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71. Use any method or combination of methods in the circuit of Figure 5.95 to determine vad. E1 = 900°, E2 = 1200° and I = 400E−3180°. 72. Consider the 2.7 kΩ + j4 kΩ combo to be the load in Figure 5.60. Determine if this value achieves maximum load power. If not, determine a new value for the load in order to achie... | ACElectricalCircuitAnalysis_Page_201_Chunk701 |
75. Find the Thévenin equivalent looking into nodes a and b for the circuit of Figure 5.98. I = 40°. 76. Find voltage vbc in the circuit of Figure 5.99 through the use of one or more delta-Y conversions. E = 1000°, R1 = R2 = 2 kΩ, R3 = 3 kΩ, R4 = 10 kΩ, R5 = 5 kΩ, XC1 = XC2 = −j2 kΩ. 77. Given the circuit of Figure 5... | ACElectricalCircuitAnalysis_Page_202_Chunk702 |
78. As mentioned earlier, it is possible that certain AC Y and delta networks cannot be converted. Consider the circuit of Figure 5.101. Can this circuit be converted with a practical outcome? Why/why not? Simulation Simulation 79. Verify the voltage computed for problem 1 by running a transient analysis. 80. Verify th... | ACElectricalCircuitAnalysis_Page_203_Chunk703 |
6 6 Nodal and Mesh Analysis Nodal and Mesh Analysis 6.0 Chapter Learning Objectives 6.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Utilize nodal analysis techniques to solve for voltages in multi-source series-parallel RLC networks. • Utilize mesh analysis techniques to solve fo... | ACElectricalCircuitAnalysis_Page_204_Chunk704 |
Along with nodal and mesh, we shall also introduce the concept of dependent AC sources. Dependent sources do not exhibit a fixed value, but rather the current or voltage is dependent on some other current or voltage in the circuit. What makes this interesting is that this controlling current or voltage may itself be af... | ACElectricalCircuitAnalysis_Page_205_Chunk705 |
One node is chosen as the reference. This is the point to which all other node voltages are measured against. Typically, the reference node is ground, although it does not have to be. We now write a current summation equation for each summing node, except for the reference node. In this circuit there is only one node w... | ACElectricalCircuitAnalysis_Page_206_Chunk706 |
Node a: I1 = i3 + i4 Node b: i3 + I2 = i5, and rearranging in terms of the fixed source, Node b: I2 = −i3 + i5 The currents are then described by their Ohm's law equivalents: Node a: I 1 = va −vb R2 + v a R1 Node b: I 2 = −va −vb R2 + vb jX L Expanding and collecting terms yields: Node a: I 1 =( 1 R1 + 1 R2)va −( 1 R2)... | ACElectricalCircuitAnalysis_Page_207_Chunk707 |
Example 6.1 Determine the voltage across the inductor in the circuit of Figure 6.4. Source one is 50° volts RMS and source two 290° volts RMS. Other than ground, there is only one current summing node in this circuit, and that's the junction at the top of the inductor. We will refer to this junction as node a. Follow... | ACElectricalCircuitAnalysis_Page_208_Chunk708 |
Example 6.2 In the circuit of Figure 6.5, determine va and vb. E is 200° volts peak while I is 0.10° amps peak. The system frequency is 2 kHz. There are two nodes of interest here, other than ground. This means we will generate two equations with two unknowns (va and vb). Using the standard reactance formulas, the in... | ACElectricalCircuitAnalysis_Page_209_Chunk709 |
Computer Simulation Computer Simulation To verify the results of the preceding example, the circuit of Figure 6.5 is captured in a simulator as shown in Figure 6.6. A transient analysis is performed on the circuit. Node voltages 1, 3 and 4 are plotted, corresponding to the voltage source and nodes a and b, respectively... | ACElectricalCircuitAnalysis_Page_210_Chunk710 |
The amplitudes are just as computed. Node voltage a appears to be nearly in phase with the voltage source, as expected. Node voltage b lags the source by between one-quarter to one-third of a division, or some 30 microseconds. For a 2 kHz source, this translates to around −22 degrees, verifying the calculated result. E... | ACElectricalCircuitAnalysis_Page_211_Chunk711 |
Computer Simulation Computer Simulation To verify the results of the preceding example, the circuit of Figure 6.8 is captured in a simulator as shown in Figure 6.9. A transient analysis is performed on the circuit. Node voltages 1 and 2 (i.e., nodes a and b, respectively) are plotted in Figure 6.10. 212 Figure 6.9 The ... | ACElectricalCircuitAnalysis_Page_212_Chunk712 |
