text stringlengths 1 7.76k | source stringlengths 17 81 |
|---|---|
Review Questions Review Questions 1. How is the equivalent impedance for a group of parallel connected resistors, inductors and capacitors computed? 2. How is the equivalent value for parallel connected AC current sources computed? 3. Is the product-sum rule still applicable for AC analysis? Can it be used for reactanc... | ACElectricalCircuitAnalysis_Page_101_Chunk601 |
4. Determine the effective impedance of the network shown in Figure 3.24 at 20 kHz. 5. Determine the effective impedance of the network shown in Figure 3.25. 6. Determine the effective impedance of the network shown in Figure 3.25 if the frequency is halved and if the frequency is doubled. 7. For the network shown in F... | ACElectricalCircuitAnalysis_Page_102_Chunk602 |
13. Determine the four branch currents for the circuit shown in Figure 3.28 and draw their phasor diagram. 14. Determine all of the branch currents for the circuit shown in Figure 3.29 assuming E is a 1 volt RMS sine. 15. Determine all of the branch currents for the circuit shown in Figure 3.30 given E = 10 volt peak s... | ACElectricalCircuitAnalysis_Page_103_Chunk603 |
17. Determine the component currents for the circuit shown in Figure 3.32. Draw phasor diagram of the source and branch currents. 18. Determine the resistor and capacitor voltages for the circuit shown in Figure 3.32. 19. Determine the resistor and inductor voltages for the circuit shown in Figure 3.33. 20. Determine t... | ACElectricalCircuitAnalysis_Page_104_Chunk604 |
24. Determine the source voltage for the circuit shown in Figure 3.35. I is 20 mA at 0 degrees. 25. Determine the source voltage for the circuit shown in Figure 3.36. Assume I1 is 1 mA at 0 degrees and I2 is 2 mA at +90 degrees. 26. Determine the capacitor and inductor currents in the circuit of Figure 3.36. Assume I1 ... | ACElectricalCircuitAnalysis_Page_105_Chunk605 |
30. For the network shown in Figure 3.39, determine a value of L such that the impedance magnitude of the circuit is 2 kΩ. The source is a 2 MHz sine and R is 3.3 kΩ. 31. For the circuit shown in Figure 3.38, determine a value for C such that the magnitude of the source current is 1 mA. E is a 2 volt 10 kHz sine and R ... | ACElectricalCircuitAnalysis_Page_106_Chunk606 |
37. Given the circuit shown in Figure 3.38, determine a value for C such that the impedance angle is −45 degrees. The source a 1 volt peak sine at 600 Hz and R = 680 Ω. 38. Given the circuit shown in Figure 3.39, determine a value for L such that the impedance angle is 45 degrees. The source a 10 volt peak sine at 100 ... | ACElectricalCircuitAnalysis_Page_107_Chunk607 |
45. Assume you are troubleshooting a circuit like the one shown in Figure 3.43. I is a 10 mA peak sine at 2 kHz, R = 390 Ω, C = 200 nF and L = 25 mH. The measured resistor voltage is a little under 2.5 volts. What is the likely culprit? 46. Given the circuit shown in Figure 3.43, find the values for C and L if the sour... | ACElectricalCircuitAnalysis_Page_108_Chunk608 |
Notes Notes ♫♫ ♫♫ 109 | ACElectricalCircuitAnalysis_Page_109_Chunk609 |
4 4 Series-Parallel RLC Circuits Series-Parallel RLC Circuits 4.0 Chapter Learning Objectives 4.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Identify series-only and parallel-only sub-groups in series-parallel RLC networks. • Compute complex equivalent impedance for series-paral... | ACElectricalCircuitAnalysis_Page_110_Chunk610 |
4.2 The Series-Parallel Connection 4.2 The Series-Parallel Connection Consider the circuit of Figure 4.1. This circuit is neither just a series circuit nor just a parallel circuit. If it was a series circuit then the current through all components would have to be same, that is, there would no nodes where the current c... | ACElectricalCircuitAnalysis_Page_111_Chunk611 |
Let's begin with a relatively simple series-parallel RLC network where the reactance values have already been found. Example 4.1 Determine the equivalent impedance of the network shown in Figure 4.2. Looking in from the left side, we note that the inductor and 33 kΩ resistor are in parallel as they are both tied to the... | ACElectricalCircuitAnalysis_Page_112_Chunk612 |
Example 4.2 Determine the equivalent impedance of the network shown in Figure 4.3. Looking in from the right side, we see immediately the 750 Ω resistor. This is in series with the sub-circuit comprised of the remaining five components. This sub-circuit can be seen as the −j800 Ω capacitor in parallel with another sub-... | ACElectricalCircuitAnalysis_Page_113_Chunk613 |
