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CHAP. 4]

53

ANALYSIS METHODS

Then, I ¼ V1 =10 ¼ 0:118 A.

4.11

Find the voltage Vab in the network shown in Fig. 4-25.

Fig. 4-25 The two closed loops are independent, and no current can pass through the connecting branch. I1 ¼ 2 A Vab ¼ Vax þ Vxy þ Vyb

4.12

30 ¼ 3A 10 ¼ I1 ð5Þ  5 þ I2 ð4Þ ¼ 3 V I2 ¼

For the ladder network of Fig. 4-26, obtain the transfer resistance as expressed by the ratio of Vin to I4 .

Fig. 4-26 By inspection, the network equation is 2

15 6 5 6 4 0 0

5 20 5 0

32 3 2 3 I1 Vin 0 0 7 6 7 6 5 07 76 I2 7 ¼ 6 0 7 20 5 54 I3 5 4 0 5 I4 0 5 5 þ RL

R ¼ 5125RL þ 18 750

N4 ¼ 125Vin

N Vin I4 ¼ 4 ¼ ðAÞ R 41RL þ 150 and

4.13

Rtransfer;14 ¼

Vin ¼ 41RL þ 150 ðÞ I4

Obtain a The´venin equivalent for the circuit of Fig. 4-26 to the left of terminals ab. The short-circuit current Is:c: is obtained from the three-mesh circuit shown in Fig. 4-27.

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An  
Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An  
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