is 2 Value of m is 2 and value of n is 3 [etc., etc.] Value of m is 3 and value of n is 4 ©2009 by L. Lagerstrom Nested Loop Execution From the loop code, and/or from the displayed results on the previous slide, we see that the loops execute as follows: 1. Matlab sets m to a value of 1 and starts the outer loop. 2. Once inside the outer loop, Matlab finds the inner loop. So it sets n to 1 and starts the inner loop. 3. The inner loop executes completely, meaning, in this case, that it iterates 4 times, displaying the "Value of. .." message each time. The value of the inner loop variable (n) changes each time, but the value of m (from the outer loop) does not. 4. When the inner loop is done, Matlab proceeds and finds that it is at the end of the first iteration of the outer loop. So it returns to the beginning of the outer loop, sets m to 2, and starts the next iteration. During this second iteration of the outside loop, the inner loop will
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## This note was uploaded on 02/23/2010 for the course ENG 42325 taught by Professor Lagerstrom during the Spring '10 term at UC Davis.
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32p. Nested loops _printable_ - Nested Loops For an...
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Ask a homework question - tutors are online | 718 | 2,699 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2017-26 | longest | en | 0.865355 |
https://www.examveda.com/if-logx-9-1-then-x-is-equal-to-16-2-18890/ | 1,571,194,531,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986661296.12/warc/CC-MAIN-20191016014439-20191016041939-00066.warc.gz | 912,616,359 | 6,962 | Examveda
# If $${\log _x}\left( {\frac{9}{{16}}} \right) = - \frac{1}{2},$$ then x is equal to:
A. $$- \frac{3}{4}$$
B. $$\frac{3}{4}$$
C. $$\frac{{81}}{{256}}$$
D. $$\frac{{256}}{{81}}$$
\eqalign{ & {\log _x}\left( {{9 \over {16}}} \right) = - {1 \over 2} \cr & \Rightarrow {x^{ - {1 \over 2}}} = {9 \over {16}} \cr & \Rightarrow {1 \over {\sqrt x }} = {9 \over {16}} \cr & \Rightarrow \sqrt x = {{16} \over 9} \cr & \Rightarrow x = {\left( {{{16} \over 9}} \right)^2} \cr & \Rightarrow x = {{256} \over {81}} \cr} | 228 | 524 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2019-43 | latest | en | 0.361095 |
http://mathhelpforum.com/calculus/54778-example-limit-function.html | 1,524,505,851,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946120.31/warc/CC-MAIN-20180423164921-20180423184921-00008.warc.gz | 208,468,549 | 9,853 | Thread: Example of limit of a function
1. Example of limit of a function
Find an example of a function $\displaystyle f0,1) \rightarrow \mathbb {R}$ that is differentiable with $\displaystyle f'(x)$ not bounded, and that $\displaystyle \lim _{x \rightarrow 0^+ } f(x)$ do not exist.
Umm... I can't really think of any examples. Since $\displaystyle f(x)= \frac {1}{x}$ has a bounded derivative.
$\displaystyle f(x)= \frac {1}{x}$ has a bounded derivative.
3. the derivative is $\displaystyle -x^{(-2)} = - \frac {1}{x^2}$, yeah, it is not bounded below, hell. Thanks. | 166 | 571 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2018-17 | latest | en | 0.820435 |
https://www.webqc.org/balance.php?reaction=Zn%28s%29+%2B+H2SO4%28aq%29+%3D+ZnSO4%28aq%29+%2B+H2%28g%29 | 1,571,687,773,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987787444.85/warc/CC-MAIN-20191021194506-20191021222006-00551.warc.gz | 1,159,537,875 | 5,416 | #### Balance Chemical Equation - Online Balancer
Balanced equation:
Zn(s) + H2SO4(aq) = ZnSO4(aq) + H2(g)
Reaction type: single replacement
Reaction stoichiometry Limiting reagent
CompoundCoefficientMolar MassMolesWeight
Zn(s)165.38
H2SO4(aq)198.07848
ZnSO4(aq)1161.4426
H2(g)12.01588
Units: molar mass - g/mol, weight - g.
Direct link to this balanced equation:
Instructions on balancing chemical equations:
• Enter an equation of a chemical reaction and click 'Balance'. The answer will appear below
• Always use the upper case for the first character in the element name and the lower case for the second character. Examples: Fe, Au, Co, Br, C, O, N, F. Compare: Co - cobalt and CO - carbon monoxide
• To enter an electron into a chemical equation use {-} or e
• To enter an ion specify charge after the compound in curly brackets: {+3} or {3+} or {3}.
Example: Fe{3+} + I{-} = Fe{2+} + I2
• Substitute immutable groups in chemical compounds to avoid ambiguity.
For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced,
but PhC2H5 + O2 = PhOH + CO2 + H2O will
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https://www.jiskha.com/display.cgi?id=1160160540 | 1,503,285,579,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886107487.10/warc/CC-MAIN-20170821022354-20170821042354-00695.warc.gz | 913,557,882 | 3,956 | # Calculus
posted by .
A rectangle is bounded by the x-axis and the semicircle y=ã(25-x^2).
Question is, what length and width should the rectangle have so that its area is a maximum, and what is the maxuimum area?
Area= length*width
= 2x*y= 2x*sqrt(25-x^2)
Now, take that, differentiate it, set to zero, and solve for x,y. Length = 2x, width (or height) is y.
I will be happy to critique your work or thinking.
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More Similar Questions | 697 | 2,685 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2017-34 | latest | en | 0.920788 |
http://www.cr.ie.u-ryukyu.ac.jp/hg/Members/kono/Proof/ZF-in-agda/raw-rev/773e03dfd6ed | 1,606,543,458,000,000,000 | text/plain | crawl-data/CC-MAIN-2020-50/segments/1606141195069.35/warc/CC-MAIN-20201128040731-20201128070731-00256.warc.gz | 116,757,384 | 1,407 | # HG changeset patch # User Shinji KONO # Date 1592031550 -32400 # Node ID 773e03dfd6ede9a2092799a08e02413aab1f59d7 # Parent ef93c56ad3115a1ecb36197dde0848d6a4b4cd67 ... diff -r ef93c56ad311 -r 773e03dfd6ed filter.agda --- a/filter.agda Fri Jun 12 19:21:14 2020 +0900 +++ b/filter.agda Sat Jun 13 15:59:10 2020 +0900 @@ -33,32 +33,37 @@ A ＼ B = record { def = λ x → def A x ∧ ( ¬ ( def B x ) ) } +-- Kunen p.76 and p.53 record Filter ( L : OD ) : Set (suc n) where field filter : OD - ¬f∋∅ : ¬ ( filter ∋ od∅ ) - f∋L : filter ∋ L f⊆PL : filter ⊆ Power L filter1 : { p q : OD } → q ⊆ L → filter ∋ p → p ⊆ q → filter ∋ q filter2 : { p q : OD } → filter ∋ p → filter ∋ q → filter ∋ (p ∩ q) +open Filter + +proper-filter : {L : OD} → (P : Filter L ) → {p : OD} → Set n +proper-filter {L} P {p} = filter P ∋ L + +prime-filter : {L : OD} → Filter L → ∀ {p q : OD } → Set n +prime-filter {L} P {p} {q} = filter P ∋ ( p ∪ q) → ( filter P ∋ p ) ∨ ( filter P ∋ q ) + record Ideal ( L : OD ) : Set (suc n) where field ideal : OD - i∋∅ : ideal ∋ od∅ - ¬i∋L : ¬ ( ideal ∋ L ) i⊆PL : ideal ⊆ Power L ideal1 : { p q : OD } → q ⊆ L → ideal ∋ p → q ⊆ p → ideal ∋ q ideal2 : { p q : OD } → ideal ∋ p → ideal ∋ q → ideal ∋ (p ∪ q) -open Filter open Ideal -L-filter : {L : OD} → (P : Filter L ) → {p : OD} → filter P ∋ p → filter P ∋ L -L-filter {L} P {p} lt = {!!} -- filter1 P {p} {L} {!!} lt {!!} +proper-ideal : {L : OD} → (P : Ideal L ) → {p : OD} → Set n +proper-ideal {L} P {p} = ideal P ∋ od∅ -prime-filter : {L : OD} → Filter L → ∀ {p q : OD } → Set n -prime-filter {L} P {p} {q} = filter P ∋ ( p ∪ q) → ( filter P ∋ p ) ∨ ( filter P ∋ q ) +prime-ideal : {L : OD} → Ideal L → ∀ {p q : OD } → Set n +prime-ideal {L} P {p} {q} = ideal P ∋ ( p ∩ q) → ( ideal P ∋ p ) ∨ ( ideal P ∋ q ) + ultra-filter : {L : OD} → Filter L → ∀ {p : OD } → Set n ultra-filter {L} P {p} = L ∋ p → ( filter P ∋ p ) ∨ ( filter P ∋ ( L ＼ p) ) | 818 | 1,905 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-50 | latest | en | 0.502738 |
https://rdrr.io/cran/smint/man/branin.html | 1,579,722,058,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250607407.48/warc/CC-MAIN-20200122191620-20200122220620-00130.warc.gz | 624,493,870 | 14,574 | # branin: Branin-Hoo 2-dimensional test function In smint: Smooth Multivariate Interpolation for Gridded and Scattered Data
## Description
Branin-Hoo 2-dimensional test function.
## Usage
`1` ```branin(x) ```
## Arguments
`x` Numeric vector with length 2.
## Details
The Branin-Hoo function is defined here over [0, 1] x [0, 1], instead of [-5, 0] x [10, 15] as usual. It has 3 global minima at (nearly) : x1 = c(0.96, 0.15), x2 = c(0.12, 0.82) and x3 = c(0.54, 0.15).
## Value
The Branin-Hoo function's value.
## Author(s)
David Ginsbourger
## Examples
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28``` ```GD <- Grid(nlevels = c("x" = 20, "y" = 20)) x <- levels(GD)[[1]]; y <- levels(GD)[[2]] f <- apply_Grid(GD, branin) dim(f) <- nlevels(GD) contour(x = x, y = y, z = f, nlevels = 40) nOut <- 100; Xout2 <- array(runif(nOut * 2), dim = c(nOut, 2)) colnames(Xout2) <- c("x", "y") ## interpolate using default method (Lagrange) GIL <- interp_Grid(X = GD, Y = f, Xout = Xout2) ## interpolate using a natural spline GIS <- interp_Grid(X = GD, Y = f, Xout = Xout2, cardinalBasis1d = function(x, xout) { cardinalBasis_natSpline(x = x, xout = xout)\$CB }) F <- apply(Xout2, 1, branin) mat <- cbind(Xout2, fTrue = F, fIntL = GIL, errorLag = F - GIL, fIntS = GIS, errorSpline = F - GIS) apply(mat[ , c("errorLag", "errorSpline")], 2, function(x) mean(abs(x))) ## Not run: ## for the users of the "rgl" package only... library(rgl) persp3d(x = x, y = y, z = f, aspect = c(1, 1, 0.5), col = "lightblue", alpha = 0.8) spheres3d(Xout2[ , 1], Xout2[ , 2], GIS, col = "orangered", radius = 2) ## End(Not run) ```
smint documentation built on April 14, 2017, 1:49 p.m. | 669 | 1,704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-05 | latest | en | 0.45505 |
http://www.freecontentweb.com/content/blackjack-versus-hold%E2%80%99em-poker-how-these-table-games-vary-slightly-how-each-game-played-and- | 1,566,657,540,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027321140.82/warc/CC-MAIN-20190824130424-20190824152424-00341.warc.gz | 247,948,037 | 8,182 | # Blackjack Versus Hold’em Poker - How These Table Games Vary Slightly, How Each Game is Played and the Strategies Used
Blackjack Versus Hold’em Poker - How These Table Games Vary Slightly, How Each Game is Played and the Strategies Used
Thesis Statement: Although blackjack and hold’em poker are both table games, they vary slightly on how each game is played, and the strategies used.
I.The basics on how to play blackjack and hold’em poker.
A. Blackjack is not a difficult game to play.
1. Player versus dealer.
2. The player has only three cards to analyze.
3. The player has only four possible decisions to make.
B. Hold’em poker compared to blackjack is a much more complicated game to play.
1. Player versus player.
2. The player has more cards to analyze.
3. The player has many decisions to make.
II. The strategies used when playing blackjack and hold’em poker.
A. Blackjack, though not difficult to play, has basic strategies of play once the cards have been dealt.
1. Knowing when to stay.
2. Knowing when to hit.
3. Knowing when to double down.
4. Knowing when to split.
B. Hold’em poker, though more difficult to play, also has basic strategies of play once the cards have been dealt.
1. Knowing when to call.
2. Knowing when to raise.
3. Knowing when to fold.
Gambling is done all over the world, with table games being the most popular games for professional gamblers like Avery Cardoza. Two of these popular table games are blackjack and hold’em poker. Although blackjack and hold’em poker are both table games, they vary slightly on how each game is played, and the strategies used.
Blackjack is not a difficult game to play. Avery (2004) says in his book, QUICK GUIDE TO WINNING BLACKJACK, “You are playing to beat the dealer” (p.17). The statement simplifies the game to a competition between the player and the dealer. This does not mean that there is only player at the table. There can be up to six players plus the dealer at one table. The object of the game is to have a higher point count then the dealer without going over twenty one. The game begins with the dealer dealing one card at a time to the players and themselves until each has two cards. The dealer will flip up one of there cards so all the players can see. Avery (2004) states “The player goes first in blackjack” (p.17). At this point the player has four possible decisions to make. The player can stay with the two cards they have. The player can take a hit, which means the dealer will give the player another card. The player can continue to hit until they wish to stay or they go over twenty one, which is called a bust. The player can also double down and receive one additional card. The last choice the player has is, if the two card’s in the players hand are the same denomination, they can split the card’s giving them two hands to play.
Hold’em poker, compared to blackjack, is a much more complicated game to play. Avery (2006) says in his book, THE BASICS OF WINNING HOLD’EM POKER, “Hold’em is played as high poker, that is, the player with the highest five-card combination at the show-down will have the winning hand and collect the money in the pot” (p.13). This quote indicates that the game is played against the other players at the table and not against the dealer. When playing in casinos the dealer does not play a hand, they just monitor the play of the players at the table. The usual number of players is eight to ten at the start of the game, with one player being called the dealer or button. This player will have a button placed in front of them to signify the position. The button is typically a plastic disk with the word dealer on it. The player to the left of the dealer is called the small blind with the person to the left of them being called the big blind. Before the cards are dealt the small and big blinds need to put their bets on the table. The dealer begins dealing one card at a time at the small blind position continuing clockwise until each player has received two cards. Once the cards are dealt Avery (2006) says “The player to the big blind’s left acts first” (p.27). At this point this player has three choices to make. The player can call, which is matching the bet of the big blind. The player can raise, which is matching the bet of the big blind then increasing the bet. The last choice the player has is to fold the hand dealt to them, which means they are done playing for that pot. This will continue around the table clockwise until all the bets have been called, which means that the players still in the game have an equal bet on the table. This can happen up to three more times in one hand.
Blackjack, though not difficult to play, has basic strategies of play once the cards have been dealt. In his book Avery (2006) says “…there is only one mathematically correct way to play your hand against the dealer” (p.49). Depending on the cards the player has will determine the strategy to use. Should the player have a total card count of seventeen to twenty, the proper play would be to stay. The player has a couple of decisions to make should their card count be twelve to sixteen. The player should stay if the dealers shown card is two to six and hit if the dealers card is seven to eleven. If the player has ten or eleven there are many more strategies to deal with, making the dealers up card more important. Should the dealers up card be ten or less the players correct move would be to double down. This is done by placing the same value amount of chips on the table next to the player’s original chips, which will signal the dealer to deal just one more card. If the player has a card count of nine or less, this is easy to play. Since the player can not go over twenty one with one more card the players proper play is to take a hit. After the hit the player then needs to reevaluate the situation before making his next decision. The player may find that the two cards he has are the same which will require a more strategic approach. According to Avery (2004) “Whenever [the player has] two cards of the same rank, his first thought must go to splitting” (p. 65). Avery’s strategy is pretty simple to follow here. Should the player have two fours or fives he should never split them. Having two tens through kings the player has the second best hand possible and should not split them either. The player may have two eights or aces in which case they should always be split. The rest of the combinations need to be decided on by what the up card of the dealer is. If the player decides to split his cards he will need to put the same value amount of chips on the table with a noticeable gap from his original stack, which will signal the dealer that he wishes to split the cards. Once this is done the player will have two hands to use these strategies on.
Hold’em poker, though more difficult to play, also has basic strategies of play once the cards have been dealt. Avery (2006) says, “The following key concepts apply to all forms of hold’em” (p.34). In the first round after the deal there are three decisions to make. The player can call, raise or fold. The players further away from the button have the advantage of knowing what the other players have done. The player must decide whether the cards dealt are cards worth risking money on. Knowing what the other players’ trends are will improve your ability to win the pot. Avery (2006) states, “Hold’em is a game where aggression brings the best returns” (p.35). Since the goal of hold’em poker is to win the chips of the other players at the table, being aggressive can be a strategic move. When deciding whether the hand is a losing one or not, just remember it will cost you chips to find out. Lastly, Avery (2006) says, “Hold’em is a game of patience” (p.36). Patience can be a frustrating emotion to deal with when playing hold’em poker. The player may want to play even though they have not been dealt cards worth betting chips on. Once the players have called or folded the round is complete. The dealer at this point will discard the top card form the remaining cards and flip the next three cards over and the strategies will need to be gone through again. The next two rounds only one card will be turned over after the top card is discarded.
In the world of gambling, two of the more popular table games are blackjack and hold’em poker. Although both are card games, the way in which they are played is slightly different. Each of the games has their own set of rules and strategies used by the player to make decisions. However, the object each game is the same, to win each hand played.
References:
Cardoza, A. (2005). Quick guide to winning blackjack. New York, NY: Cardoza Publishing.
Cardoza, A. (2005). The basics of winning hold’em poker. New York, NY: Cardoza Publishing. | 1,961 | 8,823 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-35 | longest | en | 0.94174 |
http://sourceforge.net/p/vxl/mailman/vxl-users/thread/200110300938.KAA11454@caravaggio.esat.kuleuven.ac.be/ | 1,432,824,311,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207929418.92/warc/CC-MAIN-20150521113209-00304-ip-10-180-206-219.ec2.internal.warc.gz | 216,764,851 | 6,832 | Learn how easy it is to sync an existing GitHub or Google Code repo to a SourceForge project!
## vxl-users
Re: [Vxl-users] vnl_angle() From: Peter.V - 2001-10-30 09:38:22 ```> The current implementation of angle in vnl_vector.txx is > acos( abs( cos_angle(a,b) ) ) > > Why is the abs there? Thinking about it again, I believe this is a "remainder" of an angle implementation for lines, not vectors. For lines, it makes sense to "forget" about orientation, but for vectors it doesn't. So this "abs" should probably be removed. Peter. ```
Re: [Vxl-users] vnl_angle() From: Peter.V - 2001-11-09 12:16:57 ```> how is the angle between two complex vectors defined? > Is it acos( |a.b| / |a||b| ) ? No, it is just acos( a.b / |a||b| ), which is in general a complex number. This does not really make sense, i.e., there is no natural interpretation of this kind of angles, but it is the only sensible extension of "real" angles. acos( |a.b| / |a||b| ) cannot be correct since then the angle between (1,0) and (-1,1) would be 45 degrees instead of 135 degrees. Peter. ```
Re: [Vxl-users] vnl_angle() From: Amitha Perera - 2001-11-09 15:52:19 ```> No, it is just acos( a.b / |a||b| ), which is in general a complex > number. How is acos( complex ) defined? Such a function does not exist in vcl, so the template instantiations for vnl_vector fail. > acos( |a.b| / |a||b| ) cannot be correct since then the angle between > (1,0) and (-1,1) would be 45 degrees instead of 135 degrees. Certainly, for real-valued vectors. However, for complex-valued vectors, perhaps it is "good enough"? Either way, it doesn't really solve the problem of implementation. Amitha. ```
Re: [Vxl-users] vnl_angle() From: Peter Vanroose - 2001-11-09 17:44:49 ```> > No, it is just acos( a.b / |a||b| ), which is in general a complex > > number. > > How is acos( complex ) defined? Such a function does not exist in vcl, > so the template instantiations for vnl_vector fail. Since cos(complex) is well defined ( cos(a+ib) = cos(a)*cosh(b) - i sin(a)*sinh(b) if I did not make a mistake) also acos(complex) is well defined, e.g. by using the (Taylor) series expansion. I don't have an implementation at hand. And it's indeed a rather unusual function. We could of course decide to ensure it is defined in vcl, but that would probably not really be ANSI conformant. > > acos( |a.b| / |a||b| ) cannot be correct since then the angle between > > (1,0) and (-1,1) would be 45 degrees instead of 135 degrees. > > Certainly, for real-valued vectors. However, for complex-valued > vectors, perhaps it is "good enough"? No: it should be defined consistently for real and complex arguments. Doing it like this would not be consistent. Peter. ``` | 774 | 2,707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2015-22 | longest | en | 0.850276 |
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### Epidemic Modelling
##### Age 14 to 18 Challenge Level:
Use the computer to model an epidemic. Try out public health policies to control the spread of the epidemic, to minimise the number of sick days and deaths.
### Integration Matcher
##### Age 16 to 18 Challenge Level:
Can you match the charts of these functions to the charts of their integrals?
### What's That Graph?
##### Age 14 to 16 Challenge Level:
Can you work out which processes are represented by the graphs?
### Matrix Meaning
##### Age 16 to 18 Challenge Level:
Explore the meaning behind the algebra and geometry of matrices with these 10 individual problems.
### Brimful
##### Age 16 to 18 Challenge Level:
Can you find the volumes of the mathematical vessels?
### Square Pair
##### Age 16 to 18 Challenge Level:
Explore the shape of a square after it is transformed by the action of a matrix.
### Big and Small Numbers in the Living World
##### Age 11 to 16 Challenge Level:
Work with numbers big and small to estimate and calculate various quantities in biological contexts.
### The Wrong Stats
##### Age 16 to 18 Challenge Level:
Why MUST these statistical statements probably be at least a little bit wrong?
### Big and Small Numbers in Physics
##### Age 14 to 16 Challenge Level:
Work out the numerical values for these physical quantities.
### A Question of Scale
##### Age 14 to 16 Challenge Level:
Use your skill and knowledge to place various scientific lengths in order of size. Can you judge the length of objects with sizes ranging from 1 Angstrom to 1 million km with no wrong attempts?
### Back Fitter
##### Age 14 to 16 Challenge Level:
10 graphs of experimental data are given. Can you use a spreadsheet to find algebraic graphs which match them closely, and thus discover the formulae most likely to govern the underlying processes?
### Crystal Symmetry
##### Age 16 to 18 Challenge Level:
Use vectors and matrices to explore the symmetries of crystals.
### Constantly Changing
##### Age 14 to 16 Challenge Level:
Many physical constants are only known to a certain accuracy. Explore the numerical error bounds in the mass of water and its constituents.
### Approximately Certain
##### Age 14 to 18 Challenge Level:
Estimate these curious quantities sufficiently accurately that you can rank them in order of size
### Pdf Matcher
##### Age 16 to 18 Challenge Level:
Which pdfs match the curves?
### Counting Dolphins
##### Age 14 to 16 Challenge Level:
How would you go about estimating populations of dolphins?
### Debt Race
##### Age 16 to 18 Challenge Level:
Who will be the first investor to pay off their debt?
### Electric Kettle
##### Age 14 to 16 Challenge Level:
Explore the relationship between resistance and temperature
### Far Horizon
##### Age 14 to 16 Challenge Level:
An observer is on top of a lighthouse. How far from the foot of the lighthouse is the horizon that the observer can see?
### Building Approximations for Sin(x)
##### Age 16 to 18 Challenge Level:
Build up the concept of the Taylor series
### Differential Equation Matcher
##### Age 16 to 18 Challenge Level:
Match the descriptions of physical processes to these differential equations.
### Scale Invariance
##### Age 16 to 18 Challenge Level:
By exploring the concept of scale invariance, find the probability that a random piece of real data begins with a 1.
### Taking Trigonometry Series-ly
##### Age 16 to 18 Challenge Level:
Look at the advanced way of viewing sin and cos through their power series.
### Reaction Rates
##### Age 16 to 18 Challenge Level:
Explore the possibilities for reaction rates versus concentrations with this non-linear differential equation
### What Do Functions Do for Tiny X?
##### Age 16 to 18 Challenge Level:
Looking at small values of functions. Motivating the existence of the Taylor expansion.
### Investigating the Dilution Series
##### Age 14 to 16 Challenge Level:
Which dilutions can you make using only 10ml pipettes?
### Elastic Maths
##### Age 14 to 18
How do you write a computer program that creates the illusion of stretching elastic bands between pegs of a Geoboard? The answer contains some surprising mathematics.
### Big and Small Numbers in Biology
##### Age 14 to 16 Challenge Level:
Work with numbers big and small to estimate and calulate various quantities in biological contexts.
### Stats Statements
##### Age 16 to 18 Challenge Level:
Are these statistical statements sometimes, always or never true? Or it is impossible to say?
### Big and Small Numbers in the Physical World
##### Age 14 to 16 Challenge Level:
Work with numbers big and small to estimate and calculate various quantities in physical contexts.
### Immersion
##### Age 14 to 16 Challenge Level:
Various solids are lowered into a beaker of water. How does the water level rise in each case?
### Polygon Walk
##### Age 16 to 18 Challenge Level:
Go on a vector walk and determine which points on the walk are closest to the origin. | 1,704 | 7,975 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2019-26 | longest | en | 0.862039 |
https://electronics.stackexchange.com/questions/292200/what-does-the-c-stand-for-when-charging-li-ion-batters-ie-max-discharge-rate-2/292201 | 1,716,920,608,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059143.84/warc/CC-MAIN-20240528154421-20240528184421-00545.warc.gz | 188,154,537 | 41,229 | # what does the c stand for when charging li ion batters ie Max. Discharge rate: 2C (6.8A)
looking to charge a 4 pack of li-ion battery but dont understand what the 2c means know the rest. i.e. Max. Discharge rate: 2C (6.8A)
• There is a paragraph in this answer which explains the C notation. Mar 13, 2017 at 23:13
C in this context means the battery capacity in Ampere-Hours but divided by hours. If battery's capacity is 3.4 Ah, then C would be 3.4A, and 2C would be 6.8.
Charge and discharge rates of a battery are governed by C-rates. The capacity of a battery is commonly rated at 1C, meaning that a fully charged battery rated at 1Ah should provide 1A for one hour. The same battery discharging at 0.5C should provide 500mA for two hours, and at 2C it delivers 2A for 30 minutes. Losses at fast discharges reduce the discharge time and these losses also affect charge times.
A C-rate of 1C is also known as a one-hour discharge; 0.5C or C/2 is a two-hour discharge and 0.2C or C/5 is a 5-hour discharge. Some high-performance batteries can be charged and discharged above 1C with moderate stress. Table 1 illustrates typical times at various C-rates
"C" is the ampere-hour rating of the battery.
This would mean 2C discharge rating for the battery is 6.8 Amp, the battery would be rated at 3.4 Ampere-hours.
A battery has a capacity rating usually in mA hours or Amp Hours.
The C or C-Rate is the C harge and dis C harge rate, expressed as a ratio of the battery capacity.
A 1Ah battery should provide 1A for one hour. If it is powering something that is draining 2 Amps that is a 2C discharge rate.
The rate is used when the specifics of the battery such as voltage and capacity are not an issue. It is a generalize term applying to all batteries or the type of battery being addressed. For example: Li-Ion batteries are recommend to be charged at an 0.8C rate. This means the 1Ah battery should be charged at 800mA. A 2Ah battery would be charged at 1.6 Amp.
The C rate can be applied to any battery regardless of it capacity or voltage. | 552 | 2,058 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-22 | latest | en | 0.952355 |
https://bloxhamprimary.co.uk/classes/year-5/class-5b/term-4-week-4 | 1,632,741,310,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058415.93/warc/CC-MAIN-20210927090448-20210927120448-00517.warc.gz | 190,226,941 | 7,955 | # Term 4 Week 4
Term 4 Week 4
This week in Maths we have been revising our knowledge and learning about fractions. We have practised simplifying, adding and subtracting fractions. In Outdoor Learning, we also practised fractions. Using coloured chalks, we created our own fraction walls and played fraction snap and bingo.
In groups, we began creating our own street dance to Dance Monkey.
We have been having fun in DT making a variety of mechanisms, using levers and linkages, to create waving flags and moving dragons. Some of the mechanisms were very tricky!
For Home Learning this week we had a puzzle to solve. We had to work out how to get a chicken, a fox and a bag of grain across a river without them eating each other. We also made our own versions of the puzzle.
In Art we chose whatever we wanted to create – painting or drawing – which was great fun.
Last week we had a times tables treasure hunt in small groups. We went outside to look for the clues and some of us struggled to find them all, as they were very well hidden!
We have started reading Tom’s Midnight Garden as our class book. We are really enjoying it so far… | 250 | 1,146 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-39 | latest | en | 0.982987 |
https://community.anaplan.com/discussion/58068/nested-ifs-early-exits-vs-or-and | 1,726,323,354,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.38/warc/CC-MAIN-20240914125424-20240914155424-00897.warc.gz | 162,526,046 | 84,635 | # Nested IFs (early exits) vs. OR/AND?
I'm curious what others' thoughts are, or if there's been any testing done regarding the performance difference between nested IFs vs. OR/AND. I found a very old thread on this topic from 2014, but I expect the thinking & research has progressed a lot since then.
Let's say I have multiple criteria I need to check in order to produce a final determination. Using careful construction of nested IFs, I can allow early exits from the formula, which supports the P (Performance) rule of PLANS, however the resulting formula can be very long and convoluted. Using OR/AND results in a much simpler formula (L - Logical and A - Auditable in PLANS) but does not allow for early exits. In situations where one case is much more common than other(s), I imagine this could make a significant difference, especially over long timescales, many dimensions, etc.
What do you prefer to use?
Tagged:
• @rob_marshall To be clear, when you say break it up, do you mean into separate line items? I agree and definitely try to do that where space allows. However, I am more interested in the difference in constructing the final formula.
For instance, if I have conditions with the following results:
and I want a 1 where any of the 3 are true, I can do:
"if A OR B OR C then 1", or
"if A then 1 else if B then 1 else if C then 1"
Is there a performance difference between these? My gut reaction is that the first would be slower since it has to evaluate all 3 conditions every time, but maybe that's not the case.
@nathan_rudman Thanks for the response! I'm not sure what the real performance difference is (just my guess as above, which might be wrong) and find it very difficult to nail down, particularly in a general case like this where it's less about optimizing 1 terrible formula and more making sure I take the right approach because small differences add up. If you have any tips for measuring performance difference, I would love to hear them!
• This is really useful to know!
• Thank you @ben_speight; this is super handy to know and exactly what I was looking for.
Thanks all for the thoughtful replies! I had posted a follow-up on Friday but it looks like it was deleted somehow; not sure why that happened.
•
in your example, I believe "if A OR B OR C then 1" will be faster than "if A then 1 else if B then 1 else if C then 1" because the 2nd one has to go through the three different IF statements whereas the first does not.
Rob | 579 | 2,484 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-38 | latest | en | 0.955223 |
https://www.clutchprep.com/chemistry/practice-problems/90865/is-the-use-of-significant-figures-in-each-of-the-following-statements-appropriat-1 | 1,607,208,796,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141750841.83/warc/CC-MAIN-20201205211729-20201206001729-00534.warc.gz | 620,864,709 | 32,230 | # Problem: Is the use of significant figures in each of the following statements appropriate?On July 1, 2005, the population of Cook County, Illinois, was 5,303,683.
###### FREE Expert Solution
We’re being asked to determine if the use of significant figures in the following statement appropriate.
Recall that the significant figures in a measured quantity include one estimated digit, the last digit of the measurement. The significant figures indicate the extent of the uncertainty of the measurement
Certain rules must be followed so that a calculation involving measured quantities is reported with the appropriate number of significant figures.
Given: On July 1, 2005, the population of Cook County, Illinois, was 5,303,683.
83% (391 ratings)
###### Problem Details
Is the use of significant figures in each of the following statements appropriate?
On July 1, 2005, the population of Cook County, Illinois, was 5,303,683. | 200 | 935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-50 | latest | en | 0.915133 |
https://id.scribd.com/document/359350815/Basic-ele-engg-pdf | 1,563,691,831,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526931.25/warc/CC-MAIN-20190721061720-20190721083720-00554.warc.gz | 425,267,476 | 153,480 | Anda di halaman 1dari 198
# VTU
## Common to All Branches
1 & 2 Semester
Physics Cycle
2010 Scheme
Basic
Electrical Engineering
[10ELE15]
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Branch Name: Common to all branches
SEM: 1/2
University: VTU
Syllabus: 2010
Basic Electrical Engineering (10ELE15):
## Sl. No. Units
1 (a) D.C. Circuits
(b) Electromagnetism
2 Single phase A.C. Circuits
3 Three phase circuits
4 (a) Measuring instruments
(b) Domestic Wiring
5 DC Machines
6 Transformers
7 Synchronous generators
8 Three phase induction motors
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## BASIC ELECTRICAL ENGINEERING
PART A
Unit-I
1a) D. C. Circuits: Ohm's Law and Kirchhoffs Laws, analysis of series, parallel and
series- parallel circuits excited by independent voltage sources. Power and Energy.
Illustrative examples. 04Hours
.in
Ib) Electromagnetism: Faradays Laws, Lenz's Law, Fleming's Rules, Statically and
dynamically induced emfs. Concept of self inductance, mutual inductance and
sy
coefficient of coupling. Energy stored in magnetic field. Illustrative examples.
03Hours
ea
Unit-II
## 2. Single-phase A.C. Circuits: Generation of sinusoidal voltage, definition of average
y
value, root mean square value, form factor and peak factor of sinusoidally varying
voltage and current, phasor representation of alternating quantities. Analysis, with phasor
ud
diagrams, of R, L, C, R-L, R-C and R-L-C circuits, real power, reactive power, apparent
power and power factor. Illustrative examples involving series, parallel and series-
parallel circuits 07 Hours
st
Unit-III
3 Three Phase Circuits: Necessity and advantages of three phase systems, generation of
three phase power, definition of Phase sequence, balanced supply and balanced load.
Relationship between line and phase values of balanced star and delta connections. Power
in balanced three-phase circuits, measurement of power by two-wattmeter method.
Illustrative examples. 06 Hours
Unit-IV
## 4a) Measuring Instruments: Construction and Principle of operation of dynamometer
type wattmeter and single-phase induction type energy meter (problems excluded).
03 Hours
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4b) Domestic Wiring: Service mains, meter board and distribution board. Brief
discussion on Cleat, Casing & Capping and conduit (concealed) wiring. Two-way and
three-way control of a lamp. Elementary discussion on fuse and Miniature Circuit
Breaker (MCBs). Electric shock, precautions against shock Earthing: Pipe and Plate.
03 Hours
PART B
Unit-V
## 5. DC Machines: Working principle of DC machine as a generator and a motor. Types
and constructional features. emf equation of generator, relation between emf induced and
.in
terminal voltage enumerating the brush drop and drop due to armature reaction.
Illustrative examples.
DC motor working principle, Back emf and its significance, torque equation. Types of
D.C. motors, characteristics and applications. Necessity of a starter for DC motor.
sy
Illustrative examples on back emf and torque. 07 Hours
Unit-VI
ea
6. Transformers: Principle of operation and construction of single-phase transformers
(core and shell types). emf equation, losses, efficiency and voltage regulation (Open
Circuit and Short circuit tests, equivalent circuit and phasor diagrams are excluded).
y
ud
Unit-VII
## 7. Synchronous Generators: Principle of operation. Types and constructional features.
st
emf equation. Concept of winding factor (excluding derivation of distribution and pitch
factors). Illustrative examples on emf. equation. 06 Hours
Unit-VIII
## 8. Three Phase Induction Motors: Concept of rotating magnetic field. Principle of
operation. Types and Constructional features. Slip and its significance. Applications of
squirrel - cage and slip - ring motors. Necessity of a starter, star-delta starter. Illustrative
examples on slip calculations. 06 Hours
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TEXT BOOKS
2009.
REFERENCE BOOKS:
## 2. Basic Electrical Engineering, Abhijit Chakrabarti, Sudiptanath, Chandan Kumar
Chanda, TMH, First reprint 2009.
.in
3. Problems in Electrical Engineering, Parker Smith, CBS Publishers and Distributors,
9th Edition, 2003.
sy
y ea
ud
st
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Compiled by www.studyeasy.in
Contents
## Sl.NO UNITS Page No.
1 D. C. CIRCUITS 5 - 27
2 SINGLE-PHASE CIRCUITS 28 74
.in
75 91
5
sy 119 139
D.C. MACHINES.
ea
y
ud
st
## 8 THREE PHASE INDUCTION MOTOR 168 - 196
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PART A
Unit-I
1a) D. C. Circuits: Ohm's Law and Kirchhoffs Laws, analysis of series, parallel and
series- parallel circuits excited by independent voltage sources. Power and Energy.
Illustrative examples 04Hours
Ib) Electromagnetism: Faradays Laws, Lenz's Law, Fleming's Rules, Statically and
dynamically induced emfs. Concept of self inductance, mutual inductance and
coefficient of coupling. Energy stored in magnetic field. Illustrative examples.
03Hours
.in
D. C. Circuits
## Ohms law and state its limitations. sy
Ohms Law : the current flowing through the electric the electric circuit is directly
proportional to the potential difference across the circuit and inversely proportional to the
ea
resistance of the circuit, provided the temperature remains constant.
Limitations of Ohms Law
The limitations of the Ohms law are,
y
ud
## voltage regulators ect.
2) It does not hold good for non-metallic conductors such as silicon carbide.
The law for such conductors is given by,
V = K Im
st
## where k, m are constants.
( I ) Current is what flows on a wire or conductor like water flowing down a river.
Current flows from negative to positive on the surface of a conductor. Current is
measured in (A) amperes or amps.
( E ) Voltage Ohm's Law defines the relationships between (P) power, (E) voltage, (I)
current, and (R) resistance. One ohm is the resistance value through which one volt will
maintain a current of one ampere is the difference in electrical potential between two
points in a circuit. It's the push or pressure behind current flow through a circuit, and is
measured in (V) volts.
( R ) Resistance determines how much current will flow through a component. Resistors
are used to control voltage and current levels. A very high resistance allows a small
amount of current to flow. A very low resistance allows a large amount of current to
flow. Resistance is measured in ohms.
.
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.in
sy
y ea
ud
To make a current flow through a resistance there must be a voltage across that
resistance. Ohm's Law shows the relationship between the voltage (V), current (I) and
st
V V
V=IR or I= or R=
R I
## where: V = voltage in volts (V) or: V = voltage in volts (V)
I = current in amps (A) I = current in milliamps (mA)
R = resistance in ohms ( ) R = resistance in kilohms (k )
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## State and explain Kirchhoffs laws.
V
Kirchhoffs current law
I R
The law can be stated as,
## equal to the total current flowing away from
that junction
.in
point.
sy Another way to state the law is,
## The algebraic sum of all the current meeting at a
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junction point is always zero.
y
## The word algebraic means considering the signs of various currents.
ud
at junction point = 0
st
## positive whie currents flowing away from a junction point assumed to be
negative.
e.g. Refer to Fig. 1, currents I1 and I2 are positive while I3 and I4 are negative.
## Applying KCL, at junction 0 = 0
I1 + I2 - I3 - I4 = 0 i.e. I1 + I2 = I3 + I4
## Kirchhoffs voltage law :
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In any network, the algebraic sum of the voltage drops across the circuit
elements of any closed path (or loop or mesh) is equal to the algebraic sum of the
## e.m.f s in the path
In other words, the algebraic sum of all the branch voltages, around any closed
## Around a closed path =0
The law states that if one starts at a certain point of a closed path and goes on
.in
tracing and noting all the potential changes (either drops or rises), in any one
particular direction, till the starting point reached again, he must be at the same
sy
potential with which he started tracing a closed path.
Sum of all the potential rises must be equal to sum of all the potential drops while
ea
tracing any closed path of the circuit. The total change in potential along a closed
y
ud
st
## respectively connected in series with 15 resistor is 3 A,
.If current through 15
Find : i) Current in 20 and 30 resistors ii) The voltage across the whole
circuit iii) The total power and power consumed in all resistors. (8)
## Sol. : The arrangement is shown in the Fig. 2.
Total current I = 3 A
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=
+ 15
= 27
I=
## V = 81 V Voltage across each circuit
.in
I1 = I
=3 = 1.8 A Current through 20
I2 = I
sy
= 3 = 1.2 A Current through 30
ea
P = V I = 81 3 = 243 W Current total power
y
P20
20 = (1.8)2 20 = 64.8 W
ud
## P30 = 30 = (1.2)2 30 = 43.2 W
st
2 2
P15 = I 15 = (3) 15 = 135 W
## Cross check is P = P20 + P30 + P15
Resistance
Resistance is the property of a component which restricts the flow of electric current.
Energy is used up as the voltage across the component drives the current through it and
this energy appears as heat in the component.
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## When resistors are connected in
series their combined resistance is equal to the individual resistances added
together. For example if resistors R1 and R2 are connected in series their
combined resistance, R, is given by:
R = R1 + R2
## This can be extended for more resistors: R = R1 + R2 + R3 + R4 + ...
Note that the combined resistance in series will always be greater than any of the
.in
individual resistances.
## Resistors connected in Parallel
sy
When resistors are connected in parallel their
combined resistance is less than any of the individual
resistances. There is a special equation for the
ea
combined resistance of two resistors R1 and R2:
y
Combined resistance of R1 R2
two resistors in parallel: R =
ud
R1 + R2
For more than two resistors connected in parallel a more difficult equation must be used.
This adds up the reciprocal ("one over") of each resistance to give the reciprocal of the
st
combined resistance, R:
1 1 1 1
= + + + ...
R R1 R2 R3
The simpler equation for two resistors in parallel is much easier to use!
Note that the combined resistance in parallel will always be less than any of the
individual resistances.
( P ) Power is the amount of current times the voltage level at a given point measured in
wattage or watts.
electrical energy - energy made available by the flow of electric charge through a
conductor; "they built a car that runs on electricity" measured in k Watt Hour
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Energy=VItKWhour
A 8 ohm resistor is in series with a parellel combination of two resistors 12 ohm and 6
ohm. If the current in the 6 ohm resistor is 5 A, determine the total power dissipated in
the circuit. (6)
## Ans. : The arrangement is shown in the fig.
VAB = I1 6 = 5 6 = 30 V
I2 = = = 2.5 A
I = I1 + I2 = 7.5 A
.in
P12 = 12 = 12 = 75 W
P6 = 6= sy
6 = 150 W
P8 = 8= 8 = 450 W
PT = P12 + P6 +P8 = 675 W.
ea
2) A coil consists of 600 turns and a current of 10 A in the coil gives rise to a magnetic
flux of 1 milli Weber. Calculate i) Self induction ; ii) The e.m.f. induced and iii) The
energy stored when the current is reversed in 0.01 second . (5)
y
## Ans.: N = 600, I = 10 A , = 1 mWb
ud
i) L= = = 0.06H
ii) Current is reversed in 0.01 sec i.e. I 2 = - 10 A, I1 = 10 A
st
e =- L =-L = = 120v
iii) E=L = 0.06 =3J
## A current of 20 A flows through two ammeters A and B in series. The potential
difference across A is 0.2 V and across B is 0.3 V.Find how the same current will divide
between A and B when they are in parallel.
Ans : RA = resistance of ammeter A
= = = 0.01
RB = resistance of ammeter B
= = = 0.015
Now the ammeters are connected in parallel as shown in the Fig. 1(b)
Req = RARB =
= 0.006
V = voltage across both the ammeters
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V = IARA = IBRB
While V = IReq = 200.006 = 0.12 V
IA =
and IB = =8A
Note: This can be verified using current division rule as,
IA = I = 12 A and IB = I =8A
## A parallel circuit comprises a resistor of 20 ohm in series with an inductive reactance
of 15 ohm in one in one branch and a resistor of 30 ohm in series with a capacitive
reactance of 20 ohm in the other branch. Determine the current and power dissipated in
.in
each branch if the total current drawn by the parallel circuit is 10 -300 Amp(8)
## Ans.: The arrangement is shown in the Fig.2.
sy Z1= 20 + j 15 = 25 36.8690
Z2= 30 j 20 =36.055 -33.690
ea
By current division rule,
y
I1= IT =
ud
st
=
=7.1752 A
And =
= = 4.9752 +
A
## Only resistive part of each branch consumes the power given by R
= = 20 = 1029.67 W.
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1st law: Whenever magnetic flux linking with a coil changes with time an emf is
induced in that coil or whenever a moving conductor cuts the magnetic flux, an
emf is induced in the conductor.
2nd law: The magnitude of the induced emf is equal to the product of the number
of turns of the coil and the rate of change of flux linkage.
Lenzs law :
It states that the direction of an induced emf produced by the electromagnetic
induction is such that it sets up a current which always opposes the cause that is
.in
responsible for inducing the emf.
In short the induced emf always opposes the cause producing it which is represented
sy
by negative sign, mathematically in its expression
Consider a solenoid as shown in Fig.1. Let a bar magnet is moved towards coil such that
N-pole of magnet is facing a coil which will circulate the current through the coil.
ea
According to Lenzs law, the direction of current due to induced emf is so as to
oppose the cause. The cause is motion of bar magnet towards coil So emf will set up a
y
current through coil in such a way that the end of solenoid facing bar magnet will become
N-pole. Hence two like poles will face each other experiencing force of repulsion which
ud
## is opposite to the motion of bar magnet as shown in the above .
st
Flemings rules:
1. Flemings Right hand rule: This rule helps in deciding
the direction of the induced emf.