This group of four is in parallel with the second capacitor of −j800 Ω. Finally, we arrive at the equivalent total value by placing the resulting group of five in series with the 750 Ω resistor. Z left5 = 1 1 X C2 + 1 Z left4 Zleft5 = 1 1 −j800Ω + 1 822.530.2°Ω Zleft5 = 813.4−31.3° Ω Ztotal = Z left5 +R3 Ztotal = 813... | ACElectricalCircuitAnalysis_Page_114_Chunk614 |
One path would be to find the total impedance seen by the voltage source, Ztotal. Dividing the source voltage by this impedance gives us the source current. We could then perform a current divider between the capacitor and inductor-resistor branches to find the inductor current. Once that current is found, it can be mu... | ACElectricalCircuitAnalysis_Page_115_Chunk615 |
A time domain plot of vb and the source voltage is shown in Figure 4.6. Computer Simulation Computer Simulation To verify the results of the prior example, the circuit of Figure 4.5 is entered into a simulator as shown in Figure 4.7. A time domain or transient analysis is run, examining vb and the source voltage. Node ... | ACElectricalCircuitAnalysis_Page_116_Chunk616 |
The plot is delayed one full cycle in order to get past the initial turn-on transient. The resulting amplitudes and phase shift line up perfectly with the plot of theoretical values in Figure 4.6. Example 4.4 For the circuit of Figure 4.5, determine vab. This circuit can be analyzed as a pair of voltage dividers. By de... | ACElectricalCircuitAnalysis_Page_117_Chunk617 |
vb = e source −j X C −jX C+jX L vb = 1000°V −j800Ω −j800Ω + j1 kΩ vb = 400180° V This may also be written as −4000°. Now we subtract the two voltages to find vab. v ab = va −vb v ab = 80 0°V −400180° V v ab = 4800°V Note that vab is nearly five times larger than the source voltage. This is mostly due to the fact ... | ACElectricalCircuitAnalysis_Page_118_Chunk618 |
Example 4.5 In the circuit of Figure 4.11, determine the current flowing down through the inductor. Use the source as the reference (0°). One possible approach for this is to find the equivalent total impedance that the source drives in order to find the source current. A current divider can then be used between the i... | ACElectricalCircuitAnalysis_Page_119_Chunk619 |
Once va is found, the inductor current can be found using Ohm's law. Each of these solution paths requires about as much work as the other so there is no clear preference. As we just used the voltage divider rule in the prior example, let's use the current divider rule this time. We are going to need the combined capac... | ACElectricalCircuitAnalysis_Page_120_Chunk620 |
iinductor = isource X Ctotal X Ctotal+X L iinductor = 15.6E-3−38.7° A −j 2667Ω −j 2667Ω+j 1000Ω iinductor = 24.99E-3−38.7°A Once again we see a branch current that is larger in magnitude than the source current. This current should produce an inductor voltage of vinductor = iinductor× X L vinductor = 24.99E-3−38.7 °... | ACElectricalCircuitAnalysis_Page_121_Chunk621 |
vb = isource×Zb vb = 20° A×11.7 27.9°Ω vb = 23.4 27.9° V The voltage across the capacitor is vab. We can find this through Ohm's law. Given the reference direction of the current source, the capacitor's voltage reference polarity is + to − from left to right. v ab = i source×X C v ab = 20°A×18−90°Ω v ab = 36−90° ... | ACElectricalCircuitAnalysis_Page_122_Chunk622 |
Example 4.7 For the circuit of Figure 4.14, determine vb if the 1 amp source is used as the reference (0°) and the 3 amp source has a 30° lagging phase angle. The two current sources are in parallel and can be combined together. We must be a little careful regarding polarity, though. First of all, a “30° lagging phase... | ACElectricalCircuitAnalysis_Page_123_Chunk623 |
Z total = 1 1 X C + 1 R1+X L + 1 R2 Ztotal = 1 1 −j80Ω + 1 22 Ω+j50 Ω+ 1 33Ω Ztotal = 26.386.45°Ω v a =−itotal×Ztotal v a =−2.192136.8° A× 26.386.45°Ω v a = 57.8−36.8° V If we had reversed the reference direction of the current source, using 2.192−43.2° amps instead, the leading minus sign would not be required an... | ACElectricalCircuitAnalysis_Page_124_Chunk624 |
Example 4.8 Consider the circuit shown in Figure 4.15. Assume the source is a one volt peak sine wave. Determine voltages va, vb and vc if the source frequency is 10 kHz. Repeat this for an input frequency of 10 Hz. If we treat E as the input and vc as the final output, this circuit behaves as a series of cascading fre... | ACElectricalCircuitAnalysis_Page_125_Chunk625 |
