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AC Electrical Circuit Analysis AC Electrical Circuit Analysis A Practical Approach A Practical Approach James M. Fiore James M. Fiore | ACElectricalCircuitAnalysis_Page_1_Chunk501 |
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AC Electrical Circuit Analysis AC Electrical Circuit Analysis A Practical Approach A Practical Approach by James M. Fiore Version 1.1.10, 30 March 2023 3 | ACElectricalCircuitAnalysis_Page_3_Chunk503 |
This AC Electrical Circuit Analysis, by James M. Fiore is copyrighted under the terms of a Creative Commons license: This work is freely redistributable for non-commercial use, share-alike with attribution Device data sheets and other product information are copyright by their respective owners and have been obtained t... | ACElectricalCircuitAnalysis_Page_4_Chunk504 |
Introduction Introduction Welcome to the AC Electrical Circuit Analysis, an open educational resource (OER). The goal of this text is to introduce the theory and practical application of analysis of AC electrical circuits. It assumes familiarity with DC circuit analysis. If you have not studied DC circuit analysis, it ... | ACElectricalCircuitAnalysis_Page_5_Chunk505 |
For the Have-nots and the Has-beens “Well, the people, I would say. There is no patent. Could you patent the sun?” — Jonas Salk, inventor of the polio vaccine, when asked who owns the patent to it. 6 | ACElectricalCircuitAnalysis_Page_6_Chunk506 |
Table of Contents Table of Contents Chapter 1: Fundamentals . . . . . . . 10 1.0 Chapter Objectives . . . . . . . . 10 1.1 Introduction . . . . . . . . . 10 1.2 Sinusoidal Waveforms . . . . . . . 11 1.3 Basic Fourier Analysis . . . . . . . . 25 1.4 Complex Numbers . . . . . . . . 29 1.5 Reactance and Impedance . . . . ... | ACElectricalCircuitAnalysis_Page_7_Chunk507 |
Chapter 5: Analysis Theorems and Techniques . . . . 144 5.0 Chapter Objectives . . . . . . . . 144 5.1 Introduction . . . . . . . . . 144 5.2 Source Conversions . . . . . . . . 145 5.3 Superposition Theorem . . . . . . . 153 5.4 Thévenin's and Norton's Theorems . . . . . . 160 5.5 Maximum Power Transfer Theorem . . . .... | ACElectricalCircuitAnalysis_Page_8_Chunk508 |
Chapter 9: Polyphase Power . . . . . . 338 9.0 Chapter Objectives . . . . . . . . 338 9.1 Introduction . . . . . . . . . 338 9.2 Polyphase Definition . . . . . . . . 339 9.3 Three-Phase Connections . . . . . . . 341 9.4 Power Factor Correction . . . . . . . 354 Summary . . . . . . . . . 364 Exercises . . . . . . . . . ... | ACElectricalCircuitAnalysis_Page_9_Chunk509 |
1 1 Fundamentals Fundamentals 1.0 Chapter Learning Objectives 1.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Describe sinusoidal waveforms in mathematical terms. • Express complex waveforms using basic Fourier analysis. • Perform mathematical operations using complex numbers. • ... | ACElectricalCircuitAnalysis_Page_10_Chunk510 |
1.2 Sinusoidal Waveforms 1.2 Sinusoidal Waveforms AC, or alternating current, is so named because the current alternates or flips back and forth between two polarities. In other words, the current (and consequently the voltage) is a function of time. This is fundamentally different from direct current that is fixed in ... | ACElectricalCircuitAnalysis_Page_11_Chunk511 |
Sine waves exhibit quarter wave symmetry. That is, each quarter (in time) of the wave is identical to any other if you simply flip it around the horizontal axis and/or rotate it upright about its peak. The time it takes to complete one cycle is called the period and is denoted with the symbol T (for Time). The reciproc... | ACElectricalCircuitAnalysis_Page_12_Chunk512 |
Combining these parameters, consider the voltage waveform shown in Figure 1.4. Here we see two cycles of an AC voltage waveform. 13 Variation in Amplitude Percentage of Peak -200 -150 -100 -50 0 50 100 150 200 Time 0 0.2 0.4 0.6 0.8 1 One half amplitude Initial Double amplitude Figure 1.3 Sine wave amplitude variation.... | ACElectricalCircuitAnalysis_Page_13_Chunk513 |
The peak value is 4 volts and the peak-to-peak value is 8 volts (typically abbreviated as “8 V pp”). The period of one cycle is 0.2 seconds, or T = 200 milliseconds. Further, the frequency, f = 1/200 milliseconds, or 5 Hz (5 cycles in one second). AC waveforms may also be combined with a DC offset. Adding a positive DC... | ACElectricalCircuitAnalysis_Page_14_Chunk514 |
