Estimate single- or balanced three-phase bolted symmetrical RMS fault current from transformer rating and %Z or known R+jX. An optional interrupting-rating comparison gives an arithmetic margin only.

Use total three-phase rating and secondary line-to-line voltage. Infinite primary bus; transformer impedance only.

Estimates bolted symmetrical RMS current. Not a line-to-ground, arc-flash, peak-current or equipment-approval study.

Enter nameplate percent: 5 means 5%, not 0.05%.

Interrupting rating comparison (optional)

Blank omits comparison. Use the rating applicable at the entered voltage; an arithmetic margin is not approval.

Fault Current Formula

The known-impedance method uses the driving voltage and equivalent impedance magnitude. For a single-phase fault loop, I = V / |Z|. For a balanced three-phase bolted fault, I = VLL / (√3 |Z1|), where Z1 is the positive-sequence per-phase Thevenin impedance. The Eaton short-circuit guide describes transformer and network contributions.

Single phase: I = V / |Z| Balanced three phase: I = VLL / (√3 |Z1|) Transformer: Isc = IFL / (Z% / 100)
  • I or Isc is the bolted symmetrical RMS fault-current estimate (amps)
  • V is the voltage across a single-phase loop; VLL is three-phase line-to-line voltage (volts)
  • |Z| = √(R² + X²); use complete single-phase loop impedance or three-phase positive-sequence per-phase impedance, in ohms

The transformer method first finds full-load current: VA / V for single phase, or VA / (√3 VLL) for three phase, then divides by %Z / 100. See Schneider Electric’s transformer-current formulas. Use matching full-winding rating and voltage for single phase. This simplified infinite-primary-bus estimate excludes upstream/conductor impedances, other source contributions and tolerance adjustments; it is not a center-tapped line-to-neutral or system ground-fault calculation. The optional rating margin is not a device-selection or certification check.

Fault Current Definition

Fault current is the current that flows when an electrical fault (such as a short circuit) occurs. The available (or prospective) fault current at a point is the maximum current the system can deliver into a fault at that location under stated conditions, and it depends on the source and network impedance and the type of fault.

During a bolted fault the fault-path impedance can be very low, which can make the current very large. In most real power systems, however, the fault current is primarily limited by the total Thevenin/source impedance of the supply (transformers, generators, and conductors), not just the fault’s own resistance.

How to calculate fault current?

Example Problem #1.

Select Known R + jX · single-phase. For this example, use 120 V across the complete fault loop.

Use a complete loop resistance of 0.25 Ω and reactance of 0 Ω, giving |Z| = 0.25 Ω. Leave the optional interrupting rating blank unless a separate arithmetic comparison is needed.

Finally, use Ohm's Law to calculate the fault current.

I = V/Z

= 120/0.25

= 480 amps.

Example Problem #2.

In this next example, we will take a look to see how much a smaller impedance affects the current.

Again, we will use the same voltage as above, 120 V.

Keep the single-phase loop model and set R = 0.05 Ω, X = 0 Ω, so the impedance magnitude is 0.05 Ω.

Using the formula as in example 1:

I = V/Z

= 120/0.05

= 2,400 amps.

This is 5 times the current of example 1.

Fault Current Calculator screenshot