Calculate single-phase or three-phase short-circuit current from voltage and impedance or transformer kVA and percent impedance, or solve for equivalent impedance.

Three-phase: line-to-line voltage and impedance per phase. Single-phase: applied voltage and complete fault-loop impedance.

Estimate only. This does not calculate peak or arc-flash current or approve protective-device ratings.

Short Circuit Current Formula

Short-circuit current is fault current limited by the equivalent impedance of the source and fault path. For a single-phase circuit, use applied RMS voltage and complete loop impedance. For a balanced bolted three-phase fault, use line-to-line RMS voltage and equivalent impedance per phase. Resistance alone is sufficient only when reactance is negligible.

Isc = (V) / (k Z) k = 1 (single - phase), k = √(3) (three - phase)

Where:

  • Isc = symmetrical RMS short-circuit current
  • V = applied RMS voltage for single-phase, or line-to-line RMS voltage for three-phase
  • Z = complete loop impedance magnitude for single-phase, or equivalent impedance magnitude per phase for three-phase

If you are solving for a different variable, the same relationship can be rearranged:

V = k Isc Z
Z = (V) / (k Isc)

How to Use the Calculator

  1. Select the quantity or transformer method to solve, then select the system phase.
  2. For the system method, enter the two known quantities using the voltage and impedance definitions above. For the transformer method, enter the secondary voltage, matched nameplate kVA and percent impedance.
  3. Select Calculate. Expand Calculation details for the formula and assumptions. After editing inputs, select Calculate again to update the previous result.

The calculator can also solve for applied voltage or equivalent impedance from the other two quantities. Unit changes convert the entered quantity. The transformer method uses IFL = 1000 × kVA ÷ (k × V) and Isc = IFL × 100 ÷ %Z. It estimates a bolted fault at the secondary terminals with negligible upstream source impedance; it does not include feeder impedance, motor contribution or impedance tolerance.

Why Short Circuit Current Gets So Large

Short-circuit current rises as fault-path impedance falls. At the same voltage and phase configuration, current is inversely proportional to equivalent impedance.

For example, if equivalent impedance drops from 1 Ω to 0.01 Ω, current becomes 100 times larger at the same voltage and phase configuration. That is why fault current is often much higher than normal operating current.

Example 1

For a single-phase circuit with 500 V and complete fault-loop impedance of 0.03 Ω:

Isc = (500) / (0.03) ≈ 16666.67 A

The short circuit current is 16,666.67 A.

Example 2

For a single-phase circuit with 50 V and complete fault-loop impedance of 0.25 Ω:

Isc = (50) / (0.25) = 200

The short-circuit current is 200 A.

Interpreting the Result

The calculated value is a symmetrical RMS estimate under the selected fault model and entered equivalent impedance. It is not peak current or an arc-flash calculation, and the single-phase model does not represent an earth fault on a three-phase network. In real electrical systems, actual fault current can also be affected by:

  • Source impedance
  • Transformer impedance
  • Conductor length and size
  • Connection quality
  • Motor contribution
  • System voltage drop under fault conditions

Because of this, the calculator is best used for quick estimation, concept checks, and educational calculations. Protective device selection and fault-duty verification should be based on a full system study when required.

Common Unit Checks

  • 1 kV = 1,000 V
  • 1 mΩ = 0.001 Ω
  • 1 kA = 1,000 A

Common Mistakes

  • Using normal load resistance instead of equivalent fault-path impedance
  • Mixing volts and kilovolts without converting properly
  • Entering milliohms as ohms
  • Assuming the estimate includes all real-world impedance effects

Practical Notes

A higher short-circuit current generally means more thermal stress, stronger magnetic forces, and greater demands on fuses, breakers, bus bars, and conductors. Even small errors in impedance can produce large current differences, so use verified system data.

Frequently Asked Questions

Is short-circuit current the same as normal operating current?
No. Operating current is drawn by the intended load. Fault current is limited by the source and fault-path impