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or more commonly, solved for the voltage and expressed as: V = I×R (3.3) This is called Ohm's law, and is named in honor of Georg Ohm, a researcher from the early nineteenth century. It is, along with Kirchhoff's laws that we shall see shortly, one of the most important and useful equations available for the analysis o... | DCElectricalCircuitAnalysis_Page_79_Chunk1001 |
Computer Simulation Computer Simulation To verify the results of the preceding example, the circuit is entered into a simulator, as shown in Figure 3.12. Multisim™ is used here, although any quality circuit simulator will do. Virtual DMMs are used to measure the current and voltage. Remember, current is the rate of cha... | DCElectricalCircuitAnalysis_Page_80_Chunk1002 |
Example 3.4 Refer to the battery and lamp circuit shown back in Figure 3.3. Determine the resistance of the lamp if the current flowing through it is 300 mA and the battery voltage is 6 volts. Also determine the power dissipated by the lamp. R = V I R = 6 V 0.3A R =20Ω From power law, P = I 2×R P = 0.3A 2×20Ω P =1.8wat... | DCElectricalCircuitAnalysis_Page_81_Chunk1003 |
Example 3.6 Determine the resistor value required in the circuit of Figure 3.14 in order for the source to generate 24 watts of power. First, note that power generated must equal power dissipated. Thus, the power generated by the source must equal the power dissipated by the resistor. From power law, P = I 2×R R = P I ... | DCElectricalCircuitAnalysis_Page_82_Chunk1004 |
supplied voltage times the ratio of the resistor of interest to the total resistance: VRx = E ∙ RX /RTOTAL (3.5) In fact, Equation 3.5 is just a combination of two Ohm's law calculations into a single formula. The circulating current is equal to E/RTOTAL. This current is then multiplied by the resistor of interest (RX)... | DCElectricalCircuitAnalysis_Page_83_Chunk1005 |
Often, there will be more than one way to solve a given problem. We shall refer to these as solution paths. No particular path is “more correct” than any other, although some paths may be shorter or easier in a given case, as will be illustrated next. Example 3.7 Determine the voltage Vb (i.e., the voltage across the 1... | DCElectricalCircuitAnalysis_Page_84_Chunk1006 |
Computer Simulation Computer Simulation Let's verify the preceding example using a computer simulation. Instead of using virtual instruments, we shall use a more direct route, namely a listing of node voltages. A node voltage is the voltage at a given point with respect to some other point, normally ground, as in Vb in... | DCElectricalCircuitAnalysis_Page_85_Chunk1007 |
Example 3.8 Determine the voltage Va (i.e., the voltage across the current source) in the circuit of Figure 3.18. Given the direction of the current source, the current will flow counterclockwise. This produces voltage drops (plus to minus) left to right across the 10 Ω, bottom to top across the 3 Ω, and right to left ... | DCElectricalCircuitAnalysis_Page_86_Chunk1008 |
in a simulator and it is usually called Monte Carlo Analysis or something similar. Basically, the simulator computes a series of simulation runs, each with component values chosen randomly within the specified tolerances of the chosen components, mimicking the effect of pulling the components out of a parts bins. To il... | DCElectricalCircuitAnalysis_Page_87_Chunk1009 |
The Importance of Polarity The Importance of Polarity A key item of importance when analyzing series circuits, or indeed any electrical circuit, is noting the polarity of the voltages. This was apparent in Example 3.8. Determining polarity for voltage sources is easy as their polarity is fixed (positive is always at th... | DCElectricalCircuitAnalysis_Page_88_Chunk1010 |
to point a, or Vca, we will wind up with the same voltage magnitude but the sign will flip. In the laboratory, the first point letter (the a in Vac) is where the red lead of the voltmeter is connected and the second letter is where the black lead is connected. As a reminder, if a single connection point is used, as in ... | DCElectricalCircuitAnalysis_Page_89_Chunk1011 |
source, it is still considered a voltage drop because the polarity is + to −. The polarity on the 400 Ω resistor is the same, so we add in its 8.4 volts for 9.9 volts total. That is, node b is 9.9 volts above ground. For Vca we proceed similarly. Starting at node c we move through the 1.5 volt source, but this time the... | DCElectricalCircuitAnalysis_Page_90_Chunk1012 |