The simulation results agree nicely with the computed values in terms of both amplitude and phase. Inspection Method Inspection Method The system of equations can be obtained directly through inspection if the circuit contains current sources and no voltage sources. Let's take another look at the equations developed in... | ACElectricalCircuitAnalysis_Page_213_Chunk713 |
The huge advantage of the inspection method is that it cuts out a time consuming and error prone section of the process, namely converting the original KCL summations into a set of simplified equations with coefficients for each unknown. The inspection method generates the equations directly. To further speed the proce... | ACElectricalCircuitAnalysis_Page_214_Chunk714 |
Example 6.4 Write the system of equations for the circuit of Figure 6.12. I1 = 100° A and I2 = 490° A. We begin at node a, the first node of interest. Find all of the current sources connected to this node. All we have is I1. It is exiting, and thus negative. −10 0°A = ... Now find all of the items connected to this... | ACElectricalCircuitAnalysis_Page_215_Chunk715 |
Find all of the current sources connected to this node. There are none. 0 = ... Find all of the items connected to this node and create a sum of admittances. 0 = ...+( 1 1Ω + 1 2Ω + 1 −j5Ω)v b... Include the terms that are common between node a and node b. This is negative and goes into the lead (a before b) to keep ev... | ACElectricalCircuitAnalysis_Page_216_Chunk716 |
10.7721.8°A =−( 1 8Ω)va −( 1 1Ω)v b +( 1 1Ω + 1 8Ω + 1 j10 Ω)vc The third and final equation is finished. The completed set of equations is: −100° A =( 1 4Ω + 1 2Ω + 1 8Ω)va −( 1 2Ω)vb −( 1 8Ω)vc 0 =−( 1 2Ω)va +( 1 1Ω + 1 2Ω + 1 −j5Ω)vb −( 1 1Ω)v c 10.7721.8° A =−( 1 8Ω)va −( 1 1Ω)vb +( 1 1Ω + 1 8Ω + 1 j 10Ω)vc Note... | ACElectricalCircuitAnalysis_Page_217_Chunk717 |
Supernode Supernode From time to time you may see a circuit utilizing an ideal voltage source like the one shown in Figure 6.13. That is, this voltage source does not have a series impedance associated with it. Without that impedance, it becomes impossible to create an expression for the current passing through the sou... | ACElectricalCircuitAnalysis_Page_218_Chunk718 |
exiting we see i4. At this point we'll create an expression where all of currents entering the super node are on the left side of the equals sign and all of the exiting currents are on the right: Σiin = Σiout I x+ I y+i1+i2 = i1+i2+i3+i4 This can be simplified to: I x+ I y = i3 +i4 Writing this in terms of Ohm's law we... | ACElectricalCircuitAnalysis_Page_219_Chunk719 |
We also know that va − vb = 160° volts. Therefore vb = va − 160° volts. Substituting this into the prior equation yields: 0.269368.2°A = j10 mSv a −j2 mS(va −160°V) 0.269368.2°A = j10 mSv a −j2 mSva +32E-3 90 °A 0.239467.9° A = j8 mSva va = 29.92−22.1° V We know that vb is 160° volts below va, and thus after s... | ACElectricalCircuitAnalysis_Page_220_Chunk720 |
Example 6.6 Find node voltages va, vb and vc for the circuit of Figure 6.17. The sources are: E = 200° volts and I = 245° amps. Once again we have a situation of a voltage source lacking a series impedance which makes a source conversion impossible. Without having to short it and thus treating nodes a and c as an exp... | ACElectricalCircuitAnalysis_Page_221_Chunk721 |
And finally node c: i4 = i1 +i5 i1 = i4 −i5 i1 = 0.1S(vb −v c)−(−j0.2S)vc i1 = 0.1S(vb −(v a−20 0° V)) +j 0.2S(va −20 0°V) i1 = 0.1S(vb −v a +20 0°V) +j0.2S(va −200° V) i1 = 0.1S(vb −v a) +j 0.2Sva +(2 −j 4)A i1 = (−0.1 +j0.2)Sv a +0.1Svb +4.472 −63.4°A The equations for nodes a and c both equal i1, thus they equa... | ACElectricalCircuitAnalysis_Page_222_Chunk722 |
Consider the circuit of Figure 6.19. This circuit has two voltage sources and cannot be simplified further, although it can be solved using either superposition or nodal analysis. For mesh analysis, we begin by designating a set of current loops. These loops should be minimal in size and together cover all components a... | ACElectricalCircuitAnalysis_Page_223_Chunk723 |
As the component values and source voltages are known, we have two equations with two unknowns. These can be solved for i1 and i2 using the simultaneous equation solution techniques of your choice. Example 6.7 For the circuit of Figure 6.21, determine vb and vc. The sources are: E1 = 90° volts and E2 = 12−90° volts. ... | ACElectricalCircuitAnalysis_Page_224_Chunk724 |
E2 = (−jX C+ jX L2)i2+R2(i2 −i1) E 2 =−R2i1+(R2 −jX C+ jX L2)i2 12−90 °V = −20Ωi1 +(20Ω −j75+ j 50Ω)i2 12−90 °V = −20Ωi1 +(20Ω −j 25Ω)i2 The two loop equations are: 1553.1°V = (100Ω+ j 20Ω)i1 −20Ωi2 12−90 °V = −20Ωi1 +(20Ω−j 25Ω)i2 The equations show diagonal symmetry. The currents are i1 = 0.178519.9° amps and i2... | ACElectricalCircuitAnalysis_Page_225_Chunk725 |
