PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

266 Practical MATLAB® Applications for Engineers

Then, I 1 = I 12 − I 31

I 1 
 12 ∠∠ ∠ 90 12 30
12 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