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Hi could someone check my answer to the question below. THIS IS NOT HOMEWORK.
Question:
My Answer:
\( I_{fA} = 3I_{1A} = 3I_{2A} = 3I_{0A} \Rightarrow I_{fA} = \frac{3E}{Z_1 + Z_2 + Z_0} = \frac{33kV}{20+20+30} = 471.43A\)
Then using the same method I got \( I_{fB} = 942.86A\)
Question:
My Answer:
\( I_{fA} = 3I_{1A} = 3I_{2A} = 3I_{0A} \Rightarrow I_{fA} = \frac{3E}{Z_1 + Z_2 + Z_0} = \frac{33kV}{20+20+30} = 471.43A\)
Then using the same method I got \( I_{fB} = 942.86A\)
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