We start by finding the reactance values. Using the standard reactance formulas we find that XL = j628.3 Ω and XC = −j318.3 Ω. After substituting these into the original circuit and defining the loops, we have Figure 6.24. We have three loops with three unknown currents, and therefore three equations. We'll number the ... | ACElectricalCircuitAnalysis_Page_226_Chunk726 |
For loop 3: 0 = v R1+v R2+v R3 0 = R1(i3 −i1)+R2(i3 −i2)+R3i3 0 =−R1i1 −R2i2+(R1+R2+R3)i3 0 =−1 kΩi1 −500Ωi2+(1k Ω+500Ω+2k Ω)i3 0 =−1 kΩi1 −500Ωi2+3.5k Ωi3 The final set of equations is: 100° V = (1k Ω −j318.3Ω)i1+ j318.3Ωi2 −1 kΩi3 0 = j318.3Ω i1+(500Ω+ j 310Ω)i2 −500Ωi3 0 =−1 kΩi1 −500Ωi2+3.5k Ωi3 The system of equa... | ACElectricalCircuitAnalysis_Page_227_Chunk727 |
A transient analysis is run on the circuit, plotting node 2 which corresponds to vb, and node 1, the input voltage, which is handy for phase reference. Examining the plot, we can see that the node 2 voltage is just above 7.5 volts peak, as calculated. Further, this waveform leads the the input waveform by just over a q... | ACElectricalCircuitAnalysis_Page_228_Chunk728 |
capacitor). As usual, the set of equations produced must exhibit diagonal symmetry. While it is possible to extend this technique to include current sources, usually it is easier and less error-prone to convert the current sources into voltage sources. Then the process can continue with the direct inspection method out... | ACElectricalCircuitAnalysis_Page_229_Chunk729 |
We continue the process by determining the components that are in common between this loop and the next loop. Remember, this coefficient is negative. 100° V = (20Ω+6Ω+j8Ω)i1 −(6Ω+j 8Ω)i2... We repeat the process by determining the common components with the next loop. This coefficient is negative. In this situation, n... | ACElectricalCircuitAnalysis_Page_230_Chunk730 |
Supermesh Supermesh On occasion you may find a current source which has no associated internal impedance, such as the one in the circuit of Figure 6.29. This is similar to the situation discussed previously with nodal analysis where a voltage source does not have a specified internal impedance. As with nodal, there are... | ACElectricalCircuitAnalysis_Page_231_Chunk731 |
We now perform a KVL summation around the supermesh loop, similar to what we have done in prior work. The difference this time around is that we need to recognize that components each see one of the original mesh currents; namely i1 or i2 here. We do not solve for a supermesh current. Instead, we just use the supermesh... | ACElectricalCircuitAnalysis_Page_232_Chunk732 |
By inspection we can see that: 10E-30° A = i2 −i1 or i2 = i1 +10E-30° A We can substitute this expression into the prior supermesh expression and solve for i1: 26.9 −42° V = 1kΩi1+ j 400Ωi2 26.9 −42° V = 1kΩi1+ j 400Ω(i1+10E-30° A) 26.9 −42° V = 1kΩi1+ j 400Ωi1+490° V 29.7−44.7°V = (1 kΩ+ j 400Ω)i1 i1 ≈27.6E-3... | ACElectricalCircuitAnalysis_Page_233_Chunk733 |
hand, if you need to find voltages in a circuit that contains numerous current sources, nodal would be more effective. 6.4 Dependent Sources 6.4 Dependent Sources A dependent source is a current or voltage source whose value is not fixed. Instead, the value depends on some other circuit current or voltage. The general ... | ACElectricalCircuitAnalysis_Page_234_Chunk734 |
Dependent sources are not “off-the-shelf” items in the same way that a battery or signal generator are. Rather, dependent sources are used to model the behavior of more complex devices. For example, a bipolar junction transistor commonly is modeled as a CCCS while a field effect transistor may be modeled as a VCCS. Sim... | ACElectricalCircuitAnalysis_Page_235_Chunk735 |
In this example it should be obvious that the voltage from the dependent source can affect the voltage at node a, and it is this very voltage that defines ix, which in turn sets up the value of the dependent source. As far as analysis is concerned, either mesh or nodal can be used. The dependent source(s) will contribu... | ACElectricalCircuitAnalysis_Page_236_Chunk736 |
Collecting terms and simplifying yields: Node a : E R1 =( 1 R1 + 1 R2 + 1 −jX C)va − 1 −jX C vb Nodeb: 0 =−(k + 1 R2)va +( 1 R2 + 1 jX L)vb At this point, the component values and independent source value would be inserted into the equations and the system solved. Finally, referring back to the prior chapter, it is pos... | ACElectricalCircuitAnalysis_Page_237_Chunk737 |
