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How do you calculate the maximum possible Available Fault Current at the secondary of a Transformer?
Issue:
Transformer Available Fault Current
Product Line
LV Transformers
Environment:
Applies to LV 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 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).
Transformer Available Fault Current
Product Line
LV Transformers
Environment:
Applies to LV 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 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).
Released for:Schneider Electric Canada
