Alternating Current Analysis
313
X c = −j3 Ω
X L = j 5 Ω
R = 3 Ω
30 90°
FIGURE 3.92
Network diagram of P.3.18.
V 1 = 100 0° V
+
–
V 2 = 100 90° V
+
–
Z 1 = j 25 Ω
Z 2 = j 25 Ω
Z 3 = −j 50 Ω
Z 5 = −j 50 Ω
Z 6 = −j 50 Ω
Z 4 = −j 25 Ω
I 1
I 3
I 2
FIGURE 3.91
Network diagram of P.3.16.
CRC_47760_CH003.indd 313
CRC_47760_CH003.indd 313
7/23/2008 1:27:55 PM
7/23/2008 1:27:55 PM
313
X c = −j3 Ω
X L = j 5 Ω
R = 3 Ω
30 90°
FIGURE 3.92
Network diagram of P.3.18.
V 1 = 100 0° V
+
–
V 2 = 100 90° V
+
–
Z 1 = j 25 Ω
Z 2 = j 25 Ω
Z 3 = −j 50 Ω
Z 5 = −j 50 Ω
Z 6 = −j 50 Ω
Z 4 = −j 25 Ω
I 1
I 3
I 2
FIGURE 3.91
Network diagram of P.3.16.
CRC_47760_CH003.indd 313
CRC_47760_CH003.indd 313
7/23/2008 1:27:55 PM
7/23/2008 1:27:55 PM