Frequently asked questions
How do you calculate the maximum possible Available Fault Current at the secondary of a Transformer?
29 September 2021Issue:
Transformer Available Fault Current
Applies to LV Transformers by SquareD/Schneider Electric
Transformers are often connected to equipment that is rated for certain AIC, Average Interrupting Current, or Available Fault Current
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 75kVA Transformer with a 208Y/120 secondary and 5% impedance has a capacity of 208 amps 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).