Ray77

Just studying for an exam and came across this question: Calculate the available fault current on the secondary side of a 2500kVA, 4160-480V utility transformer with an impedance of 5.75%. Assume an infinite available fault current on the primary side of the transformer. Assume there are no motors that will contribute to the fault current.

I'm just not quite sure how to approach this, do I need the full load current first? Any guidance would be greatly appreciated.

studiot

Was this the whole question?
It seems a bit short on information.

Impedance is measured in Ohms, not percentages.

If 5.75% was mentioned I would guess this referred to the % regulation at fault conditions.

Without further information I further guess that the fault is a secondary short circuit.

In a secondary short circuit, the secondary output voltage is zero so the whole of the primary applied voltage is used in sending current through both the primary and secondary windings.