The dependent source is 30 times this value, or 15E−390° amps. Given the reference direction of this source, the current is flowing upwards through the 65 kΩ resistor and parallel −j50 kΩ capacitor. This establishes vc as negative. Multiplying ix by the parallel impedance yields the desired voltage. Z RC = R3× jX C2 R... | ACElectricalCircuitAnalysis_Page_238_Chunk738 |
The KCL summation is: Σiin = Σiout i1+i2 = i x This is expanded using Ohm's law (in several steps, for clarity). ix = i1+i2 va R2 = E −v a R1+ jX L + 1000(Ω)i x−va R3 va 2k Ω = 20 0°V−v a 1k Ω+ j1257Ω + 1000(Ω)i x−va 3k Ω 20 0°V 1k Ω+ j1257Ω = va 1 kΩ+ j1257Ω + v a 2 kΩ + va 3k Ω −1000(Ω)i x 3k Ω 12.45−51.5° A = v a... | ACElectricalCircuitAnalysis_Page_239_Chunk739 |
Note the connection to sense the current ix. It is inserted just like an ammeter. As mentioned previously, the constant for the dependent source is a transresistance and has units of ohms. A transient analysis is run the circuit, plotting the independent source, E, as node 1 (blue), and va as node 3 (red). Both the amp... | ACElectricalCircuitAnalysis_Page_240_Chunk740 |
and combination of terms to arrive at the end equations. The second approach is referred to as the inspection method. If the circuit contains only current sources (or if the voltage sources are converted to equivalent current sources), this method allows direct generation of the system of equations without the need for... | ACElectricalCircuitAnalysis_Page_241_Chunk741 |
6.6 Exercises 6.6 Exercises Analysis Analysis (All source values are in amps or volts unless specified otherwise) 1. Given the circuit in Figure 6.43, use nodal analysis to determine vc. I1 = 30°, I2 = 0.90°. 2. Use nodal analysis to find the current through the 120 Ω resistor in the circuit of Figure 6.44. I1 = 0.5... | ACElectricalCircuitAnalysis_Page_242_Chunk742 |
6. Use nodal analysis to find the current through the j45 Ω inductor in the circuit of Figure 6.45. I1 = 20°, I2 = 1.560°. 7. Use nodal analysis to find the current through the 4 Ω resistor in the circuit of Figure 6.46. I1 = 145°, I2 = 245°. 8. Given the circuit in Figure 6.46, use nodal analysis to determine vc. ... | ACElectricalCircuitAnalysis_Page_243_Chunk743 |
12. Given the circuit in Figure 6.48, use nodal analysis to determine vc. I1 = 490°, I2 = 10120°, I3 = 50°. 13. Given the circuit in Figure 6.49, use nodal analysis to determine vc. I1 = 3E−30°, I2 = 10E−30°, I3 = 2E−30°. 14. Use nodal analysis to find the current through the −j2 kΩ capacitor in the circuit of Fi... | ACElectricalCircuitAnalysis_Page_244_Chunk744 |
18. Use nodal analysis to find the current through the 2.2 kΩ resistor in Figure 6.51. E = 2400°, I = 100E−30°. 19. Use nodal analysis to find vbc in the circuit of Figure 6.52. 20. Use nodal analysis to find the current through the 2.7 kΩ resistor in the circuit of Figure 6.53. 21. Given the circuit in Figure 6.54, ... | ACElectricalCircuitAnalysis_Page_245_Chunk745 |
22. Given the circuit in Figure 6.55, use nodal analysis to determine vad. E1 = 90°, E2 = 540°. 23. Use nodal analysis to find vcb in the circuit of Figure 6.56. E1 = 10−180°, E2 = 250°. 24. Given the circuit in Figure 6.57, use nodal analysis to determine vbc. E = 200°, R1 = 10 kΩ, R2 = 30 kΩ, R3 = 1 kΩ, XC = −j1... | ACElectricalCircuitAnalysis_Page_246_Chunk746 |
25. Given the circuit in Figure 6.58, write the mesh loop equations and determine vb. 26. Use mesh analysis to find the current through the 2.7 kΩ resistor in the circuit of Figure 6.58. 27. Use mesh analysis to find the current through the 75 Ω resistor in the circuit of Figure 6.52. 28. Given the circuit in Figure 6.... | ACElectricalCircuitAnalysis_Page_247_Chunk747 |
35. Use mesh analysis to find the current through the −j200 Ω capacitor in the circuit of Figure 6.59. E1 = 180°, E2 = 1290°. 36. Given the circuit in Figure 6.59, use mesh analysis to determine vac. E1 = 10°, E2 = 500E−30°. 37. Given the circuit in Figure 6.56, use mesh analysis to determine vc. E1 = 10−180°, E2 ... | ACElectricalCircuitAnalysis_Page_248_Chunk748 |
43. Use mesh analysis to find the current through resistor R3 in Figure 6.61. E = 600°, R1 = 1 kΩ, R2 = 2 kΩ, R3 = 3 kΩ, XC = −j10 kΩ, XL = j20 kΩ. 44. Given the circuit in Figure 6.61, use mesh analysis to determine vbc. E = 12090°, R1 = 100 kΩ, R2 = 20 kΩ, R3 = 10 kΩ, XC = −j5 kΩ, XL = j20 kΩ. 45. Given the circuit... | ACElectricalCircuitAnalysis_Page_249_Chunk749 |