Z right3 = 1 1 X C2 + 1 Z right2 Zright3 = 1 1 −j31.83Ω + 1 5000Ω −j79.58Ω Zright3 = 31.83−89.6°Ω Zright4 = R2+Z right3 Zright4 = 2000Ω +31.83−89.6°Ω Zright4 = 2000−0.91° Ω Zright5 = 1 1 X C1 + 1 Z right4 Z right5 = 1 1 −j15.92Ω + 1 2000−.91°Ω Zright5 = 15.92−89.5°Ω At last we come to va: v a = esource Z right5 R1... | ACElectricalCircuitAnalysis_Page_126_Chunk626 |
Finally, to find vc we perform a voltage divider between the 5 kΩ resistor and 200 nF capacitor using vb as the input. v c = vb X C3 R3+X C3 v c = 253.3−177.3°μ V −j79.58Ω 5000Ω−j79.58Ω v c = 4.0393.6°μ V Obviously, only a tiny percentage of the source signal is found at node c at this frequency. Repeating this proce... | ACElectricalCircuitAnalysis_Page_127_Chunk627 |
The top graph plots the voltages at nodes a, b and c across frequency. It is obvious that, as the frequency increases, the voltage at each node decreases. The lower graph plots the phase shift at each of the nodes and it is apparent that the phase shift increases in the negative direction as frequency is increased. Thi... | ACElectricalCircuitAnalysis_Page_128_Chunk628 |
parallel combination appear more capacitive, and approaching −90 degrees each. A quick check of the voltage magnitudes and phases at 10 Hz indicates that very little signal is lost at the three node and that the phase shifts are close to zero. Further, at 10 kHz, there is considerable signal loss through each section, ... | ACElectricalCircuitAnalysis_Page_129_Chunk629 |
The second capacitor is three times as large and therefore its reactance will be one-third as much, or −j4.08 Ω. For the inductor, X L = j 2π f L X L = j 2π2.6 kHz 360μ H X L≈j 5.88Ω Now that we have the reactances, the loudspeaker voltage can be computed via a pair of voltage dividers. In order find the loudspeaker vo... | ACElectricalCircuitAnalysis_Page_130_Chunk630 |
Computer Simulation Computer Simulation In order to get a better sense of the loudspeaker voltage as a function of frequency, the circuit of Figure 4.18 is captured in a simulator as shown in Figure 4.19. An AC analysis simulation is performed with the output shown in Figure 4.20. 131 Figure 4.19 The circuit of Figure ... | ACElectricalCircuitAnalysis_Page_131_Chunk631 |
Both the magnitude and phase plots corroborate the calculated loudspeaker voltage at 2.6 kHz. The magnitude plot shows that the loudspeaker voltage is very close to the input level at frequencies above about 3 kHz. Below this frequency, the loudspeaker voltage rolls off considerably. Down at 100 Hz, well into the bass ... | ACElectricalCircuitAnalysis_Page_132_Chunk632 |
Review Questions Review Questions 1. In general, describe the process of reducing an AC series-parallel RLC network down to a single equivalent impedance. 2. Do Ohm's law, KVL and KCL still apply in AC series-parallel RLC networks? Why? 3. Is there a finite number of variations of AC series-parallel RLC networks? Why/w... | ACElectricalCircuitAnalysis_Page_133_Chunk633 |
3. Determine the impedance of the circuit of Figure 4.23 at frequencies of 300 Hz, 30 kHz and 3 MHz. 4. Determine the impedance of the circuit of Figure 4.24 at frequencies of 1 kHz, 20 kHz and 1 MHz. 5. Determine the impedance of the circuit of Figure 4.25. 6. Determine the impedance of the circuit of Figure 4.26. 134... | ACElectricalCircuitAnalysis_Page_134_Chunk634 |
7. For the circuit of Figure 4.27, determine the source current and the current through each of the components. 8. For the circuit of Figure 4.27, determine voltages vab and vb. 9. For the circuit of Figure 4.28, determine voltages across R, L and C if the source is 7 volts RMS. 10. For the circuit of Figure 4.28, dete... | ACElectricalCircuitAnalysis_Page_135_Chunk635 |
13. For the circuit of Figure 4.30, determine voltages vab and vb if the source is 20 volts peak. 14. For the circuit of Figure 4.30, determine the source current and the current through each of the four components if the source is 20 volts peak. 15. For the circuit of Figure 4.31, determine the source current and the ... | ACElectricalCircuitAnalysis_Page_136_Chunk636 |
19. For the circuit of Figure 4.33, determine the currents each of the three components. 20. For the circuit of Figure 4.33, determine voltages va and vb. 21. For the circuit of Figure 4.34, determine voltages va and vb. 22. For the circuit of Figure 4.34, determine the middle and right branch currents and draw a phaso... | ACElectricalCircuitAnalysis_Page_137_Chunk637 |
