- #1
lam58
- 29
- 0
Hi could someone check my answer to the question below.
Question:
My Answer:
[tex] I_{fA} = 3I_{1A} = 3I_{2A} = 3I_{0A} \Rightarrow I_{fA} = \frac{3E}{Z_1 + Z_2 + Z_0} = \frac{33kV}{20+20+30} = 471.43A[/tex]
Then using the same method I got [tex] I_{fB} = 942.86A[/tex]
Question:
My Answer:
[tex] I_{fA} = 3I_{1A} = 3I_{2A} = 3I_{0A} \Rightarrow I_{fA} = \frac{3E}{Z_1 + Z_2 + Z_0} = \frac{33kV}{20+20+30} = 471.43A[/tex]
Then using the same method I got [tex] I_{fB} = 942.86A[/tex]