49. Given the circuit in Figure 6.64, use nodal analysis to determine vab. 50. Use nodal analysis to find the current through the 100 mH inductor in the circuit of Figure 6.64. 51. Use nodal analysis to find the current through the 330 Ω resistor in the circuit of Figure 6.65. 52. Given the circuit in Figure 6.65, writ... | ACElectricalCircuitAnalysis_Page_250_Chunk750 |
56. Determine vc in the circuit of Figure 6.67 if the source E = 3120°. 57. Determine the current through the 5 kΩ resistor in the circuit of Figure 6.67 if E = 100°. 58. In the circuit of Figure 6.68, determine the capacitor current if the source E = 120°. 59. In the circuit of Figure 6.68, determine vc if the sour... | ACElectricalCircuitAnalysis_Page_251_Chunk751 |
62. In the circuit of Figure 6.70, determine the current flowing into the 600 Ω resistor if I1 = 1E−3180°. 63. Determine va and vb in the circuit of Figure 6.70 if the source I1 = 2E−30°. 64. Determine va in the circuit of Figure 6.71 if the source E = 20°. 65. Given the circuit in Figure 6.71, determine the current... | ACElectricalCircuitAnalysis_Page_252_Chunk752 |
68. In the circuit of Figure 6.73, determine vd. 69. Given the circuit in Figure 6.73, determine the current flowing through the 1 kΩ resistor. 70. Given the circuit in Figure 6.74, determine the current flowing through the 100 Ω resistor. 71. Determine vd in the circuit of Figure 6.74. 72. Determine vab in the circuit... | ACElectricalCircuitAnalysis_Page_253_Chunk753 |
Challenge Challenge 73. Given the circuit in Figure 6.76, write the node equations. E1 = 500°, E2 = 35120°, I = 500E−390°. 74. Given the circuit in Figure 6.76, use either mesh or nodal analysis to determine ved. E1 = 90°, E2 = 120°, I = 50E−30°. 75. Given the circuit in Figure 6.77, use mesh analysis to determin... | ACElectricalCircuitAnalysis_Page_254_Chunk754 |
76. Find voltage vbc in the circuit of Figure 6.78 using either mesh or nodal analysis. E = 1000°, R1 = R2 = 2 kΩ, R3 = 3 kΩ, R4 = 10 kΩ, R5 = 5 kΩ, XC1 = XC2 = −j2 kΩ. 77. Given the circuit in Figure 6.79, use nodal analysis to find vac. I1 = 8E−30°, I2 = 12E−30°, E = 500°. 255 Figure 6.78 Figure 6.79 | ACElectricalCircuitAnalysis_Page_255_Chunk755 |
78. Given the circuit in Figure 6.80, use nodal analysis to determine vad. I1 = 0.10°, I2 = 0.20°, I3 = 0.30°. 79. Given the circuit in Figure 6.81, determine vad. E1 = 150°, E2 = 60°, I = 100E−30°. 80. Given the circuit in Figure 6.82, determine vad. E1 = 220°, E2 = −100°, I = 2E−30°. 256 Figure 6.80 Figure 6... | ACElectricalCircuitAnalysis_Page_256_Chunk756 |
81. Given the circuit in Figure 6.83, determine vab. I1 = 1.20°, I2 = 2120°, E = 750°. 82. Given the circuit in Figure 6.84, determine vad. I1 = 0.80°, I2 = 0.2180°, I3 = 0.10°, E = 150°. Simulation Simulation 83. Perform a transient analysis simulation on the circuit of problem 25 (Figure 6.58) to verify the re... | ACElectricalCircuitAnalysis_Page_257_Chunk757 |
87. Investigate the variation of vb due to frequency in problem 28 (Figure 6.52) by performing an AC simulation. Run the simulation from 1 Hz up to 10 kHz. 88. Investigate the variation of vb due to component tolerance in problem 28 (Figure 6.52) by performing a Monte Carlo simulation. Apply a 10% tolerance to the resi... | ACElectricalCircuitAnalysis_Page_258_Chunk758 |
Notes Notes ♫♫ ♫♫ 259 | ACElectricalCircuitAnalysis_Page_259_Chunk759 |
7 7 AC Power AC Power After completing this chapter, you should be able to: • Describe current, voltage and power relationships in AC RLC networks. • Plot and make use of the power triangle to determine real, apparent and reactive power components in an AC power system. • Compute the power factor of an RLC network. • D... | ACElectricalCircuitAnalysis_Page_260_Chunk760 |
7.2 Power Waveforms 7.2 Power Waveforms Computation of power in AC systems is somewhat more involved than the DC case due to the phase between the current and voltage. It has been stated in prior work that power dissipation is characteristic of resistors, and that ideal inductors and capacitors do not dissipate power. ... | ACElectricalCircuitAnalysis_Page_261_Chunk761 |
The power product is shown in blue. Unless the frequency is ridiculously low, the resistor's heating will respond to the average value of this waveform thanks to the device's thermal time constant. Due to the fact that sinusoids are symmetrical around zero, the effective power dissipation averaged over time will be the... | ACElectricalCircuitAnalysis_Page_262_Chunk762 |
Note that this expression does not contain a constant term and only contains a time- varying term. Consequently, without an offset, there is no net power dissipation. The result is shown in Figure 7.2. Here power is being alternately generated and dissipated (i.e., positive values indicate dissipation while negative va... | ACElectricalCircuitAnalysis_Page_263_Chunk763 |