25. For the circuit of Figure 4.36, determine voltages va and vb. isource = 25 mA. 26. For the circuit of Figure 4.36, determine the currents through the two capacitors. 27. For the circuit of Figure 4.37, determine the current through the capacitor. i1 = 10E−30° A and i2 = 3E−390° A. 28. For the circuit of Figure 4.... | ACElectricalCircuitAnalysis_Page_138_Chunk638 |
31. For the bridge circuit of Figure 4.39, determine vab. The source is 50 volts peak. 32. For the bridge circuit of Figure 4.40, determine vab. The source is 6 amps peak. Design Design 33. Determine a new value for the capacitor in Figure 4.27 such that vb is 1.5 volts. 34. Determine the required inductive reactance i... | ACElectricalCircuitAnalysis_Page_139_Chunk639 |
36. In the circuit of Figure 4.41, determine a value for L such that the magnitude of vb equals va/2 if the source frequency is 10 kHz, R = 2.7 kΩ and C = 10 nF. 37. Given the circuit of Figure 4.41, determine a value for C such that the source current is in phase with the source voltage. The source frequency is 1 kHz,... | ACElectricalCircuitAnalysis_Page_140_Chunk640 |
Challenge Challenge 40. Consider the circuit drawn in Figure 4.43. Using only the available components of 1 kΩ, 2.2 kΩ, 1 mH, 5 mH, 10 nF, 75 nF and 560 nF, is it possible to configure a circuit such that va is half the magnitude of vb for a source frequency of 1 kHz? If so, indicate which values could be used for the ... | ACElectricalCircuitAnalysis_Page_141_Chunk641 |
Simulation Simulation 44. Perform a transient analysis to verify the node voltages computed for problem 8. 45. Perform a transient analysis to verify the node voltages computed for problem 12. 46. Perform a transient analysis to verify the node voltages computed for problem 20. 47. Perform a transient analysis to verif... | ACElectricalCircuitAnalysis_Page_142_Chunk642 |
143 Figure 4.45 Figure 4.46 | ACElectricalCircuitAnalysis_Page_143_Chunk643 |
5 5 Analysis Theorems and Techniques Analysis Theorems and Techniques 5.0 Chapter Learning Objectives 5.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Find the voltage source equivalent of a current source and vice versa. • Compute voltages and currents in multi-source RLC network... | ACElectricalCircuitAnalysis_Page_144_Chunk644 |
Finally, we will examine how to find equivalent circuits for certain component arrangements that use three connecting points, in other words, RLC combinations shaped like a triangle or like the letter Y. These are known as delta and Y configurations. These configurations are difficult to address with basic series- para... | ACElectricalCircuitAnalysis_Page_145_Chunk645 |
Consider the signal source shown in Figure 5.2. This is a pickup for an electric bass guitar. This device is used to translate the motions of guitar strings into a voltage that can be fed to an amplifier. It consists of a few thousand turns of very fine magnet wire wrapped around a magnet. When the metal guitar strings... | ACElectricalCircuitAnalysis_Page_146_Chunk646 |
For a current source, the improved model adds an impedance in parallel, as shown in Figure 5.4. This impedance 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 ... | ACElectricalCircuitAnalysis_Page_147_Chunk647 |
For example, given a voltage source, the current that can be developed when the load is shorted is E/Zinternal. Under that same load condition, all of the current from the current source version must be flowing through the load (otherwise the load isn't shorted). Therefore, the value of the equivalent current source mu... | ACElectricalCircuitAnalysis_Page_148_Chunk648 |
Example 5.1 Convert the source of Figure 5.6 into its current source equivalent. E = 2 0° volts RMS. First, the existing impedance of 80 − j60 Ω does not change, it is simply moved to a parallel position. To find the value of the current source, compute the short circuit current the existing source is capable of produ... | ACElectricalCircuitAnalysis_Page_149_Chunk649 |
Computer Simulation Computer Simulation To verify this process, the circuits of Figures 5.8 and 5.9 are entered into a simulator. The original circuit simply specified an inductive reactance so a convenient frequency of 1 kHz was used and then an appropriate inductor was created that would yield the desired reactance o... | ACElectricalCircuitAnalysis_Page_150_Chunk650 |
Now let's turn our attention to using source conversion to simplify and solve a multi- source circuit. Example 5.3 For the circuit of Figure 5.12, determine vb. I = 2E−390° amps peak and E = 100° volts peak. One method of solution is to transform the voltage source into a current source. By doing so, the entire circu... | ACElectricalCircuitAnalysis_Page_151_Chunk651 |
