Raiden
Staff Engineer
The problem speaks about a transformer connected to the utilitiy at a 6.6kv and the source impedance (utilitie's) is 0.94%. If the transformer rated impedance is 4.25% and rated current is 1720A. (Secondary voltage is 480V) What is the short circuit current available at the transformer secondary?
For the answer: Isc= FLA / (%Zsource+%Ztrafo) = 33,141A
My question is: How can you add the Z% source? Is it not at a different base than the transformer?. After getting the equivalent impedance (add in series) dont we need to do 480/ Zequivalent to get the Isc?
For the answer: Isc= FLA / (%Zsource+%Ztrafo) = 33,141A
My question is: How can you add the Z% source? Is it not at a different base than the transformer?. After getting the equivalent impedance (add in series) dont we need to do 480/ Zequivalent to get the Isc?