value whose magnitude ranges from 0 up to 1. Therefore, unless θ is zero, the offset will not equal the peak value of the sinusoidal portion. This is a particularly important point which shall be amplified in a moment. Example waveforms using θ = 45° are shown graphically in Figure 7.3. The power waveform dips slightly... | ACElectricalCircuitAnalysis_Page_264_Chunk764 |
At this point we can see that resistors dissipate true power but that reactive components do not. This raises a practical problem, namely, what to call the current- voltage product for purely reactive or complex loads. That is, we can't lump together the current-voltage values for an inductor with those of a resistor a... | ACElectricalCircuitAnalysis_Page_265_Chunk765 |
v RMS = v peak √2 v RMS = 10V √2 v RMS ≈7.07V i = v Z i = 7.07 V 10+j 6.283Ω i = 0.5986−32.1° A For the power calculations, we shall only use the magnitudes of the voltage and current. Here, the symbol “| |” refers to just the magnitude of the reactance or impedance. P = i2R P = (0.5986A) 210Ω P ≈3.58 W Q = i 2| X | Q... | ACElectricalCircuitAnalysis_Page_266_Chunk766 |
The first simulation plots the circulating current (green), the resistor voltage (red) and their product (the power, in blue). This is shown in Figure 7.7. We can see that current and voltage are perfectly in phase, as expected. Also, the power waveform ranges from zero up to about 7 watts. The average of this is appro... | ACElectricalCircuitAnalysis_Page_267_Chunk767 |
A second set of plots is generated for the inductor. This is shown in Figure 7.8. Again, compare this set against the curves seen in Figure 7.2 for the general reactive case. We can see that the current (green) is 90 degrees out of phase with the inductor's voltage (red) and lagging, as expected. More importantly, we s... | ACElectricalCircuitAnalysis_Page_268_Chunk768 |
Example 7.2 Determine the power dissipated by the resistor in the circuit of Figure 7.10. Also find the apparent power drawn by the circuit and the reactive power of the capacitor. The source frequency is 1 kHz. The first item is to find the reactance of the capacitor. X C = 1 2π f C X C = 1 2π1kHz 1μ F X C =−j15.92Ω U... | ACElectricalCircuitAnalysis_Page_269_Chunk769 |
The capacitor voltage is found via KVL: vC = E −vR vC = 7.070° −3.76157.9 °V vC = 5.99−32.1° V For the powers, we just use the magnitude of the voltage. P = v 2 R P = (3.761V) 2 10Ω P ≈1.414W Q = i2 | X | Q = (5.99 V) 2 15.92Ω Q ≈2.25 VAR, capacitive S = E 2 |Z | S = (7.07 V) 2 |10Ω −j15.92Ω| S ≈2.66 VA, capacitive ... | ACElectricalCircuitAnalysis_Page_270_Chunk770 |
In Figure 7.12 we see the results of a transient analysis run on the resistor. We can see that the voltage and current are in phase. Also, the power waveform swings from zero up to around 2.8 watts or so. This corresponds to an average value of just under 1.5 watts, and this agrees nicely with the computed result. Figu... | ACElectricalCircuitAnalysis_Page_271_Chunk771 |
272 Figure 7.13 Transient analysis for the capacitor of Figure 7.11. Figure 7.14 Transient analysis for the resistor and capacitor together in Figure 7.11. | ACElectricalCircuitAnalysis_Page_272_Chunk772 |
7.3 Power Triangle 7.3 Power Triangle The prior section revealed that the phase angle between the current and voltage cannot be ignored when computing power. For example, if a 120 volt RMS source delivers 2 amps of current, it appears that it delivers 240 watts. This is only true if the load is purely resistive. For a ... | ACElectricalCircuitAnalysis_Page_273_Chunk773 |
S = v RMS×iRMS (7.4) S = √P 2+Q 2 (7.5) θ = tan −1 Q P (i.e., the impedance angle) (7.6) P = S cos θ (7.7) P = √S 2−Q 2 (7.8) Q = S sinθ (7.9) Q = √S 2−P 2 (7.10) Power Factor Power Factor As we are often interested in the true power, it is worth noting that a rearrangement of Equation 7.7 shows that the ratio of true ... | ACElectricalCircuitAnalysis_Page_274_Chunk774 |
From here we can determine the system impedance which will, in turn, allow us to determine the source current. Z = R× jX L R + jX L Z = 160Ω×( j56.55Ω) 160Ω+ j56.55Ω Z = 53.370.5°Ω isource = esource Z i source = 1200° V 53.370.5°Ω isource = 2.252−70.5° A The apparent power is the product of the magnitudes of circui... | ACElectricalCircuitAnalysis_Page_275_Chunk775 |