I = ETotal RTotal I = 26 V 13 k Ω I = 2 mA V 10k = I×R1 V 10k = 2 mA×10 k Ω V 10k = 20 volts Similarly, V2k = 4 volts and V1k = 2 volts with the polarities shown. To determine Vbe, we can start at node b and move to node e by either going to the left or right side. We shall do both and compare the results. Going to the... | DCElectricalCircuitAnalysis_Page_91_Chunk1013 |
V d = E R4 RT V d = 36 V 4.5 kΩ 12 kΩ V d = 13.5 volts V b = E R2+R3+R4 RT V b = 36V 8k Ω 12 k Ω V b = 24 volts V bd = E R2+R3 RT V bd = 36 V 3.5k Ω 12 kΩ V bd = 10.5 volts There is another way to find Vbd here, and that's to notice that by definition Vbd = Vb − Vd. Therefore, Vbd = 24 volts − 13.5 volts, or 10.5 volts... | DCElectricalCircuitAnalysis_Page_92_Chunk1014 |
A cutaway view of a rotary panel mount pot with its back casing removed is shown in Figure 3.29. Most circular potentiometers utilize a 270° rotation from end to end, also known as “3/4 turn”. Some precision pots used for calibration may use an internal worm gear to move the wiper and achieve 20 or more turns to shuttl... | DCElectricalCircuitAnalysis_Page_93_Chunk1015 |
As a result, potentiometers are very useful as controllable voltage dividers (hence the name). Sometimes it is desirable to control two signals simultaneously from a single knob (e.g., one volume knob for a hi-fi stereo controls both the left and right channels). For this application, multiple resistive elements can be... | DCElectricalCircuitAnalysis_Page_94_Chunk1016 |
V c(min) = E Ri RT V c(min) = 16V 1 kΩ 8 kΩ V c(min) = 2 volts For the midway case, the pot can be treated as two 2.5 kΩ resistors. V c(mid )= E Ri RT V c(mid )= 16 V 3.5k Ω 8 kΩ V c(mid )= 7 volts Note that the mid-position achieves a voltage midway between the extremes. Thus the pot allows a range of voltage adjustme... | DCElectricalCircuitAnalysis_Page_95_Chunk1017 |
I (min)= E RT I (min) = 50 V 11 kΩ I (min) = 4.545 mA 3.9 Summary 3.9 Summary Electrons are the charge carrier in metals. Their direction of travel, called electron flow, runs from the negative terminal of a voltage source to the positive terminal. The more commonly used conventional current flow runs from the positive... | DCElectricalCircuitAnalysis_Page_96_Chunk1018 |
are used typically to adjust a voltage level. If the wiper and only one of the end terminals is used, the device is called a rheostat. A common application for a rheostat is the control of current. Finally, computer simulation tools may be used to build virtual circuits and determine various parameters such as voltages... | DCElectricalCircuitAnalysis_Page_97_Chunk1019 |
3. For the circuit of Figure 3.34, determine the power dissipated in the resistor. 4. For the circuit of Figure 3.35, determine the power dissipated in the resistor. 5. Determine the voltage at the open terminals of Figure 3.36. 6. Determine the voltage at the open terminals of Figure 3.37. 7. Determine the voltage at ... | DCElectricalCircuitAnalysis_Page_98_Chunk1020 |
9. Determine the equivalent resistance of circuit shown in Figure 3.40. 10. Determine the equivalent resistance of circuit shown in Figure 3.41. 11. For the circuit of Figure 3.42, determine the circulating current. 12. For the circuit of Figure 3.42, determine the voltages across each resistor and find Vab. 13. Given ... | DCElectricalCircuitAnalysis_Page_99_Chunk1021 |
14. For the circuit of Figure 3.43, determine the circulating current. 15. Given the circuit of Figure 3.43, determine the voltages across each resistor and find Vba. 16. For the circuit of Figure 3.43, determine the power dissipated by each resistor and the power delivered by the source. 17. For the circuit of Figure ... | DCElectricalCircuitAnalysis_Page_100_Chunk1022 |
22. For the circuit of Figure 3.46, determine the circulating current and indicate all voltage polarities. 23. Given the circuit of Figure 3.46, determine the voltages across each resistor and find Vc, Vac, and Va. 24. For the circuit of Figure 3.46, determine the power dissipated by the 10 Ω resistor. 25. For the circ... | DCElectricalCircuitAnalysis_Page_101_Chunk1023 |
29. Given the circuit of 3.49, determine the voltages across each resistor and find Vb, Vc, and Vac. 30. Referring to the circuit of Figure 3.49, determine the circulating current and indicate all voltage polarities. 31. Given the circuit of 3.50, determine the circulating current and indicate all voltage polarities. 3... | DCElectricalCircuitAnalysis_Page_102_Chunk1024 |
35. Using the voltage divider rule, determine the voltages Vb, Vc and Vac for the circuit shown in Figure 3.52. 36. Using the voltage divider rule, determine the voltages Vb, Vc and Vbd for the circuit shown in Figure 3.53. 37. For the circuit of Figure 3.53, determine Vb if the 4 kΩ resistor is accidentally shorted. H... | DCElectricalCircuitAnalysis_Page_103_Chunk1025 |
39. Given the circuit shown in Figure 3.54, find the voltage drop across the resistor. 40. Given the circuit shown in Figure 3.55, find the voltage drops across the resistor. 41. Find the voltage drops across the resistors in the circuit of Figure 3.56. 42. Find the voltage drops across the resistors in the circuit of ... | DCElectricalCircuitAnalysis_Page_104_Chunk1026 |
43. Find the voltage drops across the resistors in the circuit of Figure 3.58. 44. Find the voltage drops across the resistors in the circuit of Figure 3.59. 45. The circuit of Figure 3.60 uses a linear taper potentiometer. Determine Vb when the wiper arm is at position a, position b, and at the halfway point. 46. What... | DCElectricalCircuitAnalysis_Page_105_Chunk1027 |
50. For the circuit of Figure 3.41, find the value of a series voltage source that would generate 1 mA of current if it was connected across the terminals. 51. Determine values for the resistors in Figure 3.61 such that R1 is four times the size of R2 and R2 is three times the size of R3, with the total resistance equa... | DCElectricalCircuitAnalysis_Page_106_Chunk1028 |
54. For the circuit shown in Figure 3.64, determine values for R1 and R2 such that Vab is 6 volts if I is a 2 mA source and the total voltage drop is 24 volts. 55. Consider the circuit of Figure 3.53. Is it possible to add a fifth resistor such that the circulating current is 0.1 mA? If so, what is that resistor value?... | DCElectricalCircuitAnalysis_Page_107_Chunk1029 |
59. Given the circuit of Figure 3.66, determine the required source voltage if R1 is 1 kΩ, R2 is 1 kΩ, the power dissipation in R1 is 4 mW and the power dissipation in R3 is 2 mW. 60. Given the circuit of Figure 3.67, determine Vc, Vdb and Vce. 61. Given the circuit of Figure 3.67, determine Vac, Veb and Vd. 62. Refer ... | DCElectricalCircuitAnalysis_Page_108_Chunk1030 |
Source: www.xkcd.com This page intentionally left not blank. 109 | DCElectricalCircuitAnalysis_Page_109_Chunk1031 |
4 4 Parallel Resistive Circuits Parallel Resistive Circuits 4.0 Chapter Learning Objectives 4.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Identify parallel resistive circuits that include a single voltage source or one or more current sources. • Compute equivalent resistance of... | DCElectricalCircuitAnalysis_Page_110_Chunk1032 |
4.2 The Parallel Connection 4.2 The Parallel Connection Consider the generic layout of Figure 4.1, each component represented by a box. Although it may be drawn oddly, there are only two common connection nodes in this configuration: one along the top and the other along the bottom (if you're having difficulty seeing t... | DCElectricalCircuitAnalysis_Page_111_Chunk1033 |
When it comes to current sources, they may simply be added together, however, like series voltage sources, polarity is important. Referring to the three parallel current sources in Figure 4.4, the middle and right sources are pumping current into the top node while the left source is feeding the bottom node. You can al... | DCElectricalCircuitAnalysis_Page_112_Chunk1034 |
Among other things, Equation 4.3 tells us that the equivalent resistance of a group of parallel resistances will always be smaller than the smallest resistance in that group. It also indicates that if we have N identical resistors of value R, the equivalent will be R/N (e.g., 3 parallel 3.6 kΩ resistors are equivalent ... | DCElectricalCircuitAnalysis_Page_113_Chunk1035 |
Thus the equivalent will be N/(N+1), or 3/4ths, of the smaller resistor, yielding 6 kΩ. Similarly, a 180 Ω in parallel with a 90 Ω would yield 2/3rds of 90 Ω, or 60 Ω. Example 4.1 A group of resistors is placed in parallel as shown in Figure 4.6. Determine the equivalent parallel value. There are several solution paths... | DCElectricalCircuitAnalysis_Page_114_Chunk1036 |
4.4 Kirchhoff's Current Law (KCL) 4.4 Kirchhoff's Current Law (KCL) Just as Kirchhoff's voltage law is a key element in understanding series circuits, Kirchhoff's current law (KCL) is the operative rule for parallel circuits. It states that the sum of all currents entering and exiting a node must sum to zero. Alternate... | DCElectricalCircuitAnalysis_Page_115_Chunk1037 |
The current through each of the resistors must be this voltage divided by the corresponding resistance. For the current through R1: I R1 = V R1 I R1 = I Total R1 R2 R1(R1+R2) I R1 = I Total R2 R1+R2 Similarly, for the current through R2: I R2 = V R2 I R2 = I Total R1 R2 R2(R1+R2) I R2 = I Total R1 R1+R2 Thus, the curre... | DCElectricalCircuitAnalysis_Page_116_Chunk1038 |