Hold the right hand thumb, fore finger and the middle
finger set at right angles to each other and the thumb points
the direction of the motion of the conductor and the fore finger
points the direction of the field and the middle finger points the
direction of the induced emf.
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2. Flemings Left hand rule: This rule helps in deciding the direction of
force acting on a conductor.
Hold the left hand thumb, fore finger and the middle finger
set at right angles to each other and the thumb points the direction
of the force acting on the conductor and the direction
of the fore finger points the direction of the magnetic field and
the middle finger points the direction of the current in the conductor
## Statically and dynamically induced emfs
Statically induced emf STATICALLY INDUCED EMF
The emf induced in a coil due to change of flux linked with it (change of flux is by the
increase or decrease in current) is called statically induced emf.
.in
Transformer is an example of statically induced emf. Here the windings are
stationary,magnetic field is moving around the conductor and produces the emf.
## DYNAMICALLY INDUCED EMF
sy
The emf induced in a coil due to relative motion of the conductor and the magnetic field
ea
is called dynamically induced emf.
## example:dc generator works on the principle of dynamically induced emf in the
y
conductors which are housed in a revolving armature lying within magnetic field
ud
## Statically induced e.m.f
The change in flux lines with respect to coil can be achieved without physically
st
moving the coil or the magnet. Such induced e.m.f. in a coil which is without physical
movement of coil or a magnet is called statically induced e.m.f.
To have an induced e.m.f there must be change in flux associated with a coil.
Such a change in flux can be achieved without any physical movement by increasing and
decreasing the current producing the flux rapidly, with time.
Consider an electromagnet which is producing the necessary flux for producing
e.m.f. Now let current through the coil of an electromagnet be an alternating one. Such
alternating current means it changes its magnitude periodically with time.
This
produces the flux which is also alternating i.e. changing with time. Thus there exists
associated with coil placed in the viscinity of an electromagnet. This is responsible
for
producing an e.m.f in the coil. This is called statically induced e.m.f.
There is no physical movement of magnet or conductor; it is the alternating
supply which is responsible for such an induced e.m.f.
Such type of an induced e.m.f. is available in transformers.
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## Dynamically induced e.m.f.
The change in the flux linking with a coil, conductor or circuit can be brought
about by its motion relative to magnetic field. This is possible by moving flux with
respect to coil conductor or circuit or it is possible by moving conductor, coil, circuit with
respect to stationary magnetic flux.
Such an induced e.m.f. which is due to physical movement of coil, conductor
with respect to flux or movement of magnet with respect with to stationary coil,
conductor is called dynamically induced e.m.f. or motional induced e.m.f.
This type of induced e.m.f. is available in the rotating machines such as
alternators, generator etc.
d) A coil resistance 150 is placed in a magnetic field of 0.1 mWb. The coil has
500 turns and a galvanometer of 450 is connected in series with it. The coil is moved in
.in
0.1 sec from the given field to another field of 0.3 mWbs. Find the average induced e.m.f.
and the average current through the coil. [5]
## Ans : N = 500, = 0.3 mWb, = 0.1mWb, dt = 0.1 sec.
2
|e| = N =
=1V
1
=
sy
ea
RT = Rcoil + Rg = 150+450 = 600
I= = = 1.6667 mA.
y
Find the currents in all the resistors of the network shown in Fig. 1. Also find the voltage
ud
## s.: The current distribution in various branches is shown in the Fig. 2.
Apply KVL to the two loops,
st
Loop ACDA,
- 10 I1 5 I 2 +20(1-I1) = 0
30I1 + 5I2 = 20 (1)
Loop CBDC,
-2.5(I1-I2) + 5(1-I1+I2) +5I2 = 0
## 7.5I - 12.5I =5 .(2)
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AC 10 0.666 A
CD 5 0A
CB 205 0.666 A
DB 5 0.333A
## Trace the path from A to B as shown in the Fig. 3.
.in
sy
ea
VAB = 6.666+1.666
= 8.3333V with A positive
y
ud
## With examples, clearly differentiate between statically induced e.m.f. and
dynamically induced e.m.f.
Ans : Statically induced e.m.f
The change in flux lines with respect to coil can be achieved without physically
st
moving the coil or the magnet. Such induced e.m.f. in a coil which is without physical
movement of coil or a magnet is called statically induced e.m.f.
To have an induced e.m.f there must be change in flux associated with a coil.
Such a change in flux can be achieved without any physical movement by increasing and
decreasing the current producing the flux rapidly, with time.
Consider an electromagnet which is producing the necessary flux for producing
e.m.f. Now let current through the coil of an electromagnet be an alternating one. Such
alternating current means it changes its magnitude periodically with time. This
produces the flux which is also alternating i.e. changing with time. Thus there exists
associated with coil placed in the viscinity of an electromagnet. This is responsible for
producing an e.m.f in the coil. This is called statically induced e.m.f.
There is no physical movement of magnet or conductor; it is the alternating
supply which is responsible for such an induced e.m.f.
Such type of an induced e.m.f. is available in transformers.
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## Dynamically induced e.m.f.
The change in the flux linking with a coil, conductor or circuit can be brought
about by its motion relative to magnetic field. This is possible by moving flux with
respect to coil conductor or circuit or it is possible by moving conductor, coil, circuit with
respect to stationary magnetic flux.
Such an induced e.m.f. which is due to physical movement of coil, conductor
with respect to flux or movement of magnet with respect with to stationary coil,
conductor is called dynamically induced e.m.f. or motional induced e.m.f.
This type of induced e.m.f. is available in the rotating machines such as
alternators, generator etc.
d) A coil resistance 150 is placed in a magnetic field of 0.1 mWb. The coil has
500 turns and a galvanometer of 450 is connected in series with it. The coil is moved in
0.1 sec from the given field to another field of 0.3 mWbs. Find the average induced e.m.f.
.in
and the average current through the coil.
[5]
Ans : N = 500, 2 = 0.3 mWb, 1 = 0.1mWb, dt = 0.1 sec.
|e| = N =
=1V
= sy
RT = Rcoil + Rg = 150+450 = 600
ea
I= = = 1.6667 mA.
y
self inductance : According to Lenzs law the direction of this induced e.m.f. will be so
ud
as to oppose the cause producing it. The cause is the current I hence self induced e.m.f
will try to set up a current which is in opposite direction to that of current I. When current
is increased, self induced e.m.f. reduces the current tries to keep to its original value. If
current is decreased, self induced e.m.f. increases the current and tries to maintain it back
st
to its original value. So any change in current through coil is opposed by the coil.
This property of the coil which opposes any change in the current passing through it
is called self inductance or only inductance.
It is analogous to electrical inertia or electromagnetic inertia.
The formula for self inductance is given by,
L=
It can be defined as flux linkages per ampere current in it. Its unit is Henry (H)
## Expressions for coefficient of self inductance (L):
L=
But =
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L =
L= henries
Now s
L=
L =
= Henries
Where l = length of magnetic circuit
a = area of cross-section of magnetic circuit which flux is
passing.
.in
Derive an Expression for energy stored in the inductor:
Let the induced e.m.f. in a coil be,
e = -L sy
This opposes a supply voltage. So supply voltage V supplies energy to overcome this,
which ultimately gets stored in the magnetic field.
ea
V = -e = - =L
Power supplied = V I = L I
y
ud
## Energy supplied in time dt is,
E = power x time = L x I x dt
st
= L di x I joules.
This is energy supplied for a change in current of dI but actually current changes from
zero to I.
Integrating above total energy stored is,
E=
## state i) Flemmings right hand rule, and
ii) Flemings left hand rule.
Mention their applications.
Flemings right hand rule : The Flemings left hand rule is used to get direction of force
experienced by conductor carrying current placed in magnetic field while Flemings right
hand rule can be used to get direction of induced e.m.f. when conductor is moving at
right angles to the magnetic field.
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According to this rule, outstretch the three fingers of right hand namely the
thumb, fore finger and the middle finger, perpendicular to each other. Arrange the right
hand so that finger point in the direction of flux lines ( from N to S ) and thumb in the
direction of motion of conductor with respect to the flux then the middle finger will point
in the direction of the induced e.m.f. ( or current ).
.in
sy
y ea
ud
st
Flemings left hand rule: The direction of the force experienced by the current
carrying conductor placed in magnetic field can be determined by a rule called
Flemings left hand rule. The rule states that outstretch the three fingers on the left
hand namely the first finger, middle finger and thumb such that they are mutually
perpendicular to each other. Now point the first finger in the direction of magnetic field
and middle finger in the direction of the current then the thumb gives the direction of the
force experienced by the conductor.
The rule is explained in the diagrammatic form in fig. 2.
Applications: Flemings right
hand rule is used to get the direction
of induced emf in case of generators
and alternators while left hand rule
is used to get the direction of torque
induced in motors.
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## Define i) self inductance, and ii) mutual inductance.
Mention their units and formula to calculate each of them. Derive an
expression for the energy stored in an inductor of self inductance L Henry carrying the
current of I amperes.
Sol. : self inductance : According to Lenzs law the direction of this induced e.m.f. will
be so as to oppose the cause producing it. The cause is the current I hence self induced
e.m.f will try to set up a current which is in opposite direction to that of current I. When
current is increased, self induced e.m.f. reduces the current tries to keep to its original
value. If current is decreased, self induced e.m.f. increases the current and tries to
maintain it back to its original value. So any change in current through coil is opposed by
the coil.
This property of the coil which opposes any change in the current passing through it
is called self inductance or only inductance.
It is analogous to electrical inertia or electromagnetic inertia.
The formula for self inductance is given by,
.in
L=
It can be defined as flux linkages per ampere current in it. Its unit is Henry (H)
sy
Expressions for coefficient of self inductance (L):
L=
ea
But =
=
y
L =
ud
L= henries
Now s
st
L=
L =
= Henries
Where l = length of magnetic circuit
a = area of cross-section of magnetic circuit through which flux
is passing.
8) Derive an Expression for energy stored in the inductor:
Let the induced e.m.f. in a coil be,
e = -L
This opposes a supply voltage. So supply voltage V supplies energy to overcome this,
which ultimately gets stored in the magnetic field.
V = -e = - =L
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Power supplied = V I = L I
## Energy supplied in time dt is,
E = power x time = L x I x dt
= L di x I joules.
This is energy supplied for a change in current of dI but actually current changes from
zero to I.
Integrating above total energy stored is,
E=
.in
=L =L
E = joules
Mutual inductance:
sy
ea
Magnitude of mutually induced e.m.f
Let N1 = Number of turns of coil A
N2 = Number of turns of coil B
I1 = Current flowing through coil A
y
ud
## 2 = Flux linking with coil B
According to Faradays law, the induced e.m.f. in coil B is,
st
e2 =
Negative sign indicates that this e.m.f will set up a current which will oppose the change
Now 2 =
## If permeability of the surroundings is assumed constant then 2 I1 and hence I1 is
constant.
Rate of change of 2 =
Rate of change of current I1
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## Here is called co efficient of mutual inductance dented by M
Volts
Coefficient of mutual inductance is defined as the property by which e.m.f gets induced
in the second coil because of change in current through first coil.
Coefficient of mutual inductance is defined as the property by which e.m.f gets induced
in the second coil because of change in current through first coil.
.in
Coefficient of mutual inductance is also called mutual inductance. It is measured in
Henries.
sy
Definitions of mutual inductance and its unit:
1) The coefficient of mutual inductance is defined as the flux linkages of the coil per
ampere current in other coil.
ea
2) It can also be defined as equal to e.m.f induced in volts in one coil when current
in other coil changes uniformly are rate of one ampere per second.
y
## Similarly its unit be defined as follows:
1) Two coils which are magnetically coupled are said to have mutual inductance of
ud
one hence when a current of one ampere flowing through one coil produces a
flux linkage of one Weber turn in the other coil.
2) Two coils which are magnetically coupled are said to have mutual inductance of
st
one Henry when a current changing uniformly at the rate of one ampere per
second in one coil, induces as e.m.f of one volts in the other coil.
## 9) Expressions of the mutual inductance (M):
1)
2) 2 is the part of the flux 1 produced due to I1. Let K1 be the fraction of 1 which is
## 3) The flux 1 can be expressed as,
1 =
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If all the flux produced by the coil A links with coil B K1=
1.
and K1 = 1
4) Now
Then
.in
sy
5) If second coil carries current I2, producing flux 2, the part of which links with coil A
i.e.1 then,
ea
1 = K2 2 and
=
y
=
ud
Now
st
coupling coefficient: The coefficient of coupling is define as the ratio of the actual
mutual inductance present between the two coils as the maximum possible value of the
mutual inductance. It gives an idea about magnetic coupling between the two coils. This
coefficient indicates the amount of linking with other coil which is produced by one
coil.
Let N1 = Number of turns of first coil
N2 = number of turns of second coil
I1 = current through first coil
I2 = current through by first coil
1 = flux produced by first coil
2 = flux produced by second coil
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M= and M =
Multiplying the two expressions,
MM=
M2 = K1K2
But = L1 = self inductance of first coil
And = L2 = self inductance of second coil
M2 = K1K2L1L2
M=
Let K= = coefficient of coupling
M=K
.in
K=
Three similar coils each having resistance of 10 ohm and reactance of 8 ohm are
sy
connected in star across a 400 V, 3 phase supply . Determine the i) Line current;
ii) Total power and iii) Reading of each of two wattmeters connected to measure
the power. (8)
ea
Ans.: R=10 , =8 , = 400 V, star
= 10 + j 8 = 12.082
y
ud
= = = 230.94 V,
star
st
i) = = = 18.0334 V,
= = 18.0334 A.
ii) = cos where =
38.6590
= 400 18.0334 cos
= 9756.2116 W
iii) = cos (30- ) = 400 18.0334 cos ( - )
= 7131.1412 W
= cos (30+ ) = 400 18.0334 cos ( + )
= 2625.0704 W
A coil of 300 turns wound on a core of non magnetic material has an inductance of 10
mH. Calculate i) flux produced by a current of 5A ii) the average value of the emf
induced when a current of 5Amps is reversed in 8 milli seconds.
Sol: Given values are N = 300, I = 5A, L = 10mH = 10x10-3H
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Self inductance, L= = =
= 0.166mWb
## Self induced emf, e = -L
= -L
= -10x10-3
= 12.5 volts
Induced emf = 12.5 volts
A coil of 300 turns wound on a core of non magnetic material has an inductance of 10
mH.Calculate i) flux produced by a current of 5A ii) the average value of the emf
.in
induced
when a current of 5Amps is reversed in 8 milli seconds.
Sol: Given values are N = 300, I = 5A, L = 10mH = 10x10 -3H
Self inductance, L=
= 0.166mWb
sy = =
ea
Self induced emf, e = -L
= -L
y
ud
= -10x10-3
= 12.5 volts
st
## Induced emf = 12.5 volts
Two coupled coils of self inductances 0.8 H and 0.20 H have a coefficient of coupling
0.9. Find the mutual indulgence and turns ratio [7]
Sol.: L1 = 0.8 H
L2 = 0.2 H
K = 0.9
## Mutual Inductance, M=K = 0.9
= 0.9 = (0.9) (0.4) = 0.36 H
M = 0.36 H
Turns ratio
Turns ratio = 4
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A coil of 1000 turns is wound on a silicon steel ring of relative permeability 1200. The
ring has a mean diameter of 10cm and cross sectional area of 12 sq.cm. When a current
of 4 amperes flows through the coil. Find
i) Flux in the core
ii) Inductance of the coil
iii) The emf induced in the coil if the flux falls to zero in 15 milii seconds.
iv) Now, if another similar coil is placed such that 70% magnetic coupling exists
between the coils. Find the mutual inductance. (Chapter-1) [10]
Ans.: Given:
N = 1000 turns
r = 1200
.in
d = 10 cm = 0.1 m
a = 12 cm2 = 12
I=4A
sy
o = 4
## i) Flux in the core:
ea
Let I = Length of magnetic circuit
=
= 0.1 m
y
= 0.3141 m
ud
S=
st
= 1.735 AT/Wb
=
Wb
= 23.05
## flux in the core = = 23.05 Wb
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L=
L = 5.763 H
=-N
.in
= - 1000
= 1536.6V
## iv) 70% of magnetic coupling i.e.
sy
ea
K1 =
Mutual Inductance =
y
ud
M=
=
st
= 4.033 H
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Unit-II
## Single-phase A.C. Circuits: Generation of sinusoidal voltage, definition of average
value, root mean square value, form factor and peak factor of sinusoidally varying
voltage and current, phasor representation of alternating quantities. Analysis, with phasor
diagrams, of R, L, C, R-L, R-C and R-L-C circuits, real power, reactive power, apparent
power and power factor. Illustrative examples involving series, parallel and series-
parallel circuits
SINGLE-PHASE CIRCUITS
## Alternating voltage may be generated:
.in
a) By rotating a coil in a magnetic field as shown in Fig.3.1.
b) By rotating a magnetic field within a stationary coil as shown in Fig.3.2.
sy
y ea
ud
st
In each case, the value of the alternating voltage generated depends upon the number of
turns in the coil, the strength of the field and the speed at which the coil or magnetic field
ritates.
The alternating voltage generated has regular changes in magnitude and direction. If a
load resistance (e.g. a light bulb) is connected across this alternating voltage, an
alternating current flows in the circuit. When there is a reversal of polarity of the
alternating voltage, the direction of current flow in the circuit also reverses.
3.2 Equation of Alternating E.M.F.
Let us take up the case of a rectangular
coil of N turns rotating in the anticlockwise
direction, with an angular velocity of radians
per second in a uniform magnetic field as shown
in Fig.3.3. let the time be measured from the instant
of coincidence of the plane of the coil with the
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X-axis. At this instant maximum flux max links with the coil. As the coil rotates, the flux
linking with it changes and hence e.m.f. is induced in it. Let the coil turn through an
angle in time t seconds, and let it assume the position as shown in Fig.3.3. Obviously
= t.
When the coil is in this position, the maximum flux acting vertically downwards can
be resolved into two components, each perpendicular to the other, namely:
a) Component max sin t, parallel to the plane of the coil. This component does not
induce e.m.f. as it is parallel to the plane of the coil.
b) Component max cos t, perpendicular to the plane of coil. This component
induces e.m.f. in the coil.
## flux linkages of coil at that instant (at 0) is
= No. of turns x flux linking
.in
= N max cos t
As per faradays Laws of Electromagnetic induction, the e.m.f. induced in a coil is
equal to the rate of change of flux linkages of the coil. So, instantaneous e.m.f. e
induced in the coil at this instant is:
sy
cos t)
ea
=- (N max
## =-N max (cos t)
= -N max (-sin t)
y
## e = + N max sin t volts . . .(1)
It is apparent from eqn.(1) that the value of e will be maximum (E m), when the coil has
ud
## rotated through 900 (as sin 900 = 1)
Thus Em = N max volts . . .(2)
Substituting the value of N max from eqn.(2) in eqn.(1), we obtain:
st
e = Em sin t . . .(3)
we know that = t
e = Em sin
It is clear from this expression of alternating e.m.f. induced in the coil that
instantaneous e.m.f. varies as the sin of the time angle ( or t).
= 2f, where f is the frequency of rotation of the coil. Hence eqn.(3) can be
written as
e = Em sin 2ft ..
.(4)
If T = time of the alternating voltage = , then eqn.(iv) may be re-written as
e = Em sin t
so, the e.m.f. induced varies as the sine
function of the time angle, t, and if e.m.f.
induced is plotted against time, a curve of
sine wave shape is obtained as shown in
Fig.3.4. Such an e.m.f. is called sinusoidal
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## when the coil moves through an angle of 2 radians.
3.3 Equation of Alternating Current
When an alternating voltage e = Em sin t is applied across a load, alternating
current flows through the circuit which will also have a sinusoidal variation. The
expression for the alternating current is given by:
i = Im sin t
in this case the load is resistive (we shall see, later on, that if the load is inductive or
capacitive, this current-equation is changed in time angle).
3.4 Important Definitions
Important terms/definitions, which are frequently used while dealing with a.c.
circuits, are as given below:
1. Alternating quantity: An alternating quantity is one which acts in alternate positive
and negative directions, whose magnitude undergoes a definite series of changes in
definite intervals of time and in which the sequence of changes while negative is
.in
identical with the sequence of changes while positive.
2. Waveform: The graph between an alternating quantity (voltage or current) and
time is called waveform, Generally, alternating quantity is depicted along the Y-
sy
axis and time along the X-axis.fig.4.4 shows the waveform of a sinusoidal voltage.
3. Instantaneous value: The value of an alternating quantity at any instant is called
instantaneous value.
ea
The instantaneous values of alternating voltages and current are represented by e
and I respectively.
4. Alternation and cycle: When an alternating quantity goes through one half cycle
y
## (complete set of +ve or ve values) it completes an alternation, and when it goes
ud
through a complete set of +ve and ve values, it is said to have completed one
cycle.
5. Periodic Time and Frequency: The time taken in seconds by an alternating quantity
st
## to complete one cycle is know as periodic time and is denoted by T.
The number of cycles completed per second by an alternating quantity is know as
frequency and is denoted by f. in the SI system, the frequency is expressed in
hertz.
## Periodic Time T Time taken in completing one cycle =
Or f =
In India, the standard frequency for power supply is 50 Hz. It means that alternating
voltage or current completes 50 cycles in one second.
6. Amplitude: The maximum value, positive or negative, which an alternating quantity
attains during one complete cycle is called amplitude or peak value or maximum
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respectively.
## 3.5 Different Forms of E.M.F. Equation
The standard form of an alternating voltage, as already mentioned in sec.3.2 is
e = Em sin = Em sin t = Em sin 2f t = Em sin t
on perusal of the above equations, we find that
a) The amplitude or peak value or maximum value of an alternating voltage is
given by the coefficient of the sine of the time angle.
b) The frequency f is given by the coefficient of time divided by 2.
## Taking an example, if the equation is of an alternating voltages is given by e = 20
sin 314t, then its maximum value is 20 V and its frequency is
f= 50 Hz
.in
In a like manner, if the equation is of the form
e = Im sin 2 t, then its maximum value is
Em = Im
sy and the frequency is
ea
or Hertz
y
## 3.6 Root-mean-square (R.M.S.) Value
The r.m.s. or effective value, of an alternating current is defined as that steady
ud
current which when flowing through a given resistance for a given time produces the
same amount of heat as produced by the alternating current, when flowing through the
same resistance for the same time.
st
Let us take two circuits with identical resistance, but one is connected to a battery
and the other to a sinusoidal voltage source. Wattmeters are employed to measure heat
power in each circuit. The voltage applied to each circuit is so adjusted that the heat
power produced in each circuit is the same. In this event the direct current I will equal
, which is termed r.m.s. value of the sinusoidal current.
The following method is used for finding the r.m.s. or effective value of sinusoidal
waves.
The equation of an alternating current varying sinusoidally is given by
i= Im sin
## let us consider an elementary strip
of thickness d in the first cycle of the
squared wave, as shown in Fig.3.5. let
i2 be mid-ordinate of this strip.
Area of the strip = i2 d
Area of first half-cycle of squared wave
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= d
= d ( I = Im sin )
=
= Im2 d ( sin2 =
=
=
.in
I=
= sy
=
ea
= = 0.707Im
Hence, for a sinusoidal current,
R.M.S. value of current = 0.707 x maximum value of current.
y
Similarly, E = 0.707 Em
ud
## 3.7 Average Value
The arithmetical average of all the values of an alternating quantity over one cycle is
called average value.
In the case of a symmetrical wave e.g. sinusoidal current or voltage wave, the
st
positive half is exactly equal to the negative half, so that the average value over the entire
cycle is zero. Hence, in this case, the average value is obtained by adding or integrating
the instantaneous values of current over one alternation (half-cycle) only.
The equation of a sinusoidally varying voltage
Is given by e = Em sin .
Let us take an elementary strip of thickness d in the first half-cycle as shown in
Fig.3.6. let the mid-ordinate of this strip be e.
Area of the strip = e d
Area of first half-cycle
= d
= d ( e = Em sin )
= Em d
= Em = 2Em
Average value, Eav =
Or Eav = 0.637 Em
In a similar manner, we can prove that, for alternating current varying sinusoidally,
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Iav = 0.637 Im
Average value of current = 0.637 x maximum value
## 3.8 Form Factor and crest or peak or Amplitude Factor (K f)
A definite relationship exists between crest value (or peak value), average value and
r.m.s. value of an alternating quantity.
1. Form Factor: The ratio of effective value (or r.m.s. value) to average value of an
alternating quantity (voltage or current) is called form factor, i.e.
From Factor, Kf =
For sinusoidal alternating current,
Kf =
= 1.11
.in
For sinusoidal alternating voltage,
Kf = = 1.11
sy
Hence, the R.M.S. value (of current or voltage) is 1.11 times its average value.
2. Crest or Peak or Amplitude Factor (Ka): It is defined as the ratio of maximum
value to the effective value (r.m.s. value) of an alternating quantity. i.e.,
ea
Ka =
For sinusoidal alternating current,
y
Ka = = = 1.414
ud
Ka = = 1.414
st
## The knowledge of Crest Factor is particularly important in the testing of dielectric
strength of insulating materials; this is because the breakdown of insulating materials
depends upon the maximum value of voltage.
3.9 Phase
An alternating voltage or current changes in magnitude and direction at every instant.
So, it is necessary to know the condition of the alternating quantity at a particular instant.
The location of the condition of the alternating quantity at any particular instant is called
its phase.
We may define the phase of an alternating quantity at any particular instant as
the fractional part of a period or cycle through which the quantity has advanced
from the selected origin.
Taking an example, the phase of current
at point A (+ve maximum value) is T/4 second,
where T is the time period, or expressed in
terms of angle, it is /2 radians (Fig.3.7).
in other words, it means that the condition
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of the wave, after having advanced through /2 radians (900) from the selected origin
(i.e.,0) is that it is maximum value (in the positive direction).similarly, -ve maximum
value is reached after 3/2 radians (2700) from the origin, and the phase of the current at
point B is 3T/4 second.
## 3.10 Phase Difference (Lagging or Leading of Sinusoidal wave)
When two alternating quantities, say, two voltages or two currents or one voltage
and one current are considered simultaneously, the frequency being the same, they may
not pass through a particular point at the same instant.
One may pass through its maximum value at the instant when the other passes
through a value other than its maximum one. These two quantities are said to have a
phase difference. Phase difference is specified either in degrees or in radians.
The phase difference is measured by the angular difference between the points
where the two curves cross the base or reference line in the same direction.
The quantity ahead in phase is said to lead the other quantity, whereas the second
.in
quantity is said to lag behind the first one. In Fig.3.8, current I1, represented by vector
0A, leads the current I2, represented by vector 0B, by , or current I2 lags behind the
current I1 by .
sy
y ea
ud
st
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The leading current I1 goes through its zero and maximum values first and the
current I2 goes through its zero and maximum values after time angle . The two waves
representing these two currents are shown in Fig.3.8. if I 1 is taken as reference vector,
two currents are expressed as
i1 = I1m sin t and i2 = I2m sin (t- )
The two quantities are said to be in phase with each other if they pass through zero
values at the same instant and rise in the same direction, as shown in Fig.3.9. however, if
the two quantities pass through zero values at the same instant but rise in opposite, as
shown in Fig.3.10, they are said to be in phase opposition i.e., the phase difference is
1800. When the two alternating quantities have a phase difference of 90 0 or /2 radians
they are said to be in quadrature.
.in
sy
y ea
ud
## 3.11 Phasor Representation of Alternating Quantities
We know that an alternating voltage or current has sine waveform, and generators
st
are designed to give e.m.f.s. with the sine waveforms. The method of representing
alternating quantities continuously by equation giving instantaneous values (like e = Em
sin t) is quite tedious. So, it is more convenient to represent a sinusoidal quantity by a
phasor rotating in an anticlockwise direction (Fig.3.12).
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## While representing an alternating quantity by a phasor, the following points are to
be kept in mind:
i) The length of the phasor should be equal to the maximum value of the alternating
quantity.
ii) The phasor should be in the horizontal position at the alternating quantity is zero
and is increasing in the positive direction.
iii) The inclination of the line with respect to some axis of reference gives the
direction of that quantity and an arrow-head placed at one end indicates the
direction in which that quantity acts.
iv) The angular velocity in an anti-clockwise direction of the phasor should be such
that it completes one revolution in the same time as taken by the alternating
quantity to complete one cycle.
.in
Consider phasor 0A, which represents the maximum value of the alternating e.m.f.
and its angle with the horizontal axis gives its phase (Fig.3.12). now, it will be seen that
the projection of this phasor 0A on the vertical axis will give the instantaneous value of
e.m.f.
0B = 0A sin t
sy
Or e = 0A sin wt
ea
= Em sin t
Note: The term phasor is also known as vector.
y
ud
## b) -10-j7.5= < 0.75
= 12.5 < 0.75
st
This vector also falls in the third quadrant, so, following the same reasoning as
mentioned in method 1, the angle when measured in CCW direction is
0.75)
00 0 0
= 180 +36.9
(180 + =216.9
## So this expression is written as 12.5 <-143.10
So, expression (ii) is rewritten as
10 < 36.9012.5 <-143.10
125 <-106.20 which is the same as before.
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## 3.16 A.C. circuit
The path for the flow of alternating current is called on a.c. circuit.
In a d.c. circuit, the current/flowing through the circuit is given by the simple relation
I = . However, in an a.c. circuit, voltage and current change from instant to instant and
so give rise to magnetic (inductive) and electrostatic (capacitive) effects. So, in an
a.c. circuit, inductance and capacitance must be considered in addition to resistance.
## We shall now deal with the following a.c. circuits:
i) AC circuit containing pure ohmic resistance only.
ii) AC circuit containing pure inductance only.
iii) AC circuit containing pure capacitance only.
.in
3.16.1 AC circuit containing pure ohmic Resistance
When an alternating voltage is applied across a pure ohmic resistance, electrons
(current) flow in one direction during the first half-cycle and in the opposite direction
sy
during the next half-cycle, thus constituting alternating current in the circuit.
Let us consider an a.c. circuit with just a pure resistance R only, as shown in
Fig.3.31.
ea
Let the applied voltage be given by the equation
= Vm sin t --- (i)
As a result of this
= Valternating
m sin voltage, alternating
current i will flow through the circuit.
y
## The applied voltage has to supply the drop in the
ud
resistance, i.e.,
= iR
Substituting the value of from eqn.(i), we get
st
## Vm sin t = iR or i= sin t ---(ii)
The value of the alternating current i is
maximum when sin t = 1,
i.e.,
Eqn.(ii) becomes,
i= sin t --- (iii)
From eqns.(i) and (ii), it is apparent that
voltage and current are in phase with each
other. This is also indicated by the wave and
vector diagram shown in Fig. 3.32.
## Power: The voltage and current are changing at every instant.
Instantaneous power, P= Vm sin t. =
=
=
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= .cos 2
## Thus instantaneous power consists of a constant part and a fluctuating part
cos 2 of frequency double that of voltage and current waves.
## The average value of cos 2 over a complete cycle is zero.
So, power for the complete cycle is
P= =
or P = V1 watts
Where V = r.m.s. value of applied voltage
I = r.m.s. value of the current
.in
Power curve
The power curve for a purely resistive
sy
circuit is shown in Fig. 3.33. It is apparent
that power in such a circuit is zero only at
the instants a,b and c, when both voltage
ea
and current are zero, but is positive at all
other instants. in other words, power is never
negative, so that power is always lost in a resistive
a.c. circuit. This power is dissipated as heat.
y
ud
## 3.16.2 A.C. circuit containing pure Inductance
An inductive coil is a coil with or without an iron core and has negligible resistance.
In practice, pure inductance can never be had as the inductive coil has always a small
st
resistance. However, a coil of thick copper wire wound on a laminated iron core has
negligible resistance, so, for the purpose of our study, we will consider a purely inductive
coil.
## On the application of an alternating
voltage (Fig.3.34) to a circuit containing
a pure inductance, a back e.m.f. is produced
due to the self-inductance of the coil. This
back e.m.f. opposes the rise or fall o f current,
at every stage. Because of the absence of voltage
drop, the applied voltage has to overcome this
self-induced e.m.f. only.
## Let the applied voltage be =
and the self-inductance of the coil = L henry.
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## Self-induced e.m.f. in the coil, =-L
Since applied voltage at every instant is equal and opposite to the self-induced
e.m.f., i.e. = -
=-
or di =
dt
## Integrating both sides, we get
i= dt
.in
or i= (-cos )+ A
So, i= cos
sy
Where A is a constant of integration which is found to be zero from initial conditions.
ea
Or i= sin
y
ud
st
## From the expressions of instantaneous
applied voltage ( = ) and the
instantaneous current flowing through a
purely inductive coil, it is clear that the
current lags behind the voltage by as
shown in Fig. 3.35.
Inductive Reactance:
L in the expression Im = is known as inductive reactance
and is denoted by , i.e., = L. If L is in henry and is in radians per second,
then will be in ohms. So, inductive reactance plays the part the part of resistance.
## Power: Instantaneous Power,
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P = i= . Im sin
=- cos
The power measured by a wattmeter is the average value of p, which is zero since
average of a sinusoidal quantity of double frequency over a complete cycle is zero. Put in
mathematical terms,
## Power for the whole cycle, P = - dt = 0
.in
Hence, power absorbed in a pure inductive circuit is zero.
Power curve
sy
y ea
ud
st
The power curve for a pure inductive circuit is shown in Fig. 3.36. This indicates that
power absorbed in the circuit is zero. At the instants a,c and e, voltage is zero, so that
power is zero: it is also zero at points b and d when the current is zero. Between a and b
voltage and current are in opposite directions, so that power is negative and energy is
taken from the circuit. Between b and c voltage and current are in the same direction, so
that power is positive and is put back into the circuit. Similarly, between c and d, power
is taken from the circuit and between d and e it is put into the circuit. Hence, net power is
zero.
## When an alternating voltage is applied
across the plates of a capacitor, the capacitor
is charged in one direction and then in the
opposite direction as the voltage reverses. With
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## reference to Fig. 3.38,
Let alternating voltage represented by = be applied across a capacitor of
capacitance
Instantaneous charge, q = c = CVm sin
Capacitor current is equal to the rate of change of charge,
or
(CVm sin )
i=
= CVm cos
or i= sin
.in
The current is maximum when t = 0
Im =
Substituting
sy
= Im in the above expression for instantaneous current, we
get
ea
i = Im sin
y
## Capacitive Reactance: in the expression Im =
is known as capacitive reactance
ud
st
i.e., Xc =
## If C is farads and is in radians, then Xc will be in ohms.
It is seen that if the applied voltage is given by = , then the current is given
by i = Im sin this shows that the current in a pure capacitor leads its voltage
by a quarter cycle as shown in Fig. 3.39, or phase difference between its voltage and
current is with the current leading.
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## Power: Instantaneous Power,
P= i
= t. Im sin
= Im sin t cos t
= Im
Power for the complete cycle
= Im dt = 0
.in
Hence power absorbed in a capacitive circuit is zero.
## Power curves (Fig. 3.40)
sy
At the instants b,d, the current is zero, so that power is zero; it is also zero at the
instants a,c and e, when the voltage is zero. Between a and b, voltage and current are in
ea
the same direction, so that power is positive and is being put back in the circuit. Between
b and c, voltage and current are in the opposite directions, so that power is negative and
energy is taken from the circuit. Similarly, between c and d, power is put back into the
circuit, and between d and e it is taken from the circuit.
y
ud
st
## 3.17 Series R-L circuit
Let us consider an a.c. circuit containing a pure resistance R ohms and a pure
inductance of L henrys, as shown in Fig. 3.43.
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## Let V = r.m.s. value of the applied voltage
I = r.m.s. value of the current
Voltage drop across R, VR = IR (in phase with I)
Voltage drop across L, VL = IXL (leading I by 900)
## The voltage drops across these two circuit
components are shown in Fig. 3.44, where vector OA
indicates VR and AB indicates VL. The applied voltage
V is the vector sum of the two, i.e., OB.
V= =
=I
.in
I=
The term
sy
offers opposition to current flow and is called the impedance (Z)
of the circuit. It is measured in ohms.
ea
I=
## Referring to the impedance triangle ABC, (Fig. 3.45)
y
ud
or (impedance)2 + (reactance)2
Referring back to Fig. 3.44, we observe that the
applied voltage V leads the current I by an angle .
st
tan =
## The same feature is shown by means
of waveforms (Fig. 3.46). We observe that
circuit current lags behind applied voltage
by an angle .
## So, if applied voltage is expressed as =
t, the current is given by i = Im( t-
), Where Im = .
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Definition of Real power, Reactive Power, Apparent power and power Factor
Let a series R-L circuit draw a current I (r.m.s. value) when an alternating voltage of
r.m.s. value V is applied to it. Suppose the current lags behind the applied voltage by an
angle as shown in Fig. 3.47.
## Power Factor and its signifies
Power Factor may be defined as the cosine of the
angle of lead or lag. In Fig. 3.47, the angle of lag is shown.
Thus power Factor = cos .
## In addition to having a numerical value, the power
factor of a circuit carries a notation that signifies the
.in
nature of the circuit, i.e., whether the equivalent circuit
is resistive, inductive or capacitive. Thus, the p.f. might
be expressed as 0.8 lagging. The lagging and leading
refers to the phase of the current vector with respect to
sy
the voltage vector. Thus, a lagging power factor means
that the current lags the voltage and the circuit is inductive in nature. However, in the
case of leading power factor, the current leads the voltage and the circuit is capacitive.
ea
Apparent Power: The product of r.m.s. values of current and voltage, VI, is called the
apparent power and is measured in volt-amperes (VA) or in kilo-volt amperes (KVA).
y
Real Power: The real power in an a.c. circuit is obtained by multiplying the apparent
power by the factor and is expressed in watts or killo-watts (kW).
ud
## Real power (W) = volt-amperes (VA) power factor cos
or Watts = VA cos
Here, it should be noted that power consumed is due to ohmic resistance only as a
st
## pure inductance does not consume any power.
Thus, P = V I cos
## cos = (refer to the impedance triangle of Fig. 3.45)
P=VI
=
IR = I2R
or P = I2R watts
Reactive Power: It is the power developed in the inductive reactance of the circuit. The
quantity VI sin is called the reactive power; it is measured in reactive volt-amperes or
vars (VAr).
The power consumed can be represented by means of waveform in Fig. 3.48.
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## We will now calculate power in terms of instantaneous values.
Instantaneous power, P = vi = Vm sin t Im sin ( t - )
= Vm Im sin t sin( t- )
= Vm Im [cos - cos(2 t- )]
This power consists of two parts:
.in
i)
ii)
Constant part Vm Im cos sy
which contributes to real power.
Sinusoidally varying part Vm Im cos (2 t- ), whose frequency is twice that of the
ea
voltage and the current, and whose average value over a complete cycle is
zero (so it does not contribute to any power).
y
ud
= cos
= V I cos
st
## Where V and I are r.m.s. values
Power curves: The power curve for R-L series circuit is shown in Fig. 3.48. The curve
indicates that the greater part is positive and the smaller part is negative, so that the net
power over the cycle is positive.
During the time interval a to b, applied voltage and current are in opposite
directions, so that power is negative. Under such conditions, the inductance L returns
power to the circuit. During the period b to c, the applied voltage and current are in the
same direction so that power is positive , and therefore, power is put into the circuit. In a
similar way, during the period c to d, inductance L returns power to the circuit while
between d and e, power is put into the circuit. The power absorbed by resistance R is
converted into heat and not returned.
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## Consider an a.c. circuit containing resistance R ohms and capacitance C farads, as
.in
shown in the fig. 3.52(a).
## Let V = r.m.s. value of voltage
I = r.m.s. value of current
voltage drop across R, VR = IR
sy - in phase with I
ea
Voltage drop across C, VC = IXC - lagging I by
The capacitive resistance is negative, so VC is in the negative direction of Y axis, as
shown in the fig. 3.52(b).
y
We have V=
ud
Or I=
st
## The denominator, Z is the impedance of the circuit, i.e., Z =
fig. 3.52(c)
depicts the impedance triangle.
## Power factor, cos =
Fig. 3.52(b) shows that I leads V by an
angle , so that tan =
This implies that if the alternating voltage
is v = Vm sin t, the resultant current in the
R C circuit is given by i = Im sin( t + ),
such that current leads the applied voltage
by the angle . The waveforms of fig. 3.53 depict this fact.
## Power: Average power, P = v I = VI cos ( as in sec. 3.17).
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Power curves: The power curve for R C series circuit is shown in fig. 3.54. The curve
indicates that the greater part is positive and the smaller part is negative, so that the net
power is positive.
.in
sy
3.19 Resistance, Inductance and capacitance in series
Consider an a.c. series circuit
containing resistance R ohms, Inductance L henries
ea
and capacitance C farads, as shown in the fig. 3.59.
y
## I = r.m.s. value of current
voltage drop across R, VR = IR - in phase with I
ud
## voltage drop across L, VL = I.XL - lagging I by 900
Voltage drop across C, VC = IXC - lagging I by 900
st
## Referring to the voltage triangle of
Fig. 3.60, OA represents VR, AB and AC
represent inductive and capacitive drops
respectively. We observe that VL and VC are
1800 out of phase.
## Thus, the net reactive drop across the
combination is
= AB BD ( BD = AC)
= VL VC
= I(XL - XC)
OD, which represents the applied voltage V, is the vector sum of OA and AD.
OD = OR V =
=
Or I = =
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## The denominator is the impendence of the circuit.
So (impedance)2 = (resistance)2 + (net reactance)2
Or Z2 = R2 + = R2 + X2
Where the net reactance = X (fig. 3.61)
## Phase angle is given by
tan
=
power factor,
.in
cos =
Power = VI cos
If applied voltage is represented by the equation v = Vm sin t, then the resulting current
i = Im sin( t )
sy
in an R L C circuit is given by the equation
If XC > XL , then the current leads and the +ve sign is to be used in the above equation.
ea
If XL > XC, then the current lags and the ve sign is to be used.
If any case, the current leads or lags the supply voltage by an angle , so that tan = .
If we employ the j operator (fig. 3.62), then we have
y
Z = R + j (XL - XC)
ud
Z=
st
Z =Z tan-1
=Z tan-1
## 3.20 Parallel AC circuits
In a parallel a.c. circuit, the voltage across each branch of the circuit is the same
whereas current in each branch depends upon the branch impedance. Since alternating
currents are vector quantities, total line current is the vector sum of branch currents.
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The following are the three methods of solving parallel a.c. circuits:
a) Vector method.
c) Symbolic or j- method.
## 3.20.1 Vector method
In this method the total line current is found by drawing the vector diagram of the
circuit. As voltage is common, it is taken as the reference vector and the various branch
currents are represented vectorially. The total line current can be determined from the
vector diagram either by the parallelogram method or by the method of components.
Branch 1
Impedance Z1 =
.in
Current I1 =
Cos
1= or 1 = cos-1
sy (fig. 3.65).
Current I1 lags behind the applied voltage by
Branch 2
ea
Impedance Z2 =
Current I2 =
y
Cos = cos-1
ud
2
2= or
Current I2 leads V by 2 (fig. 3.65).
st
Resultant current : The total line current I is the vector sum of the branch currents I 1
and I2 and is found by using the parallelogram law of vectors, as shown in fig. 3.65.
The second method is the method of components i.e., resolving the branch
currents I1 and I2 along the x- axis and y- axis and then finding the resultant of these
components (fig. 3.66).
Let the resultant current be I and be its phase angle, as shown in fig. 3.66
(b). Then the components of I along X- axis is equal to the algebraic sum of the
components of branch currents I1 and I2 along the X- axis ( active components).
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Similarly, the component of I along Y- axis is equal to the algebraic sum of the
components of I1 and I2 along Y- axis i.e,
Component of resultant current along Y- axis
= algebraic sum of I1 and I2 along X axis
or I cos
= I1 cos 1 + I2 cos
## Component of resultant current along Y axis
= algebraic sum of I1 and I2 along Y axis
I=
or I sin
.in
= = I1 sin 1 I2 sin
2
and tan =
sy
If tan is positive, current leads and if tan is negative, then the current lags behind
applied voltage V. power factor for the entire circuit
ea
cos =
y
The reciprocal of impedance of a circuit is called its admittance. It is represented
ud
by Y.
Y=
st
So, Y =
Its unit is Siemens (S). A circuit with an impedance of one ohm has an
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## Just as impedance Z of a circuit had two rectangular components, resistance R and
reactance X, admittance Y also has two rectangular components known as conductance g
and susceptance b. fig. 3.67 shows the impendence triangle and the admittance triangle. It
is clear the admittance has two components g and b. The component g along the X- axis
is the conductance which is the reciprocal of resistance. The component b is called
susceptance, which is the reciprocal of reactance.
In fig. 3.67(a), the impedance and admittance triangles for an inductive circuit are
shown. It is apparent that susceptance b is negative, being below X axis. Hence
inductive susceptance is negative. In fig. 3.67 (b), the impedance and admittance
triangles for capacitive circuit is shown. It is evident that susceptance is positive, being
above the X axis; hence, capacitive susceptance is positive.
.in
Relations
Conductance g = Y cos
Or g=
sy
ea
Conductance is always positive.
Susceptance b = Y sin =
y
ud
inductive.
st
## 3.20.3 Application of admittance method
Let us consider a parallel circuit with
three branches, as given in fig. 3.68.
we can determine the conductors by just
adding the conductances of the three
branches. In a like manner, susceptance
is determined by the algebraic addition
of the susceptances of the different
branches.