Power Factor Correction Power Factor Correction One issue with a reactive load is that the current is higher than it needs to be in order to achieve a certain true load power. This is wasteful and would require larger conductors. To alleviate these issues, an opposite reactance can be added to the load such that the re... | ACElectricalCircuitAnalysis_Page_276_Chunk776 |
P = S cos θ P = 3.03VA cos43° P = 2.22 W To be clear, 2.22 watts is the combined power dissipation between the two resistors. Q = S sinθ Q = 3.03 VAsin43 ° Q = 2.07 VAR, inductive The power triangle for this circuit is shown in Figure 7.19. For the second part involving power factor correction, we need to add a reactiv... | ACElectricalCircuitAnalysis_Page_277_Chunk777 |
C = 1 2π f X C C = 1 2π10 kHz62.8Ω C = 253.3nF The resulting circuit is shown in Figure 7.21. Computer Simulation Computer Simulation It is useful to see the reduction in current demand caused by using power factor correction. To so so, the circuit of Figure 7.21 is captured in a simulator as shown in Figure 7.22. We w... | ACElectricalCircuitAnalysis_Page_278_Chunk778 |
The resulting current for the original circuit is shown in Figure 7.23. Note that the peak current is just over 300 milliamps, as calculated in step two of the example. The second simulation is shown in Figure 7.24, now with power factor correction. 279 Figure 7.23 Simulation results for the circuit of Figure 7.22 with... | ACElectricalCircuitAnalysis_Page_279_Chunk779 |
The amplitude here is much reduced, in the range of 160 to 170 milliamps peak. By adding the capacitor, the two reactive currents cancel, leaving a parallel impedance of just 100 Ω. This is in series with the 20 Ω resistor. Dividing 120 Ω into the 20 volt peak source yields a peak current of approximately 167 mA, in ag... | ACElectricalCircuitAnalysis_Page_280_Chunk780 |
Purely capacitive loads are not nearly as common but some everyday devices exhibit highly complex impedances that can vary widely in both amplitude and angle. A standard moving coil dynamic loudspeaker falls into this category. While you would never plug a loudspeaker directly into an AC power system14, the amplifier t... | ACElectricalCircuitAnalysis_Page_281_Chunk781 |
loudspeakers are particularly inefficient, typically converting only about 1% of the electrical input into usable acoustic output. Example 7.5 A certain 120 VAC RMS 60 Hz 1.2 HP motor has an efficiency of 90% and a lagging power factor of 0.85. Determine the apparent power and the current drawn from the system. Also dr... | ACElectricalCircuitAnalysis_Page_282_Chunk782 |
Example 7.6 For the motor described in Example 7.5, determine an appropriate component that when placed in parallel will produce unity power factor. Draw the complete power triangle for the system and determine the new current draw. In Example 7.5 the reactive power was determined to be 617 VAR and was inductive, so we... | ACElectricalCircuitAnalysis_Page_283_Chunk783 |
Example 7.7 The system shown in Figure 7.28 is supplied by a 240 VAC RMS 60 Hz source. Load 1 is 1000 watts of resistive heating elements. Load 2 is a 3 HP motor with an efficiency of 92% and a lagging power factor of 0.75. Load 3 is a capacitor bank equivalent to 75 μF. Draw the system power triangle and determine whe... | ACElectricalCircuitAnalysis_Page_284_Chunk784 |
Now for the capacitor. First, we need to determine the reactance, then we can use the v2/Z form of power law to determine Q. X C = 1 2π f C X C = 1 2 π60 Hz75μ F X C =−j35.37Ω Q = v 2 X Q = (240V) 2 35.37Ω Q = 1629VAR, capacitive There are no other reactive elements in the system to consider. We have 515 fewer capaciti... | ACElectricalCircuitAnalysis_Page_285_Chunk785 |
7.5 Summary 7.5 Summary Power is the product of current and voltage, however, the phase relationship between the two has a major impact on the result. The most simple and straightforward case is when the current and voltage are in phase, meaning the load is a pure resistance. In this case, the true power can be compute... | ACElectricalCircuitAnalysis_Page_286_Chunk786 |
Review Questions Review Questions 1. Define apparent power, real power and reactive power. 2. Describe the power triangle. 3. What is power factor, PF? 4. Describe the difference between leading and lagging power factor. 5. What is power factor correction? 6. Give examples of resistive loads and inductive loads. 7.6 Ex... | ACElectricalCircuitAnalysis_Page_287_Chunk787 |
3. For the circuit shown in Figure 7.34, determine apparent power S, real power P, reactive power Q and power factor PF. Also, draw the power triangle. The source is 120 volts. 4. For the circuit shown in Figure 7.35, determine apparent power S, real power P, reactive power Q and power factor PF. Also, draw the power t... | ACElectricalCircuitAnalysis_Page_288_Chunk788 |
