# 0195136047.pdf

(Joyce) #1

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``