Total conductance,
G = g1 + g2 + g3
Total susceptance
B = (-b1) + (-b2) + b3
Total current I = VY
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## Let us take the parallel two branch
circuit of fig. 3.69, with the same p.d. across
the two impedances Z1 and Z2 .
I1 = and I2 =
Total current I = I1 + I2 =
=V
= V(Y1 + Y2)
.in
= VY
## Where the total admittance Y = Y1 + Y2
sy
We should note that admittances are added for parallel branches, whereas impedances
are added for series branches. Both admittances and impedances are complex quantities,
so all additions have to be performed in complex form.
ea
In case of the two parallel branches of fig. 3.70,
Y1 = =
y
=
ud
st
## b1 = ----- susceptance of top branch
In similar manner,
Y2 = =
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Total admittance Y = Y1 + Y2
=
= (g1 + g2) j(b1 b2)
= G JB
Y=
= tan-1
0
In polar form, admittance Y = Y
.in
Y= tan-1
## Total current I = VY; I1 = VY1 and I2 VY2
V = V 00 and Y = Y
sy
So I = VY = V 00 Y = VY
ea
Taking a general case,
V=V and Y = Y , then
y
So I = VY = V Y = VY +
ud
current drawn by a pure capacitor of 20 F is 1.382 A from 220V AC supply. What is the
supply frequency ?
st
sol. :
Given
C = 20 F
V = 220volts
Now,
Since it is a pure capacitor, its resistance is zero
Z = -jXc |Z| = |Xc| = 159.18
For capacitor, Xc =
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An EMF whose instantaneous value is 100 sin (314t /4) volts is applied to a circuit
and the . current flowing through it is
20sin(314t-15708) amperes. Find the frequency and the values of
. circuit elements, assuming a series combination of circuit elements.
Sol. : V= 100 sin(314t /4) volts
I= 20sin (314t-15708) A
## Vm= 100volts, Vrms=
.in
Comparing V=100 sin (314t /4) with V= Vm sin(t+)
We have
= 314 sy 2f= 314
ea
y
ud
Now, Z= =
st
## As the current is lagging behind the voltage
the circuit is inductive consisting of resistance
and inductance reactance.
R= Z cos = 5 cos 450 = 3.5355
XL = Z sin = 5 sin450 = 3.5355
## c) An inductive coil draws a current of 2A, when connected to a 230V, 50 Hz supply.
The power taken by the coil is 100 watts. Calculate the resistance and inductance of the
coil.
## Sol. : Let r be resistance of coil while L
Be inductance of coil.
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## Power taken by coil is consumed in
resistive part of coil as inductance never
consumes any part.
P=I2 r r=
r = 25
Now Z= R + jXL
Z2 = R2 +
XL =
.in
But, XL = 2fL
L=
sy
L = 0.3573 H
ea
A single phase voltage of (200+jo) v at 50 Hz is applied to a circuit comprising of a
resistance of 20 inductance of 20 mH and a capacitance of 150 F connected in series.
Find,
y
ud
## ii) Current drawn from supply
iii) Power factor
iv) Power drawn
st
## v) Energy stored in inductor and capacitor
and draw voltage diagramAns.: Given:
R = 20 , L = 20 mH, C = 150 F
## i) Impedance of the circuit:
XL = 2 fL
= 25020
= 6.28
XC =
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= 21.22
Z = R+j (XL XC)
Z = 20+j (6.28-21.22)
Z = 20-j 14.94
Z = 24.96 < - 36.750
## ii) Current drawn from supply:
.in
=
=
= sy
= 8.01 < 36.750 A
ea
= 8.01 A
y
ud
## Let current I be the reference phasor
I = 8.01 < 00 A
VR = I.R. = 8.01 (20)
st
## = 160.2 < 00 volts
VL = I (jXL)
= (8.01 < 00) (j 6.28)
= (8.01 < 00) (6.28 < 900)
= 50.30 < 900 volts
VC = (8.01 < 00) (-j 21.22)
= (8.01 < 00) (21.22 < -900)
= 169.97 < - 900 volts
## As the circuit is capacitive current leads voltage by an angle of 36.75 0
p.f. = cos 36.75 = 0.801 lead
iv) Power drawn:
Power = VI cos
= (200) (8.01) (0.801)
= 1283.20 W
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## A circuit having a resistance of 12 , an inductance of 0.15 H and a capacitance of
100F in series is connected across a 100V, 50 Hz supply. Calculate the impedance,
current, the phase difference between the current and supply voltage.
Sol:
XL = 2FL = 2 x 50 x 0.15
= 100 x x 0.15
XL = 47.12
XC = =
XC = 31.83
Z = R + jXL jXC = 12 + j47.12 j31.83
= 12 + j15.29
V = 100volts = 100 00
.in
Z = 12 + j15.29 = 19.43 51.870
I= = sy
I = 5.146A
= 5.146 51.870 A
## p.f. = cos = cos51.870 = 0.6174 lag, = 51.870
ea
c) Two circuits with impedances of Z1 = 10 + j15 and Z2 = 6-j8 are connected in
parallel. If the supply current is 20A, what is the power dissipated in each branch?
Sol :
Z1 = 10 + j15 = 18.027 56.300
y
Z2 = 6-j8 = 10 53.130
ud
I = 20A
I1 = =
st
= =
I1 = 11.45A
Power consumed or dissipated in branch 1 = = (11.45)2 (10)
= 1312.11W
I2 = =
=
I2 = 20.64A
Power dissipated in branch 2 = = (20.64)2 (6)
= 2558.40W
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## A circuit consists of a resistance of 10 , an inductance of 16 mH and a capacitance of
150 F connected in series. A supply of 100 V at 50 Hz is given to the circuit. Find the
current, power factor and power consumed by the circuit. Draw the vector diagram.
Sol.: R = 10 , L = 16 mH, C = 100 F
XL = 2fL
= 2501610-3
= 5.02
XC =
= 21.22
.in
Z = R+j (XL XC) = 10+ j(5.02-21.22)
Z = 10 j 16.2 sy
Z = 19.03 < - 58.310
ea
= , = = = 5.25 A
y
## Let current I be the reference phasor
ud
I = 5.25 < 00 A
VR = I.R = (5.25) (10) = 52.5 < 00 volts
st
VL = I (jXL) (5.25 < 00) (j 5.02) = (5.25 < 00) (5.02 < 900)
## = 26.35 < 900 volts
VC = (5.25 < 00) (-j 21.22) = (5.25 < 00) (21.22 < -900)
= 111.40 < - 900 volts
The phasor diagram is shown in the Fig.4.
V=I.Z
= (5.25 <00) (19.03 <-58310)
= 100 < - 58310 volts
As the circuit is capacitive current leads voltage
by an angle of 58.310.
## p.f. = cos 58.310 = 0.525 lead
Power = VI cos
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## = (100) (5.02) (0.525)
= 263.71 W
An inductive coil draws a current of 2A, when connected to a 230V, 50 Hz supply. The
power taken by the coil is 100 watts. Calculate the resistance and inductance of the coil.
## Sol. : Let r be resistance of coil while L
Be inductance of coil.
## Power taken by coil is consumed in
.in
resistive part of coil as inductance never
consumes any part.
P=I2 r r=
r = 25
sy
ea
Now Z= R + jXL
Z2 = R2 +
y
XL =
ud
But, XL = 2fL
L=
st
L = 0.3573 H
## Obtain the form factor of a half rectified sinewave.
Ans: The following waveform represented as,
## I) R.M.S valve : Consider the first cycle,
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I R.M.S = =
= =
= = =
## ii) Average value
.in
Iav = = =
= = =
= [ -cos + cos0] =
sy=
ea
iii) Kf = = = = 1.57
iv) KP = = =2
y
ud
## b) A non inductive resistor of 10 is in series with a capacitor of 100 F across a 250
volts, 50 Hz, A.C. supply. Determine the current taken by the capacitor and p.f. of the
st
circuit. [5]
Ans. : The circuit is shown in the Fig. 4.
XC =
= 31.8309
Z = R jXC = 10 j31.8309
Let voltage applied be the reference .
I= = 7.4929 +72.559 A Current
cos = cos(72.559) = 0.3 leading Power factor
## c) A current of average value 18.019 A is following in a circuit to which a voltage of
peak value 141.42 V is applied. Determine i) Impedance in the polar form ii) Power
Assume voltage lags current by 30 . [5]
Ans. : Iav = 18.019 A, Vm = 141.42 V , V lags I by 30.
Assuming purely sinusoidal current and voltage waveforms,
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## I = Kf Iav = 1.11 18.019 = 20 A R.M.S.
value
V= = 100 V
R.M.S. value
Let voltage be reference as shown in the Fig. 5.
V = 100 0 V
I = 20 + 30 A
i) Z=
= 5 -30 Polar form
ii) P = VI cos = 100 20 cos (30)
= 1732.0508 W
d) An a.c. current is given by I = 10 sin t + 3sin 3 t + 2 sin 5 t Find the r.m.s value of
.in
the current. [5]
Ans. : When a coil carries d.c. current and more than one alternating signals then the
total heat produced is the sum of the heats produced by d.c. component and all the
alternating components.
Let R = Resistance of wire
sy
t = Time for which signals which signals are flowing
Htotal = Hdc + H1 + H2 +..
ea
Htotal = Rt where Irms = Total r.m.s value
Hdc = Rt
H1 = R t , H2 = R t .
y
Rt=
ud
Irms =
For given example, Idc = 0
Irms1 = = 7.07106 A
st
Irms2 = = 2.1213 A
Irms3 = = 1.4142 A
Irms = = 7.5166 A
## current drawn by a pure capacitor of 20 F is 1.382 A from 220V AC supply. What is
the supply frequency?
sol. :
Given
C = 20 F
V = 220volts
Now,
Since it is a pure capacitor, its resistance is zero
Z = -jXc |Z| = |Xc| = 159.18
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For capacitor, Xc =
## 2) A parallel circuit comprises a resistor of 20 ohm in series with an inductive
reactance of 15 ohm in one in one branch and a resistor of 30 ohm in series with a
capacitive reactance of 20 ohm in the other branch. Determine the current and power
.in
dissipated in each branch if the total current drawn by the parallel circuit is 10 -300
Amp(8)
Ans.: sy
The arrangement is shown in the Fig.2.
Z12= 20
30 +
= 25 36.8690
15 =36.055 -33.690
j 20
ea
By current division rule,
y
I1= IT =
ud
st
=
=7.1752 A
And =
= = 4.9752 +
A
## Only resistive part of each branch consumes the power given by R
= = 20 = 1029.67 W.
= = 30 = 742.5784 W.
EXPECTED QUESTIONS
1) ExplinGeneration of sinusoidal AC Voltage
2) Derive an Equation of Alternating E.M.F.
Define
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a)Alternating quantity:.
b)Waveform:.
c)Instantaneous value:.
d) Alternation and cycle: cycle.
e) Periodic Time and Frequency:
## 3)DefineRoot-mean-square (R.M.S.) Value
4)DefineAverage Value
6) Show that power in a AC circuit containing pure ohmic resistance only VI
7) Show that power in a AC circuit containing containing pure inductance 0.
8) Show that power in a AC circuit containing containing pure capacitance 0.
## 9) Show that power Series R-L circuit IS VIcos
.in
10) Define Real power, Reactive Power, Apparent power and power Factor
## 11) Show that power Series R C circuit is VIcos
sy
12) Show that power Series R-L C circuit IS VIcos
ea
SOLUTION TO QUESTION BANK
y
## 3) Derive an expression for the instantaneous power in a pure conductor energised by
ud
sinusoidal voltage. Draw the waveshapes of voltage, current and power signals involved.
st
## Consider a simple circuit consisting of a
pure capacitor of c-farads, connected across a
voltage given by the equation, v = Vm sint.
The circuit is shown in Fig. 4.
The current i charges the capacitor c.
The instantaneous charge q on the plates of the
Capacitor is given by
q = Cv
q = CVm sin (t)
Now, current is rate of flow of charge,
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Where Im=
Where
The above equation clearly shows that the current is purely sinusoidal and
having . phase angle + radians i.e. + 900.
.in
sy
nature of the current. If current is assumed reference, we can say that voltage across
capacitor passing through the capacitor by 900.
ea
Fig. 5 shows waveforms of voltage applied by 900. The current and the corresponding
phasor diagram. The current waveform starts earlier by 90 0 in comparison with voltage
waveform. When voltage zero, the current has positive maximum value.
y
ud
st
## 4) A parallel circuit comprises a resistor of 20 ohm in series with an inductive
reactance of 15 ohm in one in one branch and a resistor of 30 ohm in series with a
capacitive reactance of 20 ohm in the other branch. Determine the current and
powerdissipated in each branch if the total current drawn by the parallel circuit is 10 -
300 Amp (8)
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## Ans.: The arrangement is shown in the Fig.2.
Z1= 20 + j 15 = 25 36.8690
Z2= 30 j 20 =36.055 -33.690
## By current division rule,
I1= IT =
.in
=
=7.1752 A
And
A
= sy
= = 4.9752 +
ea
Only resistive part of each branch consumes the power given by R
= = 20 = 1029.67 W.
y
= = 30 = 742.5784 W.
5) With a neat sketch briefly explain how an alternating voltage is produced when a coil
ud
is
rotated in a magnetic field.
Sol: Generation of A.C. voltage :
st
The machines which are used to generate electrical voltages are called generators.
The generators which generate purely sinusoidal a.c. voltages are called alternators.
The basic principle of an alternator the principle of electromagnetic induction. We
have discussed earlier the effect of electromagnetic induction and Faradays Law of
governing this phenomenon. So, sine wave is generated according to Faradays Law of
electromagnetic induction. It says that, whenever there is a relative motion between the
conductor and the magnetic field in which it is kept, an e.m.f. gets induced in the
conductor. The relative motion may exist because of movement of conductors with
respect to magnetic field or movement of magnetic field with respect to conductor. Such
an induced e.m.f. then can be used to supply the electrical load.
Let us see how an alternator produces a sine wave, with the help of simplest form of
an alternator called single turn or single loop alternator.
## Single turn generator:
Construction: It consists of a permanent magnet of two poles. A single turn
rectangular coil is kept in the vicinity of the permanent magnet. The coil made up of
same conducting material like copper or aluminium. The coil is made up of two
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conductors namely a-b and c-d. such two conductors are connected at one end to form a
coil. This is shown in fig.2.
The coil is so placed that it can be rotated about its own axis in clockwise or
.in
anticlockwise direction. The remaining two ends C1 and C2 of the coil are connected to
the rings mounted on the shaft called slip rings are also rotating members of the
alternator. The two brushes P and Q are resting on the slip rings. The brushes are
sy
stationary and making contact with the slip rings. The slip rings and brush assembly is
necessary to collect the current induced in the rotating coil and make it available to the
stationary external resistance. The overall construction is shown in Fig.3.
y ea
ud
st
Working:
The coil is rotated in anti clockwise direction. While rotating, the conductors ab and
cd cut the lines of flux of the permanent magnet. Due to Faradays law of electromagnetic
induction, an e.m.f. gets induced in the conductors. This e.m.f. drives a current through
resistance R connected across the brushes P and Q. The magnitude of the induced e.m.f.
depends on the position of the coil in the magnetic field. Let us see the relation between
magnitude of the induced e.m.f. and the positions of the coil. Consider different instants
and the different positions of the coil.
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Instant 1 : Let the initial position of the coil be as shown in Fig. 4. The plane of the
coil is perpendicular to the direction of the magnetic field. The instantaneous component
of velocity of conductors ab and cd, is parallel to the magnetic field as shown and there
cannot be the cutting of the flux lines by the conductors. Hence, no e.m.f. will be
generated in the conductors ab and cd and not current will flow through the external
resistance R. this position can be represented by considering the front view of the Fig. 4
as shown in Fig. 4(a).
The angle is measured from plane of the magnetic flux.
.in
sy
y ea
ud
st
Instant 2 : When the coil is rotated in anticlockwise direction through some angle ,
then the velocity will have two components v sin perpendicular to flux lines and v cos
parallel to the flux lines. Due to sin component, there will be cutting of the flux and
proportionally, there will be induced e.m.f. in the conductors ab and cd. This e.m.f. will
drive a current through the external resistance R. This is shown in Fig.4(b).
## 6) Derive an expression for the instantaneous power in a pure conductor energised by
sinusoidal voltage. Draw the waveshapes of voltage, current and power signals involved.
## Sol. : A.C. through pure capacitors:
Consider a simple circuit consisting of a
pure capacitor of c-farads, connected across a
voltage given by the equation, v = Vm sint.
The circuit is shown in Fig. 4.
The current i charges the capacitor c.
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## The instantaneous charge q on the plates of
the
Capacitor is given by
q = Cv
q = CVm sin (t)
Now, current is rate of flow of charge,
.in
Where Im= sy
Where
ea
The above equation clearly shows that the current is purely sinusoidal and
having . phase angle + radians i.e. + 900.
y
ud
nature of the current. If current is assumed reference, we can say that voltage across
capacitor passing through the capacitor by 900.
st
Fig. 5 shows waveforms of voltage applied by 900. The current and the corresponding
phasor diagram. The current waveform starts earlier by 90 0 in comparison with voltage
waveform. When voltage zero, the current has positive maximum value.
In purely capacitive circuit, current leads voltage by 900.
Please refer fig. 5 on next page.
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## Concept of capacitive reactance :
We have seen while expressing current equation in the standard form that
## The term is called capacitive reactance and is measured in ohms.
So capacitive reactance is defined as the
opposition offered by the capacitance of a circuit
to the flow of an alternating sinusoidal current.
Xc is measured in ohms and it depends on the frequency
.in
of the applied voltage.
The capacitive reactance is inversely
proportional to the Frequency for constant
C.
sy
ea
Xc for constant C
The graph of Xc Vs F is a rectangular hyperbola as shown in fig. 6.
y
If the frequency is zero, which is so for d.c. voltage, the capacitive reactance is
Infinity. Therefore , it is said that the capacitance offers open circuit to the d.c. or it
ud
block d.c.
Power: The expression for the instantaneous power can be obtained by talking the
st
## product of instantaneous voltage and current.
Thus, power curve is a shine wave of frequency double that of applied voltage. The
average value sine curve over a complete cycle is always zero.
The fig. 7 shows waveforms of current, voltage and power. It can be observed from
the figure that when power curve is positive, in practice, an electrostatic energy gets
stored in the capacitor during its changing while the negative power curve represents that
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the energy stored is turned back to the supply during its discharging. The areas of positive
and negative loops are exactly the same and hence, average power consumption is zero.
.in
sy
Pure capacitance never consumes power.
ea
7) An EMF whose instantaneous value is 100 sin (314t /4) volts is applied to a circuit
and the current flowing through it is 20sin(314t-15708) amperes. Find the frequency and
the values of circuit elements, assuming a series combination of circuit elements.
y
## Sol. : V= 100 sin(314t /4) volts
ud
I= 20sin (314t-15708) A
Vm= 100volts, Vrms=
st
We have
= 314 2f= 314
## I = 14.1421 900 A as 1.5708 rad 900
Angle between V and I = 900 450
= 450, cos = 0.7071
Now, Z= =
As the current is lagging behind the voltage
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## the circuit is inductive consisting of resistance
and inductance reactance.
R= Z cos = 5 cos 450 = 3.5355
XL = Z sin = 5 sin450 = 3.5355
## 8) An inductive coil draws a current of 2A, when connected to a 230V, 50 Hz supply.
The power taken by the coil is 100 watts. Calculate the resistance and inductance of the
coil.
## Sol. : Let r be resistance of coil while L
Be inductance of coil.
.in
Power taken by coil is consumed in
sy
resistive part of coil as inductance never
consumes any part.
ea
P=I2 r r=
r = 25
y
Now Z= R + jXL
ud
Z2 = R2 +
st
XL =
But, XL = 2fL
L=
L = 0.3573 H
9) Derive expressions for average value and RMS value of a sinusoidally varying AC
voltage.
Sol : Analytical method to find rms value of alternating quantity.
Consider sinusoidally varying alternating current and square of this current as shown
in
Fig. 5.
Please refer Fig.5 on next page.
The current I = Im sin
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## while square of current
i2 =
.in
Area of curve over half a cycle be calculated by considering an interval d as shown.
sy
Area of square curve over half cycle = d and length of the base is
Average value of square of the current over half cycle
ea
= =
2
= d = sin d
y
= d
ud
= 2 =
=
Hence, root mean square value i.e. r.m.f. value can be calculated
st
as,
=
Ir.m.f. =
Ir.m.f. = 0.707 Im
The r.m.s. value of the sinusoidal alternating current is 0.707 times the maximum or
peak value or amplitude of that alternating current.
Importance of R.M.S value :
In case of alternating quantities, the values are used for specifying magnitudes of
alternating quantities.
In practice, everywhere, r.m.f. values are used to analyze alternating quantities. The
ammeters and voltmeters record the r.m.f. value of current and voltage respectively.
9)Average value :
The average value of an alternating quantity is defined as that value which is
obtained by averaging all the instantaneous values over a period of half cycle.
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For a symmetrical a.c, the average value over a complete cycle is zero as both
positive and negative half cycle are exactly identical. Hence, the average value is defined
for half cycle only.
Average value can also be expressed by that steady current which transfers across
any circuit, the same amount of charge as is transferred by that alternating current during
the same time. The average value for sinusoidally varying alternating current can be
obtained by,
1) Graphical Method and 2) Analytical method
Analytical method:
For an unsymmetrical a.c., the average value must be obtained for one complete cycle
but symmetrical a.c. like sinusoidal, it is to be obtained for half cycle.
.in
sy
y ea
ud
## Consider the elementary interval of instant d as shown in Fig.6. The average
instantaneous value of current in this interval is say, i as shown.
The average value can obtained by taking ratio of area under cure over half cycle to
st
## length of the base for half cycle.
Iav = =
= d = sind =
= =
= =
Iav =
i.e. Iav = 0.637Im
Similarly, Vav = 0.637Vm
The average value is used for application like battery charging, etc. It is rarely used in
practice.
10) A circuit having a resistance of 12 , an inductance of 0.15 H and a capacitance
of
100F in series is connected across a 100V, 50 Hz supply. Calculate the
impedance, current, the phase difference between the current and supply voltage.
Sol:
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XL = 2FL = 2 x 50 x 0.15
= 100 x x 0.15
XL = 47.12
XC = =
XC = 31.83
Z = R + jXL jXC = 12 + j47.12 j31.83
= 12 + j15.29
V = 100volts = 100 00
Z = 12 + j15.29 = 19.43 51.870
I= = = 5.146 51.870 A
I = 5.146A
p.f. = cos = cos51.870 = 0.6174 lag, = 51.870
11) Two circuits with impedances of Z1 = 10 + j15 and Z2 = 6-j8 are connected in
.in
parallel. If the supply current is 20A, what is the power dissipated in each branch?
Sol :
Z1 = 10 + j15 = 18.027 56.300
Z2 = 6-j8 = 10 53.130
I = 20A
sy
I1 = =
ea
= =
I1 = 11.45A
y
## Power consumed or dissipated in branch 1 = = (11.45)2 (10)
= 1312.11W
ud
I2 = =
=
st
I2 = 20.64A
Power dissipated in branch 2 = = (20.64)2 (6)
= 2558.40W
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Compiled by www.studyeasy.in
## Unit-3: Three Phase Circuits
Necessity and advantages of three phase systems, generation of three phase power,
definition of Phase sequence, balanced supply and balanced load. Relationship between
line and phase values of balanced star and delta connections. Power in balanced three-
phase circuits, measurement of power by two-wattmeter method. Illustrative examples
## 3.23 Necessity and Advantages of three-phase systems.
In a single phase a.c. generation, a number of armature coils are connected to from one
winding, which when rotated in a magnetic field generates a voltage called single-phase
voltage. But it is found that in many applications, the single-phase system is not very
satisfactory. For instance, a single-phase induction motor is not self-starting unless it is
fitted with an anxiliary winding. Hence it is found that a three-phase induction motor is
.in
more suitable as it is self-starting and has better efficiency and power factor than its
single-phase counterpart.
sy
The advantages of three phase systems are as follows:
1. The output of a three-phase machine is always greater than that of a single-phase
machine of the same size of frame. So, for a given size and voltage, a three-phase
ea
machine (e.g. a 3-phase alternator) occupies less space and costs less than the
single-phase machine having the same rating.
2. To transmit and distribute a given amount of power over a given distance, a
three- phase transmission line requires less copper than a single-phase line.
y
ud
## motors (except commutator motors) have a pulsating torque.
4. Single-phase motors (except commutator motors) are not self-starting. Three-
phase motors are self-starting.
5. The pulsating nature of the armature reaction in single-phase alternators causes
st
difficulty with parallel running unless the poles are fitted with exceedingly heavy
dampers. Three-phase generators work in parallel without difficulty.
6. The connection of single-phase generators in parallel gives rise to harmonics,
whereas three-phase generators can be conveniently connected with causing
generation of harmonics.
7. In the case of a three-phase star system, two different voltages can be obtained,
one between lines and the other between line and phase, whereas in the case of a
single phase system only one voltage can be obtained.
8. In a single-phase system, the instantaneous power is a function of time and hence
fluctuates with respect to time. This fluctuating power causes considerable
vibrations in single-phase motors. Hence performance of single-phase motors in
poor; Whereas instantaneous power in a symmetrical three-phase system is
constant.
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## 3.24 Generation of 3-phase E.M.F.
In the 3-phase system, there are three equal voltages of the same frequency but
displaced from one another by 1200 electrical. These voltages are produced by a
three-phase generator which has three identical windings or phases displaced 120 0
electrical apart. When these windings are rotated in a magnetic field, e.m.f. is induced
in each winding or phase. These e.m.f. s are of the same magnitude and frequency but
are displaced from one another by 1200 electrical.
Consider three electrical coils mounted on the same axis but
0
displaced from each other by 120 electrical. Let the three coils be rotated in an
anticlockwise direction in a bipolar magnetic field with an angular velocity of
radians/sec, as shown in Fig. 3.80. Here, , are the start terminals and
are the end terminals of the coils.
When the coil is in the position AB shown in Fig. 3.80, the magnitude and
.in
direction of the e.m.f. s induced in the various coils is as under:
sy
y ea
ud
st
a) E.m.f. induced in coil is zero and is increasing in the positive direction. This
is indicated by wave in Fig. 3.80 (b).
0
b) The coil is 120 electrically behind coil The e.m.f. induced in this coil
is negative and is approaching maximum negative value. This is shown by the
wave.
c) The coil is 2400 electrically behind or 1200 electrically behind coil .
The e.m.f. induced in this coil is positive and is decreasing. This is indicated by
wave .
Thus, it is apparent that the e.m.f. s induced in the three coils are of the same
magnitude and frequency but displaced 1200 electrical from each other.
Vector Diagram: The r.m.s. values of the three phase voltage are shown
vectorially in Fig. 3.80(c).
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= sin t
= sin ; = sin
## 3.25 Meaning of phase sequence
The order in which the voltages in the voltages in the phases reach their maximum
positive values is called the phase sequence. For example, in Fig. 3.80(a), the three coils
, and are rotating in anticlockwise direction in the magnetic field. The
coil is 1200 electrical ahead of coil and 2400 electrical ahead of coil .
0
Therefore, e.m.f. in coil leads the e.m.f. in coil by 120 and that in coil
by 2400. It is evident from Fig. 3.80(b) that attains maximum positive first, then
and . In other words, the order in which the e.m.f. s in the three phases
, and attain their maximum positive values is a,b,c. Hence, the phase sequence
.in
is a,b,c.
## 3.25.1 Naming the phases
sy
The 3 phases may be numbered (1,2,3) or lettered (a,b,c) or specified colours (R Y
B). By normal convention, sequence RYB is considered positive and R B Y negative.
ea
3.26 25 Meaning of phase sequence
It is necessary to employ some systematic notation for the solution of a.c. circuits and
y
systems containing a number of e.m.f. s. acting and currents flowing so that the process
of solution is simplified and less prone to errors.
ud
## It is normally preferred to employ double-subscript notation while dealing with a.c.
electrical circuits. In this system, the order in which the subscripts are written indicates
the direction in which e.m.f. acts or current flows.
st
## For example, if e.m.f. is expressed as
it indicates that e.m.f. acts from a to b; if it is
expressed as , then the e.m.f. acts in a direction
opposite to that in which acts. (Fig. 3.81)
i.e., =- .
Similarly, Iab indicates that current flows in the direction from a to b but I ba indicates
that current flows in the direction from b to a; i.e., I ba = -Iab.
## 3.27 Balanced Supply and Load
When a balanced generating supply, where the three phase voltages are equal, and
the phase difference is 1200 between one another, supplies balanced equipment load,
where the impedance of the three phases or three circuit loads are equal, then the current
flowing through these three phases will also be equal in magnitude, and will also have a
phase difference of 1200 with one another. Such an arrangement is called a balanced load.
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3.28 Obtaining Relationship between Line & Phase Values & Expression for power
for Balanced Star Connection
This system is obtained by joining together
similar ends, either the start or the finish; the other
ends are joined to the line wires, as shown in
Fig.3.82(a). the common point N at which similar
(start or finish) ends are connected is called the
neutral or star point. Normally, only three wires
are carried to the external circuit, giving a 3-phase,
3-wire, star-connected system; however, sometimes
a fourth wire known as neutral wire, is carried to the
neutral point of the external load circuit, giving a
3-phase, 4-wire connected system.
The voltage between any line and the neutral
point, i.e., voltage across the phase winding, is
.in
called the phase voltage; while the voltage between
any two outers is called line voltage. Usually,
the neutral point is connected to earth.
sy
In Fig.3.82(a), positive directions of e.m.f.s. are
taken star point outwards. The arrow heads on
e.m.f.s. and currents indicate the positive direction.
ea
Here, the 3-phases are numbered as usual: R,Y and
B indicate the three natural colours red, yellow
and blue respectively.
y
## By convention, sequence RYB is taken as positive and RYB as negative.
In Fig.3.82(b), the e.m.f.s induced in the three phases, are shown vectorially. In a
ud
star-connection there are two windings between each pair of outers and due to joining of
similar ends together, the e.m.f.s induced in them are in opposition.
Hence the potential difference between the two outers, know as line voltage, is the
st
## vector difference of phase e.m.f.s of the two phases concerned.
For example, the potential difference between outers R and Y or
Line voltage ERY, is the vector difference of phase e.m.f.s E R and EY or vector sum of
phase e.m.f.s ER and (-EY).
i.e. ERY = ER- EY (vector difference)
or ERY = ER + (-EY) (vector sum)
as phase angle between vectors ER and (-EY) is 600,
from vector diagram shown in Fig.3.82(b),
ERY =
Let ER = EY = EB = EP (phase voltage)
Then line voltage ERy = = EP
Similarly, potential difference between outers Y and B or line. Voltage E YB = EY EB
= EP and potential difference between outers B and R, or line voltage E BR = EB ER =
EP.
In a balanced star system, ERY, EYB and EBR are equal in magnitude and are called line
voltages.
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EL = EP
Since, in a star-connected system, each line conductor is connected to a separate
phase, so the current flowing through the lines and phases are the same.
i.e. Line current IL = phase current IP
if the phase current has a phase difference of with the voltage,
power output per phase = EPIP cos
total power output, P = 3EPIP cos
= 3 IP cos
= EL IL cos
i.e. power = x line voltage x line current x power factor
apparent power of 3-phase star-connected system
= 3 x apparent power per phase
= 3EPIP = 3x x IL = E L IL
.in
3.29 Obtaining Relationship between Line and Phase Values and Expression for
Power for Balanced Delta Connection
sy
When the starting end of one coil is connection to the finishing end of another coil,
as shown in Fig.3.83(a), delta or mesh connection is obtained. The direction of the e.m.f.s
is as shown in the diagram.
ea
From Fig.3.83 it is clear that line current is the vector difference of phase currents of
the two phases concerned. For example, the line current in red outer I R will be equal to
the vector difference of phase currents I YR and IRB. The current vectors are shown in
Fig.3.83(b).
y
ud
st
## Referring to Fig.3.83(a) and (b),
Line current, IR = IYR IRB (vector difference)
= IYR + (-IRB) (vector sum)
As the phase angle between currents IYR and IRB is 600
IR =
For a balanced load, the phase current in each winding is equal and let it be = I P.
Line current, IR = = IP
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## Similarly, line current, IY = IBY IYR = IP
And line current, IB = IRB IBY = IP
In a delta network, there is only one phase between any pair of line outers, so the
potential difference between the outers, called the line voltage, is equal to phase voltage.
i.e. Line voltage, EL = phase voltage, EP
Power output per phase = EPIP cos ,
where cos is the power factor of the load.
Total power output, P = 3EPIP cos
= 3EL cos
= ELIL cos
i.e. Total power output = x Line voltage x Line current x p.f.
apparent power of 3-phase delta-connected system
= 3 x apparent power per phase
.in
= 3 EPIP = 3EL ELIL
Each of the two wattmeters connected to measure the input to a three phase reads 20
sy
kW. What does each instrument reads, when the load p.f. is 0.866 lagging with the total
three
phase power remaining unchanged in the altered condition?
ea
[6]
## sol. : Each wattmeters reads 20 kW total=power
00 = 40 kW
y
As both
Now p.f.wattmeters
ud
=1
Cos new = 0.866 new = 300
st
## We have, = here = new
As the total power in 3ph circuit remains same W1 + W2 = 410 ---
(i)
300 =
= W1 W2
W1 W2 = 13.33
(i) + (ii) gives,
2W1 = 53.33
W1 = 26.66 kW
From (i), W2 = 40 W1 = 40 26.66 = 13.33
W2 = 13.33 kW
) A balanced star connected load of (8 + jb) is connected to a 3 phase, 230 V supply.
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Find the line current, power factor, power, reactive voltamperes and total
voltamperes.
Sol: VL = 230 volts
Zph = 8+j6 = 10 36.860
Vph = = = 132.79V
Iph = = = 13.27A
IL Iph = 13.270A
p.f. = cos = cos36.860 = 0.8lag
p = 3Vph Iph cos = 3x32.79x13.27x0.8 = 4229.09W
Reactive power, Q = 3Vph Iph sin = 3x132.79x13.27x0.6
Q = 3171.82VAR
Total voltamperes, S = 3Vph Iph = 3x132.79x13.27
.in
S = 5286.36
sy
A star connected load consists of 6 resistances in series with an 8 inductive reactance
in each phase. A supply voltage of 440 V at 50 Hz is applied to the load. Find the line
current, power factor and power consumed by the load. [8]
ea
Sol.: VL = 440 volts, f=50Hz
RPh = 6
y
XPh = 8
ud
st
VPh =
IPh = IL =
= 25.4 A
## Power factor = cos 53.130 = 0.6 lagging
P= cos = 44025.40.6 = 11.614 KW
L = 0.3573 H
Each of the two wattmeters connected to measure the input to a three phase reads 20 kW.
What does each instrument reads, when the load p.f. is 0.866 lagging with the total three
phase power remaining unchanged in the altered condition? [6]
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## sol. : Each wattmeters reads 20 kW total power = 40 kW
As both wattmeters reads same cos =1 = 00
Now p.f. is 0.866 lag
Cos new = 0.866 new = 300
## We have, = here = new
As the total power in 3ph circuit remains same W1 + W2 = 410 ----
(i)
300 =
.in
= W1 W2
W1 W2 = 13.33
(i) + (ii) gives,
sy
2W1 = 53.33
W1 = 26.66 kW
ea
From (i), W2 = 40 W1 = 40 26.66 = 13.33
W2 = 13.33 kW
y
A 4 pole generator with wave wound armature has 51 slots, each having 24 conductors.
The flux per pole is 0.01 Weber. At what speed must the armature rotate to give an
ud
induced emf of 220 V? What will be the voltage developed if the voltage developed if the
winding is lap and the armature rotates at the same speed? [6]
st
## sol.: P = 4, for wave winding A = 2
Number of slots = 51, conductors/slot = 24
= 0.01 wb
E = 220 volts
Z = Number of slots conductors/slots
= 51 24
= 1224
E=
N=
=
N = 539.21 rpm
Now speed is same but winding is lap A=P
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E=
E = 110 volts
A 4 pole, 220 V. lap connected DC shunt motor has 36 slots, each slot containing 16
conductors. It draws a current of 40 A from the supply. The field resistance and armature
resistance are 110 , 0.1 respectively. The motor develops an output power of 6 kW.
The flux per pole is 40 mwb. Calculate
a) the speed
b) the torque developed by the armature and
c) the shaft torque [7]
## Sol.: P = 4, V = 220 volts, slots = 36, A=P=4
.in
Conductors/slot = 16, Z = 36 16 = 576
IL = 40 A
R sh = 110 = 0.1
P output = 6 KW, = 40 wb
Ish =
sy
ea
IL = Ia+Ish
Ia = IL - Ish
= 40 2 = 38A
Eb = V - IaRa
y
ud
= 220-3.8
Eb = 216.2 volts
st
N= =
Now, Eb =
N = 563 rpm
Ta = 0.159 Z
= 0.159 ZIa
T a = 139.20 N-m
P out = T sh .
## Tsh = = 101.76 N-m
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Compiled by www.studyeasy.in
T sh = 101.76 N-m
A 120V DC shunt motor has an armature resistance of 60 . It runs at 1800 RPM, when it
takes full load current of 40A. Find the speed of the motor while it is operating at half
Sol.:
,
IL1 = Ia1 + If
Ia1 = IL1- If = 40 2 = 38A
## = 120 7.6 = 112.4V
N1 = 1800rpm
.in
Now for shunt machine
T Ia as remains constant.
T2 = T1
sy
y ea
ud
## Ia2 = Ia1 = (38) = 19A
Eb2 = V Ia2 Ra = 120 (19)(0.2) = 120 3.8
st
Eb2 = 116.2V
Now, Eb N for
constant
N2 = 1860.85 rpm
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RECOMMENDED QUESTIONS
## 1) Necessity and Advantages of three-phase systems.
2) Explain Generation of 3-phase E.M.F
3) Define phase sequence
4) Obtain Relationship between Line & Phase Values & Expression for power for
Balanced Star Connection
5) Obtain Relationship between Line and Phase Values and Expression for Power
for Balanced Delta Connection
6) Explain Measurement of power in 3-Phase Circuits Two Wattmeter Method
7) Explain Measurement of Power Factor Balanced 3-phase Load
1) What are the advantages of a three phase system over a single phase system?
.in
Sol: Advantages of three phase system:
In the three phase system, the alternator armature has three windings and it
sy
produces three independent alternating voltages. The magnitude and frequency of all of
them is equal but they have a phase difference of 1200 between each other. Such a three
phase system has following advantages over single phase system:
ea
1) The output of three phase machine is always greater than single phase machine of
same size, approximately 1.5 times. So for a given size and voltage a three phase
alternator occupies less space and has less cost too than single phase having same
y
rating.
2) For a transmission and distribution, three phase system needs less copper or less
ud
conducting material than single phase system for given volt amperes and voltage
rating so transmission becomes very much economical.
3) It is possible to produce rotating magnetic field with stationary coils by using
st
three phase system. Hence three phase motors are self starting.
4) In single phase system, the instantaneous power is a function of time and hence
fluctuates w.r.t. time. This fluctuating power causes considerable vibrations in
single phase motors. Hence performance of single phase motors is poor. While
instantaneous power in symmetrical three phase system is constant.
5) Three phase system give steady output.
6) Single phase supply can be obtained from three phase but three phase can not be
obtained
From single phase.
7) Power factor of single phase motors is poor than three phase motors of same
rating.
8) For converting machines like rectifiers, the d.c. output voltage becomes smoother
if number of phases are increased.
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But it is found that optimum number of phases required to get all above said
advantages is three. Any further increase in number of phases cause a lot of
complications. Hence three phase system is accepted as standard polyphase system
throughout the world.
2) With a neat circuit diagram and a vector diagram prove that two wattmeters are
sufficient to measure total power in a 3 phase system.
.
c) A balanced star connected load of (8 + jb) is connected to a 3 phase, 230 V
supply.
Find the line current, power factor, power, reactive voltamperes and total
voltamperes.
Sol: VL = 230 volts
Zph = 8+j6 = 10 36.860
Vph = = = 132.79V
.in
Iph = = = 13.27A
IL Iph = 13.270A
p.f. = cos = cos36.860 = 0.8lag sy
p = 3Vph Iph cos = 3x32.79x13.27x0.8 = 4229.09W
Reactive power, Q = 3Vph Iph sin = 3x132.79x13.27x0.6
ea
Q = 3171.82VAR
Total voltamperes, S = 3Vph Iph = 3x132.79x13.27
S = 5286.36
3)obtain the relationship between line and phase values of current in a three phase
y
ud
## Sol. : Relation for delta connected load:
Consider the balanced delta connected load as shown in the fig. 10
st
## Fig. 10 Delta connected load
Lines voltage VL = VRY = VYB = VBR
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Lines currents I L = I R = IY = I B
Phase voltages Vph = VRY = VYB = VBR
Phase currents IPh = IRY = IYB = IBR
As seen earlier, Vph = VL for delta connected load. To derive the relation between I L
and IPh apply the KCL at the node R of the load shown in the fig. 10.
IR + IBR = IRY
IR = IRY IBR
Apply KCL at node Y and B we can write equations for line currents I Y and IB as,
## The phasor diagram to obtain line
Current IR by carrying out vector subtraction
.in
Of phase currents IRY and IYB is shown in the fig. 11.
## The perpendicular AC drawn on
vector OB, bisects the vector OB which
represents IL.
sy
Fig. 11
ea
0
similarly OB bisects angle between - IYB and IRY which is 60
BOA = 300
And OC = CB =
y
ud
st
IL=
## for delta connection
Again Zph decides wheather Iph has to lag, lead to remain in phase with Vph. Angle
between Vph and Iph is .
The complete phasor diagram for cos lagging power factor load is shown in the
fig. 12.
Fig. 12
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## Zph = Rph + jXLph = |Zph|
Each Iph lags respective Vph by the angle
Power: Power consumed in each phase is single phase power given by,
Pph = Vph Iph cos
Total power p = 3 Pph = 3 Vph Iph cos = 3VL
4) Show that in a three phase, balanced circuit, two wattmeters are sufficient to
measure the total three phase power and power factor of the circuit.
Sol. : Two wattmeter method: The current coils of the two wattmeters are connected in
.in
any two lines while the voltage coil of each wattmeters is connected between its own
current coil terminal and line without current coil. Consider star connected balanced load
and two wattmeters connected as shown in fig. 13. Let us consider the rms values of the
sy
currents and voltages to prove that sum of two wattmeter gives total power consumed by
y ea
ud
st
VRB = VR VB
VYB = VY VB
VR^IR =
VR = VY = VB = Vph
VRB = VR VB , IR = IY = IL = Iph
VYB = VY VB = VRB = VL
From fig. 14, IR^VRB = 30 and IR^VRB = 30 +
W1 = IR VRB cos ( 30 ) = VL IL cos ( 30 )
W2= IY VYB cos ( 30 + ) = VL IL cos ( 30 + )
W1 + W2 = VL IL[ cos ( 30 ) + cos ( 30 + ) ]
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## = 2VLIL cos30 cos = 2VLIL cos
= VLIL cos
= total power
5) Each of the two wattmeters connected to measure the input to a three phase reads 20
kW. What does each instrument reads, when the load p.f. is 0.866 lagging with the total
three phase power remaining unchanged in the altered condition? [6]
## sol. : Each wattmeters reads 20 kW total power = 40 kW
As both wattmeters reads same cos = 00
Now p.f. is 0.866 lag =1
.in
Cos new = 0.866 new = 300
## We have, = here = new
(i)
sy
As the total power in 3ph circuit remains same W1 + W2 = 410 -------
ea
300 =
y
= W1 W2
ud
W1 W2 = 13.33
(i) + (ii) gives,
st
2W1 = 53.33
W1 = 26.66 kW
From (i), W2 = 40 W1 = 40 26.66 = 13.33
W2 = 13.33 kW
6) Three similar coils each having resistance of 10 ohm and reactance of 8 ohm are
connected in star across a 400 V, 3 phase supply . Determine the i) Line current; ii) Total
power and iii) Reading of each of two wattmeters connected to measure the power. (8)
## Ans.: R=10 , = 8 , = 400 V, star
= 10 + j 8 = 12.082
= = = 230.94 V,
star
i) = = = 18.0334 V,
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Compiled by www.studyeasy.in
= = 18.0334 A.
ii) = cos where =
38.6590
= 400 18.0334 cos
= 9756.2116 W
iii) = cos (30- ) = 400 18.0334 cos ( - )
= 7131.1412 W
= cos (30+ ) = 400 18.0334 cos ( + )
= 2625.0704 W
## 7) A balanced star connected load of (8 + jb)
is connected to a 3 phase, 230 V supply.
Find the line current, power factor, power, reactive voltamperes and total
voltamperes.
.in
Sol: VL = 230 volts
Zph = 8+j6 = 10 36.860
Vph = = = 132.79V
Iph = = = 13.27A
sy
IL Iph = 13.270A
ea
p.f. = cos = cos36.860 = 0.8lag
8) obtain the relationship between line and phase values of current in a three phase
y
ud
## Sol. : Relation for delta connected load:
Consider the balanced delta connected load as shown in the fig. 10
st
## Fig. 10 Delta connected load
Lines voltage VL = VRY = VYB = VBR
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Lines currents I L = I R = IY = I B
Phase voltages Vph = VRY = VYB = VBR
Phase currents IPh = IRY = IYB = IBR
As seen earlier, Vph = VL for delta connected load. To derive the relation between I L
and IPh apply the KCL at the node R of the load shown in the fig. 10.
IR + IBR = IRY
IR = IRY IBR
Apply KCL at node Y and B we can write equations for line currents IY and IB as,
## The phasor diagram to obtain line
Current IR by carrying out vector subtraction
.in
Of phase currents IRY and IYB is shown in the fig. 11.