6. For the circuit shown in Figure 7.37, determine apparent power S, real power P, reactive power Q and power factor PF. Also, draw the power triangle. The source is 240 volts, XC = −j200 Ω, R = 75 Ω. 7. For the circuit shown in Figure 7.38, determine apparent power S, real power P, reactive power Q and power factor PF... | ACElectricalCircuitAnalysis_Page_289_Chunk789 |
13. Consider the system shown in Figure 7.39. E is a standard 120 V input. If the three loads are 45 W, 60 W and 75 W incandescent light bulbs, respectively, determine the apparent power delivered to the system, the source current, the reactive power and the real power. 14. Given the system shown in Figure 7.39, determ... | ACElectricalCircuitAnalysis_Page_290_Chunk790 |
20. Given the system shown in Figure 7.39, determine the apparent power delivered to the system, the source current, the real power, the reactive power and the power factor. E is 60 V. Load 1 is 90 W of incandescent lighting, load 2 is 800 W of heating elements from a dryer and load 3 200 VAR inductive. 21. A 120 V 3 H... | ACElectricalCircuitAnalysis_Page_291_Chunk791 |
29. For the system shown in Figure 7.41, E is 240 V. Load 1 is 1.2 kW resistive heating, load 2 is 400 W resistive lighting, load 3 is a 0.5 HP motor that is 80% efficient with a 0.7 lagging power factor, and load 4 is a 1 HP motor that is 85% efficient with a 0.8 lagging power factor. For the system, determine the app... | ACElectricalCircuitAnalysis_Page_292_Chunk792 |
Challenge Challenge 38. A power distribution system for a concert has the following specifications: Ten class D audio power amplifiers rated at 2 kW output each with 90% efficiency and unity power factor, 10 kW worth of resistive stage lighting to illuminate the musicians alongside a troupe of trained dancing kangaroos... | ACElectricalCircuitAnalysis_Page_293_Chunk793 |
8 8 Resonance esonance 8.0 Chapter Learning Objectives 8.0 Chapter Learning Objectives After completing this chapter, you should be able to: • Describe series resonance and plot impedance, voltage and current as a function of frequency. • Describe parallel resonance and plot impedance, voltage and current as a function... | ACElectricalCircuitAnalysis_Page_294_Chunk794 |
8.2 Series Resonance 8.2 Series Resonance Let's begin with the simplest RLC circuit; one consisting of a single voltage source in series with a single resistor, inductor and capacitor, as shown in Figure 8.1. Of particular interest is how the total impedance varies across the frequency spectrum and what impact this has... | ACElectricalCircuitAnalysis_Page_295_Chunk795 |
The dip in the center corresponds to an impedance equal to R. At this frequency the capacitive and inductive reactances are equal in magnitude and effectively cancel each other. All that's left is the resistive component, R. This frequency is known as the resonant frequency and is denoted by f0. The series resonant fre... | ACElectricalCircuitAnalysis_Page_296_Chunk796 |
X L = X C 2π f 0 L = 1 2π f 0C f 0 2 = 1 (2π) 2 LC f 0 = 1 2π√LC (8.2) Note that a particular resonant frequency can be obtained through a variety of LC pairs. This, along with the value of R, will alter the specific shape of the impedance curve in terms of how narrow or broad the dip is. These components will also aff... | ACElectricalCircuitAnalysis_Page_297_Chunk797 |
At this point we can more precisely define Q. Specifically, the “sharpness” of the curve is related to the half-power or “−3 dB” frequencies, f1 and f2.17 These are the frequencies at which the current (assuming voltage source drive) falls off to 0.707 of the maximum value at resonance. Therefore, they represent the fr... | ACElectricalCircuitAnalysis_Page_298_Chunk798 |
The resonant frequency, f0, in general is not located evenly between f1 and f2. It is, in fact, located at their geometric mean. In other words, f 0 = √f 1 f 2 (8.5) From Equation 8.5 we may derive: f 0 f 1 = f 2 f 0 (8.6) To find accurate values for f1 and f2 we can define a factor, k0. The derivation of k0 is found i... | ACElectricalCircuitAnalysis_Page_299_Chunk799 |
For higher Q circuits (Qcircuit ≥ 10), we can approximate symmetry, and thus f 1 ≈f 0 −BW 2 (8.10) f 2 ≈f 0+ BW 2 (8.11) As mentioned previously, the Q can be a function of either R or the L/C ratio. In Figures 8.6 and 8.7 we have impedance curves for the two cases. The frequency axis is normalized to f0 (i.e., f0 is u... | ACElectricalCircuitAnalysis_Page_300_Chunk800 |
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