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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 reactance and/or complex impedance? 4. Define Kirchhoff's current law for AC circuit analysis. 5. Is it possible for a branch current to be larger than the source current in an AC parallel circuit? Explain why/why not. 3.6 Exercises 3.6 Exercises Analysis Analysis 1. Determine the effective impedance of the network shown in Figure 3.21 at 10 MHz. 2. Determine the effective impedance of the network shown in Figure 3.22 at 100 Hz. 3. Determine the effective impedance of the network shown in Figure 3.23 at 5 kHz. 101 Figure 3.21 Figure 3.22 Figure 3.23
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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 Figure 3.21, determine the frequency below which the impedance is mostly resistive. 8. For the network shown in Figure 3.22, determine the frequency below which the impedance is mostly inductive. 9. Draw phasor impedance plot for problem 1. 10. Draw phasor impedance plot for problem 2. 11. Determine the three branch currents for the circuit shown in Figure 3.26 and draw their phasor diagram. 12. Determine the three branch currents for the circuit shown in Figure 3.27 and draw their phasor diagram. 102 Figure 3.24 Figure 3.25 Figure 3.26 Figure 3.27
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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 sine, R = 220, XC = −j500. and XL = j1.5 k. 16. Determine all of the branch currents for the circuit shown in Figure 3.31 given E = 2 volt peak sine, R = 1 k, XC = −j2 k. and XL = j3 k. 103 Figure 3.28 Figure 3.29 Figure 3.30 Figure 3.31
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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 the component currents for the circuit shown in Figure 3.33. Draw phasor diagram of the source and branch currents. 21. Determine the source voltage for the circuit shown in Figure 3.34. 22. Determine the component currents for the circuit shown in Figure 3.34. Draw the phasor diagram of the source and branch currents. 23. Determine the component currents for the circuit shown in Figure 3.35. I is 20 mA at 0 degrees. 104 Figure 3.32 Figure 3.33 Figure 3.34 Figure 3.35
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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 is 1 mA at 0 degrees and I2 is 2 mA at +90 degrees. 27. Determine the resistor and capacitor currents in the circuit of Figure 3.37. Assume I1 is 20° amps and I2 is 0.545°. 28. Determine the source voltage for the circuit shown in Figure 3.37. Assume I1 is 20° A and I2 is 0.545°. Design Design 29. For the network shown in Figure 3.38, determine a value of C such that the impedance magnitude of the circuit is 1 kΩ. The source is a 50 Hz sine and R is 2.2 kΩ. 105 Figure 3.36 Figure 3.37 Figure 3.38
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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 = 8 kΩ. 32. For the network shown in Figure 3.39, determine a value for L such that the magnitude of the source current is 10 mA. E is a 25 volt 100 kHz sine and R = 4 kΩ. 33. For the network shown in Figure 3.40, determine a value of C such that the impedance magnitude of the circuit is 10 k Ω. The source is a 440 Hz sine and R is 33 kΩ. 34. For the network shown in Figure 3.41, determine a value of L such that the impedance magnitude of the circuit is 200 Ω. The source is a 60 Hz sine and R is 680 Ω. 35. For the circuit shown in Figure 3.40, determine a value for C such that the magnitude of the circuit voltage is 200 volts. The source current is a 100 mA 1200 Hz sine and R = 15 kΩ. 36. For the circuit shown in Figure 3.41, determine a value for L such that the magnitude of the circuit voltage is 50 volts. The source current is a 2.3 A 60 Hz sine and R = 330 Ω. 106 Figure 3.39 Figure 3.40 Figure 3.41
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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 kHz and R = 1.2 kΩ. 39. Determine a value for C such |XC| = |XL| for the circuit shown in Figure 3.42. The source frequency is 1 kHz, R = 200 Ω and L = 50 mH. 40. Determine a value for L such |XC| = |XL| for the circuit shown in Figure 3.43. The source frequency is 22 kHz, R = 18 kΩ and C = 5 nF. 41. Add one or more components in parallel with the circuit of Figure 3.22 such that the resulting impedance at 20 Hz is 10 Ω with a phase angle of at least +30°. Challenge Challenge 42. Determine a value for C such that the impedance angle for the circuit shown in Figure 3.42 is purely resistive (0 degrees). The source frequency is 1 kHz, R = 200 Ω and L = 50 mH. 43. Is it possible to change the value of the resistor in Figure 3.34 so that the system voltage is 4 volts? If so, what is the value? If not, why not? 44. Is it possible to change the value of the inductor and/or capacitor in Figure 3.34 so that the system voltage is 4 volts? If so, what is/are the values? If not, why not? 107 Figure 3.42 Figure 3.43
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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 source is a sine wave at 1 kHz, R = 4 kΩ, iSource = 3 mA, iR = 2 mA, iL = 5 mA, Simulation Simulation 47. Using a transient analysis simulation, verify that the source current magnitude is 1 mA using the capacitor value determined in design problem 31. 48. Using a transient analysis simulation, verify that the source current magnitude is 10 mA using the inductor value determined in design problem 32. 49. Using a transient analysis simulation, verify that the source voltage magnitude is 200 volts using the capacitor value determined in design problem 35. 50. Using a transient analysis simulation, verify that the source voltage magnitude is 50 volts using the inductor value determined in design problem 36. 51. Using a transient analysis simulation, verify the design solution for problem 39. This can be checked by seeing if the current magnitudes in C and L are identical. 52. Using a transient analysis simulation, verify the design solution for problem 40. This can be checked by seeing if the current magnitudes in C and L are identical. 53. Impedance magnitude as a function of frequency can be investigated by driving the circuit with a fixed amplitude current source across a range of frequencies. The resulting voltage will be proportional to the effective impedance. Investigate this effect by performing an AC analysis on the circuits shown in Figures 3.32 and 3.33. Use a frequency range of 10 Hz to 1 MHz. Before running the simulations, sketch your expected results. 54. Following the idea presented in the previous problem, investigate the impedance as a function of frequency of the circuit shown in Figure 3.43. Use R = 1 k, C = 10 nF, and L = 1 mH. Run the simulation from 100 Hz to 10 MHz. Make sure to sketch your expected results first. 108
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Notes Notes ♫♫ ♫♫ 109
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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-parallel RLC circuits. • Simplify an entire RLC network into a simple series or parallel equivalent comprised of complex impedances. • Utilize KVL, KCL and other techniques to find various voltages and currents in series-parallel RLC networks driven by a single effective voltage or current source. • Interpret phasor diagrams and time domain plots for component voltages and/or currents in series- parallel RLC circuits. 4.1 Introduction 4.1 Introduction Having completed our examination of strictly series and strictly parallel AC circuits, we turn our attention now to somewhat more complex circuits, namely those circuits comprised of components in mixed series-parallel arrangements. This chapter deals with a subset of series-parallel RLC circuits, specifically those that are driven by a single effective current or voltage source, and which may be simplified using series and parallel component combinations. The rules and techniques explored for strictly series networks are still applicable to series connected subsections of larger circuits. The same is true for the rules and techniques established for strictly parallel circuits regarding parallel subsections. Thus, the key to analyzing series-parallel circuits is in recognizing those portions of the circuit that form series or parallel sub-circuits, and then applying the series and parallel analysis rules to those sections. Ohm's law, KVL and KCL may be used in turn to solve portions of the problem until all currents and voltages are found. As individual voltages and currents are determined, this makes it easier to apply these rules to determine other values. It is often useful to determine the effective impedance of individual sections at the outset in order to facilitate circuit analysis. Indeed, continuing the process until the entire network is reduced to a series-only or parallel- only simplified version is a good starting point. That is, each of the complex impedances that make up the series-only or parallel-only simplified equivalent is made up from a sub-circuit which in turn potentially is made up of other sub-circuits, and so on. The art of examining a complex series-parallel network and being able to immediately determine which elements constitute a series connection and which constitute a parallel connection is an essential skill and worthy of practice. 110
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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 could divide. This is clearly not the case as the current flowing through the capacitor can divide at node b, with one portion flowing down through the resistor and the remainder through the inductor. On the other hand, if it was strictly parallel, then all of the components would have to exhibit the same voltage and therefore there would be only two connection points in the circuit. This is also not the case as there are three such points: a, b and ground. What is true for the circuit of Figure 4.1 is that the resistor and the inductor are in parallel. We know this because both components are attached to the same two nodes; b and ground, and must exhibit the same voltage, vb. As such, we can find the equivalent impedance of this pair and treat the result as a single value, let's call it ZP. In this newly simplified circuit, ZP is in series with the capacitor and the source. We have simplified the original circuit into a series circuit and thus the series circuit analysis rules may be applied. 4.3 Series-Parallel Impedance 4.3 Series-Parallel Impedance The rules for combining resistors, capacitors and inductors in AC series-parallel circuits are similar to those established for combining resistors in DC circuits. Obviously, the first item is to determine the reactances of the capacitors and inductors. At that point, simple series and parallel combinations can be identified. These combinations are each reduced to a complex impedance. Once this is completed, the network is examined again to see if these new complex impedances can be identified as parts of new series or parallel sub-circuits, and simplified. This process is repeated until we are left with a single complex impedance. Again, it is useful to remember that the phase angles of the reactive components can sometimes lead to surprising results, such as a series sub-circuit having an impedance magnitude smaller than its largest component — something that would never happen with a network comprised of just resistors. The importance of using vector computations cannot be over stressed. 111 Figure 4.1 A simple series-parallel RLC circuit.
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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 same two nodes. Also, we can see that the capacitor is in series with the 8.2 kΩ resistor. This series combination is, in turn, in parallel with the other two parallel components. Thus, it would make sense to find the series combination first. Z series = R+(−jX C ) Z series = 8.2 kΩ −j 2k Ω Z series = 8440−13.7° Ω We now place this new complex impedance in parallel with the inductor and the 33 kΩ resistor. Z total = 1 1 Z 1 + 1 Z 2 + 1 Z 3 Ztotal = 1 1 j550Ω + 1 33 kΩ + 1 8440−13.7°Ω Ztotal = 556.885.4°Ω Clearly, the inductor dominates here. The parallel resistor is roughly two orders of magnitude larger than the inductive reactance and has minimal impact on a parallel combination. Further, the complex impedance derived from the capacitor/resistor combination is also considerably larger, and given that it has a negative (capacitive) phase angle, it partly cancels the inductive reactance. This leaves us with a magnitude a little higher than that of the inductive reactance alone, and with a phase angle shifted toward the resistive side. The series and parallel combinations can be much more complicated than that of the prior network. Ladder networks, for example, feature a set of sections that load other sections, resulting in repeated series and then parallel simplifications. In this situation, it is best to start work at the end farthest from the nodes of interest. The following example will illustrate this on a modest scale. 112 Figure 4.2 Network for Example 4.1.