Figure 1.6 illustrates the effect of phase shift. Note that in this plot, t = 0 has been moved to the center of the horizontal axis. The middle curve is the initial, or reference, wave (green). To the left (red) is a wave leading the initial wave by one- eighth cycle, or 45°. To the right (blue), is a lagging wave of h... | ACElectricalCircuitAnalysis_Page_15_Chunk515 |
To compute a phase shift, first determine the time differential between the waveform and the reference, which we'll call Δt. The reference may be a fixed point in time (e.g., t = 0) or another waveform. Generally, the easy way to do this is to measure the difference at the zero-crossings, assuming there is no DC offset... | ACElectricalCircuitAnalysis_Page_16_Chunk516 |
f = 1 T f = 1 10 ms f = 100Hz The second issue is the DC offset. Note that the positive peak occurs at 4 amps while the negative peak occurs at −2 amps. This indicates a peak-to- peak value of 6 amps. Without an offset, the positive peak would be at 3 amps, therefore there is a +1 amp DC offset. The vertical center of ... | ACElectricalCircuitAnalysis_Page_17_Chunk517 |
The waveform is shifted to the left which indicates a positive or leading phase shift. If we examine the second cycle, we see that it hits zero volts at 1.8 milliseconds. Therefore the shift is 0.2 milliseconds. Expressed in degrees this is: θ = 360° Δt T θ = 360° 0.2ms 2 ms θ = 36° The final expression is: v(t)=2sin(2... | ACElectricalCircuitAnalysis_Page_18_Chunk518 |
Laboratory Measurements Laboratory Measurements In the laboratory, a function generator is used to generate sines and other waveshapes. These devices will allow precise control over both the amplitude and frequency of the wave along with adding a DC offset, if desired. An example is shown in Figure 1.10. The correspond... | ACElectricalCircuitAnalysis_Page_19_Chunk519 |
Schematic Symbols Schematic Symbols As far as schematics are concerned, the symbols for AC voltage and current sources are shown in Figure 1.12. The polarity and direction markings are not absolute; after all, these are AC sources whose polarity and directions flip back and forth. The markers are instead used to establ... | ACElectricalCircuitAnalysis_Page_20_Chunk520 |
Example 1.4 Assume an oscilloscope displays two waves as depicted in Figure 1.14. Determine the phase shift of the smaller 20 volt peak (blue) waveform relative to the larger 25 volt peak (red) waveform. First, note that neither wave exhibits a DC offset. If one or both them had an offset, the wave(s) would have to be ... | ACElectricalCircuitAnalysis_Page_21_Chunk521 |
Sines and Cosines Sines and Cosines There are a handful of specific phase shifts that are worth a closer look. If a sine wave is inverted, that is, flipped upside down, it is indistinguishable from a sine wave that has been shifted either +180 or −180 degrees. In other words, such a wave can be written three different ... | ACElectricalCircuitAnalysis_Page_22_Chunk522 |
amplitude is also zero. It is also true that the sine wave is the slope of the negative cosine wave, the negative cosine is the slope of the negative sine, and the negative sine is the slope of the cosine. Moving in the reverse direction, we can say that the anti-derivative (indefinite integral) of a cosine wave is a s... | ACElectricalCircuitAnalysis_Page_23_Chunk523 |
The first step is to square this waveform. A useful trigonometric identity is (sin x) 2=1 2 −1 2 cos 2 x Applying this to our wave yields: v(t) 2= 1 2 −1 2 cos(2π2 f t) This expression describes an inverted cosine wave at twice the original frequency and half of the original amplitude, riding on a DC offset equal to it... | ACElectricalCircuitAnalysis_Page_24_Chunk524 |
Wavelength Wavelength Another item of interest is the speed of propagation of the wave. This varies widely. In the case of light in a vacuum (or to a close approximation, an electrical current in a wire), the velocity is about 3E8 meters per second (i.e., 300,000 km/s) or about 186,000 miles per second. Given a velocit... | ACElectricalCircuitAnalysis_Page_25_Chunk525 |
A triangle wave is similar: ) 2 )1 2 cos(( )1 2 ( 1 ) ( 1 2 ft n n t v n (1.5) Thus a triangle wave of frequency f is made up of an infinite series of cosines (sines with a 90 degree or one quarter cycle phase shift) at odd integer multiples of f, with an inverse square amplitude characteristic. For exampl... | ACElectricalCircuitAnalysis_Page_26_Chunk526 |
27 Building a Square Wave Voltage -1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 Time 0 0.2 0.4 0.6 0.8 1 Fundamental plus third, fifth and seventh harmonics Fundamental Third harmonic Fifth harmonic Seventh harmonic Figure 1.18 Building a square wave, part 2. Figure 1.19 Building a square wave, part 3. Building a Sq... | ACElectricalCircuitAnalysis_Page_27_Chunk527 |
The triangle sequence begins with a fundamental and the first harmonic as shown in Figure 1.20. The resulting combination is already trending away from a simple sine shape. The second and final graph, Figure 1.21, shows a total of seven harmonics. The result is very close to a triangle, the only obvious deviation is th... | ACElectricalCircuitAnalysis_Page_28_Chunk528 |
1.4 Complex Numbers 1.4 Complex Numbers In AC circuits, parameters such as voltage and current are vectors, that is, they have both a magnitude and a phase shift or angle. For example, a voltage might be “12 volts at an angle of 30 degrees” (or more compactly, 1230º). This is known as polar form or magnitude-angle for... | ACElectricalCircuitAnalysis_Page_29_Chunk529 |
Converting from one form to another relies on basic trigonometric relations. For convenience, the relationships between the magnitude, angle, real component and imaginary component are reproduced below: Magnitude = √Real 2+Imaginary 2 θ = tan−1 Imaginary Real Real = Magnitude cosθ jImaginary = Magnitude sinθ (1.6a – d)... | ACElectricalCircuitAnalysis_Page_30_Chunk530 |
Real = 10 cos 45º = 7.07 jImaginary = 10sin 45º = j 7.07 Real = 0.2cos−30º = 0.173 jImaginary = 0.2sin−30º =−j0.1 The answers for the second part are 7.07 + j7.07 and 0.173 − j0.1. Example 1.6 Perform the following: a) Add 7 + j8 to + j6, b) Subtract 553.1º from 10−45º, c) Divide 2090º by 4−50º, d) Multiply 90 + j7... | ACElectricalCircuitAnalysis_Page_31_Chunk531 |
ic = C dvc dt ic = C d sin(2 π f t) dt ic = C sin(2π f t+90º) Substituting this result into Equation 1.7 yields: X C = vc ic X C = sin(2π f t) 2 π f C sin(2π f t+90º) X C = 1 2π f C −90º Or more directly: X C =−j 1 2π f C (1.8) In summation, if we divide the capacitor's sinusoidal voltage by its current (Ohm's law), w... | ACElectricalCircuitAnalysis_Page_32_Chunk532 |
We can see that the value of resistance does not change with frequency while the inductive reactance increases with frequency and the capacitive reactance decreases. The linear frequency scale makes the capacitor change difficult to see. If this is plotted again but using a logarithmic frequency scale as in Figure 1.25... | ACElectricalCircuitAnalysis_Page_33_Chunk533 |
The effect of both capacitor size and frequency is shown in Figure 1.26 using a log frequency axis: the smaller the capacitor, the larger the capacitive reactance at any particular frequency. Similarly, the effect of both inductor size and frequency is shown in Figure 1.27 using a linear frequency axis: the larger the ... | ACElectricalCircuitAnalysis_Page_34_Chunk534 |
It is worth noting that the plots of Figures 1.26 and 1.27 are for ideal components. In reality, all components exhibit some resistive, capacitive and inductive effects due to their construction. For example, eventually, inductive and resistive effects will cause the capacitive reactance curves of Figure 1.26 to begin ... | ACElectricalCircuitAnalysis_Page_35_Chunk535 |
For the second source, 50 kHz is 25 times larger than the original and capacitive reactance is inversely proportional to frequency. Therefore XC is 25 times smaller, or −j1.59 Ω. For the inductor at 2 kHz, X L = +j 2π f L X L = +j 2π 2kHz1 mH X L = +j 12.57Ω Inductive reactance is directly proportional to frequency. Th... | ACElectricalCircuitAnalysis_Page_36_Chunk536 |
1.6 Phasor Diagrams 1.6 Phasor Diagrams A time domain representation of sine waves as drawn earlier tells us everything we need to know about the waves, however it is not the most compact method of displaying them, nor are they easy to sketch by hand. Consider the the plot of Figure 1.28. Here, two sine waves (green an... | ACElectricalCircuitAnalysis_Page_37_Chunk537 |
1.7 Summary 1.7 Summary The most simple of AC waveforms is the sine wave. It can be thought of as the vertical displacement of a vector rotating at a constant speed, like the second hand of a clock. The length of the second hand represents the height or amplitude of the sine wave and the speed of rotation represents it... | ACElectricalCircuitAnalysis_Page_38_Chunk538 |