## The perpendicular AC drawn on
vector OB, bisects the vector OB which
represents IL.
sy
Fig. 11
ea
0
similarly OB bisects angle between - IYB and IRY which is 60
y
ud
st
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Compiled by www.studyeasy.in
UNIT-4
## Measuring Instruments: Construction and Principle of operation of dynamometer type
wattmeter and single-phase induction type energy meter (problems excluded).
03 Hours
4b) Domestic Wiring. Brief discussion on Cleat, Casing & Capping and conduit
(concealed) wiring. Two-way and three-way control of a lamp. Elementary discussion on
fuse and Miniature Circuit Breaker (MCBs). Electric shock, precautions against shock
Earthing: Pipe and Plate
UNIT-4
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DOMESTIC WIRING
Introduction sy
A network of wires drawn connecting the meter board to the various energy consuming loads
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(lamps, fans, motors etc) through control and protective devices for efficient distribution of
power is known as electrical wiring.
Electrical wiring done in residential and commercial buildings to provide power for lights,
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fans, pumps and other domestic appliances is known as domestic wiring. There are several
wiring systems in practice. They can be classified into:
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Types of wiring: Depending upon the above factors various types of wiring used in
practice are:
1. Cleat wiring
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2. Casing wiring
3. Surface wiring
4. Conduit wiring
i ) Clear wiring:
In this type V.I.R or P.V.C wires are clamped between porcelain cleats.
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The cleats are made up of two halves. One half is grooved through which wire passes
while the other fits over the first. The whole assembly is then mounted on the wall or
wooden beam with the help of screws.
This method is one of the cheapest method and most suitable for temporary work. It
can be very quickly installed and can be recovered without any damage of material.
Inspection and changes can be made very easily.
This method does not give attractive appearance. After some time due to sagging at
some places, it looks shabby. Dust and dirt collects on the cleats. The wires are directly
exposed to atmospheric conditions like moisture, chemical fumes etc. maintenance cost is
very high.
Due to these disadvantages this type is not suitable for permanent jobs.
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ii) Casing capping: This is very popularly used for residential buildings. In this
method, casing is a rectangular strip made from teak wood or new a days made up of
P.V.C. It has two grooves into which the wires are laid. Then casing is covered with a
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rectangular strip of wood or P.V.C. of the same width, called capping. The capping is
screwed into casing is fixed to the walls the help or porcelain discs or cleats.
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Good protection to the conductors from dangerous atmospheric conditions, neat and clean
appearance are the advantages of this type.
In case of wooden casing capping, there is high risk of fire along with the requirement of
skilled labour. The method is costly.
Surface wiring: in this type, the wooden battens are fixed on the surface of the wall,
by means of screws and rawl plugs. The metal clips are provided with the battens at
regular intervals. The wire runs on the batten and is clamped on the batten using the
metal clips. The wires used may lead sheathed wires or can tyre sheathed wires.
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Depending upon type of wire used surface wiring is also called lead Sheathed wiring or
cab tyre sheathed wiring. If the wire used is though rubber Sheathed then it is called
T.R.S. wiringWhile if the wire used is cab tyre SheathedThen it is called C.T.S wiring.
## Conduit wiring: In this method, metallic
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tubes called as conduits are used to run the
wires. This is the best system of wiring as
it gives full mechanical protection to the wires.
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This is most desirable for workshops and public
Buildings. Depending on whether the conduits
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are laid inside the walls or supported on the
walls, there are two types of conduit wiring
which are :
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i) Surface conduit wiring: in this method conduits are mounted or supported on the
walls with the help of pipe books or saddles. In damp situations, the conduits are
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## spaced apart from the wall by means of wooden blocks.
ii) Concealed conduit wiring: In this method, the conduit are buried under the wall
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aty the some of plastering. This is also called recessed conduit wiring.
The beauty of the premises is maintained due to conduit wiring. It is durable and
has long life. It protects the wires from mechanical shocks and fire hazards. Proper
earthing of conduits makes the method electrical shock proof. It requires very less
maintenance.
The repairs are very difficult in case of concealed conduit wiring. This method is
most costly and erection requires highly skilled labour. These are few disadvantages
of the conduit type of wiring. In concealed conduit wiring, keeping conduit at earth
potential is must.
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Compiled by www.studyeasy.in
## FACTORS AFFECTING THE CHOICE OF WIRING SYSTEM:
The choice of wiring system for a particular installation depends on technical factors and
economic viability.
## 1. Durability: Type of wiring selected should conform to standard specifications, so that
it is durable i.e. without being affected by the weather conditions, fumes etc.
2. Safety: The wiring must provide safety against leakage, shock and fire hazards for the
operating personnel.
3.Appearance: Electrical wiring should give an aesthetic appeal to the interiors.
4. Cost: It should not be prohibitively expensive.
5. Accessibility: The switches and plug points provided should be easily accessible.
There must be provision for further extension of the wiring system, if necessary.
6 Maintenance Cost: The maintenance cost should be a minimum
7. Mechanical safety: The wiring must be protected against any mechanical damage
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Specification of Wires:
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The conductor material, insulation, size and the number of cores, specifies the electrical
wires. These are important parameters as they determine the current and voltage handling
capability of the wires. The conductors are usually of either copper or aluminum. Various
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insulating materials like PVC, TRS, and VIR are used. The wires may be of single strand
or multi strand. Wires with combination of different diameters and the number of cores or
strands are available.
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## For example: The VIR conductors are specified as 1/20, 3/22,.7/20
The numerator indicates the number of strands while the denominator corresponds to the
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diameter of the wire in SWG (Standard Wire Gauge). SWG 20 corresponds to a wire of
diameter 0.914mm, while SWG 22 corresponds to a wire of diameter 0.737 mm.
A 7/0 wire means, it is a 7-cored wire of diameter 12.7mm (0.5 inch). The selection of
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the wire is made depending on the requirement considering factors like current and
voltage ratings, cost and application.
## Example: Application: domestic wiring
1. Lighting - 3/20 copper wire
2. Heating - 7/20 copper wire
The enamel coating (on the individual strands) mutually insulates the strands and the wire
on the whole is provided with PVC insulation. The current carrying capacity depends on
the total area of the wire. If cost is the criteria then aluminum conductors are preferred.
In that case, for the same current rating much larger diameter of wire is to be used.
## Two- way and Three- way Control of Lamps:
The domestic lighting circuits are quite simple and they are usually controlled from one
point. But in certain cases it might be necessary to control a single lamp from more than
one point (Two or Three different points).
For example: staircases, long corridors, large halls etc.
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## Two-way Control of lamp:
Two-way control is usually used for staircase lighting. The lamp can be controlled from
two different points: one at the top and the other at the bottom - using two- way switches
which strap wires interconnect. They are also used in bedrooms, big halls and large
corridors. The circuit is shown in the following figure.
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Switches S1 and S2 are two-way switches with a pair of terminals 1&2, and 3&4
respectively. When the switch S1 is in position1 and switch S2 is in position 4, the circuit
does not form a closed loop and there is no path for the current to flow and hence the
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lamp will be OFF. When S1 is changed to position 2 the circuit gets completed and hence
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the lamp glows or is ON. Now if S2 is changed to position 3 with S1 at position 2 the
circuit continuity is broken and the lamp is off. Thus the lamp can be controlled from two
different points.
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1 3 ON
1 4 OFF
2 3 OFF
2 4 ON
## Three- way Control of lamp:
In case of very long corridors it may be necessary to control the lamp from 3 different
points. In such cases, the circuit connection requires two; two-way switches S1and S2 and
an intermediate switch S3. An intermediate switch is a combination of two, two way
switches coupled together. It has 4 terminals ABCD. It can be connected in two ways
a) Straight connection
b) Cross connection
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In case of straight connection, the terminals or points AB and CD are connected as shown
in figure 1(a) while in case of cross connection, the terminals AB and
C D is connected as shown in figure 1(b). As explained in two way control the lamp is
ON if the circuit is complete and is OFF if the circuit does not form a closed loop.
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Compiled by www.studyeasy.in
The condition of the lamp is given in the table depending on the positions of the switches S1, S2
and S3.
## p Position of S3 Position of S1 Position of S2 Condition of the
lamp
lamp
1 3 ON
1 1 4 OFF
Straight 2 3 OFF
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connection 2 4 ON
1 3 OFF
2
Cross
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2
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3
ON
ON
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connection 2 4 OFF
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EARTHING:
The potential of the earth is considered to be at zero for all practical purposes as the
generator (supply) neutral is always earthed. The body of any electrical equipment is
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## connected to the earth by means of a wire of negligible resistance to safely discharge
electric energy, which may be due to failure of the insulation, line coming in contact with
the casing etc. Earthing brings the potential of the body of the equipment to ZERO i.e. to
the earths potential, thus protecting the operating personnel against electrical shock. The
body of the electrical equipment is not connected to the supply neutral because due to
long transmission lines and intermediate substations, the same neutral wire of the
generator will not be available at the load end. Even if the same neutral wire is running it
will have a self-resistance, which is higher than the human body resistance. Hence, the
body of the electrical equipment is connected to earth only.
Thus earthing is to connect any electrical equipment to earth with a very low resistance
wire, making it to attain earths potential. The wire is usually connected to a copper plate
placed at a depth of 2.5 to 3meters from the ground level.
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Compiled by www.studyeasy.in
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BLOCK DIAGRAM
## The earth resistance is affected by the following factors:
1.
2.
3.
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Material properties of the earth wire and the electrode
Temperature and moisture content of the soil
Depth of the pit
4. Quantity of the charcoal used
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The importance of earthing is illustrated in the following figures
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Necessity of Earthing:
1. To protect the operating personnel from danger of shock in case they come in
contact with the charged frame due to defective insulation.
2. To maintain the line voltage constant under unbalanced load condition.
3. Protection of the equipments
4. Protection of large buildings and all machines fed from overhead lines against
lightning.
Methods of Earthing:
The important methods of earthing are the plate earthing and the pipe earthing. The earth
resistance for copper wire is 1 ohm and that of G I wire less than 3 ohms. The earth
resistance should be kept as low as possible so that the neutral of any electrical system,
which is earthed, is maintained almost at the earth potential. The typical value of the
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earth resistance at powerhouse is 0. 5 ohm and that at substation is 1 ohm.
## 1. Plate earthing 2. Pipe earthing
Plate Earthing
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In this method a copper plate of 60cm x 60cm x 3.18cm or a GI plate of the size 60cm x
60cm x 6.35cm is used for earthing. The plate is placed vertically down inside the ground
at a depth of 3m and is embedded in alternate layers of coal and salt for a thickness of 15
cm. In addition, water is poured for keeping the earth electrode resistance value well
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below a maximum of 5 ohms. The earth wire is securely bolted to the earth plate. A
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cement masonry chamber is built with a cast iron cover for easy regular maintenance.
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Pipe Earthing
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Earth electrode made of a GI (galvanized) iron pipe of 38mm in diameter and length of 2m
(depending on the current) with 12mm holes on the surface is placed upright at a depth of
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4.75m in a permanently wet ground. To keep the value of the earth resistance at the desired
level, the area (15 cms) surrounding the GI pipe is filled with a mixture of salt and coal.. The
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efficiency of the earthing system is improved by pouring water through the funnel periodically.
The GI earth wires of sufficient cross- sectional area are run through a 12.7mm diameter pipe
(at 60cms below) from the 19mm diameter pipe and secured tightly at the top as shown in the
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following figure.
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Compiled by www.studyeasy.in
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When compared to the plate earth system the pipe earth system can carry larger leakage
currents as a much larger surface area is in contact with the soil for a given electrode size. The
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system also enables easy maintenance as the earth wire connection is housed at the ground
level.
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PROTECTIVE DEVICES
Protection for electrical installation must be provided in the event of faults such as short circuit,
overload and earth faults. The protective circuit or device must be fast acting and isolate the
faulty part of the circuit immediately. It also helps in isolating only required part of the circuit
without affecting the remaining circuit during maintenance. The following devices are usually
used to provide the necessary protection:
Fuses
Relays
Miniature circuit breakers (MCB)
Earth leakage circuit breakers (ELCB)
FUSE
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The electrical equipments are designed to carry a particular rated value of current under normal
circumstances. Under abnormal conditions such as short circuit, overload or any fault the
current raises above this value, damaging the equipment and sometimes resulting in fire hazard.
Fuses are pressed into operation under such situations. Fuse is a safety device used in any
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electrical installation, which forms the weakest link between the supply and the load. It is a
short length of wire made of lead / tin /alloy of lead and tin/ zinc having a low melting point
and low ohmic losses. Under normal operating conditions it is designed to carry the full load
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current. If the current increases beyond this designed value due any of the reasons mentioned
above, the fuse melts (said to be blown) isolating the power supply from the load as shown in
the following figures.
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Compiled by www.studyeasy.in
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CHARACTERISTICS OF FUSE MATERIAL
1.
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The material used for fuse wires must have the following characteristics
Low melting point
2. Low ohmic losses
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3. High conductivity
4. Lower rate of deterioration
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## Different types of fuses:
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Re-wirable or kit -kat fuses: These fuses are simple in construction, cheap and
available up-to a current rating of 200A. They are erratic in operation and their
performance deteriorates with time.
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Plug fuse: The fuse carrier is provided with a glass window for visual inspection
of the fuse wire.
Cartridge fuse: Fuse wire usually an alloy of lead is enclosed in a strong fiber
casing. The fuse element is fastened to copper caps at the ends of the casing. They are
available up-to a voltage rating of 25kV. They are used for protection in lighting
installations and power lines.
Miniature Cartridge fuses: These are the miniature version of the higher rating
cartridge fuses, which are extensively used in automobiles, TV sets, and other electronic
equipments.
Transformer fuse blocks: These porcelain housed fuses are placed on secondary
of the distribution transformers for protection against short circuits and overloads.
Expulsion fuses: These consist of fuse wire placed in hollow tube of fiber lined
with asbestos. These are suited only for out door use for example, protection of high
voltage circuits.
Semi-enclosed re-wirable fuses: These have limited use because of low breaking
capacity.
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Time-delay fuse: These are specially designed to withstand a current overload for
a limited time and find application in motor circuits.
## HRC CARTRIDGE FUSE
The high rupturing capacity or (HRC) fuse consists of a heat resistant ceramic body.
Then silver or bimetallic fuse element is welded to the end brass caps. The space
surrounding the fuse element is filled with quartz powder. This filler material absorbs the
arc energy and extinguishes it.
When the current exceeds the rated value the element melts and vaporizes. The vaporized
silver fuses with the quartz and offers a high resistance and the arc is extinguished.
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1. Fast acting
2. Highly reliable
3. Relatively cheaper in comparison to other high current interrupting device
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2. Requires replacement
3. The associated high temperature rise will affect the performance of other devices
## TERMS RELATED WITH FUSES
Rated current: It is the maximum current, which a fuse can carry without undue heating
or melting. It depends on the following factors:
1. Permissible temperature rise of the contacts of the fuse holder and the
fuse material
2. Degree of deterioration due to oxidation
Fusing current: The minimum current at which the fuse melts is known as the fusing
current. It depends on the material characteristics, length, diameter, cross-sectional area
of the fuse element and the type of enclosure used.
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Fusing Factor: It is the ratio of the minimum fusing current to the rated current. It is
always greater than unity.
## This is a measuring instrument/device which works on the principle of induction
and measures the energy consumed over a definite period.
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## 1) Upper Magnet/ shunt magnet (P.P)
2) Potential coil/ Voltage coil
4) Friction compensator
5) Aluminium disc
6) Brake magnet
7) Lower magnet/Series magnet
8) Current coil (C-C)
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## This instrument consisting two electromagnets as in fig.
1. Upper magnet or Shunt magnet: which carries the potential coil on its central
limb which also carries one or two copper shading bands for the power factor adjustment.
2. Lower magnet or Series magnet: Which carries the current coil as shown.
An aluminum disc is between the fields of the upper and lower electro magnets. There is
a friction compensator in the upper magnets for the measurement at very low loads. The
aluminum disc rotates in the field of a brake magnet whose position can be set so that the
disc rotates at proper speeds at higher loads.
This instrument works on the principle of induction that when both the shunt and series
coils are energized by ac, there will be tow alternative fluxes are in the shunt coil and one
in the series coil these time varying fluxes are cut by a stationary disc. Inducing currents
in the disc. These currents interacts with the fluxes and results in a torque which is given
by
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T k1 sh i se K 2 se i sh there by the disc rotates in a particular direction and the number
and speed of rotations depends on the energy consumed by the load.
Some times the energy meters disc rotates slowly even on no load conditions as
the potential coil is continuously energized and this effect is called the CREEP and the
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speed is called the CREEP SPEED to minimum this creep one pair of diametrically
opposite holes are made in the aluminum disc which alters the reluctance and minimizes
the creep effect.
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DYNAMOMETER WATTMETER:-
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In this type there will not be any permanent magnets and there will be a pair of
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fixed coils connected in series when energized gives the same effect as that of the
permanent magnets. In the field of these fixed coils there will be a moving coil which
when energized acted upon by a torque by which it deflects
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## F1 F2: Fixed coils
M: Moving coil
R: High resistance in series with m
I2: current through
The two fixed coils in series act as the current coil and the moving coil in series with R
act as the potential coil. The moving coil is pivoted between the two fixed coils carries a
current I2 proportional to V. This current is fed to m through two springs which also
provides the necessary controlling torque. This instrument can be used on both ac and dc
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circuits as both the coils are energized simultaneously by a common source due to which
a unidirectional torque is produced.
RECOMMENDED QUESTIONS
1. Mention the different types of wiring. With relevant circuit diagrams and
switching tables, explain the two way and the three way control of lamps.
2. Explain the different types of wiring used in practice
3. Explain in brief the following:
* Fuses
* Specification of wires
* Earthing and its necessity
4. Sketch any one type of earthing and indicate why such earthing of electrical
equipments is necessary
5. With a neat sketch explain any one method of earthing electrical appliance.
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6. With a neat circuit diagram, explain the two way control of a filament lamp.
7. Define domestic wiring. What important factors are to be considered in domestic
wiring? Mention the difference types of wiring in practice.
8. What do you understand by Earthing? With a neat diagram explain plate
earthing.
9.
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With a neat circuit diagram and a switching table, explain the two point control of
a lamp.
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12. With a neat sketch explain the pipe earthing method.
12. Explain the working principle of a fluorescent lamp when connected to an electrical
supply source, with necessary auxiliary components and their functions
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## 1) With a neat sketch explain the pipe earthing method.
Sol: pipe earthing :
In this method of earthing a G.I pipe of 38 mm diagram and 2 meter (7 feet) length
is embedded vertically into the ground. This pipe acts as an earth electrode. The depth
depends on the condition of soil.
The earth wires are fastened to the top section of the pipe above the ground level
with nut and bolts.
The pit area around the pipe is filled with salt and coal mixture for improving the
condition of the soil and earthing efficiency. The schematic arrangement of pipe earthing
system is shown in the Fig.10.
The contact surface of G.I. pipe with the soil is more as compared to the plate due
to its circuit section and hence can handle heavier leakage current for the same electrode
size.
According to Indian standard, the pipe should be placed at a depth of 4.75m.
impregnating the coke with salt decreases the earth resistance. Generally alternate layers
of salt and coke are used for best results.
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In summer season, soil becomes dry, in such case salt water is poured through the
funnel connected to the main G.I.pipe through19 mm diameter pipe. This keeps the soil
wet.
The earth wires are connected to the G.I. pipe above the ground level and can be
physically inspected from time to time. These connections can be checked for performing
continuity tests. This is the important advantage of pipe earting over the plate earthing.
The earth lead used must be G.I. wire of sufficient cross-sectional area to carry fault
current safely. It should not be less than electrical equivalent of copper conductor of
12.97 mm2 cross-sectional area.
The only disadvantage of pipe earthing is that the embedded pipe length has to be
increased sufficiently in case the soil specific resistivity is of high order. This increases
the excavation work and hence increased cost. In ordinary soil condition the range of the
earth resistance should be 2 to 5 ohms.
In the places where rocky soil earth bed exists, horizontal strip earthing is used. This
is suitable as soil excavation required for plate or pipe earthing is difficult in such places.
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For such soils earth resistance is between 5 to 8 ohms.
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2) with the help of a neat diagram. Explain the construction and principle of operation of
single phase energy meter.
## Sol. : Induction type single phase energy meter:
Induction type instruments are most commonly used as energy meters. Energy
meter is an integrating instrument which measures quantity of electricity. Induction type
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of energy meters are universally used for domestic and industrial applications. These
meters record the energy in kilo-watt-hours (k Wh).
Fig. 15 shows the inductance type single phase energy meter.
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It works on the principle of induction i.e. on the production of eddy currents in the
moving system by the alternating fluxes. These eddy currents induced in the moving
system interact with each other to produce a driving torque due to which disc rotates to
record the energy.
In the energy meter there is no controlling torque and thus due to driving torque
only, a continuous rotation of the disc is produced. To have constant speed of rotation
braking magnet is provided.
Construction:
There are four main parts of operating mechanism
1) Driving system 2) moving room 3) braking system 4) registering system.
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## 1) Driving system: It consists of two electromagnets whose core is made up of
silicon steel laminations. The coil of one of the electromagnets, called current
coil, is excited by load current which produces flux further. The coil of
another electromagnetic is connected across the supply and it carries current
proportional to supply voltage. This is called pressure coil. These two
electromagnets are called as series and shunt magnets respectively.
The flux produced by shunt magnet is brought in exact quadrature with
supply voltage with the help of copper shading bands whose position is adjustable.
2) Moving system: Light aluminium disc mounted in a light alloy shaft is the
main part of moving system. This disc is positioned in between series and
shunt magnets. It is supported between jewel bearings. The moving system
runs on hardened steel pivot. A pinion engages the shaft with the counting
mechanism. There area no springs and no controlling torque.
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3) Braking system: a permanent magnet is placed near the aluminium disc for
barking mechanism. This magnet reproduced its own field. The disc moves in
the field of this magnet and a braking torque is obtained. The position of this
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magnet is adjustable and hence braking torque is adjusted by shifting this
magnet to different radial positions. This magnet is called braking magnet.
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4) Registering mechanism: It records continuously a number which is
proportional to the revolutions made by the aluminium disc. By a suitable
system, a train of reduction gears, the pinion on the shaft drives a series of
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pointers. These pointers rotate on round dials which an equally marked with
equal division.
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## Working: since the pressure coil is carried by shunt magnet M2 which is
connected across the supply, it carries current proportional to the voltage. Series
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magnet M1 carries current coil which carries the load current. Both these coils
produced alternating fluxes 1 and 2 respectively. These fluxes are proportional to
currents in their coils. Parts of each of these fluxes link the disc and induces e.m.f. in
it. Due to these e.m.f.s eddy currents are induced in the disc. The eddy current
induced by the electromagnet M2 react with magnetic field produced by M1 react with
magnetic field produced by M2. Thus each portion of the disc experiences a
mechanical force and due to motor action, disc rotates. The speed of disc is controlled
by the C shaped magnet called braking magnet. When disc rotates in the air gap, eddy
currents are induced in disc which oppose the cause producing them i.e. relative
motion of disc with respect to magnet. Hence braking torque Tb is generated. This is
proportional to speed N of disc. By adjusting position of this magnet, desired speed of
disc is obtained. Spindle is connected for ecording mechanism through gears which
record the energy supplied.
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3) Mention the different types of wiring. With relevant circuit diagrams and
switching tables, explain two way and three way control of lamps.
Sol.: Types of wiring: Depending upon the above factors various types of wiring used in
practice are:
5. Cleat wiring
6. Casing wiring
7. Surface wiring
8. Conduit wiring
i ) Clear wiring:
In this type V.I.R or P.V.C wires are clamped between porcelain cleats.
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The cleats are made up of two halves. One half is grooved through which wire passes
while the other fits over the first. The whole assembly is then mounted on the wall or
wooden beam with the help of screws.
This method is one of the cheapest method and most suitable for temporary work. It
can be very quickly installed and can be recovered without any damage of material.
Inspection and changes can be made very easily.
This method does not give attractive appearance. After some time due to sagging at
some places, it looks shabby. Dust and dirt collects on the cleats. The wires are directly
exposed to atmospheric conditions like moisture, chemical fumes etc. maintenance cost is
very high.
Due to these disadvantages this type is not suitable for permanent jobs.
ii) Casing capping: This is very popularly used for residential buildings. In this
method, casing is a rectangular strip made from teak wood or new a days made up of
P.V.C. It has two grooves into which the wires are laid. Then casing is covered with a
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rectangular strip of wood or P.V.C. of the same width, called capping. The capping is
screwed into casing is fixed to the walls the help or porcelain discs or cleats.
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Good protection to the conductors from dangerous atmospheric conditions, neat and clean
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appearance are the advantages of this type.
In case of wooden casing capping, there is high risk of fire along with the requirement of
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skilled labour. The method is costly.
Surface wiring: in this type, the wooden battens are fixed on the surface of the
wall, by means of screws and rawl plugs. The metal clips are provided with the battens
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at regular intervals. The wire runs on the batten and is clamped on the batten using the
metal clips. The wires used may lead sheathed wires or can tyre sheathed wires.
ud
Depending upon type of wire used surface wiring is also called lead
Sheathed wiring or cab tyre sheathed
st
## wiring. If the wire used is though rubber
Sheathed then it is called T.R.S. wiring
While if the wire used is cab tyre Sheathed
Then it is called C.T.S wiring.
## Conduit wiring: In this method, metallic
tubes called as conduits are used to run the
wires. This is the best system of wiring as
it gives full mechanical protection to the wires.
This is most desirable for workshops and public
Buildings. Depending on whether the conduits
are laid inside the walls or supported on the
walls, there are two types of conduit wiring
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which are :
iii) Surface conduit wiring: in this method conduits are mounted or supported on the
walls with the help of pipe books or saddles. In damp situations, the conduits are
spaced apart from the wall by means of wooden blocks.
iv) Concealed conduit wiring: In this method, the conduit are buried under
the wall aty the some of plastering. This is also called recessed conduit wiring.
The beauty of the premises is maintained due to conduit wiring. It is durable and
has long life. It protects the wires from mechanical shocks and fire hazards. Proper
earthing of conduits makes the method electrical shock proof. It requires very less
maintenance.
The repairs are very difficult in case of concealed conduit wiring. This method is
most costly and erection requires highly skilled labour. These are few disadvantages
of the conduit type of wiring. In concealed conduit wiring, keeping conduit at earth
.in
potential is must.
5) Explain two way & three way control of lamps
Two way control of lamps:
sy
ea
Thisis also called as staircase wiring as
it is commonly used for stair cases
y
## and corridor lighting. It consist of
two way switches. A two way switch
ud
## operatesalways in one of the two
possible positions.The circuits is
shown in the fig. 20.
st
## Assume that lamp is on first floor.
Switch A In on first floor and B in
on second floor. In the position
shown, the lamp is OFF.
## When person changes position of
Switch A from (1) to (2) then lamp
gets phase through Switches A and B
it gets switched ON as shown In fig. 21.
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When person reaches on second floor, the lamp is required to be switched OFF. So
person will change switch B from (2) to (1), due to which phase connection reaching
to the lamp gets opened and lamp will switched OFF as shown in fig. 22.
.in
Switch A position
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Switch B position `Switch C position
ea
1 1 ON
2 2 ON
y
2 1 OFF
ud
1 2 OFF
3) Explain Three way control of lamps: This is also a type of stair case wiring. It
st
consist of two way switches A and B and one intermediate switch C. The circuit
used to have three way control of lamps is shown in the fig. 23.
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## The intermediate switch can have positions to connect point 1 4, 3 2, as shown or
1 2 and 3 4 shown dotted. The switch A is on the first floor and switch B is on
third floor say.
In the position shown is, fig. 23, the lamp is ON.
When person from floor 2 changes switch C position to have connections 1 2
and 3 4 then it can be seen from fig. 24 that there is open circuit in the connection
and lamp gets switched OFF.
.in
sy
y ea
Now if the person from the third floor changes the position of switch B from 1 to 2,
then again lamp gets supply through position 2 of A, 3 4 of C and 2 of B shown in
ud
## the fig. 25. The lamp gets switched ON.
st
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Compiled by www.studyeasy.in
## Sr. no. Switch A Switch B Switch C Condition of
position position position lamp
1 1 1 1 4, 3 2 OFF
2 2 2 1 4, 3 2 OFF
3 1 2 1 4, 3 2 ON
4 2 1 1 4, 3 2 ON
5 1 1 1 2, 3 4 ON
6 2 2 1 2, 3 - 4 ON
.in
7 1 2 1 2, 3 4 OFF
8 2 1 1 2, 3 4 OFF
sy
y ea
ud
st
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Compiled by www.studyeasy.in
Unit-5:
## DC Machines: Working principle of DC machine as a generator and a motor. Types and
constructional features. emf equation of generator, relation between emf induced and
terminal voltage enumerating the brush drop and drop due to armature reaction.
Illustrative examples.
DC motor working principle, Back emf and its significance, torque equation. Types of
D.C. motors, characteristics and applications. Necessity of a starter for DC motor.
Illustrative examples on back emf and torque.
UNIT-5
D.C. MACHINES.
.in
Working principle of D.C.Machine as generator and motor, constructional features,
EMF equation of generator and simple problems, back emf and torque equation of
DC motors, simple problems, types of DC motors, characteristics and applications,
necessity of starter, 3-point starter. sy
______________________________________________________________________
A machine which works on direct current is defined as a D.C.Machine.
ea
D.C.Machines are of two types. (i) D.C.Generator and (ii) D.C.Motor.
Sl. D.C. Generator D.C.Motor
y
No.
1 Definition: Definition:
ud
## A generator is a rotating A motor is a machine which
machine which converts converts electrical energy into
mechanical energy into mechanical energy
st
electrical energy
2 Principle: Principle:
Whenever a coil is rotated in a magnetic Whenever a current coil is placed
field an e.m.f. will be induced in this coil under a magnetic field the coil
and is given by experiences a mechanical force,
e=BlvSin volts/coil side and is given by F= BIlSin
where, B=The flux density in Tesla, Newtons/coil side.
l=the active length of the coil side in Where, I is the current through the
meters, v=the velocity with which the coil in ampere.
coil is moved in meters/sec and is the
angle between the direction of the flux
and the direction of rotation of the coil
side.
3 The direction of the emf induced is fixed The direction of the force acting is
by applying the Flemings right hand rule fixed by applying the Flemings
left hand rule.
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CONSTRUCTION OF A D.C.MACHINE.
## Salient parts of a D.C.machine are:
Field system (poles)
Coil arrangement (armature)
Commutator
Brushes
Yoke
Fig shows the details of a four pole D.C. machine with both shunt and series field windings.
Field system: This is made of electromagnets, wherein a iron laminated core is wound with
well insulated enameled copper wire. The core is laminated to minimize the eddy current loss.
Each lamination is dipped in varnish and dried. A pole shoe is attached to the pole face to
direct the flux to concentrate radially on to the armature thereby reducing the leakage and
.in
fringing flux. Poles are fixed to the yoke by means of bolts.
Armature: This is the rotating part of the machine made of laminated iron core cylindrical in
structure with slots on its periphery. Insulated copper coils are laid in these slots, and these
sy
coils are connected for lap or wave connection. The core laminations are firmly mounted on a
shaft fitted with smooth bearings on either side for smooth rotation.
## Comparison of lap and wave windings:
ea
LAP WAVE
Number of armature parallel paths Number of parallel paths is always
y
ud
## Preferred when large current at Preferred when large voltage with
lesser voltage is the requirement. lesser current is the requirement.
st
## alternating e.m.f. into unidirectional e.m.f.
This is cylindrical in structure made of copper segments with mica insulation between them
and is firmly fixed on to the shaft carrying the armature and the armature coil free ends are
brazed to the commutator segments.
Brushes: These are current collecting devices placed on the body of the commutator with a
holder. Brushes are made of carbon, copper or graphite.
Yoke: This is the outer most part of the machine made of cast steel which is the mechanical
enclosure for the machine to protect it from dust and moisture and also provides the return
path for the magnetic flux and carries half the flux per pole.
E.M.F. Equation:
Let the D.C. machine has P number of poles, Z number of armature conductors arranged in A
number of parallel paths. Let be the flux per pole and N is the speed of rotation in
revolutions per minute.
Consider one North Pole of the machine under which a group of armature conductors all
being connected in series. Let x be the spacing between any two neighboring conductors ant t
be the time taken to move through this distance of x.
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The total flux per pole is made of several lines and one line of flux is cut by one
conductor when it moves through a distance of x in t seconds.
Therefore the induced emf in the 1st conductor when cut by the flux of 1
is
e1 = 1/t volts
nd
Similarly in the 2 conductor e2 = 2/t volts, and so on.
Therefore the total emf induced in all the conductors under one pole is the sum of all these
emfs.
E= e1 + e2 + e3 + e4 + ..
E= 1/t + 2/t + 3/t + 4/t + ..
E= /t volts/pole.
For all the P number of poles E= P/t volts
The speed is defined as N revolutions per minute,
.in
N revolutions in one minute or 60 seconds.
1 revolutions will be in time of 60/N seconds, and as one revolution corresponds to all the Z
number of conductors the time t for a travel of distance x can be written as t= 60/NZ seconds.
sy
Therefore the induced EMF E= P/t = P/60/(NZ)=PZN/60.
As the Z number of conductors are arranged in A number of parallel paths,
The induced e.m.f per parallel path is
ea
E= PZN/60A volts.
y
As P, Z, A are fixed the induced e.m.f is mainly dependent on the flux and the speed, and
hence we write that the induced e.m.f E is proportional to the product of the speed N and the
ud
flux .
## Types of D.C. Generators.
st
D.C. Generators are classified on the basis of the method of exciting the field coils as (i)
Separately excited generators and
(ii) Self excited generators.
In separately excited type the field coils are excited from an independent D.C. source.
In self excited type excitation of the field coils are done by feeding back a part of the output
of the generator.
Self excitation can be done in three ways:
(i) By connecting the field coils across the armature- Shunt excitation ( Fig 1.)
(ii) By connecting the field coils in series with the armature- Series. (Fig 2) excitation.
(iii) By using both the shunt and series field coils together- Compound excitation.
In compound excitation the fluxes due to the shunt field and the series field may support each
other or oppose each other and accordingly they are called Cumulatively compounded or
Differentially compounded generators. (Figs. 3 and 4.)
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There are two more ways of connecting the shunt and series field coils with the armature: (i)
Long shunt connection:- Here the armature and the series field coils are connected in series
and the shunt field circuit is connected in parallel with this combination. (Fig. 5.)
(ii) Short shunt connection: - Here the armature and the shunt field circuits are connected in
parallel and the series field coils are connected in series with this combination.( Fig. 6)
A 500V shunt motor has 4 poles and a wave connected winding with 492 conductors. The
flux per pole is 0.05 Wb. The full load current is 20 Amps. The armature and
shunt field
resistances are 0.1
and 250 respectively. Calculate the speed and the developed
torque.
Sol:
P = 4, V =500volts
= 0.05Wb
.in
Z = 492
IL = IF.L. = 20A
Ra = 0.1
Rsh = 250 ,A = 2 for wave winding
sy
Ish = = = 2A
ea
IL = Ia + Ish Ia = IL Ish = 20-2 = 18A
Now, Eb = =
y
## But, Eb = V-IaRa = 500-(18)(90.1) = 500-1.8
ud
498.2 =
N=
st
N = 607.56 rpm
Now, EbIa = Ta
Ta =
=
Ta =
Ta = 140.94N-m) A 250 KVA, 11000/415V, 50 Hz single phase
## Questions and Problems on D.C. Generators:
1. Draw the cross sectional views of a typical 4 pole D.C. machine and explain the
function of each part.
2. With usual notations derive expression for the induced emf of a D.C. machine.
3. A 4 pole generator with wave wound armature has 51 slots, each having 24
conductors. The flux per pole is 0.01 wb. At what speed must the armature be rotated to
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give an induced emf of 220 volt. What will be the voltage developed if the winding is
lap and the armature rotates at the same speed.
4. Find the useful flux per pole of a 250 V, 6pole shunt generator having two circuit
connected armature winding with 220 conductors. At normal temperature the overall
armature resistance including brushes is 0.2 ohm. The armature current is 13 A and the
speed is 910 rpm.
5. A 4 pole, 100V, shunt generator with lap connected armature winding having
a field and armature resistances of 50 and 0.1 ohm respectively supplies sixty, 100V,
40 watt lamps. Calculate the total armature current, current per parallel path and
the generated emf.
6. A long shunt cumulatively compounded D.C. generator supplies 7.5 KW
power at 250V. The shunt field, series field and armature resistances are respectively 125
ohm, 0.3 ohm and 0,5 ohm. Calculate the induced emf.
7. In a 110V compound generator the resistances of the armature, shunt
and series field coils are 0.06, 25 and 0.04 ohms respectively. The load
.in
consists of 200 lamps each rated at 55 Watt, 110 Volt. Find the
generated emf when connected for (i) Long shunt and (ii) Short shunt.
sy
D.C. MOTOR.
ea
Principle: Whenever a current coil is placed in a magnetic field the coil experiences a
mechanical force, and is given by
y
ud
## I is the current through the coil
l is the active length of the coil side
is the angle between the movement of the
st
## coil and the direction of the flux
The direction of the force acting can be decided by applying Flemings left hand rule.
## The construction of a D.C.Motor is same as the construction of a D.C.generator.
Types of D.C.Motors:
Depending on the interconnection between the armature and the field circuit D.C.Motors
are classified as (i) Shunt Motor, (ii) Series Motor and (iii) Compound motors just like
D.C.Generators.
Back EMF: Whenever a current coil is placed under a magnetic field the coil experiences
a mechanical force due to which the coil starts rotating. This rotating coil again cuts the
magnetic lines of force resulting an EMF induced in it whose direction is to oppose the
applied EMF (as per Flemings right hand rule), and hence the name BACK EMF or
Counter Emf.
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Significance of Back EMF: Back EMF is a must in a motor which helps to regulate the
armature current and also the real cause for the production of torque.
## Expression for the back Emf is given by E=V-IaRa,
Where E is the back emf, V is the applied emf, Ia is the armature current and Ra is the
armature circuit resistance. And also E= PZN/60A volts, from the machine parameters.
## Production of torque in a D.C. Motor.
The production of torque in a d.c. motor can be well explained with the help of the following
figures.
Fig (a) represents the magnetic field distribution between a bipolar magnet from North pole to
South pole.
Fig(b) shows the field set up around a current carrying coil
.in
In fig the current carrying coil is brought under the influence of bipolar magnetic field.
The resultant field around the coil due to the inter action of the main field and the coil field is
seen in fig (d) where in the flux is strengthened in the left part of the upper coil side and
sy
weakened in the right part of the upper coil side and vice-versa in the lower coil side. The
resultant flux which strengthened at one point exerts a force on the conductor as per
Flemings left hand rule and thereby the coil side experiences a mechanical force.
ea
In the construction it is seen that several coils sides are on the armature and the tangential
force acting on each of these coil sides add each other and resulting in a unidirectional
y
movement which makes the armature to rotate at a uniform speed thereby torque is
produced.
ud
TORQUE EQUATION:
st
Let P be the total number of poles, Z be the total number of armature conductors arranged in
A number of parallel paths. Let be the flux per pole, N be the speed of rotation in rpm, and
T be the torque in Nm.
## We know that the back emf E=V-IaRa
Multiplying the above equation by Ia on both sides
We get EIa=VIa-Ia2 Ra
Where VIa represents the Power input to the armature, Ia2 Ra represents the armature copper
loss and EIa represents the Total power output of the armature which is the electrical power
converted into mechanical power called the electro-mechanical power
in watts. The equivalent mechanical power is given by 2 NT/60 watts.
## Therefore, EIa=2 NT/60 watts
But E = PZN /60A, therefore the torque T= PZIa/2A Nm.
From the above equation it can be seen that the torque is directly proportional to the product
of the flux and the armature current.
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Speed of a D.C.Motor:-
## We know that for a motor in general the back emf e is given by
E= V-IaRa=PZN/60A
From which we write,
N=(V-IaRa)/PZ60A,
and the speed N is proportional to (V-IaRa)/,
From the above equation we write the speed is directly proportional to the applied voltage
V, and the armature current Ia and inversely proportional to the flux .
## Characteristics of D.C.Motors: To study the performance of a motor it is necessary to study
the variation of its speed and torque with the variations of the load on it.
There are two types of characteristics: (i) Speed v/s load characteristics
.in
(i)
sy
In a shunt motor the flux is considered to be constant because of the reason that the field
ea
circuit is connected across a constant power supply. Also as the applied voltage is constant the
speed is directly proportional to the armature current only, and also as the load is increased
the armature current also increases at the same rate and the speed becomes constant. But due
to the increased friction at the bearings with the increase of the load there is a small decrease
y
in the speed. The characteristic is shown in the fig. and is compared with the ideal
ud
characteristics. The drop in the speed can be reduced by slightly de-exciting the field flux,
there by the speed is controlled.
## (b) Series Motor:
st
In a series motor the flux is solely dependent on the armature current hence the speed
variation with load is not like shunt motor. At no load condition only residual flux is in action
which is very very small resulting in a dangerously high speed. Therefore series motors are
not to be started on no load, which result in the initial speed of dangerously high value called
RUN AWAY SPEED which severely damages the motor. Hence in series motors there is a
provision of a fly wheel fixed to the shaft which acts like a mechanical load to prevent the
motor to attain this high speed.
## characteristics of D.C. i) series and ii) shunt motors. Mention two
applications of each motor.
DC series motor :
i) N=1 characteristics :
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## Applications : Cranes, trolleys .
.in
DC shunt motor :
i) N-I characteristics :
sy
y ea
ud
st
## Applications : Lathe machine, milling machine.
b) Give reasons for the following in an alternator
i) Armature is stationary field is rotating .
## ii) Distributed winding is used instead
of concentrated winding. (6)
## Sol.:i) Armature is stationary field is rotating
a) The stationary armature can be easily insulated for high a.c. generators
voltages.
b) The armature winding can be braced easily to prevent deformation.
c) The armature winding can be directly connected to load. Thus load current
does not pass through brush contain.
d) The excitation voltage is low.
## ii) Distributed winding is used instead of concentrated winding
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In distributed type of winding all the coils belonging to a phase are well
distributed over the m slots per phase, under every pole. Distributed makes
the waveform of the induced e.m.f. more sinusoidal in nature. Also in concentrated
winding due to large number of conductors per slot, heat dissipation is poor . So in
practice, double layer, short pitched and distributed type of armature winding is
preferred for the alternators.
A 120V DC shunt motor has an armature resistance of 60 . It runs at 1800 RPM, when it
takes full load current of 40A. Find the speed of the motor while it is operating at half
Sol.:
.in
IL1 = Ia1 + If
Ia1 = IL1- If = 40 2 = 38A
## = 120 7.6 = 112.4V
N1 = 1800rpm
sy
ea
Now for shunt machine
T Ia as remains constant.
y
T2 = T1
ud
st
## Ia2 = Ia1 = (38) = 19A
Eb2 = V Ia2 Ra = 120 (19)(0.2) = 120 3.8
Eb2 = 116.2V
Now, Eb N for
constant
N2 = 1860.85 rpm
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A 250 V d.c. shunt motor has an armature resistance of 0.5 and shunt field resistance
of 250 .When driving a load at 600 rpm., the torque of which is constant, the armature
takes 20 A. If it is desired to raise the speed from 600 to 800 r.p.m., what resistance must
be inserted in the field circuit? Assume the magnetization curve to be a straight line. [10]
.in
Ans. :
Ish1 = = 1A
Eb1 = V Ia1Ra = 250 20 0.5 = 240 V
sy
ea
T Ia Ish Ia
=1 Torque is constant
=1 i.e. Ia2 Ish2 = 20
y
N
ud
st
= 320 (2)
But = V Ia2 Ra = 250- 0.5Ia2
= 250 0.5 From (1)
## Using in (2) , = 320
250 Ish2 10 = 320 i.e. 320 - 250 Ish2 + 10 = 0
Solving, Ish2 = 0.7389 A, 0.0422 A
Neglecting lower value, Ish2 = 0.7389 A
But, Ish2 = i.e. 0.7389 =
RX = 88.3407 External resistance
required
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Compiled by www.studyeasy.in
## Recommended Questions and problems on D.C.Motors:
1. Fundamentally, upon what two quantities does the speed of the motor depend?
Derive the equation which gives the speed in terms of terminal voltage, armature
resistance drop, and the flux per pole.
2. When load is applied to a motor what is its first reaction? With the shunt
Motor, how does this reaction affect the counter emf? The current to the armature?
3. How does the flux in series motor vary with the load current? Show the relation
of internal torque to load current, assuming no saturation in the magnetic circuit. Show
the relation of speed to load current. What precautions should be taken when the series
motor is being installed for industrial drives?
4. In what way do the windings of a compound motor differ from those of a shunt
motor and a series motor? In what two ways, with respect to the shunt winding, may the
series winding be connected?
5. Discuss the speed characteristic and torque characteristic of the cumulative-
.in
compound motor. What is the advantage of this motor over the series motor?
6. Why is a starting resistance necessary for D.C. motors? With the shunt motor, in
what circuit the starting resistor is connected? Why should it not be connected in the
line?
7.
sy
The resistance of the armature of a 25-hp 240-volt shunt motor is 0.083 ohm.
When connected to a 240-volt supply the armature develops a counter emf of 232.8-volts.
ea
Determine: (a) armature current; (b) armature current when connected across same power
supply while stationary; (c) counter emf when armature current is 110 amperes.