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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-circuit containing the other four components. This four component sub- circuit consists of the inductor in series with yet another sub-circuit consisting of the final two resistors and capacitor. This three element sub- circuit consists of the 2.2 k Ω resistor in parallel with the series combination of the 1 k Ω resistor and the −j400 Ω capacitor. The most sensible way to approach this is to start at the left end with the simple RC series combination and then work right, toward the nodes of interest. We'll number the components from left to right for identification. Zleft2 = R1+(−jX C1) Zleft2 = 1k Ω−j 400Ω Zleft2 = 1077−21.8°Ω We now place this complex impedance in parallel with the 2.2 kΩ resistor. This creates a three element sub-circuit which is in series with the inductor. Z left3 = 1 1 R2 + 1 Zleft2 Zleft3 = 1 1 2.2 kΩ + 1 1077−21.8°Ω Zleft3 = 734.7−14.7° Ω Zleft4 = Z left3 +jX L Zleft4 = 734.7−14.7°Ω +j 600Ω Zleft4 = 822.530.2 °Ω 113 Figure 4.3 Network for Example 4.2.
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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 = 813.4−31.3°Ω +750Ω Ztotal = 1506−16.3°Ω In rectangular form this is 1443 −j422.3 Ω, meaning that this network is equivalent to a 1443 Ω resistor in series with a capacitive reactance of −j422.3 Ω. Series-parallel simplification techniques will not work for all circuits. Some networks such as delta or bridge configurations require other techniques that will be addressed in later chapters. 4.4 Series-Parallel Analysis 4.4 Series-Parallel Analysis Given the infinite variety of series-parallel configurations, there are myriad ways of solving any given circuit for a particular current or voltage. Many solution paths exist. This is good, because while you might not see a particular path, there are others that will also provide correct results. The only issue is which path is most efficient or convenient for you. Suppose we are trying to find vb in the circuit of Figure 4.4. How might we approach this problem? 114 Figure 4.4 A series-parallel RLC circuit.
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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 multiplied by the inductive reactance to find vb. Alternately, having found the total impedance, we could compute the voltage divider between the three components on the right and R1 to find va. Knowing va, a second voltage divider between XL and R2 gives us vb. A third possibility would be to find the source current and use that to find va, perhaps by finding the drop across R1 and subtracting that from the source, E. Once va is found, a voltage divider can be used to find vb. Undoubtedly there are other solution paths that will work here. Some are more “computationally expensive” than others, but as long as you can identify one of them, the answers are within reach. Remember, the larger the circuit becomes, the greater the number of possible solution paths. Don't fall into the trap of relying on the same “trick” for every circuit, though. It is useful to solve these circuits using a variety of techniques as a means of cross-checking the results and sharpening your skill set. Example 4.3 Determine vb for the circuit of Figure 4.5 if the source frequency is 100 Hz. The first thing to do is to find the capacitive reactance. X C =−j 1 2 π f C X C =−j 1 2 π100Hz 75 nF X C ≈−j 21.22kΩ This reactance is in parallel with the 27 kΩ resistor. Their combination is: Zrc = R×(−jX C) R +(−jX C) Zrc = 27 kΩ×(−j21.22 kΩ) 27 kΩ −j 21.22k Ω Zrc ≈16.68E3−51.8°Ω This impedance forms a voltage divider with the 47 kΩ resistor to create vb. vb = e source Z rc R+Z rc vb = 90 0° V 16.68E3−51.8Ω 47k Ω +16.68E3−51.8 ° Ω vb ≈25.5−38.9° V 115 Figure 4.5 Circuit for Example 4.3.
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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 2 corresponds to vb. The results are shown in Figure 4.8. 116 Time Domain Graph Voltage (volts) -100 -50 0 50 100 Time (milliseconds) 0 2 4 6 8 10 vb esource Figure 4.6 Time domain plot of voltages for the circuit of Figure 4.5. Figure 4.7 The circuit of Example 4.3 in the simulator.
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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 definition, vab = va − vb. Numbering the resistors from top to bottom gives us: v a = esource R2 R1+R2 v a = 1000° V 40 kΩ 10k Ω + 40 kΩ v a = 80 0°V 117 Figure 4.8 Transient analysis of the circuit of Example 4.3. Figure 4.9 Circuit for Example 4.4.
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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 that vb itself is four times the source magnitude. Due to the fact that XL and XC are relatively close in size, they largely cancel each other when placed in series. This produces a small net reactance which creates a large current. This considerable current then produces large voltages across these components. The closer the magnitudes of XL and XC, the higher the L and C component voltages. We will examine this effect in detail when we discuss resonance in Chapter 8. To verify this result, we can calculate the voltage across the inductor and check to see if KVL is satisfied. vinductor = esource j X L −jX C+jX L vinductor = 1000° V j 1k Ω −j800Ω+ j1k Ω vinductor = 5000° V Adding the vb of −4000° volts to vinductor does indeed yield the source voltage of 1000° volts. This can be seen graphically in Figure 4.10. First, note that the inductor voltage is in phase with the source voltage. This is because the LC branch appears to be net inductive, producing a current lagging the source voltage by 90 degrees. This same current flows through the inductor, meaning its voltage leads this current by 90 degrees, and thus the inductor voltage is in phase with the source voltage. The lagging current also flows through the capacitor which produces a further 90 degree lag for the capacitor voltage (i.e., vb) or 180 degrees total. Combining the large inductor voltage with a capacitor voltage that is nearly as large but effectively inverted yields the smaller source voltage. 118
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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 inductor and the pair of capacitors (all three being in parallel). Another option would be to find the impedance of the three reactive components and then use the voltage divider rule to find va. 119 Time Domain Graph Voltage (volts) -500 -400 -300 -200 -100 0 100 200 300 400 500 Time (cycles) 0 0.2 0.4 0.6 0.8 1 vb vinductor esource Figure 4.10 Time domain plot of the circuit of Figure 4.9. Figure 4.11 Circuit for Example 4.5.
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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 capacitive reactance for the current divider, and we will also need it to find the total impedance, so let's do that first. X Ctotal = 1 1 X C1 + 1 X C2 X Ctotal = 1 1 −j 4 kΩ + 1 −j 8k Ω X Ctotal = 2667−90°Ω This value is in parallel with the inductive reactance, and that combo is in series with the resistor, yielding the total impedance. Z CL = 1 1 X Ctotal + 1 X L ZCL = 1 1 −j2667Ω + 1 j1k Ω ZCL = 160090°Ω Ztotal = R+Z CL Ztotal = 2 kΩ+160090°Ω Ztotal = 256138.7°Ω The source current is found using Ohm's law. isource = esource Z total isource = 400° V 256138.7°Ω isource = 15.6E-3−38.7°A We now apply a current divider between XCtotal and the inductor. 120
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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 ° A×100090 ° Ω vinductor = 24.99 51.3° V The voltage across the resistor is v R = i source×R v R = 15.6E-3 −38.7° A×20000° Ω v R = 31.2−38.7°V KVL indicates that the sum of vR and vinductor should equal the source of 400° volts, and it does (within rounding limits). It is now time for some examples that use current sources. Example 4.6 Determine va, vb and vab in the circuit of Figure 4.12. Use the source as the reference angle of 0 degrees. To find vb we can determine the equivalent impedance of the two resistors and the inductor and multiply it by the source current. The rightmost resistor and inductor are in series, yielding 10 + j20 Ω. This is in parallel with the 18 Ω resistor. Z b = R1×Z LR R1 +Z LR Zb = 18Ω×(10Ω+ j 20Ω) 18Ω+(10Ω+ j 20Ω) Zb = 11.727.9°Ω 121 Figure 4.12 Circuit for Example 4.6.
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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° V Finally, va is just vab plus vb based on KVL. v a = vab+v b v a = 36 −90° V+23.4 27.9°V v a = 32.48−50.5°V A phasor diagram is shown in Figure 4.13. Graphically, it can be seen that subtracting vb from va yields vab, as expected. Remember, this is a series- parallel circuit and therefore we do not see necessarily 0 degree or 90 degree angles between the various voltages as found in simple series-only circuits. 122 Figure 4.13 Phasor voltage plot for the circuit of Figure 4.12.
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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 angle” means that the second source is 3−30° amps. Along with this, its reference direction is opposite that of the first source. This means that the second source is negative or inverted by 180 degrees relative to source one. Thus, we can treat it as a downward source of −3−30° amps, or 3150° amps, whichever we prefer. Now that they're both configured as having a downward reference direction, we simply add them together. itotal = i1+i2 itotal = 10° A+3150° A itotal = 2.192136.8° A Alternately, we could subtract 3−30° amps from the first source based on the reference directions, and note that the resulting direction of the combination is the same as that of the first source. Another option would be to reverse the reference direction of the first source. This would yield an upward direction with a value of 2.192−43.2° amps. Having simplified the circuit to a single current source, it should be obvious that the inductor is in series with the 22 Ω resistor, and that combination is in parallel with both the capacitor and the 33 Ω resistor. Finding that parallel impedance would allow us to find va. Knowing va, a voltage divider between the series inductor/resistor combo will yield vb. An important thing to note is that, given the downward reference direction of the equivalent current source, KCL indicates that the current direction through the other components must be upward, meaning that both va and vb are negative with respect to ground. 123 Figure 4.14 Circuit for Example 4.7.
