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For the network of Problem 9.25, compute the transfer admittance Ytransfer;12 and use it to obtain node voltage V2 .
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Y 0:194 55:498 0:388 12 0:50 I1 12:9 55:498 V V2 Ytransfer;12
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SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS
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[CHAP. 9
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Replace the active network in Fig. 9-34(a) at terminals ab with a Thevenin equivalent.
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Z 0 j5 5 3 j4 2:50 j6:25 5 3 j4 
The open-circuit voltage V 0 at terminals ab is the voltage across the 3 j4  impedance: V0   10 08 3 j4 5:59 8 j4 26:568 V
Fig. 9-34
For the network of Problem 9.27, obtain a Norton equivalent circuit (Fig. 9-35).
At terminals ab, Isc is the Norton current I 0 . By current division,   10 08 3 j4 0:830 41:638 A I0 j5 3 j4 3 j9 5 3 j9
Fig. 9-35
Fig. 9-36
The shunt impedance Z 0 is as found in Problem 9.27, Z 0 2:50 j6:25 .
Obtain the Thevenin equivalent for the bridge circuit of Fig. 9-36. Make V 0 the voltage of a with respect to b.
By voltage division in either branch, 12 j24 20 33 j24 30 j60 20 80 j60
Vax
Vbx
CHAP. 9]
SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS   12 j24 30 j60 08 0:326 33 j24 80 j60
Hence,
Vab Vax Vbx 20
169:48 V V 0
Viewed from ab with the voltage source shorted out, the circuit is two parallel combinations in series, and so Z0 21 12 j24 50 30 j60 47:35 33 j24 80 j60 26:818 
Replace the network of Fig. 9-37 at terminals ab with a Norton equivalent and with a Thevenin equivalent.
Fig. 9-37 By current division, 3   6 7 3 j4 10 08 7 Isc I 0 6 4 5 3 j6 0:439 j10 3 j4 10 3 j6 and by voltage division in the open circuit, Vab V 0 3 j4 10 13 j4 2
105:268 A
08 3:68
36:038 V
Then
Z0
V0 3:68 I0 0:439
36:038 8:37 105:268
69:238 
See Fig. 9-38.
Fig. 9-38
SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS
[CHAP. 9
Supplementary Problems
9.31 Two circuit elements in a series connection have current and total voltage i 13:42 sin 500t 53:48 Identify the two elements. 9.32 Ans: A v 150 sin 500t 108 V
R 5 ; L 20 mH
Two circuit elements in a series connection have current and total voltage i 4:0 cos 2000t 13:28 Identify the two elements. Ans: A v 200 sin 2000t 50:08 V
R 30 ; C 12:5 mF
A series RC circuit, with R 27:5  and C 66:7 mF, has sinusoidal voltages and current, with angular frequency 1500 rad/s. Find the phase angle by which the current leads the voltage. Ans: 208 A series RLC circuit, with R 15 , L 80 mH, and C 30 mF, has a sinusoidal current at angular frequency 500 rad/s. Determine the phase angle and whether the current leads or lags the total voltage. Ans: 60:68, leads A capacitance C 35 mF is in parallel with a certain element. Identify the element, given that the voltage and total current are v 150 sin 3000t V Ans: R 30:1   . Determine the iT 16:5 sin 3000t 72:48 A
A two-element series circuit, with R 20  and L 20 mH, has an impedance 40:0 angle  and the frequency. Ans: 608; 276 Hz
Determine the impedance of the series RL circuit, with R 25  and L 10 mH, at (a) 100 Hz, (b) 500 Hz, (c) 1000 Hz. Ans: a 25:8 14:18 ; b 40:1 51:58 ; c 67:6 68:38  Determine the circuit constants of a two-element series circuit if the applied voltage v 150 sin 5000t 458 results in a current i 3:0 sin 5000t 158 (A). Ans: V
25 ; 8:66 mH
A series circuit of R 10  and p 40 mF has an applied voltage v 500 cos 2500t 208 (V). Find the C resulting current i. Ans: 25 2 cos 2500t 258 (A) p Three impedances are in series: Z1 3:0 458 , Z2 10 2 458 , Z3 5:0 908 . Find the applied voltage V, if the voltage across Z1 is 27:0 108 V. Ans: 126:5 24:68 V For the three-element series circuit in Fig. 9-39, (a) nd the current I; (b) nd the voltage across each impedance and construct the voltage phasor diagram which shows that V1 V2 V3 100 08 V. Ans: a 6:28 9:178 A; (b see Fig. 9-40.
Fig. 9-39
CHAP. 9]
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