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SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS
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Obtain the voltage Vx in the network of Fig. 9-28, using the mesh current method.
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Fig. 9-28 One choice of mesh currents is shown on the circuit diagram, with I3 passing through the 10- resistor in a direction such that Vx I3 10 (V). The matrix equation can be written by inspection: 3 2 32 3 2 7 j3 j5 5 I1 10 08 6 7 4 j5 12 j3 2 j2 54 I2 5 4 5 308 5 I3 5 2 j2 17 j2 0 Solving by determinants, 7 j3 j5 10 08 j5 12 j3 5 308 5 2 j2 0 667:96 169:098 0:435 I3 7 j3 j5 5 1534:5 25:068 j5 12 j3 2 j2 5 2 j2 17 j2 and Vx I3 10 4:35 194:158 V.
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In the netwrok of Fig. 9-29, determine the voltage V which results in a zero current through the 2 j3  impedance.
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SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS
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Choosing mesh currents as shown on the circuit diagram, 5 j5 30 08 0 1 I2 j5 0 6 0 z 0 V 10 Expanding the numerator determinant by cofactors of the second column, j5 6 V 5 j5 0 0 whence V 35:4 30 08 j5 6 0 10
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45:08 V
Solve Problem 9.19 by the node voltage method.
The network is redrawn in Fig. 9-30 with one end of the 2 j3 impedance as the reference node. By the rule of Section 9.7 the matrix equation is 2   32 3 2 3 1 1 1 1 1 V1 30 08 6 5 j5 2 j3 7 5 j5 76 7 6 5 6 76 7 6 7   6 74 5 4 4 1 1 1 1 1 15 30 08 V 5 V2 5 j5 5 j5 4 6 5 4 For node voltage V1 to be zero, it is necessary that the numerator determinant in the solution for V1 vanish. 30 08 0:200 j0:200 5 0 from which V 35:4 458 V N1 30 08 V 0:617 j0:200 5 4
Fig. 9-30
Fig. 9-31
Use the node voltage method to obtain the current I in the network of Fig. 9-31.
There are three principal nodes in the network. The reference and node 1 are selected so that the node 1 voltage is the voltage across the j2- reactance. 21 32 3 2 3 1 1 1 V1 50 08 6 5 j2 4 76 7 6 7 4 5 6 76 7 6 7 4 1 1 1 1 54 5 4 50 908 5 V2 4 4 j2 2 2 from which
SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS 0:250 13:52 56:318 0:750 j0:500 24:76 V1 0:450 j0:500 0:546 15:948 0:250 0:250 0:750 j0:500 and I 24:76 72:258 12:38 2 908 17:758 A 10 j25
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72:258 V
Find the input impedance at terminals ab for the network of Fig. 9-32.
Fig. 9-32 With mesh current I1 selected as shown on the diagram. 8 j2 3 0 3 8 j5 5 0 5 7 j2 315:5 16:198 6:98 Zinput;1 z 8 j5 5 11 45:2 24:868 5 7 j2
8:678 
For the network in Fig. 9-32, obtain the current in the inductor, Ix , by rst obtaining the transfer impedance. Let V 10 308 V.
Ztransfer;12 z 315:5 16:198 14:45 3 5 12 0 7 j2 308 0:692 32:148 32:148 
Then
Ix I2
V 10 Ztransfer;12 14:45
2:148 A
For the network in Fig. 9-32, nd the value of the source voltage V which results in V0 5:0 08 V.
The transfer impedance can be used to compute the current in the 2 j2  impedance, from which V0 is readily obtained. Ztransfer;13 z 315:5 16:198 21:0 13 15 08 16:198 
V0 I3 2 j2
V 2 j2 V 0:135 Ztransfer;13
61:198
CHAP. 9]
SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS
Thus, if V0 5:0
08 V, V 5:0 08 37:0 0:135 61:198 61:198 V
Alternate Method The node voltage method may be used. V0 is the node voltage V2 for the selection of nodes indicated in Fig. 9-32. 1 1 1 V 5 j2 3 j5 5 j2 1 0 j5 V 0:134 61:158 V0 V2 1 1 1 1 5 j2 3 j5 j5 1 1 1 1 j5 j5 5 2 j2 For V0 5:0 08 V, V 37:3 61:158 V, which agrees with the previous answer to within roundo errors.
For the network shown in Fig. 9-33, obtain the input admittance and use it to compute node voltage V1 .
1 1 1 1 10 j5 2 2 1 1 1 1 2 2 3 j4 j10 0:311 Y 1 1 1 11 2 3 j4 j10 V1 I1 5:0 08 16:1 Yinput;1 0:311 49:978
Yinput;1
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