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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 and we would arrive at the same result. Continuing, vb = v a( R1 R1+X L) vb = 57.8−36.8° V 22Ω 22Ω+j50Ω vb = 23.29−103° V Analysis Across the Frequency Domain For the most part, we have examined the response of a circuit to a single frequency of excitation. In many electronic systems, such as in the field of communications, numerous frequencies are present simultaneously. Recall from Chapter 1 how complex wave shapes such as square waves, triangle waves or music signals can be built from a series of sine waves. In such systems, the reactive components behave as different values to the various frequencies simultaneously. For example, a capacitor may have a reactance of −j400 Ω for a 100 Hz signal while at the same time offering a reactance of −j40 Ω for a 1 kHz signal. It is this dynamic quality that allows us to design circuits to suppress or block certain frequency components, or to select specific frequencies from a large range or spectrum of frequency components. We shall introduce this concept by first analyzing the circuit at a couple of specific frequencies and then employ a simulator to perform a frequency domain analysis (sometimes called an AC analysis) to plot complex response curves of voltage versus frequency. The concept of frequency domain response will be expanded in upcoming work, particularly in Chapter 10. 124
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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 frequency-dependent voltage dividers. Generally speaking, at low frequencies the capacitive reactances will be larger than the associated resistors, and most of the input voltage will make it to node c. At high frequencies, the capacitive reactances will be small resulting in considerable voltage division at each node. Thus, only a small percentage of the input will make it to the final output. In other words, this circuit will filter out or remove high frequencies from the input with considerable effect, much more so than a single RC network. First, we need to find the three capacitive reactances at 10 kHz. Starting at the left, we find X C =−j 1 2 π f C X C =−j 1 2 π10Hz 1μ F X C ≈−j15.92Ω The other capacitive reactances work out to −j31.83 Ω and −j79.58 Ω. The voltage va can be determined by a voltage divider between the 1 kΩ resistor and the series-parallel combination of the remaining five components. First, the 5 kΩ is in series with the 200 nF. That combination is in parallel with the 500 nF, which is in turn in series with the 2 kΩ resistor. Finally, that group of four is in parallel with the 1 μF capacitor. The resistors and capacitors are numbered from left to right in the equations following. We will need each of the segment impedances for subsequent calculations. 125 Figure 4.15 Circuit for Example 4.8.
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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+Z right5 v a = 10° V 15.92−89.5° Ω 1000Ω+15.92−89.5°Ω v a = 15.92−88.6°mV To find vb we perform a voltage divider between the 2 kΩ resistor and Zright3 using va as the input. vb = v a Z right3 R2+Z right3 vb = 15.92−88.6° mV 31.83−89.6°Ω 2000Ω +31.83−89.6° Ω vb = 253.3−177.3°μ V 126
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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 process at 10 Hz yields capacitive reactances of −j15.92 kΩ, −j31.83 kΩ and −j79.58 kΩ. At this level, the amount of signal lost through each segment is inconsequential. For example, for the final segment the voltage divider ratio works out to: v c vb = X C3 R3+X C3 vc vb = −j79.58k Ω 5000Ω−j79.58k Ω v c vb = 0.998−3.6° In other words, a mere 0.2% of the signal is lost and there is a modest −3.6° phase shift. The results at the other nodes are similar and left as an exercise. Thus we see that that the low frequencies are allowed through this network while the high frequencies are attenuated. Computer Simulation Computer Simulation While the results of Example 4.8 should be convincing as to the performance of the circuit, it should also be obvious that determining the voltages for any set of source frequencies would be a tedious exercise. Fortunately, there are other techniques that may be employed, such as those examined in Chapter 10. For now, though, we will turn our attention to a simulator. Most simulators offer an AC analysis or frequency domain analysis that will create two linked graphs; one for the voltage magnitude and another for the phase. We begin by entering the circuit of Figure 4.15 into a simulator as shown in Figure 4.16. Even though the schematic shows a 1 kHz source frequency, the AC analysis will allow us to specify the starting and ending frequencies for the plots. In this case, we'll use the 10 Hz and 10 kHz points specified in the example. The results are shown in Figure 4.17. 127
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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. This is expected because, as the frequency increases, the capacitive reactance decreases, making each 128 Figure 4.16 The circuit of Figure 4.15 in a simulator. Figure 4.17 Frequency domain plots of voltage for the circuit of Figure 4.15.
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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, with each section producing nearly −90 degrees, just as calculated. Perhaps the only curious bit here is the abrupt change in phase shift shown at node c around 300 Hz (red trace). This is just an artifact of the plotting software. If an angle goes beyond ±180 degrees, the value is rotated back the other way to keep the value within ±180. For instance, −185 degrees is the same as +175 degrees. Combining reactive elements can be a very effective means of selecting out a certain range of frequencies, as further illustrated in the following example. Example 4.9 In Chapter 2 we introduced the concept of a loudspeaker crossover network. The idea was to “steer” low frequencies to the woofer (low frequency transducer) and high frequencies to the tweeter (high frequency transducer). An advancement on that simple system is to use a combination of capacitors and inductors in place of a simple RC or RL network. One possible configuration is illustrated in Figure 4.18. At high frequencies, the capacitive reactance will be small while the inductive reactance will be large. Thus, virtually all of the input signal will reach the loudspeaker. In contrast, at low frequencies the capacitive reactances will be large and the inductive reactance small, resulting in hardly any of the input signal reaching the loudspeaker. Somewhere in the middle, a significant portion of the signal will make it through. This point is referred to as the crossover frequency. If the source voltage is 1 volt peak, determine the voltage developed across an 8 Ω loudspeaker at a frequency of 2.6 kHz for this circuit. First, we need to determine the reactances at the frequency of interest. X C1 = −j 1 2π f C X C1 = −j 1 2π 2.6kHz 5μ F X C1 ≈−j12.24Ω 129 Figure 4.18 Circuit for Example 4.9.
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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 voltage we'll first find the voltage developed across the inductor. To find that, we need to find the combined impedance of the three components on the right. Zright3 = 1 1 X L + 1 Z right2 Zright3 = 1 1 j5.88Ω + 1 8Ω −j 4.08Ω Zright3 = 6.4450.3°Ω Now for the voltage divider to find vinductor. vinductor = esource Z right3 Z right3+X C1 vinductor = 10° V 6.4450.3°Ω 6.44 50.3°Ω−j12.24Ω vinductor = 0.77110.8° V And now the final voltage divider to find vloudspeaker. vloudspeaker = vinductor Zloudspeaker Z loudspeaker+X C2 vloudspeaker = 0.77110.8°μ V 8Ω 8Ω−j 4.08Ω vloudspeaker = 0.686137.9° V At this particular frequency the loudspeaker sees about 2/3rds of the source voltage. For any higher frequency, the loudspeaker will see a larger share of the 1 volt source and for any lower frequency, the loudspeaker see less. At very low frequencies, only a few microvolts may get through. 130
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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 4.18 in a simulator. Figure 4.20 Frequency domain plot of the loudspeaker voltage for the circuit of Figure 4.18.
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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 region, less than 100 microvolts, or under 0.01% of the input, reaches the loudspeaker. This circuit would make for an effective crossover network to a high frequency tweeter. In closing, it is worth noting that a loudspeaker exhibits a complex impedance instead of simple resistive value, however, modeling it as an 8 Ω resistor is sufficient to illustrate the operation of this circuit. We will take a closer look at the impedance of loudspeakers and other devices in upcoming chapters. 4.5 Summary 4.5 Summary In this chapter we have determined how to identify basic series-parallel RLC networks driven by a single effective voltage or current source. The key to this is to identify sub-circuits or subgroups of components that are comprised of either series- only or parallel-only configurations within themselves. These groupings can then be reduced to equivalent impedances using the series and parallel combination techniques examined in prior chapters. This process may be repeated until the entire circuit is simplified down to either a single series loop or parallel arrangement of components driven by a voltage or current source. Once a circuit has been simplified, series and parallel analysis techniques, and laws such as Ohm's law, Kirchhoff's voltage and current laws, and the voltage and current divider rules, may be employed to determine various voltages and currents in the simplified equivalent. Given these results, the circuit may be expanded back into its original form in stages, reapplying these rules and techniques to determine voltages and currents within the sub-circuits. The process may be iterated until every current and voltage in the original circuit is discovered, if desired. As the impedances of the individual sub-circuits can be anywhere between +90 and −90 degrees, phasor diagrams of the various component voltages or currents will no longer exhibit the strict right angles seen in series-only and parallel-only circuits. What is true is that this perpendicular relationship will still exist among the RLC components that comprise a specific series or parallel sub-circuit. There are infinite varieties of series-parallel RLC configurations and consequently no single solution technique will work for all of them. In fact, the more complex the circuit, the more solution paths that exist for said circuit. Consequently it is prudent to plan out a solution path instead of just randomly “diving in” as this will lessen the ultimate effort. 132
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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/why not? 4. Describe a general procedure to find the voltage between two arbitrary points in a series-parallel circuit. 5. In an AC series-parallel RLC circuit, will it always be the case that voltage across any resistor is in phase with that resistor's current? Why/why not? 6. In an AC series-parallel RLC circuit, will it always be the case that voltage across any inductor leads any resistor's voltage by 90 degrees? Why/why not? 4.6 Exercises 4.6 Exercises Analysis Analysis Assume the source's angle is 0 degrees unless specified otherwise. 1. Determine the impedance of the circuit of Figure 4.21 at frequencies of 100 Hz, 10 kHz and 1 MHz. 2. Determine the impedance of the circuit of Figure 4.22 at frequencies of 20 Hz, 1 kHz and 20 kHz. 133 Figure 4.21 Figure 4.22
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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 Figure 4.23 Figure 4.24 Figure 4.25 Figure 4.26
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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, determine the source current and the current through each of the three components. Also, draw a phasor diagram of E, vL and vR. 11. For the circuit of Figure 4.29, determine the source current and the current through each of the components. 12. For the circuit of Figure 4.29, determine voltages vab and vb. Also, draw a phasor diagram of E, vab and vb. 135 Figure 4.27 Figure 4.28 Figure 4.29
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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 current through each of the four components. 16. For the circuit of Figure 4.31, determine voltages vab and vb. 17. For the circuit of Figure 4.32, determine voltages vab and vb if the source is 100 volts peak. 18. For the circuit of Figure 4.32, determine the currents through the two resistors. 136 Figure 4.30 Figure 4.31 Figure 4.32
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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 phasor diagram of three circuit currents. 23. For the circuit of Figure 4.35, determine voltages va and vb. 24. For the circuit of Figure 4.35, determine the currents through the two resistors. 137 Figure 4.33 Figure 4.34 Figure 4.35