8. A 230-volt 4-pole 15 hp shunt motor has 702 conductors connected for simplex
y
wave and its resistance is 0.252 ohm. The flux per pole is 7.65 milliwebers and the
armature current is 60-Amp. Determine speed of armature.
ud
9. A four pole d.c. shunt motor has 456 surface conductors connected in simplex
wave. The flux is 2.41 milliwebers per pole (a) Determine the counter emf when the
speed is 1500 rpm. The armature resistance is 0.2 ohm. (b) Determine the terminal
st
voltage when armature current is 60-amp if the speed and flux remain constant.
10. A 60-hp, 250-volt, 1200 rpm shunt motor takes 214 amp at 250 volts. The field
current is 1.05 amp, and the combined total armature resistance is 0.039 ohm. The motor
speed when running light is 1200 rpm, and the line current is 8.6 amp. Determine the
internal power and torque developed.
11. A 25-hp, 250-volt d.c. series motor has its armature and series field resistance of
0.12 ohm and 0.10 ohm respectively. When the motor takes 85 amp, the speed is 600
rpm. Determine the speed when the current is (a) 100 amp, (b) 40 amp. Assume
saturation curve is a straight line, and neglect armature reaction.
12. A 6 pole d.c. generator runs at 850 rpm, and each pole has a flux of 0.2
milliwebers. If there are 150 conductors in series between each pair of brushes, what is
the value of the generated emf?
13. A 220-volt shunt motor has a field resistance of 400 ohm and an armature
resistance of 0.1 ohm. The armature current is 50 amps, and the speed is 900 rpm.
Assuming a straight line magnetization curve, calculate (a) the additional resistance in the
field to increase the speed to 1000 rpm for the same armature current, and (b) the speed
with the original field current and an armature current of 200 amps.
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## SOLUTION TO QUESTION BANK
1) A 500V shunt motor has 4 poles and a wave connected winding with 492 conductors.
The
flux per pole is 0.05 Wb. The full load current is 20 Amps. The armature and shunt
field
resistances are 0.1 and 250 respectively. Calculate the speed and the developed
torque.
Sol:
P = 4, V =500volts
= 0.05Wb
Z = 492
IL = IF.L. = 20A
.in
Ra = 0.1
Rsh = 250 ,A = 2 for wave winding
Ish = = = 2A
Now,
IL = Ia + Ish
Eb = =
sy
Ia = IL Ish = 20-2 = 18A
ea
But, Eb = V-IaRa = 500-(18)(90.1) = 500-1.8
498.2 =
N=
y
N = 607.56 rpm
EbIa = Ta
ud
Now,
Ta = =
Ta =
st
Ta = 140.94N-m
Tg = 0.159 Ia
= 0.1590.0518
Tg = 140.81N-m
2)A 4 pole , 1500 r.p.m d.c generator has a lap wound armature having 24
slots
With 10 conductors per slot. If the flux pole is 0.04 Wb, calculate the e.m.f
generated in the
b armature. What would be the generated e.m.f if the winding is wave
connected ? (6)
Ans: P =4, N = 1500 r.p.m, Lap i.e. A = P, = 0.04 Wb
## Z = Slots x Conductors / Slot = 24x10= 240
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Eg = = = 240 V
If winding is wave connected, A=2
Eg = = 480 v
## 3) Derive an expression for torque a d.c motor (4)
d) The current drawn from the mains by a 220 V D.C shunt motor is 4 A on
no- load The resistance and armature windings are 110 ohm and 0.2 ohm
respectively . if the line current on full load is A at speed of 1500 r.m.p. find the no-
.in
Ans: ILO = 4A,V =220 V, Rsh = 110 , Ra = 0.2, IFL = 40 A
Ish =
= sy
=2A
## Ia0 = ILO - Ish = 4-2==2A
ea
Ebo = V--IaoRa = 220-20.2=219.6V
## IaFL = IFL- Ish =40-2=38 A ISH is
y
constant
IbFL = V- IaFL Ra =220-380.2=212.4V
ud
N = Eb . is
constant
st
= i.e = NFL
=1500 r.p.m
## No = 1550.8474 r.p.m . No load speed.
4) Draw a neat sketch representing the cut section view of a D.C. machine. Explain the
important features of different parts involved there on.
## Sol.: Construction of a practical D.C. machine:
Whether a d.c. machine is a generator or motor, the basic construction remains the
same. The fig. 26 shows the cross- sectional view showing the various parts of four pole,
practical d.c. machine.
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.in
sy
ea
It consists of the following parts:
Yoke:
y
a) Functions :
i) It serves the purpose of outermost cover of the d.c. machine. So that the insulating
ud
materials get protected from harmful atmospheric elements like moisture, dust and
various gases like SO2, acidic fumes etc.
ii) It provides mechanical support to the poles.
st
iii) It forms a part of the magnetic circuit. It provides a path of low reluctance for
magnetic flux. The low reluctance path is important to avoid wastage of power to provide
same flux. Large current and hence the power is necessary if the path has high reluctance,
to produce the same flux.
## b) Choice of material: To provide low reluctance path, it must be made up of some
magnetic material. It is prepared by using cast iron because it is cheapest. For large
machines rolled steel, cast steel, silicon steel is used which provides high permeability
i.e. low reluctance and gives good mechanical strength.
As yoke does not need any machining or
good finishing as it rough, casting is the best method
of construction of yoke.
Poles:
Each pole is divided into two parts
Namely, a) pole core and b) pole shoe
This is shown in fig. 27.
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## a) Function of pole core and pole shoe:
i) pole core basically carries a field winding which is necessary to produce the flux.
ii) It directs the flux produced through air gap to armature core, to the next pole.
iii) pole shoe enlarges the area of armature core to come across the flux, which is
necessary to produce larger induced e.m.f. to achieve this, pole shoe has given a
particular shape.
b) Choice of material : It is made up of magnetic material like cast iron or cast
steel.
as it requires a definite shape and size, laminated construction is used. The
laminations of required size and shape are stamped together to get a pole, which is then
bolted to the yoke.
Field winding [F1-F2] :
The field winding is wound on the pole core with a definite direction.
.in
a) Functions: i) To carry current due to which pole core on which the winding placed
behaves as an electromagnet, producing necessary flux.
As it helps in producing the magnetic field i.e. exciting the pole as
electromagnet it is called Field winding or Exciting winding.
sy
4) Explain Choice of material: As it has to carry current hence obviously made up
of some conducting material. So aluminium or copper is the choice. But field
ea
coils are required to take any type of shape and bend about pole core and copper
has good pliability i.e. it can bend easily. So copper is the proper choice.
Filed winding is divided into various coils called bas field coils. These are
y
connected in series with each other and wound in such a direction around pole cores, such
that alternate N and S poles are formed.
ud
By using right hand thumb rule for current carrying circular conductor, it can be
easily determined that how a particular core is going to behave as N or S for a
particular winding direction around it.
st
Armature:
It is further divided into two parts namely,
I) Armature core and II) Armature winding
I) Armature core is cylindrical in shape mounted on the shaft. It consists of slots on its
periphery and the air ducts to permit the air flow through armature which serves cooling
purpose.
a) Functions:
i) Armature core provides house for armature winding i.e. armature conductors.
ii) To provide a path of low reluctance to the magnetic flux produced by the field
winding.
b) Choice of material :
A it has to provide a low
Reluctance path to the flux,
it is made up of magnetic
material like cast iron or
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steel.
It is made up of laminated Construction to keep eddy current Loss as low as possible.
A single circular Lamination used for the construction of the armature core is shown in
Fig. 28.
II) Armature winding is nothing but the interconnection of the armature conductors,
placed in the slots provided on the armature core periphery. When the armature is rotated,
in case of generator, magnetic flux gets cut by armature conductors and e.m.f. gets
induced in them.
a) Functions:
i) Generation of e.m.f. takes place in the armature winding in case of generators.
ii) To carry the current supplied in case of d.c. motors
iii) To do the useful work in the external circuit.
.in
b) Choice of material: As armature windings carries entire current which depends on
external load, it has to be made up of conducting material, which is copper.
sy
Armature winding is generally former wound. The conductors are placed in the
armature slots which are lined with tough insulating material.
ea
Commutator:
We have seen earlier that the basic nature of e.m.f induced in the armature
conductors is alternating. This needs rectifications in case of d.c. generator which is
possible by device called commutator.
y
a) Functions:
ud
## i) To facilitate the collection of current from the armature conductors.
ii) To convert internally developed alternating e.m.f. to unidirectional (d.c.) e.m.f.
iii) To produce unidirectional torque in case of motors.
st
## b) Choice of material: As it collects current from armature, it is also made up of copper
segments.
It is cylindrical in shape and is made up of wedge shaped segments of hard drawn, high
conductivity copper. Those segments are insulated from each other by thin layer of mica.
Each commutator segment is connected to the armature conductor by means of copper
lug or strip. This connection is shown in the fig. 29.
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## Brushes and brush gear:
Brushes and stationary and resting on the surface of the commutator.
a) Functions:
i) To collect current from commutator and make it available to the stationary external
circuit.
b) Choice of material: Brushes are normally made up of soft material like carbon.
Brushes are rectangular in shape. They are housed in brush holders, which are
usually of box type. The brushers are made to press on the commutator surface by means
of a spring, whose tension can be adjusted with the help of lever. A flexible copper
conductor called pig tail is used to connect the brush to the external circuit.
Bearings:
.in
Ball bearings are usually as they are more reliable. For heavy duty machines, roller
bearings are preferred.
6) Derive the expression of armature torque developed in a DC motor.
## Sol.: Torque equation of D.C. motor:
sy
It is seen that the turning or twisting force about
ea
an axis is called torque.
y
## Circumferential force of F Newtons as shown in fig. 30.
The wheel is rotating at a speed of N r.p.m.
ud
st
## W = F distance travelled in one revolution
= F 2R joules
And Pm = power developed =
Pm = T
where T = Torque in N-m = (FR)
. = angular speed in rad/sec = (2 N/60)
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Let Ta be the gross torque developed by the armature of the motor. It is also called
armature torque. The gross mechanical power developed in the armature is E b Ia, as seen
from the power equation. So if speed of the motor is N r.p.m. then,
Power in armature = Armature torque
## But Eb in a motor given by, Eb =
.in
sy
This is the torque equation of a d.c.motor.
N-m
ea
8) A 120V DC shunt motor has an armature resistance of 60 . It runs at 1800 RPM,
when it takes full load current of 40A. Find the speed of the motor while it is operating
y
ud
Sol.:
,
IL1 = Ia1 + If
st
## = 120 7.6 = 112.4V
N1 = 1800rpm
Now for shunt machine
T Ia as remains constant.
T 2 = T1
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## Ia2 = Ia1 = (38) = 19A
Eb2 = V Ia2 Ra = 120 (19)(0.2) = 120 3.8
Eb2 = 116.2V
Now, Eb N for constant
N2 = 1860.85 rpm
.in
A 4 pole generator with wave wound armature has 51 slots, each having 24 conductors.
The flux per pole is 0.01 Weber. At what speed must the armature rotate to give an
induced emf of 220 V? What will be the voltage developed if the voltage developed if the
sol.: P = 4,
Number of slots = 51,
sy
winding is lap and the armature rotates at the same speed?
for wave winding A = 2
conductors/slot = 24
[6]
= 0.01 wb
ea
E = 220 volts
Z = Number of slots conductors/slots
= 51 24
y
= 1224
ud
E=
st
N= =
N = 539.21 rpm
Now speed is same but winding is lap A=P
E=
E = 110 volts
A 4 pole, 220 V. lap connected DC shunt motor has 36 slots, each slot containing 16
conductors. It draws a current of 40 A from the supply. The field resistance and armature
resistance are 110 , 0.1 respectively. The motor develops an output power of 6 kW.
The flux per pole is 40 mwb. Calculate
a) the speed
b) the torque developed by the armature and
c) the shaft torque [7]
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## Sol.: P = 4, V = 220 volts, slots = 36, A=P=4
Conductors/slot = 16, Z = 36 16 = 576
IL = 40 A
R sh = 110 = 0.1
P output = 6 KW, = 40 wb
Ish =
IL = Ia+Ish
Ia = IL - Ish
= 40 2 = 38A
Eb = V - IaRa
Eb = 220 (38) (0.1)
= 220-3.8
.in
Eb = 216.2 volts
Now, Eb =
sy
N=
N = 563 rpm
=
ea
Ta = 0.159 Z
= 0.159 ZIa
y
ud
T a = 139.20 N-m
st
P out = T sh .
## Tsh = = 101.76 N-m
T sh = 101.76 N-m
A 120V DC shunt motor has an armature resistance of 60 . It runs at 1800 RPM, when it
takes full load current of 40A. Find the speed of the motor while it is operating at half
Sol.:
,
IL1 = Ia1 + If
Ia1 = IL1- If = 40 2 = 38A
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## = 120 7.6 = 112.4V
N1 = 1800rpm
Now for shunt machine
T Ia as remains constant.
T2 = T1
.in
Ia2 = Ia1 = (38) = 19A
Eb2 = V Ia2 Ra = 120 (19)(0.2) = 120 3.8
Eb2 = 116.2V
Now, Eb N for
sy
constant
y ea
ud
N2 = 1860.85 rpm
st
## Transformers: Principle of operation and construction of single-phase transformers
(core and shell types). emf equation, losses, efficiency and voltage regulation (Open
Circuit and Short circuit tests, equivalent circuit and phasor diagrams are excluded).
Illustrative problems on emf equation and efficiency only.
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UNIT-6
TRANSFORMERS
TRANSFORMER is a static device which transfer electric energy from one electric
circuit to another at any desired voltage with out any change in frequency.
## PRINCIPLE:- A transformer works on the principle of mutual induction. Whenever a
change in current takes place in a coil there will be an induced emf in the other coil
wound over the same magnetic core. This is the principle of mutual induction by which
the two coils are said to be coupled with each other.
.in
ea
Fig.1
y
The fig1 shows the general arrangement of a transformer. C is the iron core made
ud
of laminated sheets of about 0.35mm thick insulated from one another by varnish or thin
paper. The purpose of laminating the core is to reduce the power loss due to eddy
currents induced by the alternating magnetic flux. The vertical portions of the core are
st
called limbs and the top and bottom portions are called the yokes. Coils P and S are
wound on the limbs. Coil P is connected to the supply and therefore called as the primary,
coil S is connected to the load and is called as the secondary.
An alternating voltage applied to P drives an alternating current though P and this
current produces an alternating flux in the iron core, the mean path of the flux is
represented by the dotted line D. This flux links with the coil S and thereby induces an
emf in S.
## TYPES AND CONSTRUCTION OF TRANSFORMERS
There are two basic circuits in a transformer
## 1) Magnetic circuit 2) Electric circuit
The core forms the magnetic circuit and the electric circuit consists of two
windings primary and secondary and is made of pure copper. There are two types
of single phase transformers.
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## a) CORE TYPE b) SHELL TYPE
Figs (a) and (b) shows the details of the elevation and plan of a core type transformer.
The limbs are wound with half the L.V. and half the H.V. windings with proper
insulation between them. The whole assembly taken inside a steel tank filled with oil for
the purpose of insulation and cooling.
## CORE TYPE TRANSFORMER.
In the core type the core is surrounded by the coils but in the shell type the core is on the
either side of the coils. There are three limbs and the central limb is of large cross section
than that of outer limbs, and both the LV and HV windings are wound on the central limb
and the outer limb is only for providing the return path for the flux.
The windings are of concentric type (i.e. LV on which the HV windings) or Sandwich
.in
type.
The core is made of very thin laminations of high grade silicon steel material to
reduce the eddy current loss and Hysterisis losses in the core.
2222222222222222222222222222222222
sy
y ea
ud
st
## and the core and the core.
2- L V winding.
3- Insulation between L V and H V
winding winding. 4- End
insulation between the coils and
the yoke.
5- H V winding.
6- Limbs.
` 7- Yoke.
FIG. (b)
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.in
sy
y ea
## 1, 1 Outer limbs, 4 L V winding
2 Central limb, 5 Insulation between L V and H V windings.
ud
## 3 Insulation between L V and Core. 6 H V winding. 7 End insulation
st
In the shell type transformers the core is of different type having three limbs with the
central limb of larger cross section compared to the two outer limbs and carries both the
L V and H V windings wound over each other with proper insulation between them. The
entire assembly is immersed in a steel tank filled with oil for the cooling purpose.
EMF EQUATION:
Priciple:- Whenever a coil is subjected to alternating flux, there will be an induced emf
Nd
in it and is called the statically induced emf e
dt
Let N1, N2 be the no. of turns of the primary and secondary windings, E 1, E2 the induced
emf in the primary and secondary coils. be the flux which is sinusoidal f be the
frequency in Hz
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## Figure showing the sinusoidally varying flux of peak value m .
Whenever a coil of N no- of tunes are linked by a time varying flux , the average emf
induced in this coil is
.in
Nd
e
dt
As the flux is sinusoidal the change in flux from + m to - m is d = 2 m, and this
sy
change takes place in a duration dt = T/2 seconds.
The average induced emf in these N number of turns is
ea
Eavg = N.d /dt = N.2 m/ (T/2) = 4 mN/T = 4f mN volts (as f =1/T)
We know that the Form factor of a pure sine wave F.F. = Erms/Eavg = 1.11
y
ud
st
## In the secondary coil, N = N2 , E2 = 4.44f m N2 volts
TRANSFORMATION RATIO:
It is defined as the ratio of the secondary induced emf to the primary induced emf.
Therefore, E1 / E2 = N1 / N2 = K
For an ideal (loss free ) transformer, the input power is equal to the out put power.
Therefore E1I1 = E2I2, from which, E2/E1 = I1/I2
Also the induced emf per turn is same for both the primary and secondary turns.
If the value of the transformation ratio K>1, then it is a step up case.
If the value of the transformation ratio K<1, then it is a step down case.
If the value of the transformation ratio K=1, then it is a one:one transformer.
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LOSSES AND EFFICIENCY: There a two types of power losses occur in a transformer
## 1) Iron loss 2) Copper loss
1) Iron Loss: This is the power loss that occurs in the iron part. This loss is due to the
alternating frequency of the emf. Iron loss in further classified into two other losses.
## a) Eddy current loss b) Hysterisis loss
a) EDDY CURRENT LOSS: This power loss is due to the alternating flux linking the
core, which will induced an emf in the core called the eddy emf, due to which a current
called the eddy current is being circulated in the core. As there is some resistance in the
core with this eddy current circulation converts into heat called the eddy current power
loss. Eddy current loss is proportional to the square of the supply frequency.
.in
b) HYSTERISIS LOSS: This is the loss in the iron core, due to the magnetic reversal of
the flux in the core, which results in the form of heat in the core. This loss is directly
proportional to the supply frequency.
sy
Eddy current loss can be minimized by using the core made of thin sheets of
silicon steel material, and each lamination is coated with varnish insulation to suppress
ea
the path of the eddy currents.
Hysterisis loss can be minimized by using the core material having high
y
permeability.
ud
2) COPPER LOSS: This is the power loss that occurs in the primary and secondary
coils when the transformer is on load. This power is wasted in the form of heat due to the
resistance of the coils. This loss is proportional to the sequence of the load hence it is
st
called the Variable loss where as the Iron loss is called as the Constant loss as the supply
voltage and frequency are constants
EFFICENCY: It is the ratio of the out put power to the input power of a transformer
Input = Output + Total losses
= Output + Iron loss + Copper loss
Efficiency =
outputpower
outputpower Ironloss copperloss
V2 I 2 cos
V2 I 2 cos Weron Wcopper
Where, V2 is the secondary (out put) voltage, I2 is the secondary (out put) current and
cos is the power factor of the load.
The transformers are normally specified with their ratings as KVA,
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Therefore,
## Efficiency= (KVA)(103) cos
----------------------------------------------
(KVA)(103) cos + Wiron + Wcopper
Since the copper loss varies as the square of the load the efficiency of the transformer
at any desired load x is given by
## Efficiency= x. (KVA)(103) cos
-------------------------------------------------
x. (KVA)(103) cos + Wiron + (x)2 Wcopper
## where Wcopper is the copper loss at full load
.in
Wcopper = I2R watts
## CONDITION FOR MAXIMUM EFFICIENCY:
sy
ea
In general for the efficiency to be maximum for any device the losses must be
minimum. Between the iron and copper losses the iron loss is the fixed loss and the
copper loss is the variable loss. When these two losses are equal and also minimum
y
ud
## Iron loss = Copper loss ( which ever is minimum)
st
VOLTAGE REGULATION:
## The voltage regulation of a transformer is defined as the change in the secondary
terminal voltage between no load and full load at a specified power factor expressed
as a percentage of the full load terminal voltage.
= ---------------------------------------------------------------------------x 100
## Voltage regulation is a measure of the change in the terminal voltage of a transformer
between No load and Full load. A good transformer has least value of the regulation
of the order of 5%
A 600 KVA transformer has an efficiency of 92% at full load, uni8ty p.f. and at half load,
0.9 p.f. determine its efficiency of 75% of full load and 0.9 p.f.
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Sol. : S = 600 KVA, % = 92% on full load and half load both
0.92 =
(1)
and
.in
sy
0.92 =
.. (2)
ea
Subtracting (2) from (1),
= 28695.64
y
= 38260.86 watts
and Pi = 13913.04 watts
ud
## Now n = 0.75 i.e., 75% of full load and cos 2 = 0.9
(Pcu) new = n2 (Pcu) F.L. = (0.75)2 .
st
% =
= 91.95%
transformer has 80 turns on the secondary.
Calculate :
i) The rated primary and secondary currents
## ii) The number of primary turns
iii) The maximum value of flux
iv) Voltage induced per turn.
Sol: KVA rating = 250
V1 = 11000 volts
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V2 = 415 volts
N2 = 80
N1 =
= N2 = 80 = 2120
N1 = 2120
KVA =V1I1 = V2I2
I1 = = = 22.72A
I2 = = = 602.40A
Neglecting drops in primary,
V1 = E1 = 11000
.in
E1 = 4.44f m N1
= m
m = 0.023 Wb = 23 mWb
sy
Voltage induced per turn = = = 5.1886 volts.
ea
) In a 25 kVA, 2000/200 V Transformer, the iron and copper losses are 350 watts and 400
watts respectively, calculate the efficiency at U.P.F. at half and 3/4th full load.
y
Sol.: S = 25 k V A
P i = 350 W, (P cu )FL = 400 W
ud
## i) At half load, cos = 1
st
=
=
% Efficiency =
==
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% Efficiency = 96.525
## ii) At th load, cos = 1
% Efficiency =
.in
sy
% Efficiency = 97.02%
A 600 KVA transformer has=an efficiency of 92% at full load, uni8ty p.f. and at half load,
ea
0.9 p.f. determine its efficiency of 75% of full load and 0.9 p.f.
## Sol. : S = 600 KVA, %
y
ud
=
0.92 =
st
(1)
and
0.92 =
.. (2)
Subtracting (2) from (1),
= 28695.64
= 38260.86 watts
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## and Pi = 13913.04 watts
Now n = 0.75 i.e., 75% of full load and cos 2 = 0.9
(Pcu) new = n2 (Pcu) F.L. = (0.75)2 .
% =
= 91.95%
RECOMMENDED QUESTIONS
.in
1. The required No load ratio of a single phase 50Hz, core type transformer is
6000/150v. Find the number of turns per limb on the high and low voltage sides if
the flux is to be 0.06 wb. sy
2. A 125 KVA, single phase transformer has a primary voltage of 2000 v at 60
ea
Hz. If the number of turns of the primary is 182 and secondary has 40 turns
find (i) the secondary no load voltage (ii Flux in the core (iii) the full load
primary and secondary currents.
y
3. A 600 KVA, single phase transformer has an efficiency of 92% both at full load
ud
and half full load at unity power factor. Determine the efficiency at 75% of full
4. A 50KVA transformer has an efficiency of 98% at full load, 0.8 p.f. and an
efficiency of 96.9% at 1/4th full load, u.p.f. Determine the iron loss and full
st
5. In a 50KVA, 2000/200 v single phase transformer the iron loss and full load
copper losses area 700 w and 1000 W respectively. Calculate the efficiency of
the transformer on full load and at half full load. The p.f. of the load is 0.75
lag.
## 6. A 5 KVA, 200/100 v, single phase, 50Hz transformer has a rated secondary
voltage at full load. When the load is removed the secondary voltage is found to
be 110 v. Determine the percentage voltage regulation.
7. A single phase transformer has 400 primary and 1000 secondary turns. The net
cross sectional area of the core is 60 cm2. If the primary winding is connected to a
50Hz supply at 500v, calculate (i) the peak value of the flux density in the core
(ii) the voltage induced in the secondary winding.
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## SOLUTION TO QUESTION BANK
a) With neat sketches explain the constructional details of core type and shell type
transformers.
Sol: Core type transformer :
It has a single magnetic circuit. The core is rectangular having two limds. The
Winding encircles the core. The coils used are of cylindrical type. As mentioned earlier,
the coils are wound in helical layers with different layers insulated from each other by
paper or mica. Both the coils are placed on both the limbs. The low voltage coil is placed
inside near the core while high voltage coil surrounds the low voltage coil core is made
up of large number of thin laminations.
As the windings are uniformly distributed over the two limbs the natural cooling is
more effective. The coils can be easily removed by removing the laminations of the top
yoke, for maintenance.
.in
sy
y ea
ud
st
The fig.12(a) shows the schematic representation of the core type transformer while
the Fig.12(b) shows the view of actual construction of the core type transformer.
2) Shell type transformer:
It has a double magnetic circuit. The core has three limds. Both the windings are
placed on the central limb. The core encircles most part of the windings. The coils used
are generally multilayer disc type or sandwitch coils. As mentioned earlier, each high
voltage coil is in between two low voltage coils and low voltage coils are nearest to top
and bottom of the yokes.
The core is laminated. While arranging the laminations of the core, the core is taken
that all the joints at alternate layers are staggered. This is done to avoid narrow air gap at
the joint, right through the cross-section of the core. Such joints are called over lapped or
imbricated joints. Generally for very high voltage transformers, the shell type
construction is preferred. As the windings are surrounded by the core, the natural cooling
does not exist. For removing any winding for maintenance, large number of laminations
are required to be removed.
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The Fig.13(a) shows the schematic representation while the Fig.13(b) shows the
outaway view of the construction of the shell type transformer.
## 3) A 250 KVA, 11000/415V, 50 Hz single phase transformer has 80 turns on the
.in
secondary.
Calculate :
v) The rated primary and secondary currents
vi) sy
The number of primary turns
vii) The maximum value of flux
viii) Voltage induced per turn.
ea
Sol: KVA rating = 250
V1 = 11000 volts
V2 = 415 volts
y
N2 = 80
ud
=
N1 = = N2 = 80 = 2120
st
N1 = 2120
KVA =V1I1 = V2I2
I1 = = = 22.72A
I2 = = = 602.40A
Neglecting drops in primary,
V1 = E1 = 11000
E1 = 4.44f m N1
= m
m= 0.023 Wb = 23 mWb
Voltage induced per turn = = = 5.1886 volts.
4) Develop an expression for the efficiency of a single phase transformer and obtain the
condition for maximum efficiency.
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## Sol.: Efficiency of transformer: Explain the principle of operation of a single phase
transformer and derive its EMF equation.
## Sol.: principle of working: A transformer works on the principle of electromagnetic
induction and mutual induction between the two coils. To understand this, consider the
elementary transformer shown in fig. 32.
.in
sy
ea
It consists of two separate electrical windings which linked through common magnetic
circuit. electrically windings are isolated from each other.
y
The coil in which electrical energy is fed is called primary winding (P) while the other
ud
from which electrical energy is drawn out is called secondary winding (S). The primary
winding has N1 number of turns while secondary windings has N2 number of turns.
st
When primary winding is excited by alternating voltage say V1, it circulate alternating
current through it. This current produces an alternating flux which completes its path
through the common magnetic core as shown in fig. 32. This flux links with both the
windings.
Because of this it produces self induced e.m.f. E 1 in the primary winding while due to
mutual induction i.e. due to flux produced by primary linking with secondary, it produces
induced e.m.f. E2 in secondary winding.
## Taking ratio of the above expressions,
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## The frequency of two e.m.f.s is same.
If now secondary circuit is closed through the load, the mutually induced e.m.f. in the
secondary winding circulates current through the load. Thus electrical energy is
transferred from primary to secondary with the help of magnetic core.
A voltage V2 appears across the load.
Hence V1 is the supply voltage, while V2 is the secondary voltage when load is
connected then,
i.e.
K= transformation ratio
If k > 1 then v2 > v1 transformer is called step up transformer.
.in
If k < 1 then v2 < v1 transformer is called step down transformer
If k = 1 then v2 = v1 transformer is called one to one transformer or isolation transformer.
sy
The current flowing through primary is I 1 and when load is connected current I2 flows
through secondary winding. The powder transfer from primary to secondary remains
same. Assuming both primary and secondary power factors same we can write,
ea
Power input to primary = power output from secondary
V1I1 = V2I2
y
=
ud
## E.M.F. equation of a transformer:
Primary winding is exicted by a voltage which is alternating in nature. This circulates
current through primary which is also alternating and hence the flux produced is also
st
sinusoidal in nature.
Please refer Fig. 33 on next page.
= m sint Wb
m = Maximum flux
N1 = Primary turns
N2 = secondary turns
F = frequency of supply voltage Hz.
E1 = RMS value of primary induced e.m.f.
E2 = RMS value of secondary induced e.m.f.
From Faradays law of electromagnetic induction,
Average e.m.f. induced in each turn = Average rate of change of flux
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## = (considering magnitude only)
To find average rate of change of flux, consider 1/4 th cycle of flux as shown in the fig.
33. Complete cycle gets completed in 1/f seconds so for 1/4 th cycle time required is 1/4f
seconds. The flux reaches to m starting from zero during this time.
m Wb/ sec
## e purn turn = m volts
since is varying sinusoidally with time, the induced e.m.f. in each turn of both the
windings is also sinusoidal in nature.
For sinusoidal quantity , form factor =
## RMS value of induced e.m.f. per unit turn = 1.11e = 4.44f m
.in
RMS value of induced e.m.f. in primary winding = 4.44 f m N1 volts
RMS value of induced e.m.f. in secondary winding = 4.44 f m N2 volts
sy
Due to losses power output of transformer is always less than power input.
Power output = power input losses
ea
Hence efficiency is defined as ratio of output power to input power both being in
watts or kilowatts.
=
y
ud
=
st
=
where Pi = iron loss and is load p.f.
Pcu = copper loss
But V2I2 = V1I1= VA rating of transformer
=
reduces by same propotion i.e. for half load half of full load, for 1/4 th load 1/4th of full
load and so on. Thus output VA available which is V2I2 also reduces by same proportion.
While copper losses reduces propotional to square of current.
Thus if n is the fraction of full load applied to the transformer then efficiency is
given by,
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% =
Where n < 1 which is fraction of full load
6) Condition for maximum efficiency:
When transformer works on a constant input voltage and frequency and with a variable
load, its efficiency varies. Let us find out load at which efficiency is maximum and the
condition for the maximum efficiency.
The efficiency depends on the load current I 2. Assume the power factor of load constant
and secondary voltage V2 substantially constant. The for maximum efficiency,
as Pcu =
.in
i.e. (
( cancelling V2 cos
Pi - =0
Pi = = Pcu
sy
ea
Thus for maximum efficiency,
Iron loss = copper loss
Now, = Pi
y
## I2 = Amp for max
ud
This is the load current at which efficiency is maximum. This is denoted as I2 max. let
us find out KVA at which maximum efficiency occurs:
st
## Let (pcu) F.L. ( full load KVA)2 as Pcu I2
At max, Pi = Pcu
Let x be the KVA at which efficiency is maximum.
Pcu at max x2
but Pcu = Pi hence, Pi x2 at max
taking ratio of (1) and (2) ,
## Thus x is the new kVA at which efficiency is maximum.
Now at max, Pi = Pcu
% max =
Now the current at max, is I2 max hence max m cxanalso be expressed as,
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% max =
8) A 600 KVA transformer has an efficiency of 92% at full load, uni8ty p.f. and at half
load, 0.9 p.f. determine its efficiency of 75% of full load and 0.9 p.f.
## Sol. : S = 600 KVA, %
0.92==
(1)
.in
and
sy
ea
0.92 =
.. (2)
Subtracting (2) from (1),
y
= 28695.64
ud
= 38260.86 watts
and Pi = 13913.04 watts
Now n = 0.75 i.e., 75% of full load and cos 2 = 0.9
st
## (Pcu) new = n2 (Pcu) F.L. = (0.75)2 .
% =
= 91.95%
d) A 25 kVA transformer has an efficiency of 94% at full load unity p.f. and at half
0.9 p.f. determine the iron loss and full load copper loss (6)
Ans. : = 94% at cos = 1, = 94% at cos = 0.9
at full load , = fl
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n = 0.5
% =
0.94=
## (Pcu)FL + Pi = 1595.7446 (1)
.in
% FL = n = 0.5
0.94=
sy
0.25(Pcu)FL + Pi = 718.0851 (2)
ea
Subracting equations (1) and (2),
0.75(Pcu)FL=877.6595
y
## c) A 250 KVA, 11000/415V, 50 Hz single phase transformer has 80 turns on the
ud
secondary.
Calculate :
i) The rated primary and secondary currents
ii) The number of primary turns
st
## iii) The maximum value of flux
iv) Voltage induced per turn.
Sol: KVA rating = 250
V1 = 11000 volts
V2 = 415 volts
N2 = 80
=
N1 = = N2 = 80 = 2120
N1 = 2120
KVA =V1I1 = V2I2
I1 = = = 22.72A
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## Synchronous Generators: Principle of operation. Types and constructional features. emf
equation. Concept of winding factor (excluding derivation of distribution and pitch
factors). Illustrative examples on emf. equation.
UNIT-7
## THREE PHASE ALTERNATORS.
(SYNCHRONOUS GENERATORS)
(THREE PHASE A.C. GENERATORS).
## Electric power is generated using three phase alternators.
Principle: Whenever a coil is rotated in a magnetic field an EMF will be induced in the
.in
coil. This is called the dynamically induced EMF.
Alternators are also called as Synchronous Generators due to the reason that under
sy
normal conditions the generator is to be rotated at a definite speed called
SYNCHRONOUS SPEED, Ns R.P.M. in order to have a fixed frequency in the output
EMF wave.
ea
Ns is related with the frequency as Ns = 120f / P, where f is the frequency and P is the
total number of poles.
y
The following table gives the idea of the various synchronous speeds for various numbers
ud
## of poles for the fixed frequency of 50 Hz.
P 2 4 6 8 10 12 16 .
.
st
rpm
## Their two basic parts in an alternator: (i) Stator, (ii) Rotor.
Stator is the stationary part and Rotor is the revolving part.
There are two possibilities that (i) The armature can be the stator and the field system can
be the rotor, and (ii) The armature can be the rotor and the field system be the stator. In
practice large alternators are of the first type where in the stator is the armature and the
rotor is the field system. And this type is called the REVOLVING FIELD TYPE.
Revolving field types are preferred due to the following reasons:
(i) More conductors can be easily accommodated and with these high voltage and
higher power capacity
can be achieved.
(ii) Armature conductors can be easily braced over a rigid frame.
(iii) It is easier to insulate a stationary system.
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(iv) Cooling of the conductors will be very effective with proper cooling ducts /
vents in the stationary part.
(v) Power can be tapped easily with out any risk from the stationary part through
terminal bushings.
(vi) The armature conductors are totally free from any centrifugal force action which
tends to drag the conductors out of the slots.
ONSTRUCTION:
Revolving field type alternators are further classified into two types:
(i) Salient pole type, (ii) Non-salient pole type or Cylindrical rotor type.
Figs. (a), (b) and (c) shows the constructional features of the Alternator. Fig. (a)
represents the stator, the core of which is made of steel laminations with slots cut in
its inner periphery and all the stator stampings are pressed together and are fixed to
.in
the stator frame. Three phase windings are accommodated in these slots. These coils
are identical to each other and are physically distributed such that they are displaced
from each other by 120 degrees as shown in fig. (d).
sy
y ea
ud
st
Fig. (b) represents the structure of a salient pole rotor where the poles are of
projected type and are mounted on a spider and the field or the pole windings are
wound over the pole core as shown. This type is preferred where the running speeds
are low. Fig.(c) represents the structure of a non-salient pole rotor where the overall
structure is like a cylinder having 2 or 4 poles. This type is preferred where the
running speeds are very high.
The armature windings in the stator are made of copper and are normally arranged in
two layers and are wound for lap or wave depending on the requirements and are
usually connected in star with the neutral terminal brought out.
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EMF Equation:
## Let P be the total number of poles, Ns be the
synchronous speed, f be the frequency of the induced
EMF and the flux considered to be sinusoidally
distributed.
As we know that the induced emf is due to the rate of
change of flux cut by coils, the average induced emf in
Tph number of turns is
## For a flux change from m to m is d = 2 m in time dt= T / 2 seconds,
The average induced Emf = Tph. 2 m / ( T/2 ) = 4 Tph .f. m volts.
For a sine wave we know that the form factor is of value 1.11= Erms / Eavg.
.in
Therefore, Erms = 1.11.Eavg.
Erms = 4.44 f m Tph volts per phase. . . . . . . . . . .(1)
sy
If the armature windings are connected in star the line emf is El = 3 Ephase.
If the armature windings are connected in delta the line emf is the phase emf itself.
Equation (1) represents the theoretical value of the induced em f in each phase but in
ea
practice the
Induced emf will be slightly less than the theoretical value due to the following
reasons:
y
(i) The armature windings are distributed throughout the armature in various
ud
and is given by
st
## Kd = (Sin(m / 2) / mSin( / 2)), where m is the number of slots
per pole peer phase
and is the slot angle.
= 1800 / no. of slots per pole.
(ii) The span of the armature coil is less than a full pitch This is done
deliberately to eliminate some unwanted harmonics in the emf wave, this fact is
accounted by a factor called the coil span factor or the pitch factor, Kp and is given
by
Kp = Cos ( / 2), where is the angle by which the coils are short
chorded.
The modified Emf equation with these two factors taken into account will be
## E = 4.44 Kd.Kp..f Tph volts per phase.
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## The product of Kd and Kp is called as the winding factor Kw .which is of
value around 0.95.
VOLTAGE REGULATION:
The voltage regulation of an alternator is defined as the change in the terminal
voltage between no load and full load at a specified power factor, without any change
in the speed and excitation.
% Voltage regulation = --------------------------------------------------------------------- x
100
EV
% Voltage regulation = -------- x 100
.in
V
## The idea of voltage regulation is necessary to judge the performance of an
sy
alternator. Lesser the value of the regulation better will be the load sharing capacity at
better efficiency.
A 3 phase, 6 pole, star connected alternator has 48 slots and 12 conductors per slot on the
ea
armature. If the rotor at 1200RPM and the flux per pole is 0.3 Wb, calculate the e.m.f.
induced in the armature. The coils are full pitched and the winding factor is 0.95.
y
## sol. : P = 6, slots = 48, conductors/ slots = 12, = 0.3 wb
ud
Ns = Ns = 1200 rpm
f=
st
## for full pitched winding, Kc = 1, Kd = 0.95
Total conductors = slots conductors/slot = 4812 =
576
Zph =
Tph =
Eph = 4.44 Kc Kd f Tph
Eph = 4.4410.95600.396
= 75.9 96
Eph = 7288.70 volts
37288.70
Eline = 12.624
3 Eph =Kv
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REVIEW PROBLEMS:
1. The stator of a three phase, 8 pole, 750 rpm alternator has 72 slots, each of which
having 10 conductors. Calculate the RMS value of the emf per phase if the flux
per pole is 0.1 wb and the winding factor is 0.96. Also find the line Emf if the
windings are connected in (i) Star, and (ii) Delta.
2. A three phase star connected alternator driven at 900 rpm is required to generate a
line voltage of 460 volts at 60 Hz on open circuit. The stator has 2 slots per pole
per phase and 4 conductors per slot. Calculate (i) the number of poles, (ii) the
useful flux per pole.
3. A 4 pole, three phase, 50 Hz, star connected a.c. generator has 24 stator slots.
Find the number of conductors per slot if the flux per pole is 62 mwb and the
terminal voltage is 1100 volt. Assume full pitch coils.
.in
4. A 16 pole star connected alternator has 144 slots and 10 conductors per slot. The
flux per pole is 30 mwb and the speed is 375 rpm. Find the frequency, the phase
and line emfs.
sy
5. Find the number of armature conductors in series per phase required for the
armature of a three phase
50 Hz 10 pole alternator with 90 slots. The winding is star connected to give a
ea
line emf of 11 Kv. The flux is 0.16 wb. Also find the voltage regulation if the
terminal voltage on full load is 11.2 kv.
6. A three phase 10 pole star connected alternator runs at 600 rpm. It has 120 stator
y
slots with 8 conductors per slot. Determine the phase and line emfs if the flux
per pole is 56 mwb.
ud
7. Calculate the phase emf induced in a 4 pole, three phase, 50 Hz star connected
alternator with 36 slots and 30 conductors per slot. The flux per pole is 0.05
wb. Given the winding factor as 0.95.
st
## SOLUTION TO QUESTION PAPER
a) Explain the principle of operation of an alternator. Discuss the different types of rotor
construction of alternator mentioning their typical advantages and applications.
[8]
## Sol. : Working principle of alternator :
The alternators work on the principle of electromagnetic induction. When there is
a relative motion between the conductors and the flux, e.m.f. gets induced in the
conductors. The d.c. gemerators also work on the same principle. The only difference in
practical alternator and a d.c. generator is that in an alternator the conductors are
stationary and field is rotating. But for understanding purpose we can always consider
relative motion of conductors with respect to the flux produced with respect to the flux
produced by the field winding.
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Consider a relative motion of a single conductor under the magnetic field produced
by two stationary poles. The magnetic axis of the two poles produced by field is vertical,
shown dotted in the fig. 34.
.in
Let the conductor starts rotating from position 1. At this instant, the entire velocity
sy
ea
component is parallel to the flux lines. Hence there is no cutting of flux lines by the
conductor. So at this instant is zero and hence induced e.m.f. in the conductor is also
zero.
y
As the conductor moves from position 1 to position 2, the part of the velocity
ud
component becomes perpendicular to the flux lines and propotional to that e.m.f.
increases as the conductor moves from position 1 towards 2.
st
At position 2, the entire velocity component is perpendicular to the flux lines. Hence
there exists maximum cutting of the flux lines. And at this instant, the induced e.m.f. in
the conductor is at its maximum.
As the position of the conductor changes from 2 towards 3, the velocity component
perpendicular to the flux starts decreasing and hence induced e.m.f. magnitude also starts
decreasing. At position 3, again the entire velocity component is parallel to the flux lines
and hence at this instant induced e.m.f. in the conductor is zero.
## As the conductor moves from position 3 towards 4, the velocity component
perpendicular to the flux lines again starts increasing. But the direction of velocity
component now is opposite to the direction of velocity component existing during the
movement of the conductor from position 1 to 2. Hence induced e.m.f. in the conductor
increases but in the opposite direction.
At the position 4, it achives maxima in the opposite direction, as the entire velocity
component becomes perpendicular to the flux lines.
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Again from position 4 to 1, induced e.m.f. decreases and finally at position 1,again
becomes zero. This cycle continues as conductor rotates at a certain speed.
So if we plot the magnitudes of the induced e.m.f. against the time, we get alternating
nature of the induced e.m.f. as shown in the fig. 34 (b).
## This is the working principle of an alternator.
Construction : Most of the alternators prefer rotating filed type of construction. In case
of alternators the winding terminology is slightly different than in case of d.c. generators.
In alternators the stationary winding is called stator while the rotating winding is called
rotor. So most of alternators have stator as armature and rotor as field, in practice.
Constructional details of rotating field type of alternator are discussed below.
1) stator : The stator is a stationary armature. This consists of a core and the slots to
.in
hold the armature winding similar to the armature of a d.c. generator. The stator core uses
a laminated construction. It is built up of special steel stampings insulated from each
other with varnish or paper. The laminated construction is basically to keep down eddy
sy
currents losses. The entire core is fabricated in a frame made up of steel plates. The core
has slots om its periphery for housing the armature conductors. Frame does not carry and
flux and serves as the support to the core. Ventilation is maintained with the help of holes
ea
cast in the frame. The section of an alternator startor is shown in the fig. 35.
y
ud
st
2) Rotor :
There are two types of rotors used in alternators which are:
1) salient pole type and 2) smooth cylindrical type.
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## 3) salient pole type :
.in
sy
ea
This is also called projected pole type as all the poles are projected out from the surface
of the rotor. The poles are built up of thick steel laminations. The poles are bolted to the
rotor as shown in the fig. 36. The pole face has been given a specific shape as discussed
y
earlier in case of d.c. generators. The field winding is provided on the pole shoe. These
rotors have large diameters and small axial lengths. The limiting factor for the size of the
ud
rotor is centrifugal force acting on the rotating member of the machine. As mechinal
strength of sailent ple type is less, this is preferred for low speed alternators ranging from
125 r.p.m. to 500 r.p.m. the prime movers used to drive such rotor are generally water
st
## ii) Smooth cylindrical type :
This is also called non salient type or non projected pole type of rotor.
The rotor consists of smooth solid steel cylinder, having number of slots to
accommodate the field coil. The slots are covered at the top with the help of steel or
manganese wedges. The unslotted portion of the cylinder itself act as the poles. The poles
are not projecting out and the surface of the rotor is smooth which maintains uniform air
gap between and the rotor. These rotors have small diameters and large axial lengths.
This is to keep peripheral speed within limits. The main advantage of this type is that are
mechanically very strong and thus preffered for high speed alternators ranging between
1500 to 3000 r.p.m. such high speed alternators are called turboalternators. The prime
movers used to drive such type of rotors are generally steam turbines, electric motors.
