266 Practical MATLAB® Applications for Engineers
Then, I 1 = I 12 − I 31
I 1 12 ∠∠ ∠ 90 12 3012 3 120 A
and I 2 = I 23 − I 12
I 2 12 150∠∠ 12 90 12 3 120∠A
and I 3 = I 31 − I 23 = 12 ∠30° − 12 ∠150° A
I 3 = 12 √
__
3 ∠0° A
Part b
The system for part b is shown in Figure 3.48.
Transforming the Y load into a ∆ confi guration, and making use of the symmetry of
the system, the equivalent impedances are evaluated.
Z
ZZ ZZ ZZ
(^12) Z
12 23 31
3
and since Z 1 = Z 2 = Z 3 = j10, t hen
Z
Z
Z
12 1 Zjj
2
1
1
3
3 3 10() 30 Ω
Z
ZZ ZZ ZZ
(^23) Z
12 23 31
1
Z 23 = 30 j Ω
Z
ZZ ZZ ZZ
(^31) Z
12 23 31
2
and
Z 31 = 30 j Ω
FIGURE 3.48
3 Φ system connected to a Y load with Z 1 = Z 2 = Z 3 = 10 j.
- V −
23
V 31
V 12
Z 2
Z 3
Z 1
I 1
I (^22)
−
−
1
- •
N