Short Circuit Current Calculator

How It Works

When a short circuit occurs, an enormous current flows until a breaker or fuse clears the fault. The available fault current depends on the transformer size and impedance, reduced by cable impedance between the transformer and the panel. Every panel and breaker must have an AIC (Ampere Interrupting Capacity) rating equal to or greater than the available fault current.

The Formula

Variables

• kVA — Transformer kilovolt-ampere rating
• Z% — Transformer impedance percentage (typically 2-5%)
• FLA — Full load amps of the transformer secondary
• AIC — Ampere Interrupting Capacity -- the maximum fault current a device can safely interrupt
• Z_cable — Cable impedance = resistance per foot x length x 2 (out and return)

Example

A 50 kVA transformer at 3.5% impedance on 240V: FLA = 208A. Fault current at transformer = 208 / 0.035 = 5,952A (5.95 kA). With 50 ft of 4/0 cable, fault current at the panel drops to about 5.5 kA. A standard 10 kA rated panel is sufficient.

Tips

• Most residential panels are rated for 10 kA AIC. This is sufficient for services fed by utility transformers 50 kVA and smaller.
• Lower transformer impedance means higher available fault current -- counterintuitive but important.
• Longer cable runs reduce fault current at the panel but also increase voltage drop under normal loads.
• An undersized AIC rating is a serious safety hazard -- the breaker may not be able to clear a fault, causing an arc flash or fire.
• For commercial and industrial panels, always have a qualified engineer perform a formal short circuit study.