Review Questions Review Questions 1. What is a sine wave? Describe its constituent parameters (amplitude, frequency, offset, etc.). 2. What is the relationship between the frequency, period and phase of a sine wave? 3. What is the mathematical relationship between a sine wave and a cosine wave? 4. What is meant by the ... | ACElectricalCircuitAnalysis_Page_39_Chunk539 |
9. A 1 kHz sine wave has a phase of 72°. Determine the time delay. Repeat for a 20 kHz sine wave. 10. A 2 kHz sine wave has a phase of 18°. Determine the time delay. Repeat for a 100 kHz sine wave. 11. An oscilloscope measures a time delay of 0.2 milliseconds between a pair of 500 Hz sine waves. Determine the phase shi... | ACElectricalCircuitAnalysis_Page_40_Chunk540 |
25. Determine the capacitive reactance of a 1 μF capacitor at the following frequencies: a) 10 Hz b) 500 Hz c) 10 kHz d) 400 kHz e) 10 MHz 26. Determine the capacitive reactance of a 220 pF capacitor at the following frequencies: a) 10 Hz b) 500 Hz c) 10 kHz d) 400 kHz e) 10 MHz 27. Determine the capacitive reactance a... | ACElectricalCircuitAnalysis_Page_41_Chunk541 |
41. Which of the following have a reactance of less than 100 Ω for all frequencies below 1 kHz? a) 2 mH b) 99 mH c) 470 pF d) 10000 μF 42. Which of the following have a reactance of less than 8 Ω for all frequencies above 10 kHz? a) 10 nH b) 5 mH c) 56 pF d) 470 μF 43. Which of the following have a reactance of at leas... | ACElectricalCircuitAnalysis_Page_42_Chunk542 |
Notes Notes ♫♫ ♫♫ 43 | ACElectricalCircuitAnalysis_Page_43_Chunk543 |
2 2 Series RLC Circuits Series RLC Circuits 2.0 Chapter Learning Objectives 2.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Utilize Kirchhoff's voltage law, the voltage divider rule and Ohm's law to find node and component voltages in series RLC networks that utilize voltage sour... | ACElectricalCircuitAnalysis_Page_44_Chunk544 |
computed from the capacitor and inductor values, based on the frequency of the driving source. Consequently, if this frequency were to change, the reactances would change, and this would in turn cause changes in the circulating current and component voltages. 2.3 Series Impedance 2.3 Series Impedance Perhaps the first ... | ACElectricalCircuitAnalysis_Page_45_Chunk545 |
Example 2.2 Determine the effective impedance of the circuit shown in Figure 2.2 if the source frequency is 2 kHz. Repeat this for source frequencies of 200 Hz and 20 kHz. Finally, express the results in both rectangular and polar form. The first step is to find the reactance values at 2 kHz. X L = j 2π f L X L = j 2π2... | ACElectricalCircuitAnalysis_Page_46_Chunk546 |
reactive sum is then copied and shifted right to join the resistive component to show the final result. This is 511.2−12° Ω (green), as expected. Example 2.3 Determine the impedance of the network shown in Figure 2.4. If the input frequency is 1 kHz, determine the capacitor and inductor values. The reactance values ar... | ACElectricalCircuitAnalysis_Page_47_Chunk547 |
And now for the capacitor: X C =−j 1 2 π f C C = 1 2π f | X C| C = 1 2π1 kHz 200Ω C = 796nF A plot of the impedance vector summation is shown in Figure 2.5. Note how the three components combine graphically to arrive at Z. 2.4 Series Circuit Analysis 2.4 Series Circuit Analysis The techniques employed for series AC cir... | ACElectricalCircuitAnalysis_Page_48_Chunk548 |
KVL states that the sum of voltages around a loop must be zero, or that the sum of voltage rises around a loop must equal the sum of voltage drops: ∑ v↑ = ∑ v↓ (2.2) The voltage divider rule for AC states that the voltage across any component or group of components is proportional to the ratio of the impedance of said ... | ACElectricalCircuitAnalysis_Page_49_Chunk549 |
Now use Ohm's law to find the component voltages. v R = i×R v R = 188.665.5°μ A×22000°Ω v R = 0.414965.5° V vC = i× X C vC = 188.665.5° μA×4823−90°Ω vC = 0.9098−24.5° V To complete this Example, we shall generate a series of plots: First off, an impedance plot showing the vector combination of R and XC leading to... | ACElectricalCircuitAnalysis_Page_50_Chunk550 |
For further clarity, Figure 2.9 plots the voltages in the time domain. Note how the component voltages add graphically resulting in the input voltage. 51 Figure 2.8 Voltage phasor diagram for Example 2.4. Time Domain Graph Voltage (volts) -1.00 -0.75 -0.50 -0.25 0.00 0.25 0.50 0.75 1.00 Time (milliseconds) 0 0.25 0.5 0... | ACElectricalCircuitAnalysis_Page_51_Chunk551 |