I R1 = I Total R2 R1+R2 I 15 = 10 A 5Ω 15Ω+5Ω I 15 = 2.5 A I R2 = ITotal R1 R1+R2 I 5 = 10 A 15Ω 15Ω+5Ω I 5 = 7.5 A At this point it is worth taking a moment to verify the voltage polarities and current directions. Given the direction of the source, the currents through the two resistors must be flowing from top to bot... | DCElectricalCircuitAnalysis_Page_117_Chunk1039 |
• If the problem concerns determining resistance values, the basic idea will be to use these rules in reverse. For example, if a resistor value is needed to set a specific voltage, the equivalent parallel resistance can be determined from this voltage and the given current source. The conductance values of the other pa... | DCElectricalCircuitAnalysis_Page_118_Chunk1040 |
I Total = E RParallel I Total = 12 V 240Ω I Total = 50 mA Now CDR can be used to find the currents through the two resistors. I R1 = I Total R2 R1+R2 I 400 = 50 mA 600Ω 400Ω+600Ω I 400 = 30 mA I R2 = ITotal R1 R1+R2 I 600 = 50 mA 400Ω 400Ω+600Ω I 600 = 20 mA The voltage polarities and current directions are illustrated... | DCElectricalCircuitAnalysis_Page_119_Chunk1041 |
It turns out that a similar situation exists for voltmeters. Ideally, voltmeters present a very high internal resistance which, when placed across a resistor, has minimal impact. This effect cannot always be ignored, especially when measuring across resistors having large values, as the current divider rule will come i... | DCElectricalCircuitAnalysis_Page_120_Chunk1042 |
I R2 = I Total −I R1 I 100k = 2 mA −1.7857 mA I 100k ≈0.2143 mA The polarities and directions are shown in Figure 4.14. Note that the currents are flowing up through the resistors producing voltage drops of + to − from ground up to the top. This means that the voltage of the top node is negative with respect to ground.... | DCElectricalCircuitAnalysis_Page_121_Chunk1043 |
Depending on component values, the results could be much worse or hardly noticeable at all. As a general rule, make sure the voltmeter's internal resistance is at least 10 times (and preferably, 100 times) larger than any resistance it is placed across in order to avoid loading errors caused by unwanted current divisio... | DCElectricalCircuitAnalysis_Page_122_Chunk1044 |
I R1 = I Total R2 R1+R2 I 200 = 1.4 A 40Ω 200Ω+40Ω I 200 ≈233.33 mA As a crosscheck, Ohm's law indicates the system voltage must be 200 Ω times 233.33 mA, or approximately 46.667 volts. Via Ohm's law, the current through the 60 Ω must be 46.667 volts divided by 60 Ω, or approximately 777.78 mA. Similarly, the current t... | DCElectricalCircuitAnalysis_Page_123_Chunk1045 |
And finally, R4 = 1/G4, thus R4 = 300 Ω. The smaller the value of this fourth resistor, the more current it will siphon off from the other three resistors, thus reducing the system voltage further. Time for an example involving power. Example 4.7 A parallel network is shown in Figure 4.18. Determine the total power dis... | DCElectricalCircuitAnalysis_Page_124_Chunk1046 |
4.6 Current Limiting: Fuses and Circuit Breakers 4.6 Current Limiting: Fuses and Circuit Breakers Most electrical and electronic devices are designed to run off of a constant voltage source. Examples include battery powered devices and the numerous items that connect to standard residential and commercial electrical sy... | DCElectricalCircuitAnalysis_Page_125_Chunk1047 |
While some electronic systems use active current limiting schemes14, the more common approach is a fuse or circuit breaker. These devices are placed between the voltage source and the various loads, and see their combined current. If this current becomes too great, the device interrupts the current flow by opening the ... | DCElectricalCircuitAnalysis_Page_126_Chunk1048 |
4.7 Summary 4.7 Summary In this chapter we have introduced and examined parallel circuits using either one or more current sources or a single voltage source, along with two or more resistors. The hallmark of a parallel configuration is that all components are connected to just two nodes. This means that all elements i... | DCElectricalCircuitAnalysis_Page_127_Chunk1049 |
internal resistance that is lat east an order of magnitude (preferably two) larger than the resistors they are placed across in order to limit undesired current divider effects. Similarly, ammeters should have an internal resistance at least an order of magnitude (preferably two) smaller than the resistors with which t... | DCElectricalCircuitAnalysis_Page_128_Chunk1050 |