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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 4.37, determine voltages va and vb. I1 = 10E−30° A and I2 = 3E−390° A. 29. For the circuit of Figure 4.38, determine voltages va and vb. i1 = 245° A and i2 = 0.50° A. 30. For the circuit of Figure 4.38, determine the currents through the two resistors. i1 = 245° A and i2 = 0.50° A. 138 Figure 4.36 Figure 4.37 Figure 4.38
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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 in Figure 4.28 to shift the capacitor voltage to half of the source voltage. 35. Determine a new value for the 20 nF capacitor in Figure 4.29 such that the resistor current is 2 mA. 139 Figure 4.39 Figure 4.40
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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, R = 68 Ω and L = 22 mH. 38. Given the circuit of Figure 4.42, determine a value for L such that vb is 1 volt. The source is a 6 volt peak sine at 50 kHz, R1 = 510 Ω and R2 = 220 Ω. 39. Given the circuit of Figure 4.39, determine a new value for the inductor such that the magnitude of vb equals the magnitude of va. Assume that the source frequency is 20 kHz. 140 Figure 4.41 Figure 4.42
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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 four components. If not, explain your reasoning. 41. Given the circuit of Figure 4.29, determine the frequency at which vb is half of the source voltage. 42. For the circuit of Figure 4.44, determine voltages va, vb , and vc. i1 = 50° A and i2 = 390° A. 43. Given the circuit of Figure 4.39, is it possible to change the values of the two resistors such that the phase angle of va is the same as that of vb? If so, what are the new values, and if not, explain why it is not possible. 141 Figure 4.43 Figure 4.44
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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 verify the node voltages computed for problem 21. 48. Consider the circuit of problem 17. Assuming the source frequency is 10 kHz, determine values for the capacitors and inductors. Then, use a transient analysis to verify the results of problem 17. 49. Perform a transient analysis on the result of problem 33 to verify the accuracy of the design. 50. Perform a transient analysis on the result of problem 34 to verify the accuracy of the design. 51. Use an AC frequency response simulation to verify the results of problem 41. 52. The concept of a loudspeaker crossover network was presented originally in Chapter 2, Series RLC Circuits. In this chapter, we noted that by adding more components, it is possible to increase the rate of attenuation. In doing so the undesired signals are further reduced in amplitude. The circuits of Figure 4.45 and 4.46 (following page) show improved crossovers for a woofer and tweeter, respectively. Assuming standard 8 Ω loudspeakers, use an AC frequency domain simulation to determine the crossover frequency of each network. Also, compare the curves at node a to those at node b. Finally, compare the attenuation slopes to those generated by the simpler crossover network presented at the end of Chapter 2. Component values for the woofer: L1 = 760 μH, L2 = 250 μH, C = 10.6 μF. Component values for the tweeter: C1 = 5.3 μF, C2 = 16 μF, L = 380 μH. 142
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143 Figure 4.45 Figure 4.46
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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 networks using superposition. • Simplify RLC networks using Thévenin's and Norton's theorems. • Determine conditions for maximum power transfer and compute the maximum power. • Utilize delta-Y and Y-delta conversions for circuit simplification. 5.1 Introduction 5.1 Introduction In this chapter we shall examine a number of theorems and techniques to help us analyze complex circuits and address specialized applications. We will begin by examining the concept of source impedance in order to make more accurate models of our idealized constant voltage and current sources. This will be a step beyond using a simple resistance as found in the DC case. From there we will investigate how to convert from one type of source to another, such as creating a voltage source that is the functional equivalent of a current source. A functional equivalent is a source that can be swapped out for another while leaving all of the other circuit currents and voltages intact. In other words, all of the circuit's component voltage drops and branch currents will be identical to those found in the original configuration. This technique is useful in a number of ways, particularly in that it can help reduce more complex circuits to simplify analysis. The concept of equivalence can be extended beyond just a single source to an entire network. For this we will examine Thévenin's and Norton's theorems. Using these theorems, entire circuits utilizing dozens of components can be modeled as a single source with an associated complex impedance. When coupled with the maximum power transfer theorem, these tools will allow us to determine component values that produce the maximum amount of load power. We will also address a method of analyzing circuits that contain multiple current and/or voltage sources that are connected in a non-trivial fashion (i.e., not just series voltage sources or parallel current sources). This is called the superposition theorem and it can be applied to any circuit or parameter that meets certain requirements, including circuits that have a mix of current sources and voltage sources. Superposition can also be used to determine voltages and currents when sources use different frequencies. In fact, one of way of imagining a complex waveshape is to treat it as a series of connected sources, each with a unique frequency, phase and amplitude. Superposition we give us a means to handle this new situation. 144
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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- parallel simplification techniques. Converting from one configuration to the other will help solve that issue. 5.2 Source Conversions 5.2 Source Conversions In DC analysis, we noted that real world sources have practical limits: voltage sources cannot produce infinite current and current sources cannot produce infinite voltage. A simple way of creating a more accurate model for independent sources is to include an internal resistance. For DC voltage sources, a resistance is added in series with the source, and for DC current sources a resistance is added in parallel with the source. While this works well enough for typical DC sources, the AC situation is a little more complicated. Models for AC Sources Models for AC Sources Just as we added a simple resistance to the DC sources to make improved models, we can add a complex impedance to AC sources to do likewise. Once again, it is possible to make even more involved models that will be more accurate, but for most work, this addition will suffice. Generally, there are wider variations in values for the AC case than the DC case. We can think of AC sources as belonging in one of two broad categories. First, there are power generators, that is, systems designed to generate and deliver power for other electrical devices. This would include the AC mains system in a residence or a portable power generator. At the other end of the spectrum are signal sources such as transducers and sensors. These devices are generally low power and are not designed to produce particularly high currents, quite the opposite of generators. The model for an AC voltage source adds an impedance in series, as shown in Figure 5.1. This impedance sets an upper limit on the source's current output. Even if the output terminals are shorted, the maximum current will be dictated via Ohm's law to be the source voltage divided by the internal impedance, or E/Zinternal. Obviously, this internal impedance will create some voltage divider effect with the attached load. To minimize this effect, the impedance should be as small as practicably possible. Thus, The ideal internal impedance of a voltage source is zero ohms (a short). It is not always possible to get close to this ideal. In fact, in some situations the AC source is far removed from the ideal. 145 Figure 5.1 Practical AC voltage source model.
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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 vibrate, they alter the surrounding magnetic field. As the magnetic field changes, it induces a current, and hence a voltage, in the coil. The resulting signal is an electrical analog of the string vibrations. It is, quite literally, an AC signal source. But what of its internal impedance? The coil is made of wire between AWG 40 and 45, and may be in excess of 1000 feet (300 meters) in length. Consulting an AWG table3, we find that this will yield 1000 or more ohms of resistance. Of course, a few thousand turns of wire will also create a hefty inductance and pickups such as this may exhibit an inductance of a few henries. Putting these pieces together, the equivalent internal impedance might look something like that shown in Figure 5.3. Clearly, values like these are a far cry from the fractional ohm values found in power generation sources, and they present unique challenges. In the circuit of Figure 5.3, the component labeled Zin represents the input impedance of the associated amplifier. Typically, it would be highly resistive but the magnitude can vary quite a bit depending on the application. A guitar or bass amplifier may exhibit 1 MΩ or more, while an auxiliary input on a home stereo might be 10 kΩ. What happens if you plug a guitar into your hi-fi? Obviously, this system creates a voltage divider between the internal impedance of the pickup and the input impedance of the amplifier, but more importantly, the divider is a function of frequency. At low frequencies like 100 Hz, the inductive reactance is small, around j600 Ω, and the signal loss to a 10 kΩ input is not that great. On the other hand, at a high frequency such as 10 kHz, the inductive reactance will be over j60 kΩ resulting in considerable signal loss. Thus, the high frequencies are reduced relative to the low frequencies. The sonic effect is akin to turning down the treble control — everything will seem muffled. If the input impedance is increased considerably, say by a factor of 100, then the voltage divider effect at all audible frequencies will be negligible and the the instrument will sound true to form. Finally, although this example shows a substantial inductance, it is possible for AC sources to have an associated capacitance, or even have negligible reactance (i.e., be purely resistive). 3 Such as the one found in Chapter 2 of the companion text, DC Electrical Circuit Analysis, a companion free OER text by the author. 146 Figure 5.2 An electric bass guitar pickup (cover removed). Figure 5.3 Example of internal impedance of guitar pickup.
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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 times the internal impedance, or I∙Zinternal. This internal impedance will create some current divider effect with the attached load. To minimize this effect, the internal impedance should be as large as practicably possible. Thus, The ideal internal impedance of a current source is infinite ohms. From here on, whenever we deal with practical voltage and current sources, we understand that these sources have some associated internal resistance, even if they're not shown explicitly in a schematic diagram. Further, whenever we talk about ideal sources, we simply use a short for the internal impedance of a voltage source and an open for the internal impedance of a current source. Source Equivalences Source Equivalences For any voltage source consisting of an ideal voltage source with a series internal impedance, an equivalent current source may be created. Similarly, for any current source consisting of an ideal current source with a parallel internal impedance, an equivalent voltage source may be created. By “equivalent”, we mean that both circuits will produce the same voltage and current to identical loads. Consider the simple voltage source on the left side of Figure 5.5. Its equivalent current source is shown on the right. For reasons that will become apparent under the section on Thévenin's theorem following, the internal impedances of these two circuits must be identical if they are to behave identically. Knowing that, it is a straightforward process to find the required values of the other source. The current/voltage characteristic is linear for these circuits, and a straight plot line can be defined by just two points. The two obvious points to use are the opened and shorted load cases. In other words, if the circuit is equivalent for these two situations, it must work for any load. The shorted load case produces a large load current with zero load voltage, and the opened load case produces a large load voltage with zero load current. It would not make sense if the equivalent source could produce a greater current or voltage under the same extreme conditions as the original. 147 Figure 5.4 Practical AC current source model. Figure 5.5 A simple AC voltage source (left) and corresponding current source (right).