The time shifts may be computed for verification as follows: θ = 360° Δt T ΔtR = T θR 360° Δt R = 1 ms 65.5° 360° Δt R = 181.9μ s Δt C = T θC 360° ΔtC = 1ms−24.5° 360° ΔtC =−68μ s Thus we verify the resistor voltage leading (to the left) by 181.9 μs and the capacitor voltage lagging (to the right) by 68 μs, as seen in ... | ACElectricalCircuitAnalysis_Page_52_Chunk552 |
We will now examine a variety of different series circuits. To ease the computations, we will simply state values for the component reactances rather than specify a frequency along with inductor and/or capacitor values. Example 2.5 Determine the component voltages for the circuit shown in Figure 2.12. Given the referen... | ACElectricalCircuitAnalysis_Page_53_Chunk553 |
v R = e R Z v R = 60° V 1800° Ω 402.563.4 °Ω v R = 2.683−63.4° Vp The rectangular versions of vL and vR are 4.8 + j2.4 and 1.2 − j2.4 respectively, summing up to the source of 60°. This is illustrated in the phasor voltage plot of Figure 2.13. Compare the plot of Figure 2.13 to that of the RC circuit in Figure 2.8... | ACElectricalCircuitAnalysis_Page_54_Chunk554 |
Finally, as a source frequency was not specified for this circuit, the time scale of Figure 2.14 is calibrated in terms of cycles rather than seconds. Example 2.6 Find the voltages va and vb in the circuit of Figure 2.15. Assume the current source is the reference of 0°. Ohm's law may be used directly as the current is... | ACElectricalCircuitAnalysis_Page_55_Chunk555 |
vb = i×(R+jX L) vb = 20 0°mA×(4300+j 150Ω) vb = 86.052° V Notice the small size of the phase angle. This is expected given that the inductive reactance is so much smaller than the resistive component. Finally, these voltage are RMS as the current is assumed to be RMS (not having been labeled as peak or peak-to-peak).... | ACElectricalCircuitAnalysis_Page_56_Chunk556 |
The voltage divider yields: vbd = e Z bd Z Total vbd = 4.359−96.6°V 670090°Ω 699273.4 °Ω vbd = 4.177−80° Vp Although we have answered the question, at this point attempting to visualize the waveforms in your head to verify the results may be a little challenging. No problem! If we also find the voltage across the r... | ACElectricalCircuitAnalysis_Page_57_Chunk557 |
KVL tells us that the sum of the voltage rises must equal the sum of the voltage drops. We had used E1 as the reference and treated it as a rise. This made the reference direction clockwise for the circulating current. As a consequence, vR and vbd are voltage drops. E2 is also seen as a drop given its reference polarit... | ACElectricalCircuitAnalysis_Page_58_Chunk558 |
Figure 2.19 shows a simple way to block low frequencies from a tweeter. In this diagram, the loudspeaker is modeled as a resistor. Although this is not perfectly accurate, it is sufficient to illustrate the idea. To understand this concept, simply think of the circuit as a voltage divider between the loudspeaker and th... | ACElectricalCircuitAnalysis_Page_59_Chunk559 |
X C =−j 1 2π f C C = 1 2π f | X C| C = 1 2π 2500Hz 8Ω C = 7.96μ F To find the loudspeaker voltage at the three specified frequencies, first find the reactance at that frequency and then use the voltage divider rule. We already know that at 2.5 kHz the reactance is −j8 Ω. vloudspeaker = e Z Load ZTotal vloudspeaker = 10... | ACElectricalCircuitAnalysis_Page_60_Chunk560 |
2.5 Concerning Practical Inductors 2.5 Concerning Practical Inductors Up to this point, inductors have been treated as ideal components, that is, pure inductance. In reality, all inductors have some resistance associated with them due to the resistance of the wire used to make the coil. This is called ESR, or Equivalen... | ACElectricalCircuitAnalysis_Page_61_Chunk561 |
The Q, or quality factor, of an inductor can be defined in terms of the peak energy stored in the device versus the energy dissipated per cycle. Keeping time constant, we can relate this to the power via the relation i2R. The current necessarily is the same for both the reactive and resistive components, therefore the ... | ACElectricalCircuitAnalysis_Page_62_Chunk562 |
Rcoil = X L Qcoil Rcoil = 1257 50 Rcoil = 25.1Ω The impedance of the inductor is 25.1 + j1257 Ω , or 125788.9° Ω. The total impedance is 600 + 25.1 + j1257 Ω , or 140463.6° Ω. We can use a voltage divider to find the inductor's voltage. vind = e Z ind Z Total vind = 20 0°Vpp 125788.9°Ω 140463.6°Ω vind = 17.925.3°... | ACElectricalCircuitAnalysis_Page_63_Chunk563 |
