{}

Our Brands

Impact-Company-Logo-English Black-01-177x54

Welcome to the Schneider Electric Website

Welcome to our website.
How can we help you today?
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).

Schneider Electric New Zealand

Users group

Discuss this topic with experts

Visit our Community for first-hand insights from experts and peers on this topic and more.