The fig. 37 shows smooth cylindrical type of rotor.
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.in
Difference between salient and cylindrical type of rotor :
Salient pole type Smooth cylindrical type
sy
1 poles are projecting out from the surface. portion of the cylinder acts as poles hence
poles are non projecting.
2 Air gap is non uniform. Air gap is uniform due to smooth
ea
cylindrical periphery.
3 Diameter is high and axial length is Small diameter and large axial length is
small. the feature.
y
## 4 Mechanically weak. Mechanically robust.
5 Preferred for low speed alternators. Preffered for high speed alternators i.e.
ud
for tuboalternators.
6 Prime mover used is water turbines, I.C. Prime mover used are steam turbines,
engines. electric motors.
st
7 For same size, the rating is smaller than For same size, rating is higher than
cylindrical type. sailent pole type.
8 Separate damper winding is provided Separate damper winding is not
necessary.
## Sol. : Voltage regulation of an alternator :
Under the load condition, the terminal volatage of alternator is less than the
induced e.m.f. Eph. So if the load is disconnected, Vph will change from V ph to Eph, if flux
and speed is maintained constant. This is because when load is disconnected I a is zero
hence there are no voltage drops and no armature flux to cause armature reaction. This
change in the terminal voltage is significant in defining the voltage regulation.
The voltage regulation of an alternator is defined as the change in its
terminal voltage when full load is removed, keeping field exicitation and speed constant,
divided by the rated terminal voltage.
So if, Vph = Rated thermal voltage
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## Eph = No load induced e.m.f.
Then voltage regulation is defined as,
% Reg =
The value of the regulation not only depends on the load current but also on the power
factor of the load. For lagging and unity p.f. conditions there is always drop in the
terminal voltage hence regulation values are always positive. While for leading capacitive
load conditions , the terminal voltage increases as load current Increases. Hence
regulation is negative in such cases. The relationship between load current and the
terminal voltage is called load characteristics for various load power factor conditions are
shown in the fig. 38.
.in
sy
ea
c) A 3 phase, 6 pole, star connected alternator has 48 slots and 12 conductors per slot on
the armature. If the rotor at 1200RPM and the flux per pole is 0.3 Wb, calculate the e.m.f.
y
induced in the armature. The coils are full pitched and the winding factor is 0.95.
ud
## sol. : P = 6, slots = 48, conductors/ slots = 12, = 0.3 wb
Ns = Ns = 1200 rpm
st
f=
for full pitched winding, Kc = 1, Kd = 0.95
Total conductors = slots conductors/slot = 4812 = 576
Zph =
Tph =
Eph = 4.44 Kc Kd f Tph
Eph = 4.4410.95600.396
= 75.9 96
## Eline = 3 Eph = 37288.70
Eline = 12.624 Kv
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## Three Phase Induction Motors: Concept of rotating magnetic field. Principle of
operation. Types and Constructional features. Slip and its significance. Applications of
squirrel - cage and slip - ring motors. Necessity of a starter, star-delta starter. Illustrative
examples on slip calculations.
UNIT-8
## Construction of squirrel cage and slip ring Induction motors
Production of rotating flux
Principle of operation
.in
Necessity of a starter for 3-phase induction motor
Star Delta starter
Slip
sy
By the end of the chapter you will be able to answer questions like:
Why 3-phase induction motor cannot run with zero slip?
ea
Why 1-phase induction is motors not self starting?
Why induction motors are the most preferred ac motors for many industrial
applications?
What do you understand by revolving flux?
y
ud
INTRODUCTION
The asynchronous motors or the induction motors are most widely used ac motors in
st
industry. They convert electrical energy in AC form into mechanical energy. They work
on the principle of electromagnetic induction. They are simple and rugged in
construction, quite economical with good operating characteristics and efficiency,
requiring minimum maintenance, but have a low starting torque. They run at practically
constant speed from no load to full load condition. The 3 - phase induction motors are
self starting while the single phase motors are not self starting as they produce equal and
opposite torques (zero resultant torque) making the rotor stationary. The speed of the
squirrel cage induction motor cannot be varied easily.
CLASSIFICATION -
They are basically classified into two types based on the rotor construction
1. Squirrel cage motor
2. Slip ring motor or phase wound motor
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CONSTRUCTION
## Three phase induction motor consists of two parts
(1) stator (2) rotor
Stator
It is the stationary part of the motor supporting the entire motor assembly. This outer
frame is made up of a single piece of cast iron in case of small machines. In case of larger
machines they are fabricated in sections of steel and bolted together. The core is made of
thin laminations of silicon steel and flash enameled to reduce eddy current and hysteresis
losses. Slots are evenly spaced on the inner periphery of the laminations. Conductors
insulated from each other are placed in these slots and are connected to form a balanced 3
- phase star or delta connected stator circuit. Depending on the desired speed the stator
.in
winding is wound for the required number of poles. Greater the speed lesser is the
number of poles.
Rotor sy
Squirrel cage rotors are widely used because of their ruggedness. The rotor consists of
ea
hollow laminated core with parallel slots provided on the outer periphery. The rotor
conductors are solid bars of copper, aluminum or their alloys. The bars are inserted from
the ends into the semi-enclosed slots and are brazed to the thick short circuited end rings.
y
This sort of construction resembles a squirrel cage hence the name squirrel cage
induction motor. The rotor conductors being permanently short circuited prevent the
ud
addition of any external resistance to the rotor circuit to improve the inherent low starting
torque. The rotor bars are not placed parallel to each other but are slightly skewed which
reduces the magnetic hum and prevents cogging of the rotor and the stator teeth.
st
The rotor in case of a phase wound/ slip ring motor has a 3-phase double layer distributed
winding made up of coils, similar to that of an alternator. The rotor winding is usually
star connected and is wound to the number of stator poles. The terminals are brought out
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and connected to three slip rings mounted on the rotor shaft with the brushes resting on
the slip rings. The brushes are externally connected to the star connected rheostat in case
a higher starting torque and modification in the speed torque characteristics are required.
Under normal running conditions all the slip rings are automatically short circuited by a
metal collar provided on the shaft and the condition is similar to that of a cage rotor.
Provision is made to lift the brushes to reduce the frictional losses. The slip ring and the
enclosures are made of phosphor bronze.
.in
sy
ea
In both the type of motors the shaft and bearings (ball and roller) are designed for trouble
y
free operation. Fans are provided on the shaft for effective circulation of air. The
insulated (mica and varnish) stator and rotor windings are rigidly braced to withstand the
ud
short circuit forces and heavy centrifugal forces respectively. Care is taken to maintain a
uniform air gap between the stator and the rotor.
st
## 1. Rugged in construction and economical.
2. Has a slightly higher efficiency and better power factor than slip ring motor.
3. The absence of slip rings and brushes eliminate the risk of sparking which helps
in a totally enclosed fan cooled (TEFC) construction.
## The advantages of the slip ring rotor are:
1. The starting torque is much higher and the starting current much lower when
compared to a cage motor with the inclusion of external resistance.
2. The speed can be varied by means of solid state switching
## (a) Production of a rotating magnetic field
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Consider a 3- phase induction motor whose stator windings mutually displaced from each
other by 120 are connected in delta and energized by a 3- phase supply.
.in
sy
. The currents flowing in each phase will set up a flux in the respective phases as shown.
y ea
ud
st
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Compiled by www.studyeasy.in
.in
sy
The corresponding phase fluxes can be represented by the following equations
sin t sin
ea
R m m
Y m sin t 120
Y msin 120
y
B m sin t 240
ud
B m sin 240
The resultant flux at any instant is given by the vector sum of the flux in each of the
phases.
st
## (i) When 0 , from the flux waveform diagram ,we have
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R 0
3
Y km sin( 120 ) m
2
3
B m sin( 240 ) m
2
## The resultant flux r is given by,
3
r 2* m cos(30 ) 1..5 m
2
.in
3
B m
2
3
sy
ea
Y m
2
r 1.5 m
y
(ii) When 60 o
ud
3
R m
2
st
3
Y m
2
B 0
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3
R m
2
Y 0
3
B m
2
.in
sy
y ea
ud
st
## (iv) When 180
R 0;
3
Y m
2
3
B
2
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Compiled by www.studyeasy.in
.in
From the above discussion it is very clear that when the stator of a 3-phase induction
motor is energized, a magnetic field of constant magnitude (1.5 m) rotating at
synchronous speed
sy
(Ns) with respect to stator winding is produced.
## (b) Rotation of the rotor
ea
Consider a 3- phase stator winding energized from a 3 phase supply. As explained earlier
a rotating magnetic field is produced running at a synchronous speed N S
120 f
y
NS = ---------
ud
## Where f = supply frequency
P = Number of stator poles
st
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ANIMATION INSTRUCTION
Consider a portion of 3- phase induction motor as shown in the above figure which is
representative in nature. The rotating field crosses the air gap and cuts the initially
stationary rotor conductors. Due to the relative speed between the rotating magnetic field
and the initially stationary rotor,(change of flux linking with the conductor) an e.m.f. is
induced in the rotor conductors, in accordance with the Faradays laws of
electromagnetic induction. Current flows in the rotor conductors as the rotor circuit is
short circuited. Now the situation is similar to that of a current carrying conductor placed
in a magnetic field. Hence, the rotor conductors experience a mechanical force which
eventually leads to production of torque. This torque tends to move the rotor in the same
direction as that of the rotating magnetic field.
## CONCEPT OF SLIP (S)
.in
According to Lenzs law, the direction of rotor current will be such that they tend to
sy
oppose the cause producing it. The cause producing the rotor current is the relative
speed between the rotating field and the stationary rotor. Hence, to reduce this relative
speed, the rotor starts running in the same direction as that of stator field and tries to
ea
catch it. In practice the rotor can never reach the speed of the rotating magnetic field
produced by the stator. This is because if rotor speed equals the synchronous speed, then
there is no relative speed between the rotating magnetic field and the rotor. This makes
y
the rotor current zero and hence no torque is produced and the rotor will tend to remain
stationary. In practice, windage and friction losses cause the rotor to slow down. Hence,
ud
the rotor speed (N) is always less than the stator field speed (N S). Thus the induction
motor cannot run with ZERO SLIP. The frequency of the rotor current
fr = sf. The difference between the synchronous speed (NS) of the rotating stator field and
st
## Slip speed = NS N depends upon the load on the motor
NS - N
% Slip (s) = ---------- * 100
NS
## Squirrel cage induction motor
Squirrel cage induction motors are simple and rugged in construction, are relatively
cheap and require little maintenance. Hence, squirrel cage induction motors are preferred
in most of the industrial applications such as in
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i) Lathes
ii) Drilling machines
iii) Agricultural and industrial pumps
iv) Industrial drives.
## Slip ring induction motors
Slip ring induction motors when compared to squirrel cage motors have high starting
characteristics.
## They are used in
i) Lifts
ii) Cranes
.in
iii) Conveyors , etc.,
## Necessity of starters for 3 phase induction motor
sy
When a 3- phase motor of higher rating is switched on directly from the mains it draws a
ea
starting current of about 4 -7 times the full load (depending upon on the design) current.
This will cause a drop in the voltage affecting the performance of other loads connected
to the mains. Hence starters are used to limit the initial current drawn by the 3 phase
y
induction motors.
The starting current is limited by applying reduced voltage in case of squirrel cage type
ud
induction motor and by increasing the impedance of the motor circuit in case of slip ring
type induction motor. This can be achieved by the following methods.
st
## 1. Star delta starter
2. Auto transformer starter
3. Soft starter
## Star delta starter
The star delta starter is used for squirrel cage induction motor whose stator winding is
delta connected during normal running conditions. The two ends of each phase of the
stator winding are drawn out and connected to the starter terminals as shown in the
following figure.
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.in
sy
y ea
ud
st
## When the switch is closed on the star-start side
(1) The winding is to be shown connected in star
(2) The current I = 1/3 * (I direct switching)
(3) Reduction in voltage by1/ 3
V = V supply*1/ 3
When the switch is closed on to delta run side
## (1) the winding to be shown connected in delta
(2) application of normal voltage V supply
(3) normal current I
During staring the starter switch is thrown on to the STAR - START. In this position the
stator winding is connected in star fashion and the voltage per phase is 1/ 3 of the supply
voltage. This will limit the current at starting to 1/3 of the value drawn during direct
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switching. When the motor accelerates the starter switch is thrown on to the DELTA -
RUN side. In this position the stator winding gets connected in the fashion and the
motor draws the normal rated current.
WORKED EXAMPLES
## 1. A 12 pole, 3 phase alternator is coupled to an engine running at 500 rpm. It supplies
an Induction Motor which ahs a full load speed of 1440 rpm. Find the percentage slop
and the number of poles of the motor.
## Solution: NA = synchronous speed of the alternator
PNA 12 X 500
.in
F =------- = --------------- = 50 Hz (from alternator data)
120 120
When the supply frequency is 50 Hz, the synchronous speed can be 750 rpm, 1500 rpm,
sy
3000rpm etc., since the actual speed is 1440 rpm and the slip is always less than 5% the
synchronous speed of the Induction motor is 1500 rpm.
ea
NS N 1500 - 1440
s = --------- = ----------------- = 0.04 OR 4%
NS 1500
y
120f 120 x 50
ud
P P
P=4
st
## 2. A 6 pole induction motor is supplied by a 10 pole alternator, which is driven at 600
rpm. If the induction motor is running at 970 rpm, determine its percentage slip.
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P NA 10 X 600
From alternator date: f =------- = --------------- = 50 Hz
120 120
Synchronous speed of the induction motor
120 f 120 50
NS 1000rpm
From I.M. data: P 6
NS N 1000 970
% slip 100 3%
NS 1000
## 3. A 12 pole, 3 phase alternator is driven by a 440V, 3 phase, 6 pole Induction Motor
running at a slip of 3%. Find frequency of the EMF generated by the alternator
NS 1000rpm
.in
P 6
## N 1 s NS 1 0.03 1000 970rpm
sy
As the alternator is driven by the Induction motor, the alternator runs at 970 r.p.m.
ea
For alternator: PN 12 970
f 97 Hz
120 120
4. A three phase 4 pole, 440 V, 50Hz induction motor runs with a slip of 4%. Find the
y
## rotor speed and frequency of the rotor current.
ud
120 f 120 50
NS 1500rpm
P 4
st
NS N 1500 N
Solution: S i.e.0.04 , N 1440rpm
NS 1500
fr sf 0.04 50 2 Hz
5. A 3 phase, 50Hz 6 pole induction motor has a full load percentage slip of
3%.
Find(i) Synchronous speed and (ii) Actual Speed
120 f 120 50
NS 1000rpm
P 6
Solution: NS N 100 N
S i.e.0.03 N 970rpm
NS 1000
6. A 3 phase induction motor has 6 poles and runs at 960 RPM on full load. It is supplied
from an alternator having 4 poles and running at 1500 RPM. Calculate the full load slip
and the frequency of the rotor currents of the induction motor.
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Solution:
PN 4 1500
f 50 Hz ( from alternator data )
120 120
forInduction motor
120 f 120 50
NS 1000rpm
P 6
N S N 1000 960
S 0.04 or 4%
NS 1000
fr sf 0.04 50 2 Hz
7. The frequency of the e.m.f in the stator of a 4-pole induction motor is 50 Hz and that of
1
.in
1 Hz
the rotor is 2 . What is the slip and at what speed is the motor running?
Solution:
Given P = 4
f = 50Hz
sy
ea
fr 1.5 Hz
y
ud
fr sf
1.5 S 50
1.5
st
We have s
50
s 0.03
s 3%
## ii) To calculate the speed of the motor (N)
We have
120 f
Ns
P
120 50
Ns 1500rpm
4
We also have
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N Ns 1 s
N 1500 1 0.03
N 1455rpm
8. A 3-phase, 60Hz induction motor has a slip of 3% at full load. Find the synchronous
speed, the full- load speed and the frequency of rotor current at full load.
Solution
Given P = 6
f = 60Hz
s= 3% = 0.03
To find the synchronous speed (Ns)
We have
.in
120 f 120 60
Ns
P 6
N s 1200rpm
sy
To calculate the full load speed (N)
ea
We have
N Ns 1 s 1200(1 0.03)
y
N 1164rpm
ud
## To calculate the frequency of the rotor current ( fr)
We have
st
f r sf 0.03 60
fr 1.8 Hz
9. A 6-pole alternator running at 600 rpm. supplies a 3-phase, 4-pole induction motor. If
the induction motor induced e.m.f makes 2 alternations per second, find the speed of the
motor.
Solution
Alternator
P=6
Ns = 600
We have the frequency of induced e.m.f of an alternator given by
PN s 6 600
f 30 Hz
120 120
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## Hence, induction motor receives the supply at 30Hz frequency
Induction motor
It is given that the rotor induced e.m.f makes two alternations per second i.e. 1.0 cycle
per second
f r 1.0 Hz
fr sf
We have
f r 1.0
s
f 30
s 0.033
## The speed of the rotating magnetic field is given by
120 f
Ns Where Pm = Number of poles of induction motor
Pm
.in
120 30
Ns 900rpm
4
sy
The speed of the induction motor (N) is given by
N N s 1 s 900 1 0.033
ea
N 870rpm
y
## 10. A 10-pole induction motor is supplied by a 6-pole alternator, which is driven at
ud
1200rpm. If the motor runs with a slip of 3, what is the speed of the induction motor?
Solution
Alternator
st
P = 6 pole
Ns = 1200rpm
Therefore, the frequency of the e.m.f generated is given by
PN s 6 1200
f
120 120
f 60 Hz
Hence, the induction motor is supplied at 60Hz frequency.
Induction motor
Supply frequency = f = 60Hz
Slip = S = 3%
Stator poles = Pm =10
Speed of the rotating magnetic field,
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120 f 120 60
Ns
Pm 10
Ns 720rpm
## Speed of the motor
N Ns 1 S 720 1 0.03
N 698.4rpm
11. A 3-phase induction motor has 6-poles and runs at 960r.p.m. on full load. It is
supplied from an alternator having 4-poles and running at 1500 r.p.m. calculate the full
Solution
.in
Alternator
P = 4 poles
Ns = 1500 rpm
sy
The frequency generated e.m.f is given by
ea
PN s 4 1500
f
y
120 120
f 50 Hz
ud
## Hence, the induction motor is supplied at 50 Hz
st
Induction motor
Pm 6 poles
f 50 Hz
N 960rpm
The speed of the rotating magnetic field is given
by 120 f 120 50
Ns
Pm 6
N s 1000rpm
We have slip of an induction motor given by
N s N 1000 960
S 0.04
Ns 1000
S 4%
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12. A 4-pole, 30hp, 3-phase 400 volts, 50Hz induction motor operates at an efficiency of
0.85 with a power factor of 0.75(lag). Calculate the current drawn by the induction motor
from the mains
Solution
Given P = 4
V = 400
= 0.85
cos = 0.75
Output= 30 hp
= 22.06538 Kw ( 1metric hp=735.5 watts)
We have,
.in
Output
Input
Output 22.065
Input
Input 25.96 KW
0.85 sy
But, for a 3-phase induction motor circuit, the power input is also given by the expression
ea
P 3VL I L cos
P 25.96 1000
IL
y
## 3VL cos 3 400 0.75
ud
IL 49.94 Amperes
st
13. A 5 hp, 400V, 50Hz, 6-pole, 3-phase induction motor operating on full load draws a
line current of 7 amperes at 0.866 power factor with 2%slip. Find the rotor speed.
Solution.
Given P = 6
s = 2%
= 0.02
cos
= 0.866
f = 50 Hz Output 5hp 5 735.5watts
Output 3.677 KW
I L 7amperes
## To find the rotor speed:
Speed of the rotor magnetic field is given by
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120 f 120 50
Ns
P 6
N s 1000rpm
Speed of the rotor = N =Ns (1-s)
= 1000(1-0.02)
N = 980rpm
14. A 3-phase, 6-pole, 50 Hz induction motor has a slip of 1% at no-load and 3%at full-
1. Synchronous speed,
4. Frequency of rotor current at standstill, and
5. Frequency of rotor current at full-load
.in
Solution. Number of poles, p = 6
No- load slip, s0 = 1%
1. Synchronous speed,
sf = 3% sy
ea
120 f 120 50
Ns 1000r. p.m.( Ans)
p 60
y
ud
We know that
Ns N
s orN Ns 1 s
Ns
st
1
N0 N s 1 s0 100 1 990r. p.m.( Ans )
100
3
Nf Ns 1 s f 1000 1 970r. p.m.( Ans )
100
4. Frequency of rotor current at standstill, fr
At standstill,
s 1
f r sf 1 50 50 Hz.( Ans )
## 5. Frequency of rotor current at full load, fr =?
3
fr s f 50 1.5Hz.( Ans)
100
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15. A 3-phase, 12-pole alternator is coupled to an engine running at 500 rpm. The
alternator supplies an induction motor which has a full-load speed of 1455rpm. Find the
slip and number of poles of the motor.
Solution.
Number of poles of the alternator, pa =12
Speed of the engine, N e = 500rpm
Full-load speed of the induction motor, Nm = 1455rpm
Slip, s =?
Number of poles of the induction motor, Pm =?
Supply frequency,
N a p a 500 12
f 50 Hz
120 120
When the supply frequency is 50Hz, the synchronous speed can be 3000, 1500, 1000,
.in
750 rpm etc. since the full-load speed is 1455rpm and the full-load slip is always less
than 4%, the synchronous speed is1500rpm.
Slip,
s
Ns
sy
N s N 1500 1455
1500
0.03or 3%( Ans)
ea
Also,
120 f
Ns
y
Pm
ud
120 f 120 50
Pm 4 poles
Ns 1500
Hence, number of motor poles = 4. (Ans)
st
16. A 4- pole, 50 Hz induction motor at no-load (NNL) has a slip of 2%. When
operated at full load the slip increases to 3%. Find the change in speed of the motor
120 f
Ns
P
120 x 50
1500rpm
4
no load speed N NL N S (1 S NL )
1500(1 0.02) 1470rpm
full load speed N N S (1 S FL )
1500(1 0.03)
1455rpm
N NL N FL 1470 1455 15rpm
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REVIEW QUESTIONS
(1) What do you mean by rotating magnetic field and explain the production of
torque in a three phase induction motor?
(2) A three phase, 50 Hz 6 pole induction motor has a full load percentage slip o
3%.Find the synchronous speed and the actual speed.
(3) Explain how torque is produced in a 3-phase induction motor?
(4) Why a 3-phase induction motor cannot run with zero slip?
(5) What do you understand by slip?
(6) Why is a single phase induction motor not self starting?
(7) Why starters are necessary for starting a 3-phase induction motor?
(8) Explain with a neat diagram the working of a STAR-DELTA starter?
(9) Bring out the differences between a squirrel cage and a slip ring induction
.in
motor.
(10) Why is squirrel cage induction motor widely used for industrial applications?
(11) Mention the applications of squirrel cage and slip ring induction motors.
sy
SOLUTION TO QUESTION PAPERS
ea
What do you mean by rotating magnetic field and explain the production of torque in a
three phase induction motor.
## Sol. : Rotating magnetic field (R.M.F) :
y
The rotating magnetic field can be defined as the field or flux having constant
ud
amplitude but whose axis is continuously rotating in a plane with certain speed. So if the
arrangement is made to rotate a permanent magnet, then the resulting field is a rotating
magnetic field. But in this method it is necessary to rotate a magnet physically tp produce
rotating magnetic field.
st
But in three phase induction motors such a rotating magnetic field is produced by
supplying currents to a set of stationary windings, with the help of three phase a.c.
supply. The current carrying windings produce the magnetic field or flux . And due to
interaction of three fluxes produced due to three phase supply, resultant flux has a
constant magnitude and its axis rotating the windings. This type of field is nothing, but
rotating magnetic field. Let us study how it happens.
Production of RMF :
A three phase induction motor consists of three phase winding as its stationary part called
stator. The three phase stator winding is connected in star or delta. The three phase
windings are displaced from each other by 1200. The windings are supplied by a balanced
three phase a.c. supply. This is shown in the fig. 39. The three phase windings are
denoted as R-R, Y-Y and B- B.
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The three phase currents flow simultaneously through the windings and are displaced
.in
from each other by 1200 electrical. Each alternating phase current produces its own flux
which is sinusoidal. So all three fluxes are sinusoidal and are seprated from each other by
1200 . If the phase sequence of the windings is R Y B, then mathematical equations for
sy
the instantaneous values of the three fluxes R, Y and B can be written as,
R = m sin ( t) =
0
m sin
0
.. (1)
= sin ( t-120 ) = sin ( -120 ) . (2)
ea
Y m m
0 0
B = m sin ( t-140 ) = m sin( -240 ) .. (3)
As windings are indentical and supply is balanced, the magnitude of each flux is m.
due to phase sequence R- Y B, flux Y lags behind R by 1200 and B lags Y by
y
1200. So B ultimately lags R by 2400 . The flux R is talen as reference while writing
ud
the equations.
The fig. 40 shows the waveforms of three fluxes in space. The fig. 40 shows the phasor
diagram which clearly shows the assumed positive directions of each flux assumed
st
positive direction means whenever the flux is positive it must be represented along the
direction shown and whenever the flux is negative it must be represented along the
opposite direction to the assumed positive direction.
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Let R, Y and B be the instantaneous values of three fluxes. The resultant flux T as the
phasor addition of R , Y and B.
T = R + Y+ B
Let us find T at the instants 1, 2, 3, and 4 as shown in the fig. 40 which represents the
values of as 00 , 600, 1200 and 1800 respectively. The phasor addition can be performed
by obtaining the values of R, Y and B by substituting values of in the equation (1),
(2) and (3).
Case 1:
Substituting in the equations (1), (2) and (3) we get,
0
R = m sin0 = 0
0
Y = m sin (-120 ) = -0.866 m
0
B = m sin (- 240 ) = +0.866 m
.in
The phasor addition is shown in the fig. 41.
the positive values are shown in assumed
positive directions while negative values
are shown in opposite direction to thesy
assumed positive direction of the respective
fluxes. Refer to assumed positive directions
ea
shown in the fig. 41.
y
OD = DA =
ud
Cos 300 = =
## T= 2 0.866 m cos 300
st
= 1.5 m
So magnitude of T is 1.5 m and its position is vertically upwards at = 00.
Case 2 :
Equations (1), (2) and (3) give us,
0
R = m sin 60 = + 0.866 m
0
Y = m sin (-60 ) = -0.866 m
0
B = m sin (-180 ) = 0
So R is positive and Y is negative and
hence drawing in appropriate directions
we get phasor diagram as shown in the
fig. 42.
Doing the same construction drawing
perpendicular from B on T at D. we get
the same result as
T = 1.5 m
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But it can be seen that through its magnitude is 1.5 m it has rotated through 600 in space,
in clockwise direction, from its previous position.
Case 3:
Equations (1), (2) and (3) give us,
R = m sin 120 = + 0.866 m
Y = m sin 0 = 0
0
B = m sin (-120 ) = -0.866 m
So
R is positive and B is negative.
Showing R and B in the appropriate
directions, we get the phasor dioagram
as shown in the fig. 43.
After doing the construction same as
.in
before i.e. drawing perpendicular from
B on T ,it can be proved again that,
T = 1.5 m
sy
But the position of T is such that it has rotated further through 600 from its previous
position, in clockwise direction. And from its position at = 00, it has rotated through
1200 in space, in clock wise direction.
ea
Case 4: = 1800
Fropm equations (1), (2) and (3),
0
R = m sin (180 ) = 0
0
y
## Y = m sin (60 ) = + 0.866 m
0
B = m sin (-60 ) = - 0.866 m
ud
## So R = 0 , Y is positive and B is negative.
Drawing Y and B in the appropriate directions,
we get the phasor diagram as shown in the fig.44
st
## From phasor, it can be easily proved that,
T = 1.5 m
Thus the magnitude of T once again remains same. But it can be seen that it has further
rotated through 600 from its previous position in clock wise direction.
So for an electrical half cycle of 1800, the resultant T has also rotated through 1800. This
is applicable for winding wound for 2 poles.
## From the above discussion we have following conclusions :
a) The resultant of the three alternating fluxes, separated from each other by 1200, has
a constant amplitude of 1.5 m where m is maximum amplitude of an individual plux due
to any phase.
## b) The resulatant always keeps on rotating with a certain speed in space.
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This shows that when a three phase stationary windings are exicited by balanced
three phase a.c. supply then the resulting field produced is rotating magnetic field.
## 2) Working principle of induction motor :
Induction motor works on the principle of electro magnetic induction.
When a three phase supply is given to the three phase stator winding, a rotating
magnetic field of constant magnitude is produced as discussed earlier. The speed of this
rotating magnetic field is synchronous speed, Ns r.p.m
NS =
= speed of rotating magnetic field
f = supply frequency
P = number of poles for which stator winding is wound.
This rotating field produces an effect of rotating poles around a rotor. let direction of
rotation of this rotating magnetic field is clock wise as shown in the fig. 45.
.in
sy
y ea
ud
st
Now at this instant rotor is stationary and stator flux R.M.F. is rotating. So its pbvious
that thre exists a relative motion between the R.M.F. and rotor conductors. Now the
R.M.F. gets cut by rotor conductor as R.M.F. sweeps over rotor conductors. whenever
conductor cuts flux , e.m.f. gets induced init. so e.m.f. gets induced in the rotor
conductors called rotor induced e.m.f. this is electro magnetic induction. As rotor forms
closed circuit, induced e.m.f circulates current through rotor called rotor current as shown
in fig. 45. let direction of this current is going into paper denoted by a cross as shown in
the fig. 45.
## Any current carrying conductor
produces its own flux. so rotor produces
its flux called rotor flux. for assumed
direction of rotor current, the direction
of rotor flux is clock wise as shown in
the fig. 46. this direction can be easily
determined using right hand thumb rule.
now there are two fluxes, one R.M.F. and
other rotor flux. Both the fluxes interact
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## with each asa shown in the fig. 46. on left
of rotpr conductor two fluxes are in same direction hence add up to get high flux area. on
right side, two fluxes cancel each other to produce low flux area. As flux lines act as
stretched rubber band, high flux dendity area exerts a puch on rotor conductor towards
low flux dendity area. so rotor conductor experiences a force from left to right in this
case, as shown in the fig. 46, due to interaction of the two fluxes.
As all the rotor conductors experience a force, the overall rotor experiences a torque
and starts rotating. so interaction of the two fluxes is very essential for a motoring action.
As seen from the fig. 46, the direction of force experienced is same as that of rotating
magnetic field. hence rotor starts rotating in the same direction as that of rotating
magnetic field.
## Alternatively this can be explained as : according to LenZ law the direction of
.in
induced current in the rotor is so as oppose the cause producing it. The cause of rotor
current is the induced e.m.f which is induced because of relative motion present between
the rotating magnetic field and the rotor conductors. hence to oppose the relative motion
sy
i.e. to reduce the relative speed, the rotor experiences a torque in the same direction as
that of R.M.F. and tries to catch up the speed of rotating magnetic field.
so, NS = speed of rotating magnetic field in r.p.m
ea
N = speed of rotor i.e. motor in r.p.m.
NS N = Relative speed between the two rotating magnetic
filed and the rotor conductors.
y
## Thus rotor always rotates in same direction as that of R.M.F.
ud
3)Discuss the important features and advantages of squirrel cage and phase wound
rotor constructions in an induction motor.
st
## sol. : Rotor constructions in induction motor :
The rotor is placed inside the stator. The rotor is also laminated in construction
slotted at the periphery and cylindrical in shape. The rotor conductors are placed the rotor
slots. The two types of rotor construction are generally used for the induction motors
namely i) slip ring or wound rotor and ii) squirrel cage rotor. Let see the details of each of
these two types:
i) slip ring or wound rotor :
In this type of construction, rotor winding is exactly similar to the stator. The
rotor carries a three phase star or connected, disturbed winding, wound for same number
of poles as that of stator. The rotor construction is laminated and slotted slots contain the
rotor winding. The three ends of three phase winding, available after connecting the
windings in star or delts, are permanently connected to the slip rings. The slip rings are
mounted on the same shaft.
Compiled by www.studyeasy.in
Compiled by www.studyeasy.in
The slip rings are used to connect external stationary circuit to the internal
rotating circuit. so in this type of rotor, the external resistances can be added with the help
brushes and slip ring arrangement, in series with each phase of the rotor winding.
.in
Action of slip rings and brushes :
sy
The slip rings are mounted on the shaft. On end to each phase winding, after connecting
ea
the winding in star or delta is connected to the slip ring. Thus there are three slip rings.
The three slip rings now behave as the terminals R,Y and B of rotor winding. Now as
rotor rotates, slip rings also rotate. We want to add external resistance in series with rotor.
y
But external resistance is stationary while rotor is rotating. Thus to have connection
between rotating and stationary members, the brushes are used. The brushes are
ud
stationary and resting on slip rings to make electric contact with them. Thus slip rings are
rotating terminals of rotor while brushes are stationary terminals of rotor. Now whatever
resistance we connected across the brushes get internally added in series with each other
st
winding phase, through rotor is rotating. This is the action of slip ring and brush
assembly. The brushes are made up of carbon.
In the running condition, the slip rings are shorted. This is possible by connecting a
metal collar which gets pushed and connects all the slip rings together, shorting them. At
the same time brushes are also lifted from the slip rings. This avoids wear and tear of the
brushes due to friction. The possibility of addition of an external resistance in series with
the rotor, with the help of slip rings is the main feature of this type of rotor.
Such addition of resistance in rotor circuit only at start is used to improve starting torque
of the motor.
## 3) Squirrel cage rotor :
When there is no need of controlling starting torque of the motor, the rotor construction
can be made very simple. In this type, rotor consists of uninsulated copper or aluminium
bars, placed in the slots. These bars are permanently shorted at each end with the help of
conducting end ring. The bars are usually brazed to the end rings to provide good
Compiled by www.studyeasy.in
Compiled by www.studyeasy.in
mechanical strength. The entire structure looks like a cage hence called squirrel cage
rotor.
This is shown in the fig. 48.
.in
As the bars are permanently shorted to each other, the resistance of the entire rotor is very
sy
very small. hence rotor is also called short circuited rotor. As the rotor is shorted on its
own, no external resistance can have any effect on the rotor resistance. Hence for this
rotor, no external resistance can be added in the rotor circuit. So slip rings and brush
ea
assembly is absent in this type of rotor.
y
ud
st
In practice, the bars are slightly skewed as shown in the fig. 49. The skewing helps in two
ways :
1) It helps to reduce noise due to magnetic hum.
2) It reduces the tendency of magnetic locking between rotor and stator.
Fan blades are generally provided at the ends of the rotor core. This circuliates the
air through the machine while operation, providing the necessary cooling. The air gap
between stator and rotor is kept uniform and as small as possible.
Compiled by www.studyeasy.in
Compiled by www.studyeasy.in
The important points of comparison between the two types of rotor are given as follows.
5)Comparison of squirrel cage and wound rotor :
## Wound or slip ring rotor Squirrel cage rotor
1 Rotor consists of a three phase winding Rotor consists of bars which are shorted at
similar to the stator winding. the ends with the help of end rings
2 Construction is complicated. Construction is simple.
3 Resistance can be added externally As permanently shorted,external resistance
4 Slip rings and brushes are present to add Slip rings and brushes are absent.
external resistance.
5 The construction is delicate and due to The construction is robust and maintenance
brushes, frequent maintenance is free.
necessary.
6 The rotors are very costly. Due to simple construction, the rotors are
.in
cheap.
7 Only 5% of induction motors in industry Very common and almost 95% induction
use slip ring rotor. motors use this type of rotor.
8 sy
High starting torque can be obtained. Moderate starting torque which cannot be
controlled..
9 Rotor resistance starter can be used Rotor resistance starter cannot be used
ea
10 Rotor must be wound for the same The rotor automatically adujusts itself for
number of poles as that of stator. the same number of poles as that of stator.
11 Speed control by rotor resistance is Speed control by rotor resistance is not
y
possible. possible.
12 Rotor copper losses are high hence Rotor copper losses are less hence have
ud
## efficiency is less. higher efficiency is less.
13 Used for lifts, hoists, cranes, elevators, Used for lathes, drilling machines , fans,
compressors etc. blowers, water pumps, grinders, printing
st
machines etc.
6) A 9 phase, 50 Hz, 6 pole induction motor has a full load percentage slip of 30%. Find
i) synchronous speed and ii) Actual speed. [4]
Sol. : f = 50 Hz
P=6
Slip, S = 3% = 0.03
NS =
Slip, S =
## (0.03) (1000) = 1000 N
30 = 1000 N
N = 1000 30 = 970 rpm
Rotor speed = 970 rpm
----- -----END----- ----
Compiled by www.studyeasy.in | 70,210 | 259,468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2019-30 | latest | en | 0.765914 |
https://www.kodytools.com/units/enzyme/from/millikatal/to/microkatal | 1,709,472,050,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476374.40/warc/CC-MAIN-20240303111005-20240303141005-00333.warc.gz | 840,556,541 | 15,006 | # Millikatal to Microkatal Converter
1 Millikatal = 1000 Microkatals
## One Millikatal is Equal to How Many Microkatals?
The answer is one Millikatal is equal to 1000 Microkatals and that means we can also write it as 1 Millikatal = 1000 Microkatals. Feel free to use our online unit conversion calculator to convert the unit from Millikatal to Microkatal. Just simply enter value 1 in Millikatal and see the result in Microkatal.
Manually converting Millikatal to Microkatal can be time-consuming,especially when you don’t have enough knowledge about Enzyme units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Millikatal to Microkatal converter tool to get the job done as soon as possible.
We have so many online tools available to convert Millikatal to Microkatal, but not every online tool gives an accurate result and that is why we have created this online Millikatal to Microkatal converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
## How to Convert Millikatal to Microkatal (mkat to μkat)
By using our Millikatal to Microkatal conversion tool, you know that one Millikatal is equivalent to 1000 Microkatal. Hence, to convert Millikatal to Microkatal, we just need to multiply the number by 1000. We are going to use very simple Millikatal to Microkatal conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Millikatal} = 1 \times 1000 = \text{1000 Microkatals}$$
## What Unit of Measure is Millikatal?
Millikatal is a unit of measurement for enzyme's catalytic activity. By definition, one millikatal is the enzyme catalytic activity that raises the rate of a chemical reaction by 0.001 mole per second.
## What is the Symbol of Millikatal?
The symbol of Millikatal is mkat. This means you can also write one Millikatal as 1 mkat.
## What Unit of Measure is Microkatal?
Microkatal is a unit of measurement for enzyme's catalytic activity. By definition, one microkatal is the enzyme catalytic activity that raises the rate of a chemical reaction by 1e-6 mole per second.
## What is the Symbol of Microkatal?
The symbol of Microkatal is μkat. This means you can also write one Microkatal as 1 μkat.
## How to Use Millikatal to Microkatal Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Millikatal and in the first input field, enter a value.
• From the second dropdown, select Microkatal.
• Instantly, the tool will convert the value from Millikatal to Microkatal and display the result in the second input field.
Millikatal
1
Microkatal
1000
# Millikatal to Microkatal Conversion Table
Millikatal [mkat]Microkatal [μkat]Description
1 Millikatal1000 Microkatal1 Millikatal = 1000 Microkatal
2 Millikatal2000 Microkatal2 Millikatal = 2000 Microkatal
3 Millikatal3000 Microkatal3 Millikatal = 3000 Microkatal
4 Millikatal4000 Microkatal4 Millikatal = 4000 Microkatal
5 Millikatal5000 Microkatal5 Millikatal = 5000 Microkatal
6 Millikatal6000 Microkatal6 Millikatal = 6000 Microkatal
7 Millikatal7000 Microkatal7 Millikatal = 7000 Microkatal
8 Millikatal8000 Microkatal8 Millikatal = 8000 Microkatal
9 Millikatal9000 Microkatal9 Millikatal = 9000 Microkatal
10 Millikatal10000 Microkatal10 Millikatal = 10000 Microkatal
100 Millikatal100000 Microkatal100 Millikatal = 100000 Microkatal
1000 Millikatal1000000 Microkatal1000 Millikatal = 1000000 Microkatal
# Millikatal to Other Units Conversion Table
ConversionDescription
1 Millikatal = 0.001 Katal1 Millikatal in Katal is equal to 0.001
1 Millikatal = 1e-21 Exakatal1 Millikatal in Exakatal is equal to 1e-21
1 Millikatal = 1e-18 Petakatal1 Millikatal in Petakatal is equal to 1e-18
1 Millikatal = 1e-15 Terakatal1 Millikatal in Terakatal is equal to 1e-15
1 Millikatal = 1e-12 Gigakatal1 Millikatal in Gigakatal is equal to 1e-12
1 Millikatal = 1e-9 Megakatal1 Millikatal in Megakatal is equal to 1e-9
1 Millikatal = 0.000001 Kilokatal1 Millikatal in Kilokatal is equal to 0.000001
1 Millikatal = 0.00001 Hectokatal1 Millikatal in Hectokatal is equal to 0.00001
1 Millikatal = 0.0001 Dekakatal1 Millikatal in Dekakatal is equal to 0.0001
1 Millikatal = 0.01 Decikatal1 Millikatal in Decikatal is equal to 0.01
1 Millikatal = 0.1 Centikatal1 Millikatal in Centikatal is equal to 0.1
1 Millikatal = 1000 Microkatal1 Millikatal in Microkatal is equal to 1000
1 Millikatal = 1000000 Nanokatal1 Millikatal in Nanokatal is equal to 1000000
1 Millikatal = 1000000000 Picokatal1 Millikatal in Picokatal is equal to 1000000000
1 Millikatal = 1000000000000 Femtokatal1 Millikatal in Femtokatal is equal to 1000000000000
1 Millikatal = 1000000000000000 Attokatal1 Millikatal in Attokatal is equal to 1000000000000000
1 Millikatal = 1000000000000000000 Zeptokatal1 Millikatal in Zeptokatal is equal to 1000000000000000000
1 Millikatal = 1e+21 Yoctokatal1 Millikatal in Yoctokatal is equal to 1e+21
1 Millikatal = 0.001 Mole/Second1 Millikatal in Mole/Second is equal to 0.001
1 Millikatal = 1 Millimole/Second1 Millikatal in Millimole/Second is equal to 1
1 Millikatal = 0.000001 Kilomole/Second1 Millikatal in Kilomole/Second is equal to 0.000001
1 Millikatal = 1000 Micromole/Second1 Millikatal in Micromole/Second is equal to 1000
1 Millikatal = 0.06 Mole/Minute1 Millikatal in Mole/Minute is equal to 0.06
1 Millikatal = 60 Millimole/Minute1 Millikatal in Millimole/Minute is equal to 60
1 Millikatal = 0.00006 Kilomole/Minute1 Millikatal in Kilomole/Minute is equal to 0.00006
1 Millikatal = 60000 Micromole/Minute1 Millikatal in Micromole/Minute is equal to 60000 | 1,756 | 5,679 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-10 | latest | en | 0.86049 |
https://web2.0calc.com/questions/algebra_55795 | 1,618,512,861,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038087714.38/warc/CC-MAIN-20210415160727-20210415190727-00291.warc.gz | 701,860,023 | 5,534 | +0
# algebra
0
42
1
Julie's grades for this term tests are 85, 88, and 97.
In addition for the three tests that Julie already had, there will be just one more test. Julie needs to earn at least ___ points for the last test fo get no less than A for the term. "A" means the average of 93 or more.
What grade does Julie need to get on her last test?
Mar 28, 2021 | 106 | 368 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-17 | longest | en | 0.940617 |
https://www.acmicpc.net/problem/10452 | 1,701,337,641,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100184.3/warc/CC-MAIN-20231130094531-20231130124531-00471.warc.gz | 716,700,671 | 8,909 | 시간 제한메모리 제한제출정답맞힌 사람정답 비율
1 초 256 MB0000.000%
## 문제
In a city, there is a huge network of roads, each of which connects two districts. The network seems like a tree, in other words, for two arbitrary districts, there is exactly one path between them. Here, the path is a sequence of roads on which we can travel from one district to the other.
Many cars pass roads in the network day after day and so the roads are damaged seriously. The city department of transportation repairs roads regularly. Each road R has a cost c and a benefit b. We can repair the road R with the cost c and obtain the (social) benefit b when R is repaired.
Repairing all the roads belonging to a path, the cost and benefit of the path are the total cost and benefit of its roads, respectively. Given an upper bound of cost C, you should find a path in the network that maximizes its benefit while having a cost of at most C.
For example, a network is given in Figure 1. The circles and the solid lines represent districts and roads, respectively. The pair of integers <c, b> on a road describes that the road has the cost c and the benefit b. In this example, if the bound of cost is 8, then the red path given in Figure 2 has the benefit 13 with the cost 8 and it maximizes the benefit.
## 입력
Your program is to read from standard input. The input consists of T test cases. The number of test cases T is given in the first line of the input. Each test case starts with an integer n, the number of districts in the network, where 2 ≤ n ≤ 22,000. The districts are numbered from 1 to n. There are always exactly n − 1 roads in the network. Each of the following n − 1 lines contains four integers α, β, c and b, representing that the road connecting districts α and β has the cost c and the benefit b. Here, 1 ≤ α,β ≤ n and 1 ≤ c, b ≤ 1,000. Finally, the last line contains an integer C to represent the upper bound of cost, where 1 ≤ C ≤ 2 × 107.
## 출력
Your program is to write to standard output. Print exactly one line for each test case. The line should contain an integer to represent the maximum benefit which a path can obtain, while having a cost of at most C. If such a path never exists, then the line contains 0.