Kirchhoff's voltage law (KVL) states that the sum of voltage rises around any series loop must equal the sum of voltage drops around that loop. This remains true for AC analysis, however, it is imperative that a vector summation is performed and not just a simple summation of the magnitudes. If the phase angles are ign... | ACElectricalCircuitAnalysis_Page_64_Chunk564 |
3. Determine the impedance of the circuit of Figure 2.23 for a 10 kHz sine. 4. Determine the impedance of the circuit of Figure 2.23 for a 50 kHz sine. 5. Determine the impedance of the circuit of Figure 2.24 for a 1 kHz sine. 6. Determine the impedance of the circuit of Figure 2.24 for a 500 Hz sine. 7. Determine the ... | ACElectricalCircuitAnalysis_Page_65_Chunk565 |
11. Draw the voltage and current waveforms for the circuit of Figure 2.27. 12. Draw the voltage and current waveforms for the circuit of Figure 2.28 if E is a one volt peak sine at a frequency of 10 kHz and C = 3.3 nF. 13. Draw the voltage and current waveforms for the circuit of Figure 2.29 if E is a two volt peak-pea... | ACElectricalCircuitAnalysis_Page_66_Chunk566 |
15. Draw the voltage and current waveforms for the circuit of Figure 2.31 if I is a two amp peak-peak sine at a frequency of 40 Hz and L = 33 mH. 16. Determine the impedance of the circuit of Figure 2.32. 17. Determine the impedance of the circuit of Figure 2.32 using a frequency of 10 kHz. 18. For the circuit of Figur... | ACElectricalCircuitAnalysis_Page_67_Chunk567 |
22. For the circuit of Figure 2.33, determine the circulating current and the voltages across each component. Also find the time delay between the voltages of the components. 23. For the circuit of Figure 2.33 with a frequency of 3 kHz, determine the circulating current and the voltages across each component. Also find... | ACElectricalCircuitAnalysis_Page_68_Chunk568 |
29. For the circuit of Figure 2.36, determine the circulating current and the voltages across each component. 30. For the circuit of Figure 2.37, determine the circulating current and the voltages across each component. 31. For the circuit of Figure 2.38, determine the applied voltage and the voltages across each compo... | ACElectricalCircuitAnalysis_Page_69_Chunk569 |
33. For the circuit of Figure 2.40, determine the circulating current and the voltages across each component. 34. Repeat the previous problem using an input frequency of 10 kHz. 35. For the circuit of Figure 2.41, determine the circulating current and the voltages across each component. The source is a 10 volt peak sin... | ACElectricalCircuitAnalysis_Page_70_Chunk570 |
39. For the circuit of Figure 2.43, determine the circulating current and the voltages across each component. E is a 1 volt peak 2 kHz sine. Also, draw a phasor diagram of the four component voltages. 40. For the circuit of Figure 2.43, find vb and vca. E is a 1 volt peak 2 kHz sine. 41. For the circuit of Figure 2.44,... | ACElectricalCircuitAnalysis_Page_71_Chunk571 |
43. For the circuit of Figure 2.46, determine the circulating current and the voltages across each component. E is a 2 volt RMS 1 kHz sine. Also, draw a phasor diagram of the four component voltages. 44. For the circuit of Figure 2.47, determine vb, vc and vac. E is a 1 volt peak 25 kHz sine. 45. For the circuit of Fig... | ACElectricalCircuitAnalysis_Page_72_Chunk572 |
46. For the circuit of Figure 2.49, determine vac, vb and vc. The source is a 10 mA peak-peak sine at 50 kHz, R = 2 kΩ, C = 10 nF and L = 800 μH. 47. For the circuit of Figure 2.50, determine the voltages across each component. The source is a 2 mA RMS sine at 1 kHz, R = 1.2 kΩ, C = 750 nF and L = 6.8 mH. 48. For the c... | ACElectricalCircuitAnalysis_Page_73_Chunk573 |
49. For the circuit of Figure 2.52, determine the voltages across each component. 50. For the circuit of Figure 2.53, determine the voltages across each component. Further, draw a phasor diagram of the four component voltages. 51. For the circuit of Figure 2.54, find vac, vb and vc. The source is 5 mA peak at 8 kHz. 74... | ACElectricalCircuitAnalysis_Page_74_Chunk574 |
52. For the circuit of Figure 2.55, determine the voltages across each component. The source is 20 mA peak at 100 kHz. 53. For the circuit of Figure 2.56, determine the voltages across each component. 54. For the circuit of Figure 2.57, determine the voltages vb and vdb. E1 = 20º and E2 = 590º. 55. Determine the indu... | ACElectricalCircuitAnalysis_Page_75_Chunk575 |
