Fault Current Calculator

Calculate the available short-circuit current at the secondary terminals of a transformer using the infinite-bus method. Enter the transformer kVA rating, secondary voltage and percentage impedance to find the maximum available fault current for equipment AIC ratings.

How It Works

The infinite-bus method

The infinite-bus method assumes the primary supply has zero impedance (unlimited fault capacity), so the transformer impedance is the only limiting factor. This gives the worst-case (maximum) available fault current at the secondary terminals -- a conservative figure used to verify equipment AIC ratings.

The calculation proceeds in two steps:

  1. Transformer full-load amps (FLA): the rated current at the secondary terminals.
  2. Available fault current (AFC): FLA divided by the per-unit impedance.

Formulas

For a three-phase transformer:

FLA = kVA × 1000 / (√3 × V)
AFC = FLA / (%Z / 100)

For a single-phase transformer:

FLA = kVA × 1000 / V
AFC = FLA / (%Z / 100)

Worked example

Three-phase transformer: 75 kVA, 208 V secondary, 5 % impedance.

  • FLA = 75,000 / (√3 × 208) = 75,000 / 360.2 ≈ 208.2 A
  • AFC = 208.2 / 0.05 ≈ 4,164 A

All overcurrent protective devices (breakers, fuses) and switchgear downstream must have an AIC rating equal to or greater than this value.

Transformer %Z Reference

Typical percent-impedance values by transformer kVA rating (ANSI/IEEE C57.12.00). Always use the nameplate value when available.

Transformer size (kVA) Typical %Z (single-phase) Typical %Z (three-phase)
Up to 252.0 %2.0 -- 4.0 %
25 -- 502.0 -- 3.0 %4.0 %
51 -- 1003.0 %4.0 -- 5.0 %
101 -- 5004.0 %5.0 -- 5.75 %
501 -- 10005.0 %5.75 %
1001 and above6.0 %5.75 -- 7.5 %

Reference values only. Always use the transformer nameplate %Z for accurate results. The applicable code and equipment ratings govern.

Frequently Asked Questions

What is available fault current?

Available fault current (also called available short-circuit current) is the maximum current that can flow during a bolted short circuit at a given point in an electrical system. Equipment such as breakers and switchgear must have an ampere interrupting capacity (AIC) rating equal to or greater than this value to safely interrupt the fault.

What is the infinite-bus method and why is it conservative?

The infinite-bus method assumes the utility supply has zero impedance, meaning it can deliver unlimited fault current. Only the transformer impedance limits the fault current. In reality, the primary supply, cables and connections all add impedance, so the actual fault current is lower. Using the infinite-bus result ensures equipment is rated for the worst-case scenario.

What does transformer %Z mean?

Percent impedance (%Z) is the percentage of rated voltage needed to circulate full-load current through the transformer with the secondary short-circuited. A 5 % impedance means 5 % of the rated voltage produces 100 % of the rated current into a short circuit. A lower %Z means higher available fault current; a higher %Z limits it. The value is printed on the nameplate.

Why does fault current matter for AIC ratings?

If a protective device attempts to interrupt a fault current that exceeds its AIC rating, it can fail catastrophically, causing fires, explosions and arc flash events. NEC 110.9 and 110.10 require that all protective devices be rated for the available fault current at their location. The fault current value calculated here is used to select breakers, fuses, switchgear and busbar with a sufficient interrupting rating.

How does single-phase fault current differ from three-phase?

For a three-phase transformer the full-load amps formula includes the square root of 3 (approximately 1.732) in the denominator: FLA = kVA x 1000 / (1.732 x V). A single-phase transformer uses FLA = kVA x 1000 / V directly. Because the three-phase formula produces a lower FLA for the same kVA and voltage, the resulting fault current is also lower than the single-phase equivalent, even though three-phase systems deliver more power.

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