2. Determine the effective resistance of the network shown in Figure 4.24. 3. Determine the effective resistance of the network shown in Figure 4.25. 4. Find the effective source current of the network shown in Figure 4.26. 5. Determine the effective source current of the network shown in Figure 4.27. 129 Figure 4.24 F... | DCElectricalCircuitAnalysis_Page_129_Chunk1051 |
6. Find the source and resistor currents for the circuit of Figure 4.28. 7. Determine the source and resistor currents for the circuit of Figure 4.29. 8. Find the source and resistor currents for the circuit of Figure 4.30. 9. For the circuit of Figure 4.31, determine the source and resistor currents. 130 Figure 4.28 F... | DCElectricalCircuitAnalysis_Page_130_Chunk1052 |
10. Determine the current through each resistor in the circuit of Figure 4.32. Also determine the total power generated by the source. 11. Consider the circuit shown in Figure 4.32. Assume that the 100 kΩ is replaced with another resistor ten times as large. Will this have a major impact on the current exiting source? ... | DCElectricalCircuitAnalysis_Page_131_Chunk1053 |
15. Determine the current through each resistor in the circuit of Figure 4.35. 16. For the circuit shown in Figure 4.36, determine the current through each resistor and the source voltage. 17. For the circuit shown in Figure 4.37, determine the current through each resistor and the source voltage. 18. For the circuit s... | DCElectricalCircuitAnalysis_Page_132_Chunk1054 |
19. For the circuit shown in Figure 4.39, determine the current through each resistor and the source voltage. 20. For the circuit shown in Figure 4.40, determine the current through each resistor and the source voltage. 21. For the circuit shown in Figure 4.41, determine the current through each resistor and the source... | DCElectricalCircuitAnalysis_Page_133_Chunk1055 |
23. Referring to the circuit of Figure 4.42, determine the resistor currents if the right-most 75 kΩ resistor is accidentally opened (i.e., unconnected). How do these results compare to those of problem 22? 24. Referring to the circuit of Figure 4.42, determine the resistor currents if the right-most 75 kΩ resistor is ... | DCElectricalCircuitAnalysis_Page_134_Chunk1056 |
Design Design 30. For the network shown in Figure 4.46, determine a for values for R1 given that R2 is 12 kΩ and the equivalent combination is 8 kΩ. 31. Add a third parallel resistor to the circuit of Figure 4.30 such that the source current is 10 mA. 32. Add a fourth parallel resistor to the circuit of Figure 4.32 suc... | DCElectricalCircuitAnalysis_Page_135_Chunk1057 |
37. For the circuit shown in Figure 4.49, determine values for resistors R1 and R2 such that the current through R2 is twice the current through R1. The source is 10 mA and R1 is 6 kΩ. Challenge Challenge 38. For the circuit shown in Figure 4.34, determine a new value for the 11 kΩ resistor such that the supply current... | DCElectricalCircuitAnalysis_Page_136_Chunk1058 |
46. Given three current sources with values of 1 mA, 2 mA and 7 mA; how would they need to be connected in order to deliver 4 volts across a 1 kΩ load resistor? 47. Consider the circuit of Figure 4.50. Assume I is a 4 mA source. Using only 5% standard resistor values (see Appendix A), pick values for the three resistor... | DCElectricalCircuitAnalysis_Page_137_Chunk1059 |
5 5 Series-Parallel Resistive Circuits Series-Parallel Resistive Circuits 5.0 Chapter Learning Objectives 5.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Identify series-parallel resistive circuits that include a single effective voltage or current source. • Compute component and... | DCElectricalCircuitAnalysis_Page_138_Chunk1060 |
circuits can be solved in a diverse number of ways is a strength, not a weakness. After all, you only need to recognize one of those ways, not all of them, in order to be successful. In this chapter's examples, various competing methods will be explored, but not every solution path will be spelled out for each one. Fle... | DCElectricalCircuitAnalysis_Page_139_Chunk1061 |
situation is handled will become apparent shortly. To illustrate, let us consider the resistor configuration presented in Figure 5.2. Imagine that an ohmmeter is connected to the two open terminals. What would it read? The ohmmeter would apply a small current to the circuit to determine the effective resistance of the ... | DCElectricalCircuitAnalysis_Page_140_Chunk1062 |