57. For the circuit of Figure 2.58, determine the voltages across each component. E1 = 10º and E2 = 860º. Design Design 58. Redesign the circuit of Figure 2.32 using a new capacitor such that the current magnitude from the source is 100 μA. 59. Redesign the circuit of Figure 2.33 using a new frequency such that the c... | ACElectricalCircuitAnalysis_Page_76_Chunk576 |
66. The circuit of Figure 2.59 can be used as part of a loudspeaker crossover network. The goal of this circuit is to steer the low frequency tones to the low frequency transducer (labeled here as “Loudspeaker” and often referred to as a woofer). A similar network substitutes a capacitor for the inductor to steer the h... | ACElectricalCircuitAnalysis_Page_77_Chunk577 |
71. Simulate the solution of challenge problem 62 and determine if the new frequency produces the required results. Hint: if the reactance magnitudes are the same, then the voltage magnitudes will be identical. Further, their phases will cause these voltages to cancel, leaving the resistor voltage equal to the source v... | ACElectricalCircuitAnalysis_Page_78_Chunk578 |
Notes Notes ♫♫ ♫♫ 79 | ACElectricalCircuitAnalysis_Page_79_Chunk579 |
3 3 Parallel RLC Circuits Parallel RLC Circuits 3.0 Chapter Learning Objectives 3.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Utilize Kirchhoff's current law, the current divider rule and Ohm's law to find branch currents in parallel RLC networks that utilize current sources or... | ACElectricalCircuitAnalysis_Page_80_Chunk580 |
3.2 The Parallel Connection 3.2 The Parallel Connection A parallel connection is always characterized by having all components share just two nodes. There are no other junctions from which current can flow out of or into, or for voltage to split. Consequently, The voltage is the same everywhere in a parallel connection... | ACElectricalCircuitAnalysis_Page_81_Chunk581 |
Each of the individual impedances presented in Equation 3.4 (i.e., Z1, Z2, etc) can represent a simple resistance, a pure reactance or a complex impedance. Further, the product-sum rule shortcut for two components also remains valid for AC components: Ztotal = Z 1×Z 2 Z 1+Z 2 (3.5) There is one special case where Equat... | ACElectricalCircuitAnalysis_Page_82_Chunk582 |
The individual component values are: BL = 1 j 12 kΩ ≈−j83.33E-6S BC = 1 −j 48k Ω ≈j 20.83E-6S G = 1 20 kΩ = 50E-6S Y total = 1 12.49E351.3°Ω ≈80.1E-6−51.3°S In rectangular form Ytotal = 50E−6 − j62.5E−6 S. 83 Figure 3.2 Admittance diagram for the network of Figure 3.1. | ACElectricalCircuitAnalysis_Page_83_Chunk583 |
Example 3.2 Determine the impedance of the network shown in Figure 3.3 at a frequency of 10 kHz. Repeat this for a frequency of 1 kHz. First, find the reactances at 10 kHz. For the inductor we find: X L = j 2π f L X L = j 2π10 kHz680μ H X L≈j 42.73Ω And for the capacitor: X C =−j 1 2 π f C X C =−j 1 2 π10kHz 470 nF X C... | ACElectricalCircuitAnalysis_Page_84_Chunk584 |
Ztotal = 1 1 Z1 + 1 Z 2 + 1 Z 3 Ztotal = 1 1 j 4.273Ω + 1 1.8k Ω + 1 −j338.6Ω Ztotal = 4.32889.9°Ω In rectangular form this is 10.4E−3 + j4.328 Ω. The result is inductive, the opposite of what we saw at 10 kHz. Using the inductive reactance formula, it can be shown that at 1 kHz this parallel network has the same impe... | ACElectricalCircuitAnalysis_Page_85_Chunk585 |
This rule is convenient in that the parallel equivalent impedance need not be computed, but remember, it is valid only when there are just two components involved. When analyzing a parallel circuit, if it is being driven by a voltage source, then this same voltage must appear across each of the individual components. O... | ACElectricalCircuitAnalysis_Page_86_Chunk586 |
ZT = Z 1×Z 2 Z 1+Z 2 ZT = 1k Ω× j 2k Ω 1k Ω+j 2k Ω ZT = 89426.6°Ω As a side note, in rectangular form this is 800 + j400, meaning that this parallel combination is equivalent to a series combination of an 800 ohm resistance and a 400 ohm inductive reactance. Continuing with Ohm's law, we have: iSource = v Z T iSource ... | ACElectricalCircuitAnalysis_Page_87_Chunk587 |
whatever phase angle we see on a resistor's voltage, its current phase angle must be the same. The idea is to simply insert a resistor into the branch of which we wish to measure the current. As long as the resistance is much smaller than the impedance seen in the rest of that branch, it will have little impact on the ... | ACElectricalCircuitAnalysis_Page_88_Chunk588 |