## 예제 입력 1
2
11
1 2 2 2
2 3 1 4
3 4 3 6
3 5 2 2
5 6 1 4
5 8 3 3
6 7 5 1
8 9 2 4
9 10 2 1
9 11 3 2
8
18
1 9 1 2
9 14 2 3
10 14 6 2
2 9 5 2
3 10 1 3
4 11 2 6
11 15 3 3
12 15 4 4
5 12 1 5
6 12 2 6
17 18 3 4
16 18 2 5
7 13 2 3
13 16 1 2
8 16 1 2
15 17 4 1
14 17 2 3
10
## 예제 출력 1
13
18 | 790 | 2,495 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-50 | latest | en | 0.952358 |
https://numberworld.info/20011021101 | 1,624,129,961,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487649688.44/warc/CC-MAIN-20210619172612-20210619202612-00073.warc.gz | 402,036,998 | 3,943 | # Number 20011021101
### Properties of number 20011021101
Cross Sum:
Factorization:
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
4a8bff32d
Base 32:
ikbvspd
sin(20011021101)
-0.25095157481184
cos(20011021101)
-0.96799964209676
tan(20011021101)
0.25924759049317
ln(20011021101)
23.719549013775
lg(20011021101)
10.301269249916
sqrt(20011021101)
141460.31634702
Square(20011021101)
4.0044096550467E+20
### Number Look Up
Look Up
20011021101 which is pronounced (twenty billion eleven million twenty-one thousand one hundred one) is a amazing figure. The cross sum of 20011021101 is 9. If you factorisate the figure 20011021101 you will get these result 3 * 3 * 1171 * 1898759. 20011021101 has 12 divisors ( 1, 3, 9, 1171, 3513, 10539, 1898759, 5696277, 17088831, 2223446789, 6670340367, 20011021101 ) whith a sum of 28929507360. The number 20011021101 is not a prime number. The number 20011021101 is not a fibonacci number. 20011021101 is not a Bell Number. The number 20011021101 is not a Catalan Number. The convertion of 20011021101 to base 2 (Binary) is 10010101000101111111111001100101101. The convertion of 20011021101 to base 3 (Ternary) is 1220122121021012100200. The convertion of 20011021101 to base 4 (Quaternary) is 102220233333030231. The convertion of 20011021101 to base 5 (Quintal) is 311440310133401. The convertion of 20011021101 to base 8 (Octal) is 225057771455. The convertion of 20011021101 to base 16 (Hexadecimal) is 4a8bff32d. The convertion of 20011021101 to base 32 is ikbvspd. The sine of the number 20011021101 is -0.25095157481184. The cosine of the number 20011021101 is -0.96799964209676. The tangent of the number 20011021101 is 0.25924759049317. The square root of 20011021101 is 141460.31634702.
If you square 20011021101 you will get the following result 4.0044096550467E+20. The natural logarithm of 20011021101 is 23.719549013775 and the decimal logarithm is 10.301269249916. that 20011021101 is unique number! | 747 | 2,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-25 | latest | en | 0.691221 |
https://financestrategists.com/wealth-management/x-rate/ | 1,675,304,100,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499954.21/warc/CC-MAIN-20230202003408-20230202033408-00743.warc.gz | 257,090,718 | 24,167 | # X Rate
### Written byTrue Tamplin, BSc, CEPF® | Reviewed by Editorial Team
Updated on December 14, 2022
“X Rate” stands for “Exchange Rate,” and is a metric by which the value of one currency is compared to another.
It can be thought of as a measure of how much of the second currency can be “bought” with an amount of the first currency.<h2>What Does X Rate Mean in Finance?</h2> The purpose of an X Rate is to figure out the buying power of a currency as compared to another currency.
Not all currencies have the same buying power per unit, so knowing the exchange rate helps determine the “value” of an amount of money.
For example, the exchange rate of US dollars (USD) to Pound sterling (GBP) is 1.24.
This means that it takes \$1.24 to equal £1.
Put another way, if a US citizen traveled to the UK with \$10,000 to exchange into pounds, they would get about £8,034.
## X Rate Example
Several factors influence exchange rates.
Most exchange rates are free-floating, meaning they rise and fall in accordance with fluctuations in supply and demand in the foreign exchange market.
Other currencies may be pegged, meaning that their value is tied to the value of another currency within a certain margin; the Hong Kong dollar, for example, is pegged to the US dollar at a range of 7.75 to 7.85, so its value will stay within this range of the USD.
Some currencies have different exchange rates depending on whether or not the funds are held inside or outside a country’s borders.
The Chinese Yuan (CNY), for instance, has an exchange rate of ¥7.05 per \$1 outside its borders, but ¥7.04 per \$1 inside.
## X Rate Definition FAQs
### What does X Rate stand for?
X Rate stands for Exchange Rate in finance.
### What is the X Rate?
X Rate is a metric by which the value of one currency is compared to another. It can be thought of as a measure of how much of the second currency can be bought with an amount of the first currency.
### What is the purpose of the X Rate?
The purpose of an X Rate is to figure out the buying power of a currency as compared to another currency.
### Why should investors know the X Rate of their currnecy?
Not all currencies have the same buying power per unit, so knowing the exchange rate helps determine the value of an amount of money. | 539 | 2,296 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-06 | latest | en | 0.952232 |
http://mdct.ru/tag/annuity/ | 1,521,740,793,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647901.79/warc/CC-MAIN-20180322170754-20180322190754-00513.warc.gz | 181,495,292 | 13,690 | ## Future Value of an Annuity Calculator
The future value of annuity calculator is used to calculate what a cash sum received at the end of each period for n periods is worth at the end of period n, allowing for a discount rate i.
Future Value of an Annuity Calculator November 6th, 2016
## Present Value Annuity Calculator
The present value annuity calculator is used to calculate what a cash sum received at the end of each period for n periods is worth today, taking into account a discount rate i.
Present Value Annuity Calculator November 6th, 2016
## Future Value of an Annuity
The future value of an annuity due formula is used to calculate the future value of a series of periodic payments. The payments are for the same amount, made at the start of each period, and a discount rate i% is applied.
Future Value of an Annuity March 14th, 2017
## Growing Annuity Payment Formula FV
This growing annuity payment formula FV calculates the initial annuity payment required to provide a given future value FV using a growing annuity. The growing annuity payment formula assumes payments are made at the end of each period for n periods and are growing or declining at a constant rate (g), and a discount rate i is applied.
Growing Annuity Payment Formula FV November 6th, 2016
## Loan Balance Formula
The loan balance annuity formula is used to find the balance outstanding on a loan by calculating the present value of the remaining loan installments. The payments are for the same amount, made at the end of each period, and a discount rate i% is applied.
Loan Balance Formula November 6th, 2016
## How to Calculate a Mortgage Payment
A mortgage is an interest in a property that is transferred from a borrower (the mortgagor) to a lender (mortgagee) to as security for a mortgage loan. If the lender does not repay the loan then the lender can under certain circumstances take the property.
As we have a series of periodic payments from the lender to the borrower and a periodic compounding interest rate, the mortgage payment can be regarded as an annuity.
How to Calculate a Mortgage Payment November 6th, 2016
## Growing Annuity Payment Formula PV
This growing annuity payment formula PV calculates the initial annuity payment required to provide a given value today PV (present value) using a growing annuity. The growing annuity payment formula assumes payments are made at the end of each period for n periods and are growing or declining at a constant rate (g), and a discount rate i is applied.
Growing Annuity Payment Formula PV November 6th, 2016
## Number of Periods Annuity Formula FV
This number of periods annuity formula FV calculates the number (n) of annuity payments required to provide a given future value (FV). The annuity formula assumes payments (Pmt) are made at the end of each period, and a discount rate i is applied.
Number of Periods Annuity Formula FV November 6th, 2016
## Annuity Payment Calculator
The annuity payments calculator is used to calculate the regular sum to be received for n periods, based on an initial lump sum investment (PV), and a discount rate (i).
Annuity Payment Calculator November 6th, 2016
## Number of Periods Annuity Formula PV
This number of periods annuity formula PV calculates the number (n) of annuity payments required to provide a given value today PV (present value). The annuity formula assumes payments (Pmt) are made at the end of each period, and a discount rate i is applied.
Number of Periods Annuity Formula PV November 6th, 2016
## Present Value of a Perpetuity Calculator
The present value of a perpetuity calculator is used to calculate what an infinite stream of identical cash flows made at regular intervals over time is worth today.
Present Value of a Perpetuity Calculator November 6th, 2016
## Future Value of a Growing Annuity Formula
The future value of a growing annuity formula is used to calculate the value at the end of period n of a series of periodic payments which increase or decrease at a constant rate each period. The payments made at the end of each period, and a discount rate i% is applied.
Future Value of a Growing Annuity Formula November 6th, 2016
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https://markhneedham.com/blog/2013/08/05/javajblas-calculating-eigenvector-centrality-of-an-adjacency-matrix/ | 1,560,999,275,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999130.98/warc/CC-MAIN-20190620024754-20190620050754-00474.warc.gz | 519,049,154 | 13,372 | # Java/JBLAS: Calculating eigenvector centrality of an adjacency matrix
I recently came across a very interesting post by Kieran Healy where he runs through a bunch of graph algorithms to see whether he can detect the most influential people behind the American Revolution based on their membership of various organisations.
The first algorithm he looked at was betweenness centrality which I've looked at previously and is used to determine the load and importance of a node in a graph.
This algorithm would assign a high score to nodes which have a lot of nodes connected to them even if those nodes aren't necessarily influential nodes in the graph.
If we want to take the influence of the other nodes into account then we can use an algorithm called eigenvector centrality.
Eigenvector centrality is a measure of the influence of a node in a network. It assigns relative scores to all nodes in the network based on the concept that connections to high-scoring nodes contribute more to the score of the node in question than equal connections to low-scoring nodes. Google's PageRank is a variant of the Eigenvector centrality measure.
Both PageRank and Eigenvector centrality give us a probability which describes how often we'd end up visiting each node on a random walk around the graph.
As far as I can tell there are a couple of differences between PageRank and Eigenvector centrality (but I'm happy to be corrected as I'm still learning this stuff):
1. PageRank introduces a 'dampening factor' to simulate the idea that some percentage of the time we might decide not to follow any of a node's relationships but instead pick a random node in the graph.
2. PageRank makes sure that the elements in each column of the adjacency matrix add up to one. Therefore, if our node had a relationship to every other one in the graph then each would only contribute a value of 1/n rather than 1.
In this instance since Healy wanted to analyse the influence of people rather than web pages eigenvector centrality makes more sense.
Over the past few days I've been trying to understand this topic area a bit better and found the following resources useful:
I calculated a few made up matrices by hand but found it became too difficult after a 3x3 matrix so I wanted to find a Java based library which I could use instead.
These were the ones that I came across:
• JBLAS - Linear Algebra for Java
• JAMA - A Java Matrix Package
• Colt - Advanced Computing for Science
• Commons Math - The Apache Commons Mathematics Library
• la4j - Linear Algebra for Java
• MTJ - Matrix Toolkits Java
I'd heard of JBLAS before so I thought I'd give that a try on one of the adjacency matrices described in Murphy Waggoner's post about the Gould Index and see if I got the same eigenvector centrality values.
The first step was to define the matrix which can be represented as an array of arrays:
``````
DoubleMatrix matrix = new DoubleMatrix(new double[][] {
{1,1,0,0,1,0,0},
{1,1,0,0,1,0,0},
{0,0,1,1,1,0,0},
{0,0,1,1,1,0,0},
{1,1,1,1,1,1,1},
{0,0,0,0,1,1,1},
{0,0,0,0,1,1,1},
});
``````
Our next stop is to work out the eigenvalues which we can do using the following function:
``````
ComplexDoubleMatrix eigenvalues = Eigen.eigenvalues(matrix);
for (ComplexDouble eigenvalue : eigenvalues.toArray()) {
System.out.print(String.format("%.2f ", eigenvalue.abs()));
}
``````
``````
4.00 2.00 0.00 1.00 2.00 0.00 0.00
``````
We want to get the corresponding eigenvector for the eigenvalue of 4 and as far as I can tell the Eigen#eigenvectors function returns its values in the same order as the Eigen#eigenvalues function so I wrote the following code to work out the principal eigenvector :
``````
List<Double> principalEigenvector = getPrincipalEigenvector(matrix);
System.out.println("principalEigenvector = " + principalEigenvector);
private static List<Double> getPrincipalEigenvector(DoubleMatrix matrix) {
int maxIndex = getMaxIndex(matrix);
ComplexDoubleMatrix eigenVectors = Eigen.eigenvectors(matrix)[0];
return getEigenVector(eigenVectors, maxIndex);
}
private static int getMaxIndex(DoubleMatrix matrix) {
ComplexDouble[] doubleMatrix = Eigen.eigenvalues(matrix).toArray();
int maxIndex = 0;
for (int i = 0; i < doubleMatrix.length; i++){
double newnumber = doubleMatrix[i].abs();
if ((newnumber > doubleMatrix[maxIndex].abs())){
maxIndex = i;
}
}
return maxIndex;
}
private static List<Double> getEigenVector(ComplexDoubleMatrix eigenvector, int columnId) {
ComplexDoubleMatrix column = eigenvector.getColumn(columnId);
List<Double> values = new ArrayList<Double>();
for (ComplexDouble value : column.toArray()) {
}
return values;
}
``````
In getMaxIndex we work out which index in the array the largest eigenvalue belongs to so that we can look it up in the array we get from Eigen#eigenvectors. According to the documentation the eigenvectors are stored in the first matrix we get back which is why we choose that on the second line of getPrincipalEigenvector.
This is the output we get from running that:
``````
principalEigenvector = [0.3162277660168381, 0.3162277660168376, 0.316227766016838, 0.316227766016838, 0.6324555320336759, 0.316227766016838, 0.316227766016838]
``````
Finally we normalise the values so that they all add together to equal 1 which means our result will tell the % of time that a random walk would take you to this node:
``````
System.out.println("normalisedPrincipalEigenvector = " + normalised(principalEigenvector));
private static List<Double> normalised(List<Double> principalEigenvector) {
double total = sum(principalEigenvector);
List<Double> normalisedValues = new ArrayList<Double>();
for (Double aDouble : principalEigenvector) {
}
return normalisedValues;
}
private static double sum(List<Double> principalEigenvector) {
double total = 0;
for (Double aDouble : principalEigenvector) {
`````` | 1,445 | 5,852 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2019-26 | latest | en | 0.956549 |
http://www.thecodingforums.com/threads/straight-line-detection.349650/ | 1,464,266,257,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049275836.20/warc/CC-MAIN-20160524002115-00076-ip-10-185-217-139.ec2.internal.warc.gz | 838,768,789 | 13,302 | # Straight line detection
Discussion in 'Python' started by PyPK, Sep 28, 2005.
1. ### PyPKGuest
Does anyone know of a simple implementation of a straight line
detection algorithm something like hough or anything simpler.So
something like if we have a 2D arary of pixel elements representing a
particular Image. How can we identify lines in this Image.
for example:
ary =
[[1,1,1,1,1],
[1,1,0,0,0],
[1,0,1,0,0],
[1,0,0,1,0],
[1,0,0,0,1]]
So if 'ary' represents pxl of an image which has a horizontal line(row
0),a vertical line(col 0) and a diagonal line(diagonal of ary). then
basically I want identify any horizontal or vertical or diagonal line
anywhere in the pxl array.
Thanks.
PyPK, Sep 28, 2005
2. ### Tim RobertsGuest
"PyPK" <> wrote:
>
>Does anyone know of a simple implementation of a straight line
>detection algorithm something like hough or anything simpler.So
>something like if we have a 2D arary of pixel elements representing a
>particular Image. How can we identify lines in this Image.
>for example:
>
>ary =
>[[1,1,1,1,1],
> [1,1,0,0,0],
> [1,0,1,0,0],
> [1,0,0,1,0],
> [1,0,0,0,1]]
>So if 'ary' represents pxl of an image which has a horizontal line(row
>0),a vertical line(col 0) and a diagonal line(diagonal of ary). then
>basically I want identify any horizontal or vertical or diagonal line
>anywhere in the pxl array.
If all you want is horizontal, vertical, or 45 degree diagonal, it's pretty
easy to do that just be checking all of the possibilities.
But what if your array is:
[[1,1,1,1,1],
[1,1,1,1,1],
[1,1,1,1,1],
[1,1,1,1,1],
[1,1,1,1,1]]
Would you say there were 12 lines there?
--
- Tim Roberts,
Providenza & Boekelheide, Inc.
Tim Roberts, Sep 30, 2005
3. ### Juho SchultzGuest
PyPK wrote:
> Does anyone know of a simple implementation of a straight line
> detection algorithm something like hough or anything simpler.So
> something like if we have a 2D arary of pixel elements representing a
> particular Image. How can we identify lines in this Image.
> for example:
>
> ary =
> [[1,1,1,1,1],
> [1,1,0,0,0],
> [1,0,1,0,0],
> [1,0,0,1,0],
> [1,0,0,0,1]]
> So if 'ary' represents pxl of an image which has a horizontal line(row
> 0),a vertical line(col 0) and a diagonal line(diagonal of ary). then
> basically I want identify any horizontal or vertical or diagonal line
> anywhere in the pxl array.
>
> Thanks.
>
I would recommend using a module for computing, my choice would be
numarray: www.stsci.edu/resources/software_hardware/numarray
You could even write your own version of hough, should not be too complex.
A fwee things you need to consider:
1) Are all the lines through the image, or would a row with
[0,0,1 ...(a few dozen ones in here) ... 1,0] be a line?
2) Do you also need edge detection? Then you might need to convolve
the image with a Laplacian or something like that, e.g.
new[i,j] = (4*old[i,j])-old[i-1,j]-old[i+1,j]-old[i,j-1]-old[i,j+1]
3) How "full" are the images?
It is much easier if only a small fraction of your image is lines,
in your example more than half of image pixels are lines.
4) How big images are you processing? I always have at least
one million pixels, so the rest may not work for small images.
To do some quicklook checks you can of course go through each row/column
and check if the values are different enough, something like
mat = numarray.array(ima)
x = mat.mean()
dx = mat.stddev()
then check if some rows are different from others, maybe
(mat[:,i].mean() > (x + N*dx)) for "white" lines or
(mat[:,i].mean() < (x - N*dx))) for "black" lines
you probably need do a few tests to get a good value of N.
repeat for columns (mat[j,:]) and diagonals:
numarray.diagonal(mat,o) where
o is offset from mat[0,0]
and if you need non-diagonal elements, say
ima = [[1 0 0 0 0]
[0 0 1 0 0]
[0 0 0 0 1]]
would contain a line of ones, then
vect = ima.flat
gives the image as a rank-1 array and you can then take strides
(every nth element) just like with normal lists, array[a:b:n]
takes every nth element in array[a:b], so vect[::7] would be [1 1 1]
I hope this helps a bit.
Juho Schultz, Sep 30, 2005
4. ### Nigel RoweGuest
Tim Roberts wrote:
> "PyPK" <> wrote:
>>
>>Does anyone know of a simple implementation of a straight line
>>detection algorithm something like hough or anything simpler.So
>>something like if we have a 2D arary of pixel elements representing a
>>particular Image. How can we identify lines in this Image.
>>for example:
>>
>>ary =
>>[[1,1,1,1,1],
>> [1,1,0,0,0],
>> [1,0,1,0,0],
>> [1,0,0,1,0],
>> [1,0,0,0,1]]
>>So if 'ary' represents pxl of an image which has a horizontal line(row
>>0),a vertical line(col 0) and a diagonal line(diagonal of ary). then
>>basically I want identify any horizontal or vertical or diagonal line
>>anywhere in the pxl array.
>
> If all you want is horizontal, vertical, or 45 degree diagonal, it's
> pretty easy to do that just be checking all of the possibilities.
>
> But what if your array is:
>
> [[1,1,1,1,1],
> [1,1,1,1,1],
> [1,1,1,1,1],
> [1,1,1,1,1],
> [1,1,1,1,1]]
>
> Would you say there were 12 lines there?
Actually I'd say 24.
5 vertical,
5 horizontal,
7 diagonal downward to the right (lengths 2,3,4,5,4,3,2)
7 diagonal downward to the left (lengths 2,3,4,5,4,3,2)
--
Nigel Rowe
A pox upon the spammers that make me write my address like..
rho (snail) swiftdsl (stop) com (stop) au
Nigel Rowe, Oct 15, 2005
## Want to reply to this thread or ask your own question?
It takes just 2 minutes to sign up (and it's free!). Just click the sign up button to choose a username and then you can ask your own questions on the forum. | 1,749 | 5,603 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2016-22 | latest | en | 0.814896 |
https://mathoverflow.net/questions/21916/linear-programming-piecewise-linear-objective | 1,696,105,963,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510707.90/warc/CC-MAIN-20230930181852-20230930211852-00112.warc.gz | 413,383,498 | 27,469 | # Linear programming piecewise linear objective
I am fairly new at linear programming/optimization and am currently working on implementing a linear program that is stated like this:
max $\sum_{i=1}^{k}{p(\vec \alpha \cdot \vec c_i)}$
$s.t.$
$|\alpha_j| \le 1$
Where p(x) = 2x if x < 0, x otherwise, and $\vec c$ is a constant
The p(x) function is what's troubling me, since one can only determine x's sign after an assignment of $\vec \alpha$. How can I remove the function p from the objective and express this objective equivalently as a linear combination of the variables?
Thank you!
• oh, I forgot to index $\vec c$ by i, so the alpha vector in the sum is the same for each term, but each scalar is different. Also, the i subscript on the alpha constraints is unrelated to the index in the sum. Sorry for any confusion.
– AFJ
Apr 20, 2010 at 1:04
• I've corrected the formula for you but I'm still totally puzzled by the $|\alpha_i|\le 0$ condition. What is that really supposed to be? Apr 20, 2010 at 2:01
• wow, original post was full of mistakes, sorry about that. It should be $|\alpha_i| \le 1$.
– AFJ
Apr 20, 2010 at 2:12
• The last question: do you want to maximize the sum or to minimize it? Apr 20, 2010 at 2:24
• Maximize, although I don't think it matters in answering my question, but you're right that it should be have been included.
– AFJ
Apr 20, 2010 at 2:27
Sorry for making you wait 14 hours unnecessarily but you are partially guilty yourself: if you posted a correct and full version of the question from the beginning, you would get the answer in 5 minutes. Keep it in mind when you ask a question on a public forum next time.
Your problem is equivalent to maximizing the linear expression $\sum_i y_i$ under the linear restrictions $\alpha_j\ge -1$, $\alpha_j\le 1$, $y_i\le c_i\cdot\alpha$, $y_i\le 2c_i\cdot\alpha$. It is as simple as that but it is crucial that your $p$ is concave and that you maximize.
• Thank you very much. That transformation makes intuitive sense to me, I'm going to try it out. Would you mind briefly explaining how you came about that solution (both for my benefit and for others that come across this later)?
– AFJ
Apr 20, 2010 at 16:22
• @AFJ: Note that your function $p$ is actually $p(x)=\operatorname{min}(x,2x)$. By setting $y_i:=p(\alpha\cdot c_i)=\operatorname{min}(\alpha \cdot c_i,2\alpha\cdot c_i)$, your problem becomes: max $\sum y_i$ s.t. $|\alpha_j|\leq 1$ and $y_i=\operatorname{min}(\alpha\cdot c_i,2\alpha\cdot c_i)$. But clearly you will get the same maximum by relaxing $y_i=\cdots$ to $y_i\leq\cdots$. Now use the fact the $y_i\leq \operatorname{min}(a,b)$ iff $y_i\leq a$ and $y_i\leq b$ to get fedja's conclusion. Apr 20, 2010 at 21:44
The easiest way to deal with this is to split it into two problems: one obtained by adding the inequality $x \ge 0$ and using the objective function $2x$, and the other by adding the inequality $x \le 0$ and using the objective function $x$. Incidentally, I don't think that you mean the inequality $|\alpha_i| \le 0$. Perhaps you mean to omit the absolute value?
I think that the best way of dealing with this is to solve the dual problem. I'll have to think about the details, but that should get the number of vertices in the polytope down to a manageable level.
• It seems to me that it can't be split into the maximum of just two problems, since it is a sum over $p(x_i)$ where the $x_i$ can be different, resulting in $p(x_i)$ being $2 x_i$ or $x_i$ depending on the sign of $x_i$, for a total of $2^k$ ways that p can play a role in the objective. An exponentially growing number of separate problems to solve won't be doable, since $k$ can be $25$ or more. Please correct me if I'm wrong.
– AFJ
Apr 20, 2010 at 2:24 | 1,079 | 3,752 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2023-40 | latest | en | 0.933208 |
https://groups.yahoo.com/neo/groups/primenumbers/search/messages?advance=true&am=CONTAINS&at=email:Andrey_601@&dm=IS_ANY&fs=false&count=10 | 1,427,895,429,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131304598.61/warc/CC-MAIN-20150323172144-00137-ip-10-168-14-71.ec2.internal.warc.gz | 927,503,209 | 16,728 | Author
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## 394 results from messages in primenumbers
• ### Re: [PrimeNumbers] Suggested (?) 2-bits per prime storage scheme
Is there any interest in a collaboration to build a truly massive database? No. It's faster to generate primes than to download them. A good generator is http://primesieve.org/ for example. Best regards, Andrey
Andrey Kulsha Aug 2, 2014
• ### Re: [PrimeNumbers] The probability of occurrence of 3 consecutive primes
But this sequence has 3 consecutive primes {23,29,31}. What is the probability of such a occurrence? 02:03:05 03:05:07 05:07:11 07:11:13 11:13:17 13:17:19 17:19:23 19:23:29 23:29:31 pi(24)/86400 = 1/9600 Best regards, Andrey
Andrey Kulsha Dec 28, 2013
• ### Re: [PrimeNumbers] RE: p-tight integers
Here is the extra step. Let a = 1 + r*p. Then S := (a^p-1)/(a-1) = sum(k=0,p-1,a^k) = sum(k=0,p-1,1+k*r*p) mod p^2 = p mod p^2, since sum(k=0,p-1,k)=p*(p-1)/2 and p is odd. Hence S/p = 1 mod p. Happy now? That was so simple. Thanks. Andrey :-)
Andrey Kulsha Dec 23, 2013
• ### Re: [PrimeNumbers] RE: p-tight integers
> Conjecture: > if p divides a-1, then (a^p-1)/(a-1)/p is p-tight. This follows from Theorem 48 of Daniel Shanks' delightful book: https://archive.org/details/SolvedAndUnsolvedProblemsInNumberTheory Thanks for the book. Maybe I misunderstand something, but it's quite obvious that p divides (a^p-1)/(a-1) when a = 1 (mod p), because (a^p-1)/(a-1) = sum(a^k,{k,0,p-1}). The...
Andrey Kulsha Dec 21, 2013
• ### p-tight integers
Given an odd prime p, let's call the positive integer "p-tight" if all its prime factors are 1 (mod 2p). Theorem: if p does not divide a-1, then (a^p-1)/(a-1) is p-tight. (trivial proof follows from FLT) Conjecture: if p divides a-1, then (a^p-1)/(a-1)/p is p-tight. Any suggestions for a proof? Thanks a lot, Andrey
Andrey Kulsha Dec 20, 2013
PrimeSieve by Kim Walisch finds all quadruplets up to 1e12 just in one minute: http://code.google.com/p/primesieve/ Best regards, Andrey
Andrey Kulsha Dec 6, 2013
• ### Re: prime gap of 1368
No one interested? :( --- In primenumbers@^\$1, Andrey Kulsha wrote: > > The gap of 1368 is known between primes > 4105079953458040849 and 4105079953458042217: > http://www.trnicely.net/gaps/gaplist.html > > To prove the first occurence, we need to check > a relatively small range from 4e18 to 4.10508e18 > because 4e18 is already searched: > http://sweet.ua.pt/tos/goldbach.html...
andrey_601 Jul 30, 2013
• ### prime gap of 1368
The gap of 1368 is known between primes 4105079953458040849 and 4105079953458042217: http://www.trnicely.net/gaps/gaplist.html To prove the first occurence, we need to check a relatively small range from 4e18 to 4.10508e18 because 4e18 is already searched: http://sweet.ua.pt/tos/goldbach.html It seems the search could be optimized if we focus exactly on 1368. I'd provide some...
Andrey Kulsha Jul 27, 2013
• ### Re: [PrimeNumbers] Kulsha's prime-gap conjecture
> Summing the geometric series in Andrey Kulsha's conjecture > http://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind0107&L=nmbrthry&P=3477 > we arrive at a rather attractive conjectural heuristic. That conjecture seems to be wrong. BTW, the new correct link to it is https://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind0107&L=NMBRTHRY&P=21148 [there's a typo, f(g,x) should be instead of f(g,p...
Andrey Kulsha Aug 29, 2012
• ### the best merit for prime gaps
If there are k composites preceding Nth prime, then the proposed merit is k / (log N)^2 There are 17 known gaps with merit > 1: merit k N 1.15817 3 5 1.13821 33 218 1.13709 1131 49749629143527 1.10242 13 31 1.07487 71 3386 1.05228 209 1319946 1.05033 455 1094330260 1.03633 905 6822667965941 1.03522 147 149690 1.03436 111 31546 1.03257 765 662221289044 1.02282 1441 20004097201301080...
Andrey Kulsha May 9, 2012 | 1,331 | 3,867 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2015-14 | longest | en | 0.812593 |
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# management scieThe manager of a small firm is considering whether to produce a new product that would require leasing some special equipment at a cost of \$20,000 per monthnce
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1) The manager of a small firm is considering whether to produce a new product that would require leasing some special equipment at a cost of \$20,000 per month. In addition to this leasing cost, a production cost of \$10 would be incurred for each unit of the product produced. Each unit sold would generate \$20 in revenue.
Develop a mathematical expression for the monthly profit that would be generated by this product in terms of the number of units produced and sold per month. then determine how large this number needs to be each month to make it profitable to produc the product.
I got profit=\$20Q-(\$20,000+\$10Q) = -\$20,000+\$10Q, IF Q>0
SO \$10Q=\$20,000 = \$20,000/\$10 = Q=2,000 | 256 | 1,088 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2021-25 | latest | en | 0.936696 |
https://cs.nyu.edu/pipermail/fom/2014-August/018132.html | 1,709,366,473,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475757.50/warc/CC-MAIN-20240302052634-20240302082634-00542.warc.gz | 193,274,883 | 3,638 | # [FOM] An argument for V = L
Timothy Y. Chow tchow at alum.mit.edu
Tue Aug 26 12:17:38 EDT 2014
```Here's an argument for V = L, from the point of view of the "ordinary
mathematician," that I haven't seen articulated in quite this way before.
A fundamental and widespread activity in mathematics is to identify the
precise hypotheses needed to arrive at certain desirable conclusions. For
example, a mathematician studying doohickeys might want to know if every
doohickey with Property A necessarily also has Property B. If the answer
turns out to be yes, then a natural followup question will be whether
Property A can be weakened to Property A_0; if the answer is no, then a
natural followup question will be whether Property A can be strengthened
to Property A_1. The motivating idea is that we are seeking to understand
exactly what structural properties of doohickeys cause Property B.
For example, in a slightly fictionalized history of general topology, we
could imagine that continuous functions are initially defined in terms of
epsilons and deltas in real analysis. Later it is recognized that you can
generalize the concept of continuity to metric spaces, and even later it
is recognized that all you need to carry out continuity arguments are the
axioms for a topological space. The mental picture is of a topology as
being a "structure" on an otherwise structureless underlying set. By
stripping away everything but this structure, one has clarified that
certain kinds of properties of real functions depend only on a small
number of abstract properties of the reals; other idiosyncratic properties
of the reals are revealed to be irrelevant for pure continuity arguments.
Now let's look at the trickier example of the axiom of choice. The
Hahn-Banach theorem is a useful tool in functional analysis and therefore
one would like to understand what properties a function space needs to
have in order for the Hahn-Banach theorem to hold. Inspecting the
argument, one finds that in addition to properties that make specific
mention of relevant mathematical structure (linear maps, subadditivity),
one also needs an additional principle---Zorn's lemma, or at least
ultrafilters. On the face of it, this doesn't look like a *property of
linear spaces* per se. The analyst is therefore tempted *not* to phrase
these findings in the form, "the Hahn-Banach theorem holds only for
`choosable' real or complex vector spaces," but instead wants to say that
the Zorn's lemma argument is a (logical?) principle that holds in general,
leaving it out of the list of *structural assumptions* that are needed for
the Hahn-Banach theorem to go through.
Trouble in paradise begins with the following
Unpleasant Surprise: Infinite sets have structure.
This is news that the ordinary mathematician doesn't want to hear. Sets
are supposed to be the structureless substratum upon which mathematical
structure is built. The Unpleasant Surprise means that the mathematician
now has to worry about the distinction between set theory and logic, and
has to decide what kinds of set-theoretical structure to accept. These
are questions that the mathematician doesn't want to think about.
The argument for V = L, then, is that it does a tolerably good job of
making the Unpleasant Surprise go away. For example, confronted with the
axiom of measurable cardinals for the first time, the average
mathematician's instinct is that it is asserting the existence of some
kind of structure on an infinite set, and therefore it makes no sense to
accept the existence of such structure unless it is forced on us by
logical necessity. If anything, the fact that V = L rules out measurable
cardinals is a point in favor of V = L.
Set theorists, of course, have a different attitude. For them, the
Unpleasant Surprise is a Pleasant Surprise. They make their living
studying the structure of infinite sets. They quite reasonably object to
the ostrich-head-in-the-sand attitude of the ordinary mathematician who
wants, irrationally, to deny the existence of structure on infinite sets,
and who sometimes seems to make contradictory demands. For them, V = L is
just one option among many, and they don't see any particular reason to
choose it. Certainly its role as a magic spell to wave away undesirable
structure on infinite sets carries no weight with them. First of all,
they understand that V = L doesn't completely eliminate questions about
set-theoretic structure, and secondly, they don't see anything undesirable | 1,017 | 4,521 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-10 | latest | en | 0.925148 |
https://www.ruby-forum.com/t/arbitrary-symbol-rate-transmission-by-usrp2/181051 | 1,721,601,375,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517796.21/warc/CC-MAIN-20240721213034-20240722003034-00081.warc.gz | 837,118,725 | 6,475 | # Arbitrary symbol rate transmission by USRP2
Arbitrary symbol rate transmission by USRP2.
I would like to transmit dqpsk 1,536,000sps with using benchmark_tx.py
and
benchmark_rx.py thru USRP2.
Since USRP2 does have 100MHz as clock source.
With the clock, seems decimation and interpolation only allow me at
1,538,461sps(100MHz/65) is the nearest.
I thought I need to make the clock which familiar to the desired symbol
rate(1,536,000sps).
Which I modify FPGA’s DCM clock manager to prescale 100MHz by 3125
(100,000,000/3125=32KHz) and
multiply it by 384(12.288Msps = 8x 1,536,000) and feed it to the
I feel decimation and interpolation number is too huge for DCM.
Would someone please advice me out where I should tackling to at first?
Should I still go for modify DCM clock manager?
Yoshiya K…
On 01/04/2010 07:29 AM, åŠ è—¤ç¾©ä¹Ÿ wrote:
Which I modify FPGA’s DCM clock manager to prescale 100MHz by 3125
(100,000,000/3125=32KHz) and
multiply it by 384(12.288Msps = 8x 1,536,000) and feed it to the DAC/ADC.
I feel decimation and interpolation number is too huge for DCM.
Would someone please advice me out where I should tackling to at first?
Should I still go for modify DCM clock manager?
Do not modify the DCM. That will not do what you want. What you will
need to do is make the software output samples at a rate which the USRP2
can handle. There are 2 ways to do that:
1 - Create a modulator which will output a non-integer number of samples
per symbol.
2 - Use the existing modulator, and add a rational resampler to your
transmit chain to get to a useful rate. The ratio between 1.536M and
100 M is 3125/48 or (5^5) / (2^4 * 3). So you would need to interpolate
by a total of 5^5 and decimate by 2^4*3. You can get 5^2 in the USRP2
interpolator by setting the main interpolator to 25. This means you
would be supplying 4 MS/s over the ethernet to the USRP2.
You would then generate you sample stream at 5 samples per symbol,
giving you 7.68 MS/s. Now the ratio between 7.68 MS/s and 4 MS/s is
only 25/48. So you would use the GNU Radio rational resampler block to
do this sample rate conversion, which is not hard.
Matt
Matt. Thank you very much for very prompt and highly accurate reply.
To have maximum number of options for considering with my limited
resources.
-Change 100MHz crystal to 76.8MHz or 122.88MHz(lined up on Crystek
CVHD950)
can not be a option?
Will this idea give significant impact to USRP2?
maximum frequency for each cores.
Why I thought this may be a option was because Scott tried “re-clocking”
motherboard on below.
scott scott
Mon, 26 Oct 2009 19:49:56 -0700
In my understand 100MHz(CVHD950) is simply used for the drive DAC/ADC
and is
able to replace.
Unfortunately, I could not reach to the document address specified in
above
mail.
Best Regards.
2010年1月5日6:14 Matt E. [email protected]:
I thought I need to make the clock which familiar to the desired symbol
total of 5^5 and decimate by 2^4*3. You can get 5^2 in the USRP2
interpolator by setting the main interpolator to 25. This means you would
be supplying 4 MS/s over the ethernet to the USRP2.
You would then generate you sample stream at 5 samples per symbol, giving
you 7.68 MS/s. Now the ratio between 7.68 MS/s and 4 MS/s is only 25/48.
So you would use the GNU Radio rational resampler block to do this sample
rate conversion, which is not hard.
Matt
Yoshiya K. [email protected]
On 01/04/2010 07:17 PM, e\$B2CF#5ALie(B wrote:
Matt. Thank you very much for very prompt and highly accurate reply.
To have maximum number of options for considering with my limited resources.
-Change 100MHz crystal to 76.8MHz or 122.88MHz(lined up on Crystek
CVHD950) can not be a option?
Will this idea give significant impact to USRP2?
You should be able to use the 76.8 MHz option without any trouble. You
would not be able to meet timing in the FPGA with the 122.88 MHz one.
You will also need to change the firmware in the FPGA which sets the PLL
tuning if you are using an external reference.
It will be MUCH easier to use the rational resampler I mentioned in my
previous email. I would strongly advise that you try that first, as it
shouldn’t take you more than an hour or two.
Matt | 1,202 | 4,211 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-30 | latest | en | 0.909253 |
https://rdrr.io/cran/nnspat/man/EV.Nii.html | 1,713,930,669,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818999.68/warc/CC-MAIN-20240424014618-20240424044618-00238.warc.gz | 450,981,169 | 7,820 | # EV.Nii: Expected Values of the Self Entries in a Species... In nnspat: Nearest Neighbor Methods for Spatial Patterns
EV.Nii R Documentation
## Expected Values of the Self Entries in a Species Correspondence Contingency Table (SCCT)
### Description
Returns a `vector` of length k of expected values of the self entries (i.e. first column) in a species correspondence contingency table (SCCT) or the expected values of the diagonal entries N_{ii} in an NNCT. These expected values are valid under RL or CSR.
The argument `ct` can be either the NNCT or SCCT.
### Usage
```EV.Nii(ct)
```
### Arguments
`ct` The NNCT or SCCT
### Value
A `vector` of length k whose entries are the expected values of the self entries (i.e. first column) in a species correspondence contingency table (SCCT) or of the diagonal entries in an NNCT.
Elvan Ceyhan
### References
\insertAllCited
`scct` and `EV.nnct`
### Examples
```n<-20 #or try sample(1:20,1)
Y<-matrix(runif(3*n),ncol=3)
ipd<-ipd.mat(Y)
cls<-sample(1:2,n,replace = TRUE) #or try cls<-rep(1:2,c(10,10))
ct<-nnct(ipd,cls)
ct
EV.Nii(ct)
ct<-scct(ipd,cls)
EV.Nii(ct)
#cls as a factor
na<-floor(n/2); nb<-n-na
fcls<-rep(c("a","b"),c(na,nb))
ct<-nnct(ipd,fcls)
EV.Nii(ct)
#############
n<-40
Y<-matrix(runif(3*n),ncol=3)
ipd<-ipd.mat(Y)
cls<-sample(1:4,n,replace = TRUE) #or try cls<-rep(1:2,c(10,10))
ct<-nnct(ipd,cls)
EV.Nii(ct)
ct<-scct(ipd,cls)
EV.Nii(ct)
```
nnspat documentation built on Aug. 30, 2022, 9:06 a.m. | 487 | 1,484 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-18 | latest | en | 0.477262 |
https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Electronics/AC_Electrical_Circuit_Analysis%3A_A_Practical_Approach_(Fiore)/09%3A_Polyphase_Power/9.1%3A_Introduction_to_Polyphase_Power | 1,718,388,529,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861568.20/warc/CC-MAIN-20240614173313-20240614203313-00645.warc.gz | 212,179,623 | 29,043 | # 9.1: Introduction to Polyphase Power
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
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$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
In this chapter we shall introduce the concept of polyphase systems. Polyphase systems can be visualized as a group of individual sources of the same magnitude that are separated by a certain phase angle such that they are evenly divided across a single period. The polyphase load is similarly divided into individual sections or legs. By dividing the sources, the application of power can be much more smooth. Further, for the same total load power, the current delivered by each of the segments is reduced compared to a single phase system. For an analogy we could look at a bicycle. A single phase system is like pedaling with only one leg. That is, power is applied in a single burst per revolution of the pedal. Having two pedals is like a two-phase system; power is delivered twice per revolution, once for the right leg and once for the left. Because there are two pedals, it makes sense to separate them physically by 180 degrees or one half of a revolution, otherwise the power delivery will not be smooth. It should obvious to anyone who has pedaled a bike that you must pump a single pedal much harder using only one leg to achieve the same speed obtained when pumping with both legs.
Polyphase loads can be balanced or unbalanced. A balanced load means that all legs or sections of the load exhibit the same impedance. Consequently, the currents supplied by the sources will be the same except for the phase shifts between them. It is possible to create a polyphase system using any number of phases, however, the more phases we add, the more complex the construction of the polyphase source and load. Also, the number of required conductors between the source and load increases (one per phase). These all increase construction, installation and maintenance costs. Polyphase systems using three sections (hereafter simply referred to as three-phase systems) are popular because they deliver the benefits of polyphase while limiting the complexity and minimizing the costs. Therefore, we shall our limit our discussion to three-phase systems utilizing balanced loads.
Three-phase systems can be wired in either delta or Y configurations, or a combination. These are reminiscent of the delta and Y constructs seen in earlier chapters. We shall investigate all of the combinations to determine system parameters such as line voltage, line current and load power. We will also investigate power factor correction for balanced loads that have a non-negligible phase angle.
This page titled 9.1: Introduction to Polyphase Power is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by James M. Fiore via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. | 1,212 | 4,471 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-26 | latest | en | 0.354689 |
http://research.stlouisfed.org/fred2/series/RGDPESBSA666NRUG | 1,427,722,353,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131299339.12/warc/CC-MAIN-20150323172139-00250-ip-10-168-14-71.ec2.internal.warc.gz | 229,451,234 | 19,632 | # Expenditure-side Real GDP at Chained Purchasing Power Parities for Bahamas
2011: 6,723.62305 Millions of 2005 U.S. Dollars (+ see more)
Annual, Not Seasonally Adjusted, RGDPESBSA666NRUG, Updated: 2013-08-12 12:06 PM CDT
Click and drag in the plot area or select dates: Select date: 1yr | 5yr | 10yr | Max to
Source ID: rgdpe
When using these data in your research, please make the following reference: Feenstra, Robert C., Robert Inklaar and Marcel P. Timmer (2013), "The Next Generation of the Penn World Table" available for download at www.ggdc.net/pwt
For more information, see http://www.rug.nl/research/ggdc/data/pwt/.
Source: University of Groningen
Release: Penn World Table 8.0
Restore defaults | Save settings | Apply saved settings
w h
Graph Background: Plot Background: Text:
Color:
(a) Expenditure-side Real GDP at Chained Purchasing Power Parities for Bahamas, Millions of 2005 U.S. Dollars, Not Seasonally Adjusted (RGDPESBSA666NRUG)
Source ID: rgdpe
When using these data in your research, please make the following reference: Feenstra, Robert C., Robert Inklaar and Marcel P. Timmer (2013), "The Next Generation of the Penn World Table" available for download at www.ggdc.net/pwt
For more information, see http://www.rug.nl/research/ggdc/data/pwt/.
Expenditure-side Real GDP at Chained Purchasing Power Parities for Bahamas
Integer Period Range: to copy to all
Create your own data transformation: [+]
Need help? [+]
Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
Create segments for min, max, and average values: [+]
Graph Data
Graph Image
Suggested Citation
``` University of Groningen and University of California, Davis, Expenditure-side Real GDP at Chained Purchasing Power Parities for Bahamas [RGDPESBSA666NRUG], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/RGDPESBSA666NRUG/, March 29, 2015. ```
Retrieving data.
Graph updated.
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Name: Email: | 701 | 2,647 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2015-14 | latest | en | 0.757736 |
https://unimath.github.io/agda-unimath/graph-theory.matchings.html | 1,726,512,544,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651710.86/warc/CC-MAIN-20240916180320-20240916210320-00840.warc.gz | 540,096,577 | 38,767 | # Matchings
Content created by Egbert Rijke, Jonathan Prieto-Cubides, Fredrik Bakke and Eléonore Mangel.
Created on 2022-03-28.
module graph-theory.matchings where
Imports
open import foundation.contractible-types
open import foundation.coproduct-types
open import foundation.dependent-pair-types
open import foundation.identity-types
open import foundation.propositions
open import foundation.unit-type
open import foundation.universe-levels
open import foundation.unordered-pairs
open import graph-theory.undirected-graphs
open import univalent-combinatorics.standard-finite-types
## Idea
A matching in a undirected graph is a type of edges without common vertices.