In the newly reduced network, the 50 Ω and 30 Ω resistances are in series, yielding 80 Ω. Without redrawing, the process may be repeated again, this time the 240 Ω being in parallel with the new 80 Ω equivalent. 240 || 80 is equal to 60 Ω. This sequence has reduced the three right-most resistances down to one: the new ... | DCElectricalCircuitAnalysis_Page_141_Chunk1063 |
How then do we find the equivalent resistance of this network? We begin at the end farthest from the open terminals. We have already noted that R7 and R8 are in series. This pair, if treated as a single resistance, is in parallel with R6, or R6||(R7+R8). This group of three is in series with R5, yielding R5+R6||(R7+R8)... | DCElectricalCircuitAnalysis_Page_142_Chunk1064 |
with the 10 kΩ resistor yields 14.193 kΩ. This is in parallel with the 20 kΩ yielding 8.302 kΩ. Finally, the 8.302 kΩ equivalent is in series with the 5 kΩ, yielding a final equivalent resistance of approximately 13.3 kΩ. 5.4 Series-Parallel Analysis 5.4 Series-Parallel Analysis Analysis of series-parallel networks inv... | DCElectricalCircuitAnalysis_Page_143_Chunk1065 |
Another solution path would be to apply the voltage divider rule to R1 and Rx in order to derive the two voltage drops (or the rule can be applied to find just one of the drops and the other voltage may be found by subtracting that from the source, an application of KVL). Once the voltages are determined, Ohm's law can... | DCElectricalCircuitAnalysis_Page_144_Chunk1066 |
And for the second resistor we see: I 3k = V b 1 kΩ I 3k = 4.5V 3 kΩ I 3k = 1.5mA Obviously, these sum to the entering current of 6 mA. Example 5.5 Determine Vb and the source current in the circuit of Figure 5.11. In this circuit the 3 kΩ resistor is in parallel with the series combination of the 2 kΩ and 1 kΩ. This l... | DCElectricalCircuitAnalysis_Page_145_Chunk1067 |
As the circuit grows, more and more solution paths exist. Consider the circuit of Figure 5.13. In this case, R3 and R4 are in parallel. This parallel combination is in series with R2. Finally, this set of three resistors is in parallel with R1 and E, reducing to a parallel circuit. Consequently, we know that the voltag... | DCElectricalCircuitAnalysis_Page_146_Chunk1068 |
V b = E Rx Rx+Ry V b = 16V 60 Ω 60Ω+10Ω V b ≈13.71V V c = V b Rx Rx+Ry V c = 13.71V 50Ω 50 Ω+30Ω V c ≈8.57 V V d = V b Rx Rx+Ry V d = 13.71V 200Ω 200Ω+40 Ω V d ≈11.43V Lastly, the current through the 40 Ω resistor can be found by dividing its voltage by its resistance. Its voltage is Vb − Vd. A quicker method is to not... | DCElectricalCircuitAnalysis_Page_147_Chunk1069 |
but how are those voltages determined? To assist with this minor quandary, the circuit has been redrawn in Figure 5.17. Note that the current source and associated 1 kΩ resistor have been flipped to the side, essentially trading places with the 2 kΩ and 7 kΩ series combination. This is entirely acceptable as both subgr... | DCElectricalCircuitAnalysis_Page_148_Chunk1070 |
The source voltage and the voltage across the 1 kΩ resistor are determined via Ohm's law: V source = I source×Requivalent V source = 10 mA×7k Ω V source = 70 V V 1k = I source×R V 1k = 10 mA×1k Ω V 1k = 10 V As the source produces 70 volts and the drop on the 1 kΩ is 10 volts, then by KVL, Vb must be 70 volts − 10 volt... | DCElectricalCircuitAnalysis_Page_149_Chunk1071 |
Do not let the negative sign be bothersome. All it indicates is that node a is at a lower potential than node c, i.e., that node a is 13.33 volts below node c. It is also purely coincidental that Vac and Va have the same magnitude. In the lab, if a voltmeter is connected such that the red (positive) lead is attached to... | DCElectricalCircuitAnalysis_Page_150_Chunk1072 |
It turns out that this arrangement offers something very unique: a halving of successive node voltages and branch currents. This is extremely useful in digital to analog conversion circuits; the sort of hardware that would turn the digital bits of an MP3 or WAV file into listenable analog signals to feed loudspeakers o... | DCElectricalCircuitAnalysis_Page_151_Chunk1073 |
flowing down through the second 12 kΩ must be Vc/12 kΩ, or 6 mA, once again half of the preceding current, as expected. The process continues as we move to the right, each subsequent node seeing half the voltage of the prior node and each current getting cut in half as well. Computer Simulation Computer Simulation In o... | DCElectricalCircuitAnalysis_Page_152_Chunk1074 |
Bridges Bridges A bridge is a particular series-parallel configuration consisting of two pairs of series connected elements placed in parallel. An example of a resistive bridge being driven by a voltage source is shown in Figure 5.21. In this circuit, R1 and R2 create one series connection while R2 and R3 create the ot... | DCElectricalCircuitAnalysis_Page_153_Chunk1075 |
