0195136047.pdf

82 CIRCUIT ANALYSIS TECHNIQUES

EXAMPLE 2.3.1

Determine the voltage across the 20-resistor in the following circuit of Figure E2.3.1 (a) with

the application of superposition.

(a)

V

3

6 Ω

6 A

18 V V 12 Ω 80 Ω 20 Ω

+

−

+

−

Figure E2.3.1

(b)

V′

A B

3

VA′

3

6 Ω

18 V VA′^ =^ V′ 12 Ω = 80 Ω 20 Ω

+

−

+

−

I 1 ′

(c)

V′′

A B

3

VA′′

3

6 Ω

VA′′ = V′′^12 Ω =^80 Ω^20 Ω

+

−

6 A

Solution

Let us suppress the independent sources in turn, recognizing that there are two independent

sources. First, by replacing the independent current source with an open circuit, the circuit is

drawn in Figure E2.3.1(b). Notice the designation ofV′across the 12-resistor andV′/3 as the

dependent current source for this case. At nodeB,

(

1

80

+

1

20

)

VB′=

VA′

3

or VB′=

VA′

48

For the mesh on the left-hand side,( 6 + 12 )I 1 ′=18, orI 1 ′=1 A. But,I 1 ′=VA′/12, orVA′ =12 V.

The voltage across the 20-resistor from this part of the solution is

VB′=

12

48

=

1

4

V