ZT = Z 1×Z 2 Z 1+Z 2 ZT = 220Ω×(−j1273Ω) 220Ω−j 1273k Ω ZT = 216.8−9.8°Ω iSource = v Z T iSource = 20 0°V 216.8−9.8°Ω iSource = 92.26E-39.8° A A time domain plot of the currents is illustrated in Figure 3.7 along with a phasor plot in Figure 3.8. Note that the source current is close in both amplitude and phase to ... | ACElectricalCircuitAnalysis_Page_89_Chunk589 |
Computer Simulation Computer Simulation The circuit of Figure 3.6 is captured in a simulator as shown in Figure 3.9. Individual 2 ohm resistors are used to sense the currents in the resistor and capacitor branches. These sensing resistors are inserted directly above ground for convenience of measure. In this way a diff... | ACElectricalCircuitAnalysis_Page_90_Chunk590 |
A basic transient analysis is then performed as shown in Figure 3.10. The results of the simulation are in tight agreement with the plot of Figure 3.7. Note that the plotted resistor and capacitor “currents” are, in fact, the voltages across the associated sensing resistors (nodes 2 and 3) divided by 2 (ohms), in direc... | ACElectricalCircuitAnalysis_Page_91_Chunk591 |
iSource = v Z iSource = 20°V 9.23122.6°Ω iSource ≈0.2167−22.6° A A phasor diagram showing the vector summation of the currents is illustrated in Figure 3.12. Note that iC and iL are in perfect opposition as expected. Subtracting the magnitude of iC from iL leaves us with the precise vertical component of the source ... | ACElectricalCircuitAnalysis_Page_92_Chunk592 |
If a parallel circuit is driven by a current source, as shown in Figure 3.13, there are two basic methods of solving for the component currents. The first method is to use the current divider rule. If desired, the component voltage can then be found using Ohm's law. An alternate method involves finding the parallel equ... | ACElectricalCircuitAnalysis_Page_93_Chunk593 |
iL = iSource−i R iL = 0.30° A −0.234738.5° A iL = 0.1868−51.5°A This value could also have been obtained by applying the current divider rule a second time: iL = iSource R R+X L iL = 0.30° A 7500°Ω 7500°Ω+942.490 °Ω iL = 0.1868−51.5°A A time domain plot of the current waveforms is shown in Figure 3.14. From the... | ACElectricalCircuitAnalysis_Page_94_Chunk594 |
Computer Simulation Computer Simulation The circuit of Example 3.6 is captured in a simulator as shown in Figure 3.15. A pair of 1 ohm current sensing resistors have been added. By using 1 ohm, the sensing voltage will have the same magnitude as the current, with no scaling required. A transient analysis is performed, ... | ACElectricalCircuitAnalysis_Page_95_Chunk595 |
Example 3.7 Determine the branch currents in the circuit of Figure 3.17. Use the current source as the time reference (i.e., 0°). Instead of using repeated current dividers here, we can instead find the equivalent impedance and then apply Ohm's law to determine the system voltage. Each branch current may be obtained b... | ACElectricalCircuitAnalysis_Page_96_Chunk596 |
iC = v X C iC = 9.231−157.4° V 12−90° Ω iC ≈0.7692−67.4 °A iL = v X L iL = 9.231−157.40° V 890°Ω iL = 1.154112.6° A iR = v R iR = 9.231−157.4°V 100°Ω iR = 0.9231−157.4°A The sum of these three currents is approximately 1180° amps (i.e., −10°), the value of the current source, as expected. A phasor diagram of... | ACElectricalCircuitAnalysis_Page_97_Chunk597 |
Now for the sneaky bit. In case you didn't notice, the components in this circuit are identical to those of Figure 3.11 with the exception of the source. As a consequence, the phase angles and magnitudes of the branch currents through the three components will remain unchanged relative to each other. Due to the change ... | ACElectricalCircuitAnalysis_Page_98_Chunk598 |
The next step is to find the combined impedance so that we can find the applied voltage. Ztotal = 1 1 Z1 + 1 Z 2 + 1 Z 3 Ztotal = 1 1 −j100 Ω + 1 j50Ω + 1 40 Ω Ztotal = 37.1421.8°Ω v = i source×Z total v = 2.6752.6° A×37.1421.8° v ≈99.1674.4° V We now apply Ohm's law to each of the components to determine the branc... | ACElectricalCircuitAnalysis_Page_99_Chunk599 |
iR = v R iR = 99.1674.4° V 400°Ω iR = 2.47974.4 °A The sum of these three currents is approximately 2.6752.6° amps. This balances the net entering current from the two sources, verifying KVL. 3.5 Summary 3.5 Summary In this chapter we have examined parallel circuits using either a single voltage source, or one or m... | ACElectricalCircuitAnalysis_Page_100_Chunk600 |
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