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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 must be the current of E/Zinternal. Note that the resulting source normally will not have the same phase angle as the original source due to the phase angle of the associated impedance. 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 Zinternal times the current E/Zinternal, or just E. Thus, the two behave identically at the load limits. Similarly, if we start with a current source, an open load produces a load voltage of I∙Zinternal. Therefore, the equivalent voltage source must have a value of I∙Zinternal. For the current source, a shorted load would produce a load current equal to the source value, or I. The voltage source version would produce a current of E/Zinternal, where the value of E was just found to be equal to I∙Zinternal, and thus the load current would be I∙Zinternal/Zinternal, or just I. Once again, the two versions behave identically at the load limits. Changing the source frequency results in different values for both the reactance and the converted source voltage or current, thus the equivalent is valid only for the frequency in question. To summarize the process of source conversion: • The internal impedance will be the same for both versions. • If converting from a voltage source to a current source, the value of the current source will be the short circuit current available from the voltage source (i.e., shorted load case), and is equal to E/Zinternal. • If converting from a current source to a voltage source, the value of the voltage source will be the open circuit voltage available from the current source (i.e., opened load case), and is equal to I∙Zinternal. • The equivalent is unique to the frequency of the source. If a multi-source is being converted (i.e., voltage sources in series or current sources in parallel), first combine the sources to arrive at the simplest source and then do the conversion. Do not convert the sources first and then try to combine them as you will wind up with series-parallel configurations rather than simple sources. Judicious use of source conversions can sometimes simplify multi-source circuits by allowing converted sources to be combined, resulting in a single source. 148
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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 producing. I = E Z I = 20°V 80Ω −j 60Ω I = 0.0236.9° A The result is shown in Figure 5.7. The current source is 20 mA RMS with a leading phase angle of 36.9 degrees. Example 5.2 Convert the source of Figure 5.8 into its voltage source equivalent. I = 0.10° amps peak. Again, the internal impedance of 100 + j20 Ω does not change and we simply place it in series. To find the value of the voltage source, find the open circuit voltage the existing source is capable of producing. E = I×Z E = 0.10°A×(100 +j 20Ω) E ≈10.2 11.3° V The equivalent is shown in Figure 5.9. The voltage source E is 10.2 volts peak with a leading phase angle of 11.3 degrees. 149 Figure 5.6 Circuit for Example 5.1. Figure 5.7 Current source equivalent of the source of Figure 5.6. Figure 5.8 Circuit for Example 5.2.
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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 of j20 Ω. This turned out to 3.183 mH. Further, a random load resistor (200 Ω) was picked and applied to both circuits. The result is shown in Figure 5.10. If these two circuits are equivalent, then the voltages seen across the 200 Ω load resistors should be identical. These voltages correspond to node voltages 12 and 8. A quick voltage divider computation shows that the load voltage should be approximately 6.7857.5° volts peak. A transient analysis is performed, plotting the load voltages. The results are shown in Figure 5.11. The plot itself is delayed by one millisecond in order to get past the initial turn-on transient. Further, the current source circuit is shifted by approximately 2 microseconds. Without this slight shift in time, the two voltages perfectly overlap so that it appears there is only one trace. By looking carefully at the graph it can be seen that there are indeed two traces. The amplitudes of these waveforms match each other perfectly and also match the computed result. 150 Figure 5.9 Voltage source equivalent of the source of Figure 5.8. Figure 5.10 The circuits of Figures 5.8 and 5.9 in a simulator.
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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 circuit is reduced to a parallel network. The two current sources can then be summed and the remaining three components can be combined into one equivalent parallel impedance. At that point Ohm's law can be used to find vb. 151 Figure 5.11 Simulation results for the source conversion. Figure 5.12 Circuit for Example 5.3.
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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 impedance is: Ztotal = 1 1 X C + 1 R1 + 1 R2 Ztotal = 1 1 −j10 kΩ + 1 27k Ω + 1 5.6k Ω Ztotal = 4208−24.9 ° Ω And finally we compute vb: vb = I total×Ztotal vb = 2.034E-3−79.5° A×4208−24.9°Ω vb = 8.56−104.4° V We can perform a KCL crosscheck at node b to verify the results in the original circuit. Given the reference directions, all currents are flowing out of the node except for the current supplied by the voltage source. 152 Figure 5.13 Circuit of Figure 5.12 with converted source.
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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−3−14.4° + 1.53E−3−104.4° + 2E−390° amps does indeed equal 0.544E−334.4° amps, within carried rounding error. 5.3 Superposition Theorem 5.3 Superposition Theorem Superposition allows the analysis of multi-source AC series-parallel circuits. Superposition can only be applied to networks that are linear and bilateral. Fortunately, all of components we have discussed; resistors, capacitors and inductors, fall into that category. Further, superposition cannot be used to find values for non-linear functions, such as power, directly. This is not a limitation though because power can be computed from the resulting voltage or current values. The basic idea is to determine the contribution of each source by itself, and then combine the results to get the final answer. The contributions are either all voltages or all currents, depending on need. We can state the superposition theorem as: Any voltage or current in a multi-source linear bilateral network may be determined by summing the contributions caused by each source acting alone, with all other source replaced by their internal impedance. The process generates a series of new single-source circuits, one for each source. These new circuits are then analyzed for the parameter(s) of interest. 153
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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 their ideal internal impedance: for a voltage source, that's a short; and for a current source, an open. We start by considering E1. In the new circuit E2 is replaced with a short. This leaves a fairly simple network where XC and XL are in parallel. This combination is in series with R and E1. Using basic series- parallel techniques, we can solve for desired quantities such as the current flowing through R or the voltage vb. It is important to indicate the reference current direction and voltage polarity with respect to the source being considered (here, that's left-to- right and positive, respectively). The process is then repeated for E2, shorting E1 and leaving us with R in parallel with XC, which is in turn in series with XL and E2. Note that although in this version Vb is still positive, the reference current direction for R is now right-to-left. The numerical results from this version are added to those of the E1 version (minding polarities and directions) to achieve the final result. If power is needed, it can be computed from these currents or voltages. Note that superposition can work with a mix of current sources and voltage sources. The practical downside is that for large circuits using many sources, numerous derived circuits will need to be analyzed. For example, if there are three voltage sources and two current sources, then a total of five derived circuits will be created. It is also possible to use superposition to find the resulting currents or voltages in a circuit that uses sources with different frequencies. In this instance, the equivalent circuits will have different reactance values. In fact, a single non-sinusoidal source can be analyzed using this method by treating the source as a series of superimposed sine waves with each sine source producing a new circuit with its own unique reactance values. 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 impedance. This means that all remaining voltage sources will be shorted and all remaining current sources will be opened. 154 Figure 5.14 A basic multi-source circuit.
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• 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 polarities and current directions for these items. • Add all of the contributions from each of the sub-circuits to arrive at the final values, being sure to account for current directions and voltage polarities in the process. To illustrate the superposition technique, let's reexamine the dual source circuit shown in Figure 5.12 (repeated in Figure 5.15 for ease of reference). We will solve this using superposition and compare the results to those of Example 5.3 which used source conversion. Example 5.4 For the circuit of Figure 5.15, determine vb using superposition. I = 2E−390° amps peak and E = 100° volts peak. As the circuit has two sources, it will require two sub-circuits. For the voltage source, the current source will be replaced with an open. For the second circuit utilizing the current source, the voltage source will be replaced with a short. First, using the voltage source we find: 155 Figure 5.15 Circuit for Example 5.4. Figure 5.16 Circuit of Figure 5.15 considering voltage source.
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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 Z right2 + R1 vb1 = 10 0° V 4886−29.2°Ω 4886−29.2°Ω +27 k Ω vb1 = 1.558−24.88° V We turn our attention to the current source's contribution. We short the voltage source and redraw: This is a simple parallel circuit. We can find vb by placing the resistors and capacitors in parallel, and then using Ohm's law. Note that the reference direction of the current source is downward, meaning that the component current is upward, which makes vb negative (i.e., + to − bottom to top). The parallel impedance is: Ztotal = 1 1 X C + 1 R1 + 1 R2 Ztotal = 1 1 −j10 kΩ + 1 27k Ω + 1 5.6k Ω Ztotal = 4208−24.9 ° Ω 156 Figure 5.17 Circuit of Figure 5.17 considering current source.
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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 technique in the prior example. As mentioned previously, superposition can be used to determine the results even when the sources use different frequencies. This will be explored in the next example. Example 5.5 For the circuit of Figure 5.18, determine vb. Using superposition, we derive two new circuits, each with unique reactance values. The first circuit is shown in Figure 5.19. 157 Figure 5.18 Circuit for Example 5.5. Figure 5.19 Circuit for Example 5.5, first source only.
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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° Ω 211−83.9°Ω +314.290 °Ω vb1 = 19.78−161.9°V Remember, this waveform is at a frequency of 1 kHz. We can repeat this process for the second source which uses 10 kHz. At this new frequency the inductive reactance will be ten times larger, or j3142 Ω, and the capacitive reactance will be ten times smaller, or −j21.22 Ω. The new circuit is shown in Figure 5.20. Once again, a voltage divider can be used to find this portion of vb. The parallel combo of the 2 kΩ resistor and inductor is 168732.5° Ω. vb2 = E Z left2 Zleft2 + X C vb2 = 20°V 168732.5°Ω 168732.5°Ω +21.22−90°Ω vb2 = 2.010.6°V 158 Figure 5.20 Circuit for Example 5.5, second source only.
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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 5.5, the circuit of Figure 5.18 is captured in a simulator as shown in Figure 5.22. 159 Time Domain Graph Voltage (volts) -25 -20 -15 -10 -5 0 5 10 15 20 25 Time (msec) 0 0.5 1 1.5 2 vb vb1 vb2 Figure 5.21 Voltage plot for Example 5.5. Figure 5.22 Circuit of Example 5.5 in a simulator.
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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 Theorems These theorems are related in that they allow complex linear networks to be simplified down to a single source with an associated internal impedance. They simplify analysis when checking a circuit with multiple possible loads. Thévenin's Theorem Thévenin's Theorem Thévenin's theorem is named after Léon Charles Thévenin. It states that: Any single port linear network can be reduced to a simple voltage source, Eth, in series with an internal impedance Zth. 160 Figure 5.23 Simulation results for the circuit of Example 5.5.