If we want to flip the sign, we would place the photoresistor in the position of R2 instead of R1. Further, it is possible to increase the sensitivity by using sensors at opposite corners (e.g., R1 and R4), and comparative or differential measurements are possible by using both sides (e.g., R1 with R3). Example 5.8 Det... | DCElectricalCircuitAnalysis_Page_154_Chunk1076 |
A DC operating point analysis is run and the results are shown in Figure 5.24. The voltages are precisely as expected. Final Comments Final Comments Series-parallel simplification techniques will not work for all circuits. Complex multi-source circuits and some resistive networks such as delta configurations or five-el... | DCElectricalCircuitAnalysis_Page_155_Chunk1077 |
discovered, if desired. Once these values are determined, power calculations are trivial. There is an infinite variety of series-parallel configurations and consequently no single solution technique will work for all of them. Indeed, the more complex the circuit, the more solution paths there are for that circuit. It i... | DCElectricalCircuitAnalysis_Page_156_Chunk1078 |
5.6 Exercises 5.6 Exercises Analysis Analysis 1. In the circuit of Figure 5.1, which individual resistors are strictly in series and which are in parallel? 2. In the circuit of Figure 5.2, which individual resistors are strictly in series and which are in parallel? 3. In the circuit of Figure 5.3, which individual resi... | DCElectricalCircuitAnalysis_Page_157_Chunk1079 |
4. In the circuit of Figure 5.4, which individual resistors are strictly in series and which are in parallel? 5. In the circuit of Figure 5.5, which individual resistors are strictly in series and which are in parallel? 6. In the circuit of Figure 5.6, which individual resistors are strictly in series and which are in ... | DCElectricalCircuitAnalysis_Page_158_Chunk1080 |
8. In the circuit of Figure 5.8, which individual resistors are strictly in series and which are in parallel? 9. Determine the equivalent resistance of the network shown in Figure 5.1 (i.e., as if an ohmmeter is connected to the two terminals). 10. Determine the equivalent resistance of the network shown in Figure 5.2.... | DCElectricalCircuitAnalysis_Page_159_Chunk1081 |
21. For the circuit of Figure 5.11, find voltages Va, Vb and Vab. 22. For the circuit of Figure 5.11, find the current through each resistor. 23. For the circuit of Figure 5.12, find the current through each resistor. 24. For the circuit of Figure 5.12, find voltages Va, Vb and Vab. 25. In the circuit of Figure 5.13, f... | DCElectricalCircuitAnalysis_Page_160_Chunk1082 |
28. In the circuit of Figure 5.14, find voltages Va, Vb and Vab. 29. For the circuit of Figure 5.15, find voltages Vb, Vc and Vcb. 30. For the circuit of Figure 5.15, find the current through the 470 Ω and 3.6 kΩ resistors along with the source current. 31. For the circuit of Figure 5.16, find the current through the 2... | DCElectricalCircuitAnalysis_Page_161_Chunk1083 |
33. For the circuit of Figure 5.17, find voltages Va and Vb. 34. For the circuit of Figure 5.17, find the current through the 10 kΩ and 4 kΩ resistors along with the source current. 35. For the circuit of Figure 5.18, find the current through the 30 kΩ and the left-most 36 kΩ resistors along with the source current. 36... | DCElectricalCircuitAnalysis_Page_162_Chunk1084 |
41. Given the circuit of Figure 5.20, find the current through the 47 kΩ, 5.1 kΩ, and 3.9 kΩ resistors. 42. Given the circuit of Figure 5.20, find voltages Va, Vb and Vab. 43. For the circuit of Figure 5.21, find voltages Vb, Vc and Vd. 44. For the circuit of Figure 5.21, find the current through the 6.8 kΩ, 8.5 kΩ, an... | DCElectricalCircuitAnalysis_Page_163_Chunk1085 |
47. For the circuit of Figure 5.23, find voltages Va, Vb and Vab. 48. For the circuit of Figure 5.23, find the current through each of the resistors. 49. For the circuit of Figure 5.24, find the current through each of the resistors. 50. For the circuit of Figure 5.24, find voltages Va, Vb and Vab. 51. Given the circui... | DCElectricalCircuitAnalysis_Page_164_Chunk1086 |
54. For the circuit of Figure 5.26, find voltages Va, Vb and Vab. 55. Given the circuit of Figure 5.27, find voltages Va, Vb and Vab. 56. Given the circuit of Figure 5.27, find the current through the 5 kΩ, and 6 kΩ resistors. 57. For the circuit of Figure 5.28, find the current through the 9 kΩ and right- most 82 kΩ r... | DCElectricalCircuitAnalysis_Page_165_Chunk1087 |
