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 = j 10
b. A Y impedance confi guration shown in Figure 3.48 with
Z 1 = Z 2 = Z 3 = j 10
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
(^12) Zj
12
12
120 0
10
120 0
10 90
12 90
∠∠
∠
∠ A
I V
(^23) Z
23
23
120 120
10 90
12 120 90 12 210
∠
∠
∠∠() A
I 23 = 12 ∠150°
I V
(^31) Z
31
31
120 120
10 90
12 120 90 12 30
∠
∠
∠∠() 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 (^22)
3
−
−
1