tlona
Industrial
- Jun 1, 2010
- 55
Good morning. I have a 2.5mw, 12.5kv/480-277v transformer with a %Z of 6.10. My understanding is the that maximum available fault on the secondary would be The transformers full load current divided by %z (Ifla/%z). The utility has provided a secondary symmetrical fault current of 56398amps which is greater than the transformer max fault current of 49378amps (3120amps/.061). Is this possible?. I am assuming they mean asymmetrical? Thank-you.