How do you calculate the maximum possible Available Fault Current at the secondary for single-phase and three-phase Transformers?

Issue:
Transformer Available Fault Current

Product Line
LV Transformers

Environment:
Applies to LV single phase and three phase Transformers by SquareD/Schneider Electric

Cause:
Transformers are often connected to equipment that is rated for certain AIC, Average Interrupting Current, or Available Fault Current

Resolution:
Take the secondary full load current of the Transformer and divide it by the Transformer`s impedance. This is not necessary on Transformers below 15kVA, thus why UL does not require the impedance on the nameplate of Transformers below 15kVA.

For example, a three phase 75kVA Transformer with a 208Y/120 V secondary and 5% impedance has a capacity of 208 amps (three-phase formula is I=kVA/1.732*V) on the secondary. The peak let through with an infinite source (worst case scenario) would have 208 / 0.05 = 4,160 amps of peak let through current (available fault current).

Now, a single phase 75kVA Transformer with a 120 V secondary and 5% impedance has a capacity of 625 amps (single-phase formula is I=kVA/V) on the secondary. The peak let through with an infinite source (worst case scenario) would have 625 / 0.05 = 12,500 amps of peak let through current (available fault current).

Publicado para: Schneider Electric Mexico