Alternating Current Analysis 291
ANALYTICAL SolutionThe circuit of Figure 3.71 is transformed into the phasor circuit diagram shown in
Figure 3.72.
The three loop equations are() (^1020) **
1
20
jIII 123 20 0 50 0
∠
20 20 30
1
40
1
40
Ij 12 IjI 30
- *
0
1
40
10
1
40
Ij I12 3 jI 945
- * ∠
The matrix loop equation is given by
(^) ()(/)
( (/)) (/)
(/ )
10 20 1 20 20 0
20 20 30 1 40 1 40
0 1 40 10
j
jj
jj((/ )
140
50
0
945
1
2
3
∠
∠
I
I
I
C 2 = 4 F
v 1 (t) = 5 cos (10t) V
v 2 (t) = 9 cos(10t + 45°) V
I 1 I 2 I 3
R 1 = 10 Ω R 3 = 10 Ω
R 2 = 20 Ω
C 1 = 2 F L 1 = 3 H
FIGURE 3.71
Network of Example 3.14.
5 0 ° V 9 45 ° V
Xc1 = −j(1/20) Ω XL = j 30 Ω
Xc2 = −j(1/40) Ω
I 1 I 2 I 3
R 1 = 10 Ω
R 2 = 20 Ω
R 3 = 10 Ω
FIGURE 3.72
Phasor circuit diagram of Example 3.14.