63. For the circuit of Figure 5.30, find the current through the 1 kΩ, 2.2 kΩ, and 18 kΩ resistors. 64. For the circuit of Figure 5.30, find voltages Va, Vb and Vc. 65. Given the circuit of Figure 5.31, find voltages Va, Vb and Vab. 66. Given the circuit of Figure 5.31, find the current through the 20 kΩ, 15 kΩ, and 3.... | DCElectricalCircuitAnalysis_Page_166_Chunk1088 |
Design Design 69. Determine a new value for the 33 Ω resistor in Figure 5.9 such that the source current is 10 mA. 70. Determine a new value for the 20 kΩ resistor in Figure 5.12 such that Vb is 4 volts. 71. Determine a new value for the 470 Ω resistor in Figure 5.15 such that Vb is 6 volts. 72. Determine a new value f... | DCElectricalCircuitAnalysis_Page_167_Chunk1089 |
82. Given the circuit of Figure 5.25, determine a new value for the 2 k Ω resistor such that Vb is 12 volts. 83. Given the circuit of Figure 5.32, determine new values for the resistors such that all of the node voltages are twice the value of the original circuit's node voltages. Simulation Simulation 84. Perform a DC... | DCElectricalCircuitAnalysis_Page_168_Chunk1090 |
Notes Notes ♫♫ ♫♫ 169 | DCElectricalCircuitAnalysis_Page_169_Chunk1091 |
6 6 Analysis Theorems and Techniques Analysis Theorems and Techniques 6.0 Chapter Learning Objectives 6.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Define whether or not a circuit is a linear bilateral network. • Find the voltage source equivalent of a current source and vice v... | DCElectricalCircuitAnalysis_Page_170_Chunk1092 |
Finally, we will examine how to find equivalent circuits for certain resistor arrangements that use three connecting points, in other words, resistor arrangements shaped like the letter Y or like a triangle. These are known as delta and Y configurations. These configurations are difficult to address with basic series- ... | DCElectricalCircuitAnalysis_Page_171_Chunk1093 |
For a current source, the improved model adds a resistance in parallel, as shown in Figure 6.2. This resistance sets an upper limit on the source's voltage output. If the output terminals are opened, the maximum voltage will no longer produce a huge voltage. Instead, it is dictated by Ohm's law to be the source current... | DCElectricalCircuitAnalysis_Page_172_Chunk1094 |
To continue, if we look at the open load case, for the voltage source the load current would be zero and the load voltage would be the entire source voltage of E. For the current source, the load would also see no current and its voltage would be the voltage appearing across its internal resistance which is R times the... | DCElectricalCircuitAnalysis_Page_173_Chunk1095 |
The equivalent current source is shown in Figure 6.4. We know that this will work for the shorted an opened cases, but if any doubt is left as to its universal nature, simply substitute any other resistance value and compare the outcomes of the two circuits. For no particular reason, let's try using a load of 200 Ω and... | DCElectricalCircuitAnalysis_Page_174_Chunk1096 |
For a voltage source with matched resistances we wind up with a simple 50% voltage divider, thus the load voltage will be half the source voltage, or 165 volts. The original current source sees the current split in half due to the current divider rule. Thus, the load current should be 7.5 mA. Given this current, the lo... | DCElectricalCircuitAnalysis_Page_175_Chunk1097 |
For the first source, the current will be: I s = E Rs I s = 15 V 1k Ω I s = 15mA And for the second source: I s = E Rs I s = 6V 4 kΩ I s = 1.5mA The equivalent converted circuit is shown in Figure 6.9. Before continuing, it is worth noting that connection nodes a and c no longer exist in this circuit. More on this in a... | DCElectricalCircuitAnalysis_Page_176_Chunk1098 |
6.3 Superposition Theorem 6.3 Superposition Theorem As useful as the source conversion technique proved to be in Example 6.3, it will not work for all circuits. Thus, more general approaches are needed. One of these methods is superposition. Superposition allows the analysis of multi-source series-parallel circuits. Su... | DCElectricalCircuitAnalysis_Page_177_Chunk1099 |
To summarize the superposition technique: • For every voltage or current source in the original circuit, create a new sub- circuit. The sub-circuits will be identical to the original except that all sources other than the one under consideration will be replaced by their ideal internal resistance. This means that all r... | DCElectricalCircuitAnalysis_Page_178_Chunk1100 |
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