The value of the current source is: I c = E Z I c = 100° V 27 k Ω I c = 0.3704E-30° A The converted circuit is shown in Figure 5.13 (in blue). Combining the two current sources and using an upward reference direction yields: I total = Ic−I I total = 0.3704E-30° A−2E-390° A I total = 2.034E-3−79.5°A The combined im... | ACElectricalCircuitAnalysis_Page_152_Chunk652 |
iEsource = E−vb R1 iEsource = 100° V−8.56−104.4°V 27 kΩ iEsource = 0.544E-334.4° A The remaining three currents should add up to this value. icapacitor = v b X C icapacitor = 8.56 −104.4°V −j 10 kΩ icapacitor = 0.856E-3−14.4° A iR2 = vb R2 iR2 = 8.56 −104.4° V 5.6k Ω iR2 = 1.53E-3−104.4° A Finally, 0.856E−3−14.... | ACElectricalCircuitAnalysis_Page_153_Chunk653 |
Consider the circuit depicted in Figure 5.14. Here we see two voltage sources, E1 and E2, driving a three element series-parallel network. As there are two sources, two derived circuits must be created; one using only E1 and the other using only E2. When considering a given source, all other sources are replaced by the... | ACElectricalCircuitAnalysis_Page_154_Chunk654 |
• Indicate the reference current directions and voltage polarities on each of the new sub-circuits, as generated by the source under consideration. • Solve each of the sub-circuits for the desired voltages and/or currents using standard series-parallel analysis techniques. Make sure to note the reference voltage polari... | ACElectricalCircuitAnalysis_Page_155_Chunk655 |
For this circuit, vb may be determined via a voltage divider. To proceed, we need the impedance of the parallel combo on the right. Z right2 = R× jX C R −jX C Zright2 = 5.6 kΩ×(−j10 kΩ) 5.6 kΩ−j 2k Ω Zright2 = 4886−29.2°Ω Now for the voltage divider to find the contribution of the first source to vb. vb1 = E Z right2 ... | ACElectricalCircuitAnalysis_Page_156_Chunk656 |
We apply Ohm's law to find this source's contribution to vb. vb2 = I×Z total vb2 =−2E-390° A×4208−24.9°Ω vb2 = 8.416−114.9° V The final result is the sum of the two parts: vb = v b1 +v b2 vb = 1.558−24.88° V +8.416−114.9°V vb = 8.558−104.4° V This is virtually the same value obtained using the source conversion t... | ACElectricalCircuitAnalysis_Page_157_Chunk657 |
The reactance values are: X L = j 2π f L X L = j 2π1kHz 50 mH X L≈j 314.2Ω X C =−j 1 2 π f C X C =−j 1 2 π1kHz 750 nF X C ≈−j 212.2Ω A voltage divider can be used to find this portion of vb. The parallel combo of the 2 kΩ resistor and capacitor is 211−83.9° Ω. vb1 = E Z right2 Z right2 + X L vb1 = 10 0° V 211−83.9° ... | ACElectricalCircuitAnalysis_Page_158_Chunk658 |
This contribution is at a frequency of 10 kHz. Thus, the combination is a relatively small 10 kHz sine at about 2 volts peak riding on a 1 kHz sine that is nearly ten times larger in amplitude. This is shown in Figure 5.21. Computer Simulation Computer Simulation In order to verify the two-component waveform of Example... | ACElectricalCircuitAnalysis_Page_159_Chunk659 |
A transient analysis is performed. The results are illustrated in Figure 5.23. The results match the computed values nicely. We can see the small amplitude high frequency sine wave effectively following the contour of the much larger 1 kHz sine wave. 5.4 Thévenin's and Norton's Theorems 5.4 Thévenin's and Norton's Theo... | ACElectricalCircuitAnalysis_Page_160_Chunk660 |
It is important to note that a Thévenin equivalent is valid only at a particular frequency. If the system frequency is changed, the reactance and impedance values will change and the resulting Eth and Zth values will be altered. Consequently, these equivalents are generally not appropriate for a circuit using multiple ... | ACElectricalCircuitAnalysis_Page_161_Chunk661 |
The second part is finding the Thévenin impedance, Zth. Beginning with the “cut” circuit, replace all sources with their ideal internal impedance (thus shorting voltage sources and opening current sources). From the perspective of the cut point, look back into the circuit and simplify to determine its equivalent impeda... | ACElectricalCircuitAnalysis_Page_162_Chunk662 |
Measuring the Thévenin Equivalent in the Laboratory Measuring the Thévenin Equivalent in the Laboratory In a laboratory situation, the Thévenin equivalent can be found quickly and efficiently with the proper tools. First, the circuit is “cut”, leaving just the portion to be Thévenized. The voltage at the cut points is ... | ACElectricalCircuitAnalysis_Page_163_Chunk663 |