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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 sources with differing frequencies4. A generic example of a Thévenin equivalent is shown in Figure 5.24. The phrase “single port network” means that the original circuit is cut in such a way that only two connections exist to the remainder of the circuit. That remainder may be a single component or a large multi-component sub-circuit. Eth is the open circuit voltage at the port and Zth is the impedance looking back into the port (i.e., the equivalent that now drives the remainder). As there are many ways to cut a typical circuit, there are many possible Thévenin equivalents. Consider the circuit shown in Figure 5.25. Suppose we want to find the Thévenin equivalent that drives R2. We cut the circuit immediately to the left of R2. That is, The first step is to make the cut, removing the remainder of the circuit. In this case the remainder is just R2. We then determine the open circuit output voltage at the cut points (i.e., at the open port). This voltage is called the Thévenin voltage, Eth. This is shown in Figure 5.26. In a circuit such as this, basic series-parallel analysis techniques may be used to find Eth. In this circuit, due to the open, no current flows through the inductor, L, and thus no voltage is developed across it. Therefore, Eth must equal the voltage developed across the capacitor, C. 4 It is possible that an equivalent can be valid across a specified range of frequencies, but it will not hold for all frequencies. 161 Figure 5.24 Generic Thévenin equivalent circuit. Figure 5.25 Circuit under consideration for a Thévenin equivalent.
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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 impedance. This is shown in Figure 5.27. Looking in from where the cut was made (right side), we see that R1 and XC are in parallel, and this combination is then in series with XL. Thus, Zth is equal to jXL + (R1 || −jXC). As noted earlier, the original circuit could be cut in a number of different ways. We might, for example, want to determine the Thévenin equivalent that drives C in the original circuit of Figure 5.25. The new port location appears in Figure 5.28. Clearly, this will result in different values for both Eth and Zth. For example, Zth is now R1 || (R2 + jXL). A common error is to find Zth from the wrong perspective, namely, finding the impedance that the source drives. This is flatly incorrect. Remember, Zth is found by looking into the port and simplifying whatever is seen from there. One way to remember this is that it is possible to create equivalents for multi-source circuits. In that instance, there isn't a single driving source, so finding its load impedance is nonsensical. 162 Figure 5.26 Eth,the open circuit output voltage. Figure 5.27 Finding Zth. Figure 5.28 An alternate port location.
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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 measured with an oscilloscope. This is Eth. All of the sources are then replaced with their internal impedance, ideally shorting voltage sources and opening current sources5. An LCR impedance meter can then be connected to the port to read Zth. If an impedance meter is not available, then the source(s) are left in place and an LCR substitution box is placed at the cut points. The box is adjusted so that the voltage across it is equal to half of Eth. By the voltage divider rule, the value of the substitution box must be equal to Zth. In this case, the substitution box will yield either an inductance or capacitance value which can then be turned into a reactance given the frequency. Example 5.6 For the circuit of Figure 5.29, determine the Thévenin equivalent that drives the 300 Ω resistor and find vc. Assume the source angle is 0°. First, let's find Eth, the open circuit output voltage. We cut the circuit so that the 300 Ω resistor is removed. Then we determine the voltage at the cut points. This circuit is shown in Figure 5.30. 5 A typical laboratory signal generator has a 50 Ω internal impedance, and using this value would be more accurate than just replacing the source with a shorting wire. 163 Figure 5.29 Circuit for Example 5.6. Figure 5.30 Circuit for finding Eth for Example 5.6.
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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, we replace the source with a short and then look back in from the cut points. The equivalent circuit is shown in Figure 5.31. The inductor is in series with the parallel combination of the resistor and capacitor. Zleft2 = R× jX C R −jX C Zleft2 = 100Ω×(−j200Ω) 100Ω−j200Ω Zleft2 = 89.44−26.6°Ω Zth = Z left2 + X L Zth = 89.44−26.6 °Ω+ j50Ω Zth = 80.627.12°Ω or 80 +j10Ω The completed equivalent is shown in Figure 5.32. 164 Figure 5.31 Circuit for finding Zth for Example 5.6. Figure 5.32 Completed equivalent for Example 5.6.
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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 divider with the 100 Ω resistor and the 10 volt source. This leads to vb (7.156−18.61° volts ). A second divider can then be used between vb, the inductor and the 300 Ω resistor to find vc, which is 7.06−28.1° volts as expected. The big advantage of using the Thévenin equivalent is that we can easily find vc for any other value of load because we need only analyze the simpler equivalent circuit rather than the original. Thévenin's theorem can also be used on multi-source circuits. The technique for finding Zth does not change, however, finding Eth is a little more involved, as illustrated in the next example. Example 5.7 Find vb for the circuit of Figure 5.33 using Thévenin's theorem. This circuit is similar to the one used in Example 5.5 (Figure 5.18). The difference here is that the second source uses the same frequency as the first source. The reactance values are: XL = j314.2 Ω XC = −j212.2 Ω 165 Figure 5.33 Circuit for Example 5.7.
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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 (the cut points), so they are not in series. That is, if a sensing current entered at node b, it could split left and right, indicating parallel paths, not a series connection. Z th = −jX C× jX L −jX C +jX L Zth = −j 212.2Ω× j314.2Ω −j 212.2Ω+ j314.2Ω Zth = −j 653.7Ω We have a few options to find Eth. Superposition could be used, each circuit requiring a voltage divider. Alternately, the equivalent is basically a series loop as far as finding vb is concerned. Thus, we could find the voltage across the inductor and subtract that from the left source (assuming a reference current direction of clockwise). Finding the inductor voltage requires either a voltage divider or finding the current. Neither approach is considerably less work than the other, and it's probably a good idea to get in a little more practice using superposition, so... Considering the left source, we short the right source and find vb. vbR = E1 X C X C +X L v bR = 10 0° V −j 212.2Ω −j 212.2Ω +j314.2Ω vbR = 20.84180°V or −20.840° V For the right source, we short the left source and find vb. Then we add the two contributions to find the final voltage. Note that both sources will produce a reference polarity of + to − from top to bottom. 166 Figure 5.34 Circuit for finding equivalents for Figure 5.33.
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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 circuit and solve. vb = Eth R R +Z th vb = 14.68180° V 2 kΩ 2k Ω +(−j653.7Ω) vb = 13.95−161.9° V Computer Simulation Computer Simulation To verify the results of the preceding example, the circuit of Figure 5.33 is captured in a simulator, as shown in Figure 5.35. Next, a transient analysis is run, plotting the voltage at node 2, which corresponds to vb in the original circuit. The result is shown in Figure 5.36. The plot is delayed by 0.1 seconds in order to get past the initial turn-on transient. 167 Figure 5.35 The circuit of Figure 5.33 captured in a simulator.
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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 internal impedance. Formally, Norton's theorem states: Any single port linear network can be reduced to a simple voltage source, In, in parallel with an internal impedance Zn. The process of finding a Norton equivalent is very similar to finding a Thévenin equivalent. First, the Norton impedance is the same as the Thévenin impedance. Second, instead of finding the open circuit output voltage, the short circuit output current is found. This is the Norton current. Due to the equivalence afforded by source conversions, if a Thévenin equivalent for a network can be created, then it must be possible to create a Norton equivalent. Indeed, if a Thévenin equivalent is found, a source conversion can be performed on it to yield the Norton equivalent. 168 Figure 5.36 Simulation results for the circuit of Figure 5.33.
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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 same as Zth. That was 80 + j10 Ω. The Norton current, In, is the short-circuit current through the cut points. We can think of this as replacing the load resistor with an ammeter. This is the same as the current through the inductor. In this situation, the capacitor and inductor are in parallel and yield an impedance of j66.67 Ω. Thus, the source current is: isource = E R +Z LC i source = 100° 100Ω +j66.67Ω isource = 83.2E-3−33.7° A This splits between the capacitor and inductor. Using the current divider rule we find: in = iinductor = I source X C X C +X L in = 83.2 −33.7°A −j 200Ω −j 200Ω+j50Ω in = 0.1109−33.7 °A A source conversion can be applied to verify this value. The resulting voltage source is 8 − j4 volts, precisely the value of the Thévenin equivalent. Ultimately, deciding between using the Thévenin or Norton equivalents is a matter of personal taste and convenience. They work equally well. 169 Figure 5.37 Circuit for Example 5.8.
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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 voltage source with internal impedance, as seen in Figure 5.38, a useful question to ask is “What value of load impedance will yield the maximum amount of power in the load?” In the DC case, it was discovered that the load resistance must equal the source resistance in order to achieve maximum load power. In the AC case things appear to be much more complicated by the possible presence of reactances in both the source and load. As a refresher of prior study, consider the basic circuit depicted in Figure 5.38 with source E, source internal resistance Zi and load impedance Z. For the moment, we shall ignore the reactive portions and just describe the load power in terms of the load's resistive portion, R. To make the job easier, we may normalize the source resistance Ri to 1 Ω. By doing this, R also becomes a normalized value, that is, it no longer represents a simple resistance value but rather represents a ratio in comparison to Ri. In this way the analysis will work for any set of source values. Note that the value of E will equally scale the power in both Ri and R, so a precise value is not needed, and thus, we may as well chose 1 volt for convenience. The power in the load can be determined by using I2R where I = E / (Ri+R), yielding P =( E Ri+R) 2 R Using our normalized values of 1 volt and 1 Ω, P =( 1 1+R) 2 R After expanding we arrive at: P = R R 2+2 R+1 (5.1) We now have an equation that describes the load power in terms of the load resistance. Before we go any further, take a look at what this equation tells you, in general. It is obvious that maximum power will not occur at the extremes. If R = 0 or R = ∞ (i.e., shorted or opened load) the load power is zero. The maximizing case occurs somewhere in the middle. To find the precise value that produces the maximum load power, the proof can be divided into two portions. The first involves graphing the function and the second requires differential calculus to solve for a 170 Figure 5.38 Defining maximum power transfer.
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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 source of 1 volt) is set along the vertical. An examination of the power curve shows that the peak occurs at R = 1. In other words, the load must be equal to the source resistance. Thus, we can say that if no reactances are involved, maximum load power occurs when the load resistance equals the source resistance. It does not matter if the source is DC or AC. The graph shown in Figure 5.39 is asymmetric but the concept of resistance ratios is key here. This is easier to see if we plot the completed power curve using a logarithmic horizontal axis and also scale the vertical axis to 100%, as shown in Figure 5.40. The peak is more apparent and the curve is symmetrical in shape rather than lopsided. This reinforces the idea that the ratio of the resistances is what matters. 171 Normalized Load Power 0.00 0.05 0.10 0.15 0.20 0.25 0.30 Normalized Load Resistance 0 0.5 1 1.5 2 2.5 R/(R2+2R+1) Figure 5.39 The resistive portion of the power equation plotted.
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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 to maximize load current for any set of source and load resistances. We can modify the original power equation by adding a new term, X, which represents the net reactance in the circuit. In other words, X is equal to the sum of the reactances in the source impedance and load impedance. The power in the load is still determined by using I2R, however, we must now include the X term when computing the current: I = E √((Ri+R)2+X 2) This leads is to a new load power expression: P =( E √((Ri+R) 2+X 2)) 2 R (5.2) A cursory look at Equation 5.2 shows that to maximize P, X must be zero. A normalized plot of this equation is shown in Figure 5.41 for R = Ri. 6 Power in AC circuits is examined in great detail in Chapter 7. 172 Load Power as a Percentage of Maximum 0 20 40 60 80 100 Normalized Load Resistance 0.01 0.1 1 10 100 R/(R2+2R +1) Figure 5.40 The load power curve with logarithmic axis showing symmetry.
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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 resistors. 173 Figure 5.41 The reactive portion of the power equation plotted using matched resistance. Load Power as a Percentage of Maximum 0 20 40 60 80 100 Normalized Total Reactance -10 -5 0 5 10 Figure 5.42 The load power surface showing variations with both load resistance and total reactance.
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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 here and coincides with X = 0 and R = 1. This is more easily seen by viewing the surface from the back as shown in Figure 5.43. Note that the highest isocontour encircles the intersection of X = 0 and R = 1 (i.e., Rload = Ri ). In sum, we have verified that the resistive portions of the source and load impedance must be identical and that the reactive portions must be of the same magnitude but of opposite sign. This configuration is also known as the complex conjugate. Finally, we can state: Maximum load power will be achieved when the load impedance is equal to the complex conjugate of the internal impedance of the driving source. No other value of load impedance will produce a higher load power. We can imagine two general cases, one with an inductive source impedance and another with a capacitive source impedance. These are shown with the proper loads in Figure 5.44. 174 Figure 5.43 An alternate view of the load power surface.
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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 load voltage and a load current that are half of their maximums. Their product, however, is at the maximum. Further, efficiency at maximum load power is only 50% (i.e., only half of all generated power goes to the load with the other half being wasted internally). Values of R greater than Ri will achieve higher efficiency but at reduced load power. Sometimes we favor efficiency over maximal load power. As any linear single port network can be reduced to something like Figure 5.44 by using Thévenin's theorem, combining the two theorems allows us to determine maximum power conditions for any impedance in a complex circuit. Example 5.9 Consider the circuit of Figure 5.45. What is the power generated in the load if it is equal to 40 Ω? Further, is that the maximum power that can be attained, and if not, what is the maximum load power and what value of load would be needed? To find the load power, first find the circulating current, then use power law. The total impedance seen by the source is 20 + j10 Ω + 40 Ω, or 60 + j10 Ω. 175 Figure 5.44 Configurations for complex conjugate loads. Figure 5.45 Circuit for Example 5.9.
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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 source impedance. The required load for maximum load power is shown in Figure 5.46. We shall repeat the process to find the new load power. i = E Z total i = 70 V 40Ω i = 1.75 0° A Pload = i 2×Rload Pload = (1.75 A) 2×20 Ω Pload = 61.25W An alternate method notes that the new circuit's total impedance is purely resistive and that the source and load resistances are identical. Therefore the voltage source must split evenly across them. In this case that's 35 volts RMS each. 176 Figure 5.46 Circuit of Figure 5.45 with proper load configuration.
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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 that XL = j2πfL, this works out to j157 Ω. We now need to find the Thévenin equivalent. To find Zth we open the current source and look back in from the load. We see the 50 Ω resistor in series with the parallel combination of the 200 Ω resistor and the inductor. The parallel combination is: Z = R × jX L R +jX L Z = 200Ω×( j157Ω) 200Ω +j157Ω Z = 76.3+j 97.2Ω Therefore, Zth = 126.3 + j97.2 Ω. The complex conjugate is 126.3 − j97.2 Ω. The capacitive reactance formula may be used to determine the appropriate capacitance value to achieve −j97.2 Ω. C = 1 2 π f X C C = 1 2π50 kHz97.2Ω C = 32.8 nF 177 Figure 5.47 Circuit for Example 5.10.
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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 can be found via Ohm's law as we already know the impedance of the parallel branch. Eth = i×Z Eth = 0.10°A×(76.3+j97.2Ω) Eth = 12.451.9° V Again, using the complex conjugate, the voltage source splits evenly between the resistive components. As the current source was specified as RMS, so too will be the equivalent voltage. Pload = v R 2 Rload Pload = (6.2 V) 2 126.3Ω Pload = 304.4mW This represents the maximum load power that can be achieved in this circuit. Do not forgot, though, that an equal amount of power is dissipated by the source. This produces an efficiency of just 50%. In summation, we can say that maximum load power is achieved when the load impedance is equal to the complex conjugate of the internal impedance of the circuitry driving the load. Usually, this requires the application of either a Thévenin or Norton equivalent. Finally, although maximum power transfer is a desired outcome in some situations, it is not desirable in all situations. The reason is one of efficiency. At the maximum load power, efficiency is only 50%. In contrast, for load impedances that are greater than the source impedance, the load power will decrease, however, the efficiency will increase. Increased efficiency is particularly important when striving to minimize heat and extend battery life. 178 Figure 5.48 Thévenin equivalent of the circuit of Figure 5.47 with appropriate load.
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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 in question are networks with three external connection points. Due to the manner in which they drawn, they are referred to as delta networks and Y networks7. These configurations are shown in Figure 5.49. Note that the terminal designation of the delta version are upside down compared to those of the Y configuration. These networks can be redrawn without angles. In this form they are known as pi (also called “π”) networks and T (also called “tee”) networks. These configurations are shown in Figure 5.50. It is possible to convert back and forth between delta and Y networks in many cases. That is, for a given delta network, there may exist a Y network such that the impedances seen between the X, Y and Z terminals are identical, and vice versa. Consequently, one configuration can replace another in order to simplify a larger circuit. Unlike the DC version, certain AC networks cannot be converted using the following technique (see the final Challenge problem for an investigation of this). 7 In some sources the capital Greek letter delta (Δ) is used instead of spelling out “delta” and the letter Y is spelled out as “wye”. Thus, you may come across discussion of “Δ-Y ”, “Δ-wye” or “delta-wye” networks. It's all the same stuff. 179 Figure 5.49 Delta and Y (Δ-Y) networks. Figure 5.50 Alternate form: Pi and T (π-T) networks.
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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 terminals for the delta and the Y respectively are: ZXY = Za || (Zb+Zc) = Zd + Ze (5.3) ZXZ = Zb || (Za+Zc) = Zd + Zf (5.4) ZZY = Zc || (Zb+Za) = Ze + Zf (5.5) Let's assume that we have the delta network and are looking for the Y network equivalent. We start by focusing on the final set of terms for each of the three expressions (e.g., ZXY = Zd + Ze). Note that we have three equations with three unknowns (Zd, Ze and Zf). Thus, they can be solved using a term elimination process. If we subtract Equation 5.5 from Equation 5.3, we can eliminate the second impedance (Ze) and arrive at a difference between the first and third unknown impedance s (Zd − Zf). This quantity can then be added to Equation 5.4 to eliminate the third impedance (Zf), leaving just the first unknown impedance (Zd). (Zd + Ze) − (Ze + Zf) = (Zd − Zf) = Za || (Zb+Zc) − Zb || (Za+Zc) (Zd + Zf) + (Zd − Zf) = 2Zd = 2( Zb || (Za+Zc) + Za || (Zb+Zc) − Zc || (Za+Zb) ) Therefore, Zd = Zb || (Za+Zc) + Za || (Zb+Zc) − Zc || (Za+Zb) which, after simplifying8, is: Zd = Z a Zb Z a+Z b+Zc (5.6) Similarly, we can show that Ze = Z a Z c Z a+Z b+Z c (5.7) Z f = Z bZ c Z a+Z b+Z c (5.8) Note that if the magnitudes and angles of three original impedances are identical, the magnitudes of the Y equivalent impedances will all be one-third of the original magnitude, and with the original phase angle. 8 This process, though not particularly difficult, is somewhat tedious. It is, as they say, “left as an exercise for the student”. 180
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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: Z d Ze = Z a Z b Z a+Zb+Z c Z a Z c Z a+Zb+Z c Z d Ze = Z a Zb Z a Z c Z d Z e = Z b Z c Therefore, Z b Z c = Z d Z e Z b = Zc Z d Z e This process can be repeated for Equations 5.6 and 5.8 to obtain an expression for Za. The two expressions for Za and Zb can then be substituted into Equation 5.6 to obtain an expression for Zc that utilizes only Zd, Ze and Zf. A similar process is followed for Za and Zb resulting in: Za = Z d Z e+Z eZ f +Z d Z f Z f (5.9) Zb = Z d Ze+Z e Z f +Z d Z f Ze (5.10) Zc = Z d Z e+Z eZ f +Z d Z f Zd (5.11) If the Y network consists of three identical impedances, then the values of the delta equivalent will all be three times the original magnitude, the inverse of the situation when converting from delta to Y. In summation, equations 5.6, 5.7 and 5.8 can be used to convert a delta network into a Y network, and equations 5.9, 5.10 and 5.11 can be used to convert a Y network into a delta network. Examples of how to apply this technique to tame up-to-now intractable series-parallel networks follow. 181
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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 similar manner using Equations 5.10 and 5.11: Zb = 350 − j80 Ω Zc = 54.8 − j37.9 Ω The equivalent is shown in Figure 5.52. 182 Figure 5.51 Network for Example 5.11. Figure 5.52 Equivalent delta network for the Y network of Figure 5.51.
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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 can be expressed as a series combination. Example 5.12 Determine va in the circuit of Figure 5.53. Assume the source has a phase angle of zero degrees. This circuit cannot be simplified sufficiently using basic series-parallel techniques due to the bridge section. The components between and below nodes a and b comprise a delta network, as shown in Figure 5.54. If this network is replaced with a Y equivalent, the resulting circuit reduces to a simple series-parallel system. Before continuing, it would be helpful to determine the impedance of each of the parallel sections. For the leftmost pair: Zleft2 = R× jX L R + jX L Z left2 = 10Ω× j100Ω 10Ω + j 100Ω Zleft2 = 9.9Ω+ j0.99Ω 183 Figure 5.53 Circuit for Example 5.12. Figure 5.54 Delta network within the circuit of Figure 5.53.
