Alternating Current Analysis
265
R.3.86 The example shown in Figure 3.47 illustrates the general approach employed to
solve for all the currents (load and line) in a typical polyphase balanced network.
The voltage specs are
V 12 = 120 ∠0°
V 23 = 120 ∠−120°
V 31 = 120 ∠120° (in volts)
The analysis shown in the following is for the load specs for the following
structures:
a. A ∆ impedance confi guration shown in Figure 3.47 with
Z 12 = Z 23 = Z 31 = j10
b. A Y impedance confi guration shown in Figure 3.48 with
Z 1 = Z 2 = Z 3 = j10
ANALYTICAL Solution
Part a
Applying Ohm’s law to the circuit shown in Figure 3.47, the following currents can be
evaluated by
I
V
Z
j
12
12
12
120 0
10
120 0
10 90
12
90
ϭ
ϭ
Њ ϭ
Њ
Њ
ϭ
Ϫ Њ
∠
∠
∠
∠
A
I
V
Z
23
23
23
120
120
10 90
12
120
90
12
210
ϭ
ϭ
Ϫ Њ
Њ
ϭ
Ϫ ЊϪ Њ ϭ
Ϫ Њ
∠
∠
∠
∠
(
)
A
I 23 = 12 ∠150°
I
V
Z
31
31
31
120 120
10 90
12 120
90
12 30
ϭ
ϭ
Њ
Њ
ϭ
ЊϪ Њ ϭ
Њ
∠
∠
∠
∠
(
)
A
FIGURE 3.47
3 Φ system connected to a ∆ load with Z 12 = Z 23 = Z 31 = j10.
+
−
V 23
V 31
V 12
Z 12
Z 13
Z 32
I 1
I 3
I 31
I 12
I 23
I 2
2
3
+
−
−
+
1
•
•
•
CRC_47760_CH003.indd 265
CRC_47760_CH003.indd 265
7/23/2008 1:27:42 PM
7/23/2008 1:27:42 PM