## Definitions
module _
{l1 l2 : Level}
where
selected-edges-vertex-Undirected-Graph :
( G : Undirected-Graph l1 l2) →
( (p : unordered-pair-vertices-Undirected-Graph G) →
edge-Undirected-Graph G p → Fin 2) →
vertex-Undirected-Graph G → UU (l1 ⊔ l2)
selected-edges-vertex-Undirected-Graph G c x =
Σ ( vertex-Undirected-Graph G)
( λ y →
Σ ( edge-Undirected-Graph G (standard-unordered-pair x y))
( λ e → Id (c (standard-unordered-pair x y) e) (inr star)))
matching : Undirected-Graph l1 l2 → UU (lsuc lzero ⊔ l1 ⊔ l2)
matching G =
Σ ( (p : unordered-pair-vertices-Undirected-Graph G) →
edge-Undirected-Graph G p → Fin 2)
( λ c →
( x : vertex-Undirected-Graph G) →
is-prop (selected-edges-vertex-Undirected-Graph G c x))
perfect-matching : Undirected-Graph l1 l2 → UU (lsuc lzero ⊔ l1 ⊔ l2)
perfect-matching G =
Σ ( (p : unordered-pair-vertices-Undirected-Graph G) →
edge-Undirected-Graph G p → Fin 2)
( λ c →
( x : vertex-Undirected-Graph G) →
is-contr (selected-edges-vertex-Undirected-Graph G c x)) | 513 | 1,681 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-38 | latest | en | 0.46596 |
http://stackoverflow.com/questions/15118433/arrays-with-static-methods | 1,440,937,692,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644065306.42/warc/CC-MAIN-20150827025425-00302-ip-10-171-96-226.ec2.internal.warc.gz | 228,510,635 | 19,203 | # Arrays with Static Methods
I am trying to manipulate two Arrays by using two different Static Methods
1. `double dot(double[]a, double[]b)`
2. `double[][] multiply(double[][]a, double[][]b)`.
I cant seem to figure out how to multiply the two arrays using static methods and output there values to the user I believe my Dot product method is good though. I know I need to use a return method for my multiplication method but im not sure how to represent this correctly
Here is what I have so far:
``````public class LibMatrix {
public static void main(String[] args) {
double[] a = { 8, 5, 6, 3, 2, 1 };
double[] b = { 9, 8, 4, 1, 4, 7 };
}
public static double dot(double[] a, double[] b) {
double sum = 0.0;
for (int i = 0; i < a.length; i++)
sum += a[i] * b[i];
return sum;
}
public static double[][] multiply(double[][] a, double[][] b) {
int n = 6;
double[][] c = new double[n][n];
for (int i = 0; i < n; i++)
for (int j = 0; i < n; i++)
c[i][j] = a[i][j] * b[i][j];
return a;
}
}
``````
-
The second method is for matrices (bidimensional arrays), but your "static arrays" are one-dimensional... – SJuan76 Feb 27 '13 at 17:32
I'm not sure I understand the question; can't you call the methods in `main`, reference the return values, and iterate over them? – Dave Newton Feb 27 '13 at 17:32
`return a;` -> `return c;` – Mike Samuel Feb 27 '13 at 17:35
Also, `int n = 6;` -> `int n = a.length; assert n == 0 || (a[0].length == n && b.length == n && b[0].length == n);` – Mike Samuel Feb 27 '13 at 17:36
Changed the return a; to return c; in the method that is what I originally wrote I just changed it to try and get some type of output. So what you are saying is that my method of multiplication will not work because my arrays are one dimensional how can i make them 2-D? – user2092325 Feb 27 '13 at 17:38
Here is some corrections:
``````public class LibMatrixTests
{
static class LibMatrix {
public static double dot(double[] a, double[] b) {
double sum = 0.0;
for (int i = 0; i < a.length; i++)
sum += a[i] * b[i];
return sum;
}
public static double[][] mul( double[][] a, double[][] b ) {
double[][] c = new double[a.length][a[0].length];
for (int i = 0; i < a.length; i++)
for (int j = 0; j < a[i].length; j++)
c[i][j] = a[i][j] * b[i][j];
return a;
}
}
public static void main( String[] args ) {
double[] a = { 8, 5, 6, 3, 2, 1 };
double[] b = { 9, 8, 4, 1, 4, 7 };
double[][] c = { a, b };
double[][] d = { b, a };
double e = LibMatrix.dot( a, b );
double[][] f = LibMatrix.mul( c, d );
System.out.println( e );
for( double[] g : f ) {
for( double h : g ) {
System.out.print( h + ", " );
}
System.out.println();
}
}
}
``````
Outputs:
``````154.0
8.0, 5.0, 6.0, 3.0, 2.0, 1.0,
9.0, 8.0, 4.0, 1.0, 4.0, 7.0,
``````
-
Thank you so much thats exactly what I wanted the program to do now time to Trace and see how you did it:) – user2092325 Feb 27 '13 at 17:50
Use a diff utilities because you have misused the index i and j into for loops causing IndexOutOfBoundsException – Aubin Feb 27 '13 at 17:52
Don't have enough rep to comment but your return value in the second method should be c
Test:
``````double [][] a = new double [6][6];
double [][] b = new double [6][6];
for(int i = 0; i< a.length;i++){
for(int j = 0; j< a.length;j++){
a[i][j] = 3;
b[i][j] = 2;
}
}
d = multiply(a,b);
``````
This returns a 6x6 matrix filled with 6s so your method is working correctly.
``````d = [6, 6, 6, 6, 6, 6,
6, 6, 6, 6, 6, 6,
6, 6, 6, 6, 6, 6,
6, 6, 6, 6, 6, 6,
6, 6, 6, 6, 6, 6,
6, 6, 6, 6, 6, 6]
``````
-
in the multiply method, the second loop is wrong, j is always 0, and you should return c instead of a
try this
``````public static double[][] multiply(double[][] a, double[][] b) {
int n = 6;
double[][] c = new double[n][n];
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
c[i][j] = a[i][j] * b[i][j];
return c;
}
``````
- | 1,348 | 3,878 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2015-35 | latest | en | 0.815926 |
https://efnet-math.org/w/Seminar_on_Set_Theory_Basics | 1,726,011,676,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651323.1/warc/CC-MAIN-20240910224659-20240911014659-00460.warc.gz | 204,838,045 | 7,341 | # Seminar on Set Theory Basics
This seminar was held in #mathematics on Freenode (with a relay set up to EFnet) on 27th of August 2005 at 15:00 EDT by Kit.
It covered basic undergraduate set theory.
## Seminar Log
15:03 <@DRMacIver> Today I'll be introducing the basics of set theory. I'll start by giving some basic definitions, and then go on to develop some of the basic theory of cardinality of infinite sets.
15:03 <@DRMacIver> This is by no means a comprehensive introduction to set theory - it's a big subject, and there's a lot in it. This is just what I think one of the neatest bits of elementary set theory is.
15:04 <@DRMacIver> So, lets start by telling you what a set is.
15:04 <sven> !
15:04 <@DRMacIver> sven: Yes?
15:04 <sven> could you introduce yourself, or refer us to a web page of yours?
15:05 <@DRMacIver> I suppose so. :)
15:05 < South-Park> !can we also have ur mail id so that we can mail our doubts while reffering to the logs later ??
15:05 < DrChemicalX> ! i would be happy if you introduced yourself also sir
15:05 <@DRMacIver> Briefly, I'm a recent graduate of Cambridge university. I've been in EFNet's #math channel for the past few years and have given a number of online talks there.
15:06 <@DRMacIver> My website is http://www.efnet-math.org/~david/
15:06 <@DRMacIver> Some of you probably know me as Kit. I'm using a different name because someone else has already taken that on freenode.
15:07 <@DRMacIver> Now, if anyone has any more non-mathematical questions can you leave them to the end of the talk? It'll take forever if we keep getting sidetracked. :)
15:08 <@DRMacIver> Ok. Briefly, a set is a collection of objects - usually these are of some mathematical significance. For example sets of numbers.
15:08 <@DRMacIver> We use the notation {1, 2, 3} to mean the set containing precisely the numbers 1, 2 and 3.
15:09 <@DRMacIver> Important features of sets: Firstly, they have no inherent order to them.
15:09 <@DRMacIver> So {2, 3, 1} is the same as {1, 2, 3}
15:09 <@DRMacIver> Also, repetitions are ignored - {1, 2, 2, 3} is again the same as {1, 2, 3}.
15:10 <@DRMacIver> Finally, sets don't have any special 'names'. A set is determined entirely by what it contains - it doesn't matter if the definitions and names are completely different, if they contain the same things then they are the same set.
15:11 <@DRMacIver> Another notation we might use is defining a set by a formula. For example we could write { p : p is a prime } for the set of all prime numbers.
15:12 <@DRMacIver> Now I'm going to teach you how to count. :)
15:13 <@DRMacIver> Suppose we have a set such as {1, 2, 3} or {2, 7, 9}. It's obvious that these contain three elements, but what does that actually mean?
15:13 <@DRMacIver> Now, first of all, it essentially must be true by definition that {1, 2, 3} has three elements.
15:14 <@DRMacIver> There are always going to be n numbers between 1 and n. This is just inherent in the process of counting.
15:15 <@DRMacIver> Now, we can pair up the numbers in these two sets. (1, 2), (2, 7) and (3, 9).
15:15 <@DRMacIver> For each element of the first set we've given you a unique element of the second set.
15:15 <@DRMacIver> And every element is hit this way.
15:15 <@DRMacIver> So intuitively speaking these must have the same number of elements: Any way of counting the first set gives you a way of counting the second set.
15:16 <@DRMacIver> So really the numbers themself aren't important with counting (Yes, you can quote me as saying that numbers aren't important for counting. :) ). What's important is the notion of two sets having the same number of elements.
15:17 <@DRMacIver> Oh, sorry. I should say - an element of a set is just something that's contained in that set.
15:17 <@DRMacIver> We use the notation \in which is written as a little E, or an epsilon. For example 1 \in {1, 2, 3}
15:18 <@DRMacIver> For a finite set X we use the notation |X| to mean the number of elements of X.
15:18 <@DRMacIver> So |{1, 2, 3}| = |{2, 7, 9}| = 3
15:19 <@DRMacIver> Now, the above discussion is basically saying that if we have a set X and a set Y, then |X| = |Y| if we can pair up the elements of X and Y like that.
15:19 <@DRMacIver> I haven't actually proved that - what I'm really giving you here is a definition.
15:20 <@DRMacIver> This leads us onto the idea of a function, which will allow us to make the notion of that pairing up slightly more precise.
15:20 <@DRMacIver> Suppose we have a set X and a set Y. A function from X to Y is a rule that takes an element of X and gives you an element of Y.
15:21 <@DRMacIver> So if we have the sets {1, 2, 3} and {2, 3, 7} we might have the function that sends 1 to 2, 2 to 3 and 3 to 7. This is the pairing we had above.
15:21 <@DRMacIver> We often give these functions names. For example we might (imaginatively) call this function f.
15:21 <@DRMacIver> We would write f : X -> Y for a function from X to Y.
15:21 <@DRMacIver> And f(x) for the value that x is sent to.
15:22 <@DRMacIver> So in the above we have f : {1, 2, 3} -> {2, 3, 7} with f(1) = 2, f(2) = 3, f(3) = 7.
15:23 <@DRMacIver> Now, a function is called injective (or an injection) if it sends distinct things to distinct things.
15:23 <@DRMacIver> So f is injective.
15:23 <@DRMacIver> But if we defined g by g(1) = 2, g(2) = 2, g(3) = 7, this wouldn't be.
15:23 <@DRMacIver> Because 1 and 2 are different, but g(1) = g(2)
15:24 <@DRMacIver> We call a function surjective (or a surjection) if it hits every element.
15:24 <@DRMacIver> Oh, sorry. I forgot to tell you. If f : X -> Y then we call X the domain and Y the codomain.
15:24 <@DRMacIver> So in that definition I really want to say if it hits every element of the codomain.
15:25 <@DRMacIver> So f is surjective, but g isn't because it doesn't hit 3.
15:25 <@DRMacIver> Similarly if we defined f : {1, 2, 3} -> {1, 2, 3, 4} by just setting f(x) = x, then this is injective but not surjective.
15:26 <@DRMacIver> Finally, a function is bijective (or a bijection) if it is both injective and surjective.
15:26 <@DRMacIver> So what we are saying is that |X| = |Y| if and only if there is a bijection between X and Y.
15:26 <@DRMacIver> Phew. Sounds a bit heavy handed, but it really isn't once you're used to it. :)
15:26 <@DRMacIver> Any questions so far?
15:27 <@DRMacIver> ok. Guess not. :)
15:27 <@DRMacIver> Oh, we can also make sense of |X| <= |Y|, just by saying this means that there is an injection from X to Y.
15:28 <@DRMacIver> Now, if we have a function f : X -> Y and a function g : Y -> Z then we can define a new function gf : X -> Z by saying gf (x) = g(f(x))
15:29 <@DRMacIver> Exercise: If g and f are both bijections then so is gf.
15:29 <@DRMacIver> This is called composing g and f, or composition of functions.
15:30 <@DRMacIver> This lets us prove something that the notation has already suggested, which is that if |X| = |Y| and |Y| = |Z| then we have |X| = |Z|. Just take a bijection from X to Y and a bijection from Y to Z and compose them, giving a bijection from X to Z.
15:30 <@DRMacIver> Now, the clever part of all this is that our definitions don't assume the sets to be finite!
15:30 <@DRMacIver> So this will allow us to make sense of the notion of how many elements of an infinite set has.
15:30 <@DRMacIver> Other than 'lots' :)
15:31 <@DRMacIver> Our basic example of an infinite set is the set of natural numbers. That is numbers of the form 1, 2, 3, 4, ...
15:31 <@DRMacIver> We call this set N.
15:31 <@DRMacIver> We now encounter something slightly surprising - if you take away some elements from an infinite set then you still have the same number of elements left.
15:32 <@DRMacIver> For example |N| = | {2, 3, 4, ... } |
15:32 <@DRMacIver> Proof: Define f : N -> {2, 3, 4,...} by f(n) = n + 1
15:32 <@DRMacIver> Exercise: This is a bijection.
15:32 <@DRMacIver> You can even take away infinitely many elements (although of course if you take away *all* the elements, or even all but finitely many, this doesn't work).
15:32 <@DRMacIver> |N| = |{2, 4, 6, ...}|
15:33 <@DRMacIver> Proof: Define f(n) = 2n
15:33 <@DRMacIver> Same exercise.
15:33 <@DRMacIver> So there are as many even numbers as there are numbers.
15:33 <@DRMacIver> In fact, if A is any infinite subset of N then |A| = |N|
15:34 <@DRMacIver> How we prove this is as follows. Strictly speaking this is a definition by induction, but you don't really need to know what that means. :)
15:35 <Reloaded> What is Aleph, Aleph_0 and the difference between these two notations?
15:35 <@DRMacIver> I'm not going to use that notation in this talk, so I won't answer that now.
15:36 <@DRMacIver> Anyway, back to proving this.
15:36 <Reloaded> Those are some really interesting parts of set theory.
15:36 <@DRMacIver> What we do is define f : N -> A as follows.
15:36 <@DRMacIver> f(1) is the smallest element.
15:36 <@DRMacIver> f(2) is the next smallest element.
15:36 <@DRMacIver> ...
15:37 <@DRMacIver> f(n) is the smallest element greater than the ones we've previously defined.
15:37 <@DRMacIver> And there are always going to be more elements, because A is infinite, so we can keep defining it this way.
15:37 <@DRMacIver> This is injective because if m < n then we've chosen f(n) to be strictly greater than f(m).
15:38 <@DRMacIver> This is surjective because you can easily check that f(n) >= n, and because of the way we've constructed it, if something is in the image then all smaller numbers in A are also in the image.
15:38 <@DRMacIver> Sorry. Going to introduce another random definition. :)
15:38 <@DRMacIver> The image is the set of points in the codomain which get hit.
15:39 <@DRMacIver> So if f : X -> Y then the image, which we write f(X), is { f(x) : x \in X }
15:39 <@DRMacIver> So saying f is surjective is just saying that f(X) = Y
15:39 <@DRMacIver> Anyway, so f is a bijection.
15:39 <@DRMacIver> Thus |N| = |A|
15:40 <@DRMacIver> This should suggest to you that |N| is the 'smallest infinity'. In fact this is true.
15:41 <@DRMacIver> Suppose X is any infinite set. We can define an injective function f : N -> X in a very similar way. The only thing is that we need something called the axiom of choice to do this, because we dont' have any sensible way of picking elements of X.
15:41 <@DRMacIver> So we pick something for f(1), then having picked f(1), ... ,f(n-1) we pick something differet for f(n) (which we can do because A is infinite).
15:41 <@DRMacIver> I'm not going to get into the details of choice. I'm just going to occasionally mention it in this talk. :)
15:42 <@DRMacIver> i.e. for any infinite set X we have |N| <= |X|
15:42 <@DRMacIver> If |X| <= |N| then we say that X is countable. I'm now going to spend a little time showing that various sets are countable.
15:43 <@DRMacIver> First of all I'm going to give you another definition. :)
15:43 <@DRMacIver> If we have a set X then X^2 is the set of ordere dpairs of elements of X.
15:44 <@DRMacIver> i.e. { (x, y) : x, y \in X }
15:44 <@DRMacIver> X^3 is then the set of ordered triplets, etc.
15:44 <@DRMacIver> I'm now going to show that N^2 is countable.
15:45 <@DRMacIver> There's a nice explicit bijection between N and N^2, but it really needs a diagram so I'm not going to give it. :)
15:45 <@DRMacIver> Instead I'll do something slightly different.
15:45 <@DRMacIver> Define f : N^2 -> N by f(m, n) = 2^m 3^n
15:46 <@DRMacIver> This is then injective because things have unique prime factorisation (don't worry if you can't prove this. :) )
15:46 <@DRMacIver> So we've got that |N^2| <= |N|
15:46 <@DRMacIver> And then because this means we have a bijection between N^2 and f(N^2), which is an infinite subset of N, we have that |N^2| = |N|
15:47 <@DRMacIver> So there are as many pairs of numbers as there are numbers.
15:47 <@DRMacIver> Another thing. If we have a surjective map f : N -> X then X must be countable.
15:48 <@DRMacIver> Proof: Define g : X -> N by letting g(x) be the least n such that f(n) = x.
15:48 <@DRMacIver> Exercise: This is injective.
15:48 <@DRMacIver> Now, this will work if we replace N with any other countable set.
15:49 <@DRMacIver> So for example there are countably many positive rationals, because we can define f : N^2 -> {positive rationals} by f(m, n) = m / n
15:49 <@DRMacIver> And this is obviously a surjection (but not injective).
15:50 <@DRMacIver> Now I'm going to introduce another notion.
15:50 <@DRMacIver> First of all I want to note that it's ok for sets to contain other sets.
15:50 <@DRMacIver> So for example we could have { {1, 2}, {1, 2, 3} } - the set whose elements are precisely {1, 2} and {1, 2, 3}
15:51 <@DRMacIver> This leads us to the notion of a power set. If X is a set then we define P(X) = { subsets of X }
15:51 <@DRMacIver> Oh, sorry. I don't think I've defined a subset.
15:52 <@DRMacIver> But it's probably what you'd have expected. A is a subset of B if every element of A is also contained in B.
15:52 <@DRMacIver> So {1, 2, 3} is a subset of {1, 2, 3, 4}, but {1, 2, 5} is not because 5 is not in {1, 2, 3, 4}.
15:53 <@DRMacIver> Exercise: |P({1, ..., n})| = 2^n
15:53 <@DRMacIver> Because of this we often use the notation |P(X)| = 2^|X|
15:53 <@DRMacIver> This isn't really important though.
15:53 <@DRMacIver> First as a warmup we'll show that N has countably many finite subsets.
15:54 <@DRMacIver> What we do is we take n and write it as a prime factorisation. Say n = 2^1 5^7 11^1
15:54 <@DRMacIver> We now send this to the set of numbers which appear in the powers. So that number goes to {1, 7, 1} = {1, 7}
15:54 <@DRMacIver> Exercise: This is a surjective map.
15:55 <@DRMacIver> So, we've seen lots of infinite sets are countable. You're probably getting a bit bored of this by now. :)
15:55 <@DRMacIver> So as a punchline I'm finally going to give you an uncountable set.
15:55 <@DRMacIver> This is called Cantor's theorem: It says that for any set X there is no bijection between X and P(X).
15:56 <@DRMacIver> The proof is clever. Don't blink or you'll miss it. :)
15:56 <@DRMacIver> Suppose f : X -> P(X) was a bijection.
15:56 <@DRMacIver> Define the set A = { x \in X : x is not a member of f(x) }
15:57 <@DRMacIver> Now, f is surjective so we must have f(x) = A for some x
15:57 <@DRMacIver> As A is a subset of X.
15:57 <@DRMacIver> Now, x is an element of A if and only if it is not an element of A!
15:57 <@DRMacIver> Because we've defined A to be the set all x which are not contained in f(x)
15:58 <@DRMacIver> Err. Say y which are not contained in f(y)
15:58 <@DRMacIver> So, this is a contradiction. Thus our initial assumption can't have been correct.
15:58 <@DRMacIver> (This is called a proof by contradiction if you haven't seen it before. It's very useful, but occasionally feels like a bit of a rabbit out of a hat).
15:59 <@DRMacIver> So we've concluded that P(N), the set of all subsets of N, must be uncountable.
15:59 <@DRMacIver> And thus there are different sizes of infinity.
15:59 <@DRMacIver> I was going to use this to show that there are uncountably many real numbers, but I'm running behind so I'll leave it there.
15:59 <@DRMacIver> You basically just construct an injection from P(N) into the real numbers using the decimal expansion.
16:00 <@DRMacIver> Anyway, this is one of the basic ideas which got the subject started, and leads on to a lot of other useful and cool ideas.
16:00 <@DRMacIver> Hope it was interesting. :)
16:01 <@DRMacIver> Oh good. I've put everyone to sleep. :)
16:02 <Teirdes> Thank you.
16:02 <@Chandra> I will be publishing the lecture on site, would you mind?
16:02 <@Chandra> *odd site*
16:02 <@DRMacIver> Chandra: Go right ahead. We'll probably stick a copy on the EFNet #math one as well. | 5,103 | 15,691 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-38 | latest | en | 0.931202 |
https://www.ijoobi.com/math/conversion/length/micrometer-to-meter | 1,709,288,577,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475238.84/warc/CC-MAIN-20240301093751-20240301123751-00385.warc.gz | 806,555,895 | 14,455 | # Micrometer to Meters conversion
Result
m
Click Result to Copy
µm=m
How did we calculate?
To Calculate, we took the value you submtted and divided it by 1,000,000,000 to get the result.
(/1,000,000)
Our micrometer to meter(µm to m) conversion tool is a free converter that enables you to convert from micrometer to meter easily.
## How to Convert micrometer to meter
To convert a micrometer measurement(µm) to a meter measurement(m), divide the length by the conversion ratio. Since one Meter is equal to 1,000,000,000 Micrometers, you can use this simple formula to convert:
### What is the formula to convert from micrometer to meter?
meter=µm / 1,000,000
### Convert 5 micrometer to Meter
5 µm = (5 / 1,000,000) = 0.000005m
### Convert 10 micrometer to Meter
10 µm = (10 / 1,000,000) = 0.00001 m
### Convert 100 micrometer to Meter
100 µm = (100 / 1,000,000) = 0.0001 m
## Micrometer
### What is a Micrometer?
Micrometer ormicrometre, also called micron is a metric unit of measure for length equal to 0.001 mm, or about 0.000039 inch. Its symbol is μm. The micrometre is commonly employed to measure the thickness or diameter of microscopic objects, such as microorganisms and colloidal particles.
Micrometer can be abbreviated as µm; for example, 1 Micrometer can be written as 1µm.
### What is the Micrometer used for?
Micrometers are specially designed for the measurement of tiny objects. They allow for the exact measurement of any item that fits between the anvil and spindle. Standard types of micrometers can be used for the acceptable measurement of items under one inch in length, depth, and thickness.
## Meter
### What is a Meter?
A meter, or metre (m), is the length and distance base unit in the International System of Units (SI). The meter is defined as the distance traveled by light in 1/299 792 458 of a second. This definition was modified in 2019 to reflect changes in the definition of the second.
Meter can be abbreviated as m; for example, 1 Meter can be written as 1m.
### What is the Meter used for?
The meter is used worldwide in many applications, such as measuring distance, being the SI unit of length. The United States is one notable exception in that it primarily uses US customary units such as yards, inches, feet, and miles instead of meters in everyday use.
## How to use our Micrometer to Meters converter (µm to m converter)
Follow these 3 simple steps to use our micrometer to meter converter
1. Input the unit of micrometer you wish to convert
2. Click on convert and watch this result display in the box below it
3. Click Reset to reset the micrometer value
micrometermeters
µm m | 675 | 2,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-10 | latest | en | 0.871832 |
https://web2.0calc.com/questions/help_21269 | 1,716,512,856,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058675.22/warc/CC-MAIN-20240523235012-20240524025012-00741.warc.gz | 533,473,941 | 5,775 | +0
# help
-2
19
2
+1347
Help I see only one square here
Four of the points in the grid below form the vertices of a square \$S.\$ What is the sum of all the possible areas of \$S?\$ (Assume that neighboring points are one unit apart.)
Aug 16, 2023
edited by sandwich Aug 16, 2023
#1
+121
-1
```
. . . . . . . .
. . . . . . . .
. . A . . . . .
. . . . . B . .
. . . . . . . .
. . . C . . . .
. . . . . . . .
```
In this grid, points A, B, and C form a square. Let's assume that each side of the square has a length of 1 unit. Now, we can calculate the areas of all the possible squares that can be formed by these points:
1. Square ABC: Since the side length is 1 unit, the area is 1 * 1 = 1 square unit.
2. Square ABDE (formed by extending the sides of ABC): The area is 2 * 2 = 4 square units.
3. Square AFGH (formed by extending the sides in another direction): The area is 2 * 2 = 4 square units.
4. Square AIJK (formed by extending the sides in both directions): The area is 3 * 3 = 9 square units.
The sum of all the possible areas is: 1 + 4 + 4 + 9 = 18 square units.
So, the sum of all the possible areas of squares that can be formed using the given points is 18 square units.
Aug 16, 2023 | 386 | 1,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-22 | latest | en | 0.896202 |
http://mathhelpforum.com/calculus/200951-anti-derivative-please.html | 1,527,082,017,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865651.2/warc/CC-MAIN-20180523121803-20180523141803-00154.warc.gz | 183,795,202 | 9,507 | sin y dy/ cos^2 y
2. ## Re: anti derivative please
\displaystyle \displaystyle \begin{align*} \int{\frac{\sin{y}}{\cos^2{y}}\,dy} = -\int{\frac{-\sin{y}}{\cos^2{y}}\,dy} \end{align*}
Let \displaystyle \displaystyle \begin{align*} u = \cos{y} \implies du = -\sin{y}\,dy \end{align*} and the integral becomes
\displaystyle \displaystyle \begin{align*} -\int{\frac{1}{u^2}\,du} &= -\int{u^{-2}\,du} \\ &= -\left(\frac{u^{-1}}{-1}\right) + C \\ &= \frac{1}{u} + C \\ &= \frac{1}{\cos{y}} + C \\ &= \sec{y} + C \end{align*}
3. ## Re: anti derivative please
Originally Posted by sluggerbroth
sin y dy/ cos^2 y
ever wonder why you had to learn all those identities ?
$\displaystyle \frac{\sin{y}}{\cos^2{y}} = \frac{1}{\cos{y}} \cdot \frac{\sin{y}}{\cos{y}} = \sec{y} \cdot \tan{y}$
$\displaystyle \int \sec{y} \cdot \tan{y} \, dy = \sec{y} + C$ | 344 | 846 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-22 | latest | en | 0.480278 |
http://stackoverflow.com/questions/4716647/indexing-one-dimensional-numpy-array-as-matrix | 1,462,288,649,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860121561.0/warc/CC-MAIN-20160428161521-00103-ip-10-239-7-51.ec2.internal.warc.gz | 269,775,391 | 20,422 | # Indexing one-dimensional numpy.array as matrix
I am trying to index a numpy.array with varying dimensions during runtime. To retrieve e.g. the first row of a n*m array `a`, you can simply do
``````a[0,:]
``````
However, in case a happens to be a 1xn vector, this code above returns an index error:
IndexError: too many indices
As the code needs to be executed as efficiently as possible I don't want to introduce an `if` statement. Does anybody have a convenient solution that ideally doesn't involve changing any data structure types?
-
You only have 1- and 2D arrays? – Paul Jan 17 '11 at 18:50
Would simply reshaping the array to be a 2d 1xn array instead of a 1d n-length array count as "changing the data structure type"? – Josh Bleecher Snyder Jan 17 '11 at 18:51
All these are 2D arrays (mxn) theoretically, some just happend to be 1xn arrays, e.g. m=1. In fact they represent conditional probability tables and the case m=1 corresponds to a variable that doesn't have any dependencies. – Alain Jan 17 '11 at 18:56
## 2 Answers
Just use `a[0]` instead of `a[0,:]`. It will return the first line for a matrix and the first entry for a vector. Is this what you are looking for?
If you want to get the whole vector in the one-dimensional case instead, you can use `numpy.atleast_2d(a)[0]`. It won't copy your vector -- it will just access it as a two-dimensional 1 x n-array.
-
I didn't know about atleast_2d; handy. +1 – Josh Bleecher Snyder Jan 17 '11 at 19:19
I can second that, numpy.atleast_2d is very helpful and exactly what I was looking for. Thanks a lot. – Alain Jan 17 '11 at 19:33
From the 'array' or 'matrix'? Which should I use? section of the Numpy for Matlab Users wiki page:
For array, the vector shapes 1xN, Nx1, and N are all different things. Operations like A[:,1] return a rank-1 array of shape N, not a rank-2 of shape Nx1. Transpose on a rank-1 array does nothing.
Here's an example showing that they are not the same:
``````>>> import numpy as np
>>> a1 = np.array([1,2,3])
>>> a1
array([1, 2, 3])
>>> a2 = np.array([[1,2,3]]) // Notice the two sets of brackets
>>> a2
array([[1, 2, 3]])
>>> a3 = np.array([[1],[2],[3]])
>>> a3
array([[1],
[2],
[3]])
``````
So, are you sure that all of your arrays are 2d arrays, or are some of them 1d arrays?
If you want to use your command of `array[0,:]`, I would recommend actually using 1xN 2d arrays instead of 1d arrays. Here's an example:
``````>>> a2 = np.array([[1,2,3]]) // Notice the two sets of brackets
>>> a2
array([[1, 2, 3]])
>>> a2[0,:]
array([1, 2, 3])
>>> b2 = np.array([[1,2,3],[4,5,6]])
>>> b2
array([[1, 2, 3],
[4, 5, 6]])
>>> b2[0,:]
array([1, 2, 3])
``````
- | 830 | 2,672 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2016-18 | longest | en | 0.903311 |
http://wiki.nmr-relax.com/DPL94 | 1,550,414,292,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247481994.42/warc/CC-MAIN-20190217132048-20190217154048-00488.warc.gz | 289,805,556 | 9,828 | # DPL94
The Davis et al., 1994 2-site off-resonance fast exchange relaxation dispersion model for R1rho-type data. It extends the M61 model to off-resonance data, hence it collapses to this model for on-resonance data. The model is labelled as DPL94 in relax.
## Equation
$\mathrm{R}_{1\rho}= \mathrm{R}_1\cos^2\theta + \left( \mathrm{R}_{1\rho}{´} + \frac{\Phi_\textrm{ex} \textrm{k}_\textrm{ex}}{\textrm{k}_\textrm{ex}^2 + \omega_\textrm{e}^2} \right) \sin^2\theta$
## Parameters
The DPL94 model has the parameters {R', ..., Φex, kex}.
## Essentials
Note R1 should be provided in rad/s, the SI default unit for this relaxation rate.
It is essential to read in R1 values before starting a calculation:
relax_data.read(ri_id='R1', ri_type='R1', frq=cdp.spectrometer_frq_list[0], file='R1_values.txt', mol_name_col=1, res_num_col=2, res_name_col=3, spin_num_col=4, spin_name_col=5, data_col=6, error_col=7)
Where the data could be stored like
# mol_name res_num res_name spin_num spin_name value error
None 13 L None N 1.323940 0.146870
None 15 R None N 1.344280 0.140560
None 16 T None N 1.715140 0.136510
## Parameter name space in relax
Which means:
1. R' = spin.r2 (Fitted)
2. R = spin.r2eff (Back calculated)
3. Φex = spin.phi_ex (Fitted)
4. kex = spin.kex (Fitted)
5. R1 = spin.ri_data['R1'] (Loaded)
## Equation - re-written forms
• Evenäs, J., Malmendal, A. and Akke, M. (2001). Dynamics of the transition between open and closed conformations in a calmodulin C-terminal domain mutant. Structure, 9(3), 185-195. (DOI: 10.1016/S0969-2126(01)00575-5)
• Kempf, J. G. and Loria, J. P. (2004). Measurement of intermediate exchange phenomena. Methods Mol. Biol., 278, 185-231. (DOI: 10.1385/1-59259-809-9:185)
• Massi, F., Grey, M. J., Palmer, 3rd, A. G. (2005). Microsecond timescale backbone conformational dynamics in ubiquitin studied with NMR R1ρ relaxation experiments Protein science, 14(3), 735-742. (DOI: 10.1110/ps.041139505)
• Palmer, 3rd, A. G., Kroenke, C. D., and Loria, J. P. (2001). Nuclear magnetic resonance methods for quantifying microsecond-to-millisecond motions in biological macromolecules. Methods Enzymol., 339, 204-238. (DOI: 10.1016/S0076-6879(01)39315-1)
• Palmer, 3rd, A. G. and Massi, F. (2006). Characterization of the dynamics of biomacromolecules using rotating-frame spin relaxation NMR spectroscopy. Chem. Rev., 106(5), 1700-1719. (DOI: 10.1021/cr0404287)
• Trott, O. and Palmer, 3rd, A. G. (2002). R1rho relaxation outside of the fast-exchange limit. J. Magn. Reson., 154(1), 157-160. (DOI: 10.1006/jmre.2001.2466)
Different graphs.
## The R1ρ: R2 or R2,eff as function of effective field in rotating frame: ωe
### Discussion
It is clear that there is no real name for the pseudo-parameter. It looks like that Reff was Art's original way of denoting this and that he has now changed to R2 instead.
But if one look at the reference for the TP02 dispersion model, one will see yet another notation:
Here R2 does not contain the Rex contribution. Also, Reff is absent of Rex.
But in Art's Protein Science paper (Ref [5]), the definition R2 = R20 + Rex is used. The MP05 model reference also does not use Reff.
The Reff parameter name is confusing and it seems to have been dropped from 2005 onwards. The Reff name appears to be specific to Art Palmer's group and as he himself has dropped it, then it would be best to avoid it too.
Ref [2], Equation 27. Here the calculated value is noted as: R_eff: $R_{\text{eff}} = R^{0}_2 + R_{ex} = R_{1\rho}' + R_{ex} = R_{1\rho} / \sin^2(\theta) - R_1 / \tan^2(\theta)$
Ref [3], Equation 20. Figure 11+16, would be the reference. Here the calculated value is noted as: R_2: $R_{2} = R_{1\rho} / \sin^2(\theta) - R_1 / \tan^2(\theta)$.
Ref [4], Equation 43. $R_{\text{eff}} = R_{1\rho} / \sin^2(\theta) - R_1 / \tan^2(\theta)$
Ref [5], Material and Methods, page 740. Figure 4 would be the wished graphs. Here the calculated value is noted as: R_2: $R_{2} = R^{0}_2 + R_{ex}$
The following suggestions for the definition of the pseudo-parameters, which can be extracted, is then
$R_2 = R^{0}_2 + R_{ex} = R_{1\rho}' + R_{ex} = R_{1\rho} / \sin^2(\theta) - R_1 / \tan^2(\theta) = \frac{R_{1\rho} - R_1\cos^2(\theta)}{\sin^2(\theta)}$
## Reference
The reference for the DPL94 model is:
• Davis, D. G., Perlman, M. E., and London, R. E. (1994). Direct measurements of the dissociation-rate constant for inhibitor-enzyme complexes via the T1rho and T2 (CPMG) methods. J. Magn. Reson., 104(3), 266-275. (DOI: 10.1006/jmrb.1994.1084)
## Related models
The DPL94 model is simply the extension of the M61 model for off-resonance data. | 1,578 | 4,775 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-09 | latest | en | 0.624668 |
https://metanumbers.com/95997 | 1,638,110,584,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358560.75/warc/CC-MAIN-20211128134516-20211128164516-00276.warc.gz | 459,903,989 | 7,398 | # 95997 (number)
95,997 (ninety-five thousand nine hundred ninety-seven) is an odd five-digits composite number following 95996 and preceding 95998. In scientific notation, it is written as 9.5997 × 104. The sum of its digits is 39. It has a total of 3 prime factors and 8 positive divisors. There are 58,160 positive integers (up to 95997) that are relatively prime to 95997.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 39
• Digital Root 3
## Name
Short name 95 thousand 997 ninety-five thousand nine hundred ninety-seven
## Notation
Scientific notation 9.5997 × 104 95.997 × 103
## Prime Factorization of 95997
Prime Factorization 3 × 11 × 2909
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 95997 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 95,997 is 3 × 11 × 2909. Since it has a total of 3 prime factors, 95,997 is a composite number.
## Divisors of 95997
1, 3, 11, 33, 2909, 8727, 31999, 95997
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 139680 Sum of all the positive divisors of n s(n) 43683 Sum of the proper positive divisors of n A(n) 17460 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 309.834 Returns the nth root of the product of n divisors H(n) 5.49811 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 95,997 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 95,997) is 139,680, the average is 17,460.
## Other Arithmetic Functions (n = 95997)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 58160 Total number of positive integers not greater than n that are coprime to n λ(n) 14540 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 9234 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 58,160 positive integers (less than 95,997) that are coprime with 95,997. And there are approximately 9,234 prime numbers less than or equal to 95,997.
## Divisibility of 95997
m n mod m 2 3 4 5 6 7 8 9 1 0 1 2 3 6 5 3
The number 95,997 is divisible by 3.
## Classification of 95997
• Arithmetic
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
• Sphenic
## Base conversion (95997)
Base System Value
2 Binary 10111011011111101
3 Ternary 11212200110
4 Quaternary 113123331
5 Quinary 11032442
6 Senary 2020233
8 Octal 273375
10 Decimal 95997
12 Duodecimal 47679
20 Vigesimal bjjh
36 Base36 222l
## Basic calculations (n = 95997)
### Multiplication
n×y
n×2 191994 287991 383988 479985
### Division
n÷y
n÷2 47998.5 31999 23999.2 19199.4
### Exponentiation
ny
n2 9215424009 884653058591973 84924039665653632081 8152453035783751718879757
### Nth Root
y√n
2√n 309.834 45.7881 17.6021 9.91863
## 95997 as geometric shapes
### Circle
Diameter 191994 603167 2.89511e+10
### Sphere
Volume 3.70563e+15 1.15804e+11 603167
### Square
Length = n
Perimeter 383988 9.21542e+09 135760
### Cube
Length = n
Surface area 5.52925e+10 8.84653e+14 166272
### Equilateral Triangle
Length = n
Perimeter 287991 3.9904e+09 83135.8
### Triangular Pyramid
Length = n
Surface area 1.59616e+10 1.04257e+14 78381.2
## Cryptographic Hash Functions
md5 df9e078b8e81fcbb99278664413aab89 05c6744ac7ccdafd2724906e060216b55908b683 997bbf549f3a69e2de4e5647b5ded4fedc42b71cefd218cae1ff71b06422e21b cd3ba9311571bba10bf6f3c6820ff26cb729f3ecb1b4118f571804120a9cf37c08a075bebeb1ff7eb8882e39c1b12bac28724cf0f58b1c732918aeb73234f9e5 516bb456d6ede5b0f62d806980d923097af399e8 | 1,464 | 4,190 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2021-49 | latest | en | 0.798064 |
https://www.ni.com/documentation/en/labview/latest/analysis-node-ref/create-real-matrix-from-eigenvalues/ | 1,627,186,121,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151563.91/warc/CC-MAIN-20210725014052-20210725044052-00256.warc.gz | 941,279,708 | 13,882 | # Create Real Matrix from Eigenvalues (G Dataflow)
Generates a real matrix from a specified set of eigenvalues.
## eigenvalues
Eigenvalues from which to create the real matrix. Eigenvalues must be real or conjugate pairs.
## error in
Error conditions that occur before this node runs.
The node responds to this input according to standard error behavior.
Standard Error Behavior
Many nodes provide an error in input and an error out output so that the node can respond to and communicate errors that occur while code is running. The value of error in specifies whether an error occurred before the node runs. Most nodes respond to values of error in in a standard, predictable way.
error in does not contain an error error in contains an error
If no error occurred before the node runs, the node begins execution normally.
If no error occurs while the node runs, it returns no error. If an error does occur while the node runs, it returns that error information as error out.
If an error occurred before the node runs, the node does not execute. Instead, it returns the error in value as error out.
Default: No error
## matrix
Real matrix whose eigenvalues you specified.
## error out
Error information.
The node produces this output according to standard error behavior.
Standard Error Behavior
Many nodes provide an error in input and an error out output so that the node can respond to and communicate errors that occur while code is running. The value of error in specifies whether an error occurred before the node runs. Most nodes respond to values of error in in a standard, predictable way.
error in does not contain an error error in contains an error
If no error occurred before the node runs, the node begins execution normally.
If no error occurs while the node runs, it returns no error. If an error does occur while the node runs, it returns that error information as error out.
If an error occurred before the node runs, the node does not execute. Instead, it returns the error in value as error out.
## Algorithm for Calculating the Eigenvalues
This node generates the output matrix in the following form:
${\left[\begin{array}{cccccc}0& 1& 0& \cdots & \cdots & 0\\ 0& 0& 1& 0& \cdots & 0\\ ⋮& \ddots & \ddots & \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & 0& 1& 0\\ 0& \cdots & \cdots & \cdots & 0& 1\\ -{a}_{0}& -{a}_{1}& -{a}_{2}& \cdots & -{a}_{n-2}& -{a}_{n-1}\end{array}\right]}_{n×n}$
where n is the length of the input eigenvalues and a0, a1, ..., an-1 are the coefficients of the polynomial P(x).
The following equation defines P(x):
$P\left(x\right)=\left(x-{\lambda }_{0}\right)\left(x-{\lambda }_{1}\right)\dots \left(x-{\lambda }_{n-1}\right)={a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}+\dots +{a}_{n-1}{x}^{n-1}+{x}^{n}$
where λ0, λ1, ..., λn - 1 are the elements in eigenvalues.
Where This Node Can Run:
Desktop OS: Windows
FPGA: Not supported
Web Server: Not supported in VIs that run in a web application | 777 | 2,956 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-31 | latest | en | 0.768596 |
http://xlfxs.com/tally-chart-in-functions/ | 1,680,400,004,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950373.88/warc/CC-MAIN-20230402012805-20230402042805-00108.warc.gz | 97,611,933 | 18,190 | Excel functions and formulas
Tally chart in functions
Using a single function can make a graphical view to your number
Function is REPT along with IF and INT to show that number in graphical view, formula used here is:
`=IF(B4<5,REPT("|",B4),REPT(REPT("|",5)&"-",INT(B4/5))&REPT("|",B4-INT(B4/5)*5))` | 94 | 304 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-14 | longest | en | 0.830386 |
http://metamath.tirix.org/mpeuni/pm2.61ne | 1,721,888,251,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518579.47/warc/CC-MAIN-20240725053529-20240725083529-00251.warc.gz | 21,321,657 | 1,921 | # Metamath Proof Explorer
## Theorem pm2.61ne
Description: Deduction eliminating an inequality in an antecedent. (Contributed by NM, 24-May-2006) (Proof shortened by Andrew Salmon, 25-May-2011) (Proof shortened by Wolf Lammen, 25-Nov-2019)
Ref Expression
Hypotheses pm2.61ne.1 ( 𝐴 = 𝐵 → ( 𝜓𝜒 ) )
pm2.61ne.2 ( ( 𝜑𝐴𝐵 ) → 𝜓 )
pm2.61ne.3 ( 𝜑𝜒 )
Assertion pm2.61ne ( 𝜑𝜓 )
### Proof
Step Hyp Ref Expression
1 pm2.61ne.1 ( 𝐴 = 𝐵 → ( 𝜓𝜒 ) )
2 pm2.61ne.2 ( ( 𝜑𝐴𝐵 ) → 𝜓 )
3 pm2.61ne.3 ( 𝜑𝜒 )
4 3 1 syl5ibr ( 𝐴 = 𝐵 → ( 𝜑𝜓 ) )
5 2 expcom ( 𝐴𝐵 → ( 𝜑𝜓 ) )
6 4 5 pm2.61ine ( 𝜑𝜓 ) | 298 | 569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-30 | latest | en | 0.604269 |
https://www.jiskha.com/display.cgi?id=1184204897 | 1,516,278,373,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887253.36/warc/CC-MAIN-20180118111417-20180118131417-00211.warc.gz | 910,881,502 | 3,962 | # math 117/ algebra
posted by .
Fungicides account for 1/10 of the pesticides used in the US. Insecticides account for 1/4 of all pesticides used in the US. The two ratio of herbicides to insecticides used in the US can be written 1/10 รท 1/4 . Write this ratio in simplest form.
see the other post about division is the reciprocal of multiplication.
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More Similar Questions | 566 | 2,332 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2018-05 | latest | en | 0.891033 |
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