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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, we can use Equations 5.7 and 5.8 to determine Ze and Zf. Ze = 9.9 Ω + j0.99 Ω Zf = 4.95 Ω + j0.495 Ω Swapping the equivalent Y network into the original circuit leads us to the circuit of Figure 5.55 (Y network shown in blue). This circuit can be simplified directly to find va. In this equivalent circuit, va is simply the source voltage of 10° minus the voltage across the 2 Ω resistor. The immediate goal, then, is to find the current through that resistor. This can be achieved via a current divider once the source current is known. To find the source current, we need to find the total impedance of the network. On the upper left side, Zd is in series with the 2 Ω resistor for a total of 5.3 Ω + j0.33 Ω. This is in parallel with the upper right side total of 15.9 Ω + j0.99 Ω. 184 Figure 5.55 Equivalent circuit for the circuit of Figure 5.53.
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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 isource = 10° V 8.9564.76° Ω isource = 0.1117−4.76° A Now for the current divider and also Ohm's law for the 2 Ω resistor. i2Ω = isource Z upperright Zupperright+Z upperleft i2Ω = 0.1117−4.76° A 15.9+j0.99Ω (15.9+j0.99Ω)+(5.3+j 0.33Ω) i2Ω = 83.7E-3−4.76° A v 2Ω = i2Ω×R v 2Ω = 83.7E-3−4.76° A×2Ω v 2Ω = 0.1675−4.76° V Finally, we subtract that potential from the source to find va. v a = E −v 2Ω v a = 10° V −0.1675−4.76° V v a = 0.833 0.95° V 5.7 Summary 5.7 Summary In this chapter we have examined several techniques and theorems to assist with the analysis of AC electrical circuits. We began with more practical models for voltage and current sources by adding an internal impedance to set limits on the source's maximum output and make it sensitive to output frequency. For a voltage source, this impedance is in series, its ideal value being a short, just as it was for the DC case. For current sources, the impedance is in parallel, its ideal value being an open. 185
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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 into a single source, simplifying analysis. If the associated impedance does not have a zero degree phase angle, then the converted source will not be in phase with the original, but will instead be shifted by the impedance angle. The superposition theorem states that, for any multi-source linear bilateral network, the contributions of each source may be determined independent of all other sources, the final result being the summation of the contributions. This remains true in the AC case, however, care must be taken regarding phase shifts when combining the various contributions. The original circuit of N sources generates N new circuits, one for each source under consideration and with all other sources replaced by their ideal internal impedance. Thévenin's and Norton's theorems allow the simplification of complex linear single port (i.e., two connecting points) networks. The AC Thévenin equivalent consists of a voltage source with a series impedance while the AC Norton equivalent consists of a current source with a parallel impedance. These impedances can be represented in general as a resistance in series with a reactance, and given an operating frequency, the reactance can be turned into a capacitance or inductance. These equivalents, when replacing the original sub-circuit, will create the same voltage across the remainder of the circuit with the same current draw. In other words, the remainder of the circuit will see no difference between being driven by the original sub-circuit or by either the Thévenin or Norton equivalents. The maximum power transfer theorem states that for a simple voltage source with an internal impedance driving a simple load, the maximum load power will be achieved when the load impedance equals the complex conjugate of the internal impedance. The complex conjugate has the same real or resistive value, however, the reactive portion is of the opposite sign. This results in a cancellation of the reactive components, leaving just the resistive portions and maximizing load current. At this point, efficiency will be 50%. If the load impedance is higher than the internal impedance, the load power will not be as great, however, the system efficiency may improve, depending on the phase angle. Delta-Y conversions allow the generation of equivalent “three connection point” impedance networks. RLC networks with three elements in the shape of a triangle or delta (with one connection point at each corner) may be converted into a three element network in the shape of a Y or T, or vice versa. The two versions will behave identically to the remainder of the circuit. This allows the simplification of some circuits and eases analysis. 186
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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 Thévenin's and Norton's theorems state? How are they related? 5. What are the conditions to achieve maximum power transfer for AC circuits? How does this differ from the DC version? 6. What are delta and Y configurations? How are they related? 5.8 Exercises 5.8 Exercises Analysis Analysis 1. For the circuit shown in Figure 5.56, use superposition to find vb. 2. For the circuit shown in Figure 5.56, use superposition to find the current through the capacitor. 3. Use superposition to find the current through the 82 Ω resistor. For the circuit shown in Figure 5.57. 4. Use superposition to find vb and vcd for the circuit shown in Figure 5.57. 187 Figure 5.56 Figure 5.57
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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 the current through the 2.2 kΩ resistor for the circuit of Figure 5.59. E1 = 10° and E2 = 1090°. 8. Use superposition to find vab for In the circuit of Figure 5.59. E1 = 10° and E2 = 245°. 9. In the circuit of Figure 5.60, use superposition to find vb and vcd. The sources are in phase. 188 Figure 5.58 Figure 5.59 Figure 5.60
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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 circuit of Figure 5.61. Source one is 100 mV peak and source two is 1 V peak (in phase). 13. In the circuit of Figure 5.62, use superposition to find vab. 14. In the circuit of Figure 5.62, use superposition to find the current through the 15 kΩ resistor. 15. In the circuit of Figure 5.63, use superposition to find vab. 189 Figure 5.61 Figure 5.62 Figure 5.63
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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 sources are in phase. 19. Use superposition in the circuit of Figure 5.65 to find the currents through the inductor and capacitor. I1 = 145° and I2 = 245°. 20. Use superposition in the circuit of Figure 5.65 to find vab and vbc. I1 = 10° and I2 = 290°. 21. In the circuit of Figure 5.66, Use superposition to find vbc. I1 = 100° and I2 = 60°. 22. In the circuit of Figure 5.66, Use superposition to find the current flowing through the 2 Ω resistor. I1 = 4120° and I2 = 60°. 190 Figure 5.64 Figure 5.65 Figure 5.66
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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. Use superposition to determine the inductor current in the circuit of Figure 5.68. I = 4E−30° and E = 18−45°. 27. For the circuit of Figure 5.69, use superposition to determine the inductor current. I = 100E−30° and E = 260°. 28. For the circuit of Figure 5.69, use superposition to determine vab. I = 50E−30° and E = 1890°. 191 Figure 5.67 Figure 5.68 Figure 5.69
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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. Given the circuit of Figure 5.71, determine the Norton equivalent that drives the 20 nF capacitor. 33. For the circuit of Figure 5.72, determine the Thévenin and Norton equivalents that drive the 600 Ω resistor if the source E = 120°. 34. Given the circuit of Figure 5.72, determine the Thévenin equivalent that drives the j1 kΩ inductive reactance if E = 90°. 35. Given the circuit of Figure 5.72, determine the Norton equivalent that drives the j2.5 kΩ inductive reactance if E = 2445°. 192 Figure 5.70 Figure 5.71 Figure 5.72
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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, what combo will achieve maximum power and what is the resulting power? 39. Determine the Norton equivalent that drives the 500 Ω resistor in the circuit of Figure 5.73. Determine the value of component(s) that when placed in series with the 500 Ω resistor will achieve maximum load power (i.e., for the combo as the load). 40. For the circuit of Figure 5.74, determine the Thévenin and Norton equivalents that drive the combo of 36 Ω + j100 Ω. Does this combo achieve maximum load power? If not, what combo will achieve maximum power and what is the resulting power? 193 Figure 5.73 Figure 5.74
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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 the Thévenin and Norton equivalents that drive the combo of 4.7 kΩ in parallel with j300 Ω. Does this combo achieve maximum load power? If not, what combo will achieve maximum power and what is the resulting power? I = 200E−30°. 43. Determine the equivalent Y (T) network for the circuit of Figure 5.77. R1 = R2 = R3 = 10 kΩ and XL1 = XL2 = XL3 = j10 kΩ. 194 Figure 5.75 Figure 5.76 Figure 5.77
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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) network for the circuit of Figure 5.81. 195 Figure 5.78 Figure 5.79 Figure 5.80 Figure 5.81
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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
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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, XC = −j4 kΩ and XL = j2 kΩ. Design Design 53. Design an equivalent current source for Figure 5.87. E = 1290°, R = 1 kΩ and XC = − j200 Ω. The source frequency is 10 kHz. 197 Figure 5.85 Figure 5.86 Figure 5. 87
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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°, R = 10 kΩ and XL = j900 Ω. 57. Design an equivalent voltage source for Figure 5.89. I = 300E−30°, R = 4.3 kΩ and XC = −j5 kΩ. 58. Design an equivalent voltage source for Figure 5.89 if I = 100E−3120°, R = 75 Ω and L = 1 mH. The source frequency is 10 kHz. 59. Design an equivalent voltage source for Figure 5.90 if I = 10E−30°, R = 9.1 kΩ and L = 5 mH. The source frequency is 100 kHz. 60. Design an equivalent voltage source for Figure 5.90 if I = 50E−30°, R = 560 Ω and XL = j350 Ω. 198 Figure 5.88 Figure 5.89 Figure 5.90
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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. Redesign the circuit of Figure 5.56 so that it uses only current sources and produces the same node voltages as the original circuit. 64. Consider the 600 Ω resistor to be the load in Figure 5.72. Determine a new value for the load in order to achieve maximum load power. Also determine the maximum load power. 65. Using Thévenin's theorem with the circuit of Figure 5.73, determine a new value of capacitive reactance such that it cancels the Thévenin reactance. 66. Using Norton's theorem with the circuit of Figure 5.74, determine a new value of inductive reactance such that the inductor current is 1 mA in magnitude. 199 Figure 5.91 Figure 5.92
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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 circuit of Figure 5.94, use any method or combination of methods to determine vab. E1 = 50°, E2 = 1090° and E3 = 150°. 70. In the circuit of Figure 5.94, use any method or combination of methods to determine vce. E1 = 150°, E2 = 3090° and E3 = 450°. 200 Figure 5.93 Figure 5.94
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