There is no current flowing through the inductor due to the open. Therefore, the voltage across the inductor is zero. Consequently, Eth is the voltage across the capacitor, and that can be found with a voltage divider. Eth = E X C X C +R1 Eth = 10 0° V −j 200Ω −j 200Ω +100Ω Eth = 8.944−26.6°V or 8 −j 4 V To find Zth,... | ACElectricalCircuitAnalysis_Page_164_Chunk664 |
The voltage across the 300 Ω resistor can be found directly: v R2 = Eth R2 Z th + R2 v R2 = 8 −j4 V 300Ω 80Ω+ j10 Ω+ 300Ω v R2 = 7.06−28.1° V This value can be verified by following a standard series-parallel simplification. For example, the impedance of the three rightmost components (181.4−53.97° Ω) forms a voltage... | ACElectricalCircuitAnalysis_Page_165_Chunk665 |
The voltage across the 2 kΩ is vb, so we'll treat that resistor as the load in order to define the equivalent circuit. This is redrawn in Figure 5.34. To find Zth, we short the two sources. We're left with the inductor and capacitor in parallel. If this is confusing, remember that we are looking from node b to ground (... | ACElectricalCircuitAnalysis_Page_166_Chunk666 |
vbL = E2 X C X C + X L vbL = 20° V j 314.2Ω −j212.2Ω+j314.2Ω vbL = 6.160° V The sum of the two is −20.840° + 6.160°, or 14.68180° volts. The Thévenin equivalent is a source of 14.68180° volts in series with an impedance of −j653.7 Ω. To find the voltage across the 2 k Ω resistor, we apply it to the equivalent cir... | ACElectricalCircuitAnalysis_Page_167_Chunk667 |
Both the amplitude and phase of the simulation waveform match the computed results. Norton's Theorem Norton's Theorem Norton's theorem is named after Edward Lawry Norton. It is the current source version of Thévenin's theorem. In other words, complex networks can be reduced to a single current source with a parallel in... | ACElectricalCircuitAnalysis_Page_168_Chunk668 |
Example 5.8 Let's reexamine Example 5.6, this time creating a Norton equivalent circuit. For convenience, the original circuit of Figure 5.29 is repeated in Figure 5.37. Once again, the goal will be to determine the equivalent that drives the 300 Ω resistor and to find vc. As noted, the Norton impedance, Zn, is the sam... | ACElectricalCircuitAnalysis_Page_169_Chunk669 |
5.5 Maximum Power Transfer Theorem 5.5 Maximum Power Transfer Theorem The concept of maximum power transfer in DC resistive circuits was presented in earlier work. While maximizing load power is not a goal of all circuit designs, it is a goal of a portion of them and thus worth a closer look. In short, given an AC volt... | ACElectricalCircuitAnalysis_Page_170_Chunk670 |
precise value. We shall proceed with the graphing portion which will lead us to the answer. The more rigorous proof of the second portion is detailed in Appendix C. The curve of Equation 5.1 is plotted in Figure 5.39. The normalized load resistance is set along the horizontal and the normalized power (i.e., for a sourc... | ACElectricalCircuitAnalysis_Page_171_Chunk671 |
At this point we may turn our attention to the possible presence of reactances in both the source and load. It turns out that this is not nearly as complicated as it might look. The key is that only resistors dissipate power, not inductors or capacitors6. Load power is proportional to iload 2, so our immediate goal is ... | ACElectricalCircuitAnalysis_Page_172_Chunk672 |
A single peak is evident when X is 0. This can be achieved by setting the load reactance equal in magnitude to the source reactance but with the opposite sign. In this manner, the reactances will cancel out, leaving a purely resistive circuit with a minimal value, and thus producing maximal current for that set of resi... | ACElectricalCircuitAnalysis_Page_173_Chunk673 |
Two variables are involved here, so to further clarify the situation, a 3D surface plot of normalized power is shown in Figure 5.42. The vertical axis represents the percentage of maximum power while the front and side axes are the normalized total reactance and load resistance, respectively. A single peak is evident h... | ACElectricalCircuitAnalysis_Page_174_Chunk674 |
To achieve maximum load power in these circuits, Rload = Ri and |jXL| = |−jXC|. Note that XL and XC do not have to have the same magnitude as Ri. While using the complex conjugate produces the maximum load power, it does not produce the largest possible load current or load voltage. In fact, this condition produces a l... | ACElectricalCircuitAnalysis_Page_175_Chunk675 |
i = E Z total i = 70V 60Ω +j 10Ω i = 1.151−9.5° A As the voltage and current are in phase for a resistor, we can ignore the angle for the power calculation. Pload = i2×Rload Pload = (1.151A) 2×40Ω Pload ≈53 W This is not the maximum load power that can be achieved because this load is not the complex conjugate of the ... | ACElectricalCircuitAnalysis_Page_176_Chunk676 |
