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 ð Þ 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.