Alternating Current Analysis
291
ANALYTICAL Solution
The circuit of Figure 3.71 is transformed into the phasor circuit diagram shown in
Figure 3.72.
The three loop equations are
(
)
*
*
10 20
1
20
20
0
50 0
1
2
3
ϩ
Ϫ
Ϫ
ϩ
ϭ
Њ
j
I
I
I
∠
Ϫ
ϩ
ϩ
Ϫ
Ϫ Ϫ
ϭ
20
20
30
1
40
1
40
0
1
2
3
I
j
I
j
I
*
*
0
1
40
10
1
40
9 45
1
2
3
I
j
I
j
I
Ϫ Ϫ
ϩ
Ϫ
ϭ
Њ
∠
*
*
The matrix loop equation is given by
(
)
(/ )
(
( / ))
( / )
( / )
10 20
1 20
20
0
20
20
30 1 40
1 40
0
1 40
10
ϩ
Ϫ
Ϫ
Ϫ
ϩ
Ϫ
ϩ
ϩ
Ϫ
j
j
j
j
j ( ( / )
*
1 40
5 0
0
9 45
1
2
3
∠
∠
I
I
I
ϭ
Њ
Њ
C 2 = 4 F
v 1 (t ) = 5 cos (10*t) V
v 2 (t ) = 9 cos(10*t + 45°) V
I 1
I 2
I 3
R 1 = 10 Ω
R 3 = 10 Ω
R 2 = 20 Ω
L 1 = 3 H
C 1 = 2 F
FIGURE 3.71
Network of Example 3.14.
5
9
0° V
45° V
X L = j 30 Ω
X c1 = −j (1/20) Ω
X c2 = −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.
CRC_47760_CH003.indd 291
CRC_47760_CH003.indd 291
7/23/2008 1:27:50 PM
7/23/2008 1:27:50 PM
291
ANALYTICAL Solution
The circuit of Figure 3.71 is transformed into the phasor circuit diagram shown in
Figure 3.72.
The three loop equations are
(
)
*
*
10 20
1
20
20
0
50 0
1
2
3
ϩ
Ϫ
Ϫ
ϩ
ϭ
Њ
j
I
I
I
∠
Ϫ
ϩ
ϩ
Ϫ
Ϫ Ϫ
ϭ
20
20
30
1
40
1
40
0
1
2
3
I
j
I
j
I
*
*
0
1
40
10
1
40
9 45
1
2
3
I
j
I
j
I
Ϫ Ϫ
ϩ
Ϫ
ϭ
Њ
∠
*
*
The matrix loop equation is given by
(
)
(/ )
(
( / ))
( / )
( / )
10 20
1 20
20
0
20
20
30 1 40
1 40
0
1 40
10
ϩ
Ϫ
Ϫ
Ϫ
ϩ
Ϫ
ϩ
ϩ
Ϫ
j
j
j
j
j ( ( / )
*
1 40
5 0
0
9 45
1
2
3
∠
∠
I
I
I
ϭ
Њ
Њ
C 2 = 4 F
v 1 (t ) = 5 cos (10*t) V
v 2 (t ) = 9 cos(10*t + 45°) V
I 1
I 2
I 3
R 1 = 10 Ω
R 3 = 10 Ω
R 2 = 20 Ω
L 1 = 3 H
C 1 = 2 F
FIGURE 3.71
Network of Example 3.14.
5
9
0° V
45° V
X L = j 30 Ω
X c1 = −j (1/20) Ω
X c2 = −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.
CRC_47760_CH003.indd 291
CRC_47760_CH003.indd 291
7/23/2008 1:27:50 PM
7/23/2008 1:27:50 PM