Pload = v R 2 Rload Pload = (35 V)2 20Ω Pload = 61.25W Example 5.10 For the circuit of Figure 5.47, determine the value of Zload that will achieve maximum load power and also determine that power. The current source is 0.10° amps RMS at 50 kHz. The first job is determine the inductive reactance at 50 kHz. Recalling th... | ACElectricalCircuitAnalysis_Page_177_Chunk677 |
The resulting circuit is shown in Figure 5.48. To find the load power we need to find Eth. The open circuit output voltage is the potential appearing across the inductor/resistor pair in Figure 5.47. This is because there is no current flowing through the 50 Ω resistor, and therefore there is no voltage across it. Eth ... | ACElectricalCircuitAnalysis_Page_178_Chunk678 |
5.6 Delta-Y (Pi-T) Conversions 5.6 Delta-Y (Pi-T) Conversions Certain component configurations, such as bridged networks, cannot be reduced to a single impedance using basic series-parallel conversion techniques. One method for simplification involves converting sections into more convenient forms. The configurations i... | ACElectricalCircuitAnalysis_Page_179_Chunk679 |
Δ-Y Conversion Δ-Y Conversion A true equivalent circuit would present the same impedance between any two terminals as the original circuit. Consider the circuits of Figure 5.49 for the unloaded case (i.e., just these networks with nothing else connected to them). The equivalent impedances seen between each pair of term... | ACElectricalCircuitAnalysis_Page_180_Chunk680 |
Y-Δ Conversion Y-Δ Conversion For the reverse process of converting from Y to delta, start by noting the similarities of the expressions for Zd, Ze and Zf (i.e., Equations 5.6 through 5.8). If two of these expressions are divided, a single equation for Za, Zb or Zc will result. For example, using Equations 5.6 and 5.7:... | ACElectricalCircuitAnalysis_Page_181_Chunk681 |
Example 5.11 Convert the network of Figure 5.51 into its delta configuration equivalent. Referring back to Figure 5.50, use Equation 5.9 to determine Za. Za = Z d Z e+Z e Z f +Z d Z f Z f Za = (50Ω + j20 Ω)(10Ω)+(10Ω)(40Ω−j 30Ω)+(50Ω + j20 Ω)(40 Ω−j30Ω) (40Ω −j30Ω) Za = 65.6 + j 29.2Ω Zb and Zc may be determined in sim... | ACElectricalCircuitAnalysis_Page_182_Chunk682 |
Remember, a complex impedance can always be expressed in rectangular form. Rectangular form can be expressed directly as a series combination of a resistor and either an inductive or capacitive reactance. Even if the original impedances of a network are in a parallel form (or even a more complex form), the equivalent c... | ACElectricalCircuitAnalysis_Page_183_Chunk683 |
In similar manner, the top pair is determined to be 19.8 + j1.98 Ω and the rightmost pair is 29.7 + j2.97 Ω. Referring back to Figure 5.50, we can use Equation 5.6 to determine Zd. Zd = Z a Z b Za+Z b+Z c Zd = (19.8Ω+ j 1.98Ω)×(9.9Ω+ j 0.99Ω) (19.8Ω+ j 1.98Ω)+(9.9Ω+ j 0.99Ω)+(29.7Ω+ j 2.97Ω) Zd = 3.3Ω+ j0.33Ω Likewise,... | ACElectricalCircuitAnalysis_Page_184_Chunk684 |
Zupper = Z upperleft×Z upperright Z upperleft + Z upperright Zupper = (5.3Ω + j0.33Ω)×(15.9Ω + j 0.99Ω) (5.3Ω + j 0.33Ω) +(15.9Ω+ j0.99Ω) Zupper = 3.975Ω + j0.2475Ω This is in series with the lower section of 4.95 + j0.495 Ω for a total of 8.9564.76° Ω. Using Ohm's law, we find the source current: isource = E Ztotal i... | ACElectricalCircuitAnalysis_Page_185_Chunk685 |
Source conversions allow us to create an equivalent voltage source for any practical current source and vice versa. An equivalent source is one that will create the same voltage across (and current into) the remaining circuit as did the original source. In some cases, this swap allows multiple sources to be combined in... | ACElectricalCircuitAnalysis_Page_186_Chunk686 |
Review Questions Review Questions 1. What are the ideal internal impedances of AC voltage and current sources? 2. Outline the process of converting an AC voltage source into an AC current source, and vice versa. 3. In general, describe the process of using superposition to analyze a multi- source circuit. 4. What do Th... | ACElectricalCircuitAnalysis_Page_187_Chunk687 |
5. In the circuit of Figure 5.58, use superposition to find vb. Source one is one volt peak and source two is two volts peak. 6. In the circuit of Figure 5.58, use superposition to find the currents through the two inductors. Source one is two volts peak and source two is three volts peak. 7. Use superposition to find ... | ACElectricalCircuitAnalysis_Page_188_Chunk688 |
10. In the circuit of Figure 5.60, use superposition to find the current through the capacitor. The sources are in phase. 11. Use superposition to find the two source currents for In the circuit of Figure 5.61. Source one is 100 mV peak and source two is 500 mV peak (in phase). 12. Use superposition to find vcd for the... | ACElectricalCircuitAnalysis_Page_189_Chunk689 |
16. In the circuit of Figure 5.63, use superposition to find the current flowing through the resistor. 17. For the circuit of Figure 5.64, use superposition to find va and vb. The sources are in phase. 18. For the circuit of Figure 5.64, use superposition to find the currents through the inductor and capacitor. The sou... | ACElectricalCircuitAnalysis_Page_190_Chunk690 |
23. Use superposition to determine the current of source E in the circuit of Figure 5.67. E = 40180° and I = 20E−30°. 24. Use superposition to determine vac in the circuit of Figure 5.67. E = 280° and I = 8E−3−180°. 25. Use superposition to determine vb in the circuit of Figure 5.68. I = 3E−30° and E = 90°. 26. U... | ACElectricalCircuitAnalysis_Page_191_Chunk691 |
29. Use superposition to determine vab in the circuit of Figure 5.70. I = 10E−30° and E = 120°. 30. Use superposition to determine the capacitor current in the circuit of Figure 5.70. I = 5E−30° and E = 18120°. 31. For the circuit of Figure 5.71, determine the Thévenin equivalent that drives the 20 nF capacitor. 32... | ACElectricalCircuitAnalysis_Page_192_Chunk692 |
36. Use Thévenin's theorem to find vb in the circuit of Figure 5.72 if E = 180°. 37. Use Thévenin's theorem to find vb in the circuit of Figure 5.73. 38. Determine the Thévenin equivalent that drives the 3.9 kΩ + j1 kΩ combo in the circuit of Figure 5.73. Does this combo's impedance achieve maximum load power? If not,... | ACElectricalCircuitAnalysis_Page_193_Chunk693 |
41. For the circuit of Figure 5.75, determine the Thévenin and Norton equivalents that drive the combo of 300 Ω in parallel with −j1500 Ω. Does this combo achieve maximum load power? If not, what combo will achieve maximum power and what is the resulting power? E = 1200°. 42. For the circuit of Figure 5.76, determine ... | ACElectricalCircuitAnalysis_Page_194_Chunk694 |
44. Determine the equivalent Y (T) network for the circuit of Figure 5.78. 45. Determine the equivalent Y (T) network for the circuit of Figure 5.79. R1 = R2 = R3 = 4 kΩ and XC1 = XC2 = XC3 = −j3 kΩ. 46. Determine the equivalent Y (T) network for the circuit of Figure 5.80. 47. Determine the equivalent delta (pi) netwo... | ACElectricalCircuitAnalysis_Page_195_Chunk695 |
48. Determine the equivalent delta (pi) network for the circuit of Figure 5.82. 49. Determine the equivalent delta (pi) network for the circuit of Figure 5.83. 50. Determine the equivalent delta (pi) network for the circuit of Figure 5.84. 196 Figure 5.82 Figure 5.83 Figure 5.84 | ACElectricalCircuitAnalysis_Page_196_Chunk696 |
51. Find voltage vbc in the circuit of Figure 5.85 through the use of one or more delta-Y conversions. E = 100°, R1 = 1 kΩ, R2 = 2 kΩ, R3 = 3 kΩ, XC = −j4 kΩ and XL = j8 kΩ. 52. Find voltage vbc in the circuit of Figure 5.86 through the use of one or more delta-Y conversions. E = 200°, R1 = 1 kΩ, R2 = 8 kΩ, R3 = 3 kΩ... | ACElectricalCircuitAnalysis_Page_197_Chunk697 |
54. Design an equivalent current source for Figure 5.87 if E = 100°, R = 2.2 kΩ and C = 100 nF. The source frequency is 1 kHz. 55. Design an equivalent current source for Figure 5.88 if E = 10°, R = 600 Ω and L = 2 mH. The source frequency is 20 kHz. 56. Design an equivalent current source for Figure 5.88. E = 290°,... | ACElectricalCircuitAnalysis_Page_198_Chunk698 |
61. Reconfigure the circuit of Figure 5.91 so that it uses only voltage sources. Express all new component and source values in terms of the original labels. 62. Reconfigure the circuit of Figure 5.92 so that it uses only current sources. Express all new component and source values in terms of the original labels. 63. ... | ACElectricalCircuitAnalysis_Page_199_Chunk699 |
Challenge Challenge 67. Redesign the circuit of problem 7 (Figure 5.59) so that it uses only current sources and produces the same node voltages as the original circuit. 68. In the circuit of Figure 5.93, use any method or combination of methods to determine vab. I1 = 0.050°, I2 = 0.10° and I3 = 0.290°. 69. In the c... | ACElectricalCircuitAnalysis_Page_200_Chunk700 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.