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54

ANALYSIS METHODS

[CHAP. 4

Fig. 4-27 2

15 5

0

32

I1

3

2

Vin

3

7 7 6 6 76 4 5 20 5 54 I2 5 ¼ 4 0 5 0 Is:c: 0 5 15    5 20    Vin  0 5  Vin Is:c: ¼ ¼ R 150 The open-circuit voltage Vo:c: is the voltage across the 5- resistor indicated in Fig. 4-28.

Fig. 4-28 2

15 5 6 4 5 20 0 5

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

I3 ¼

25Vin Vin ¼ ðAÞ 5125 205

Then, the The´venin source V 0 ¼ Vo:c: ¼ I3 ð5Þ ¼ Vin =41, and RTh ¼

Vo:c: 150 ¼  Is:c: 41

The The´venin equivalent circuit is shown in Fig. 4-29. current is I4 ¼

With RL connected to terminals ab, the output

Vin =41 Vin ¼ ðAÞ ð150=41Þ þ RL 41RL þ 150

agreeing with Problem 4.12.

4.14

Use superposition to find the current I from each voltage source in the circuit shown in Fig. 4-30. Loop currents are chosen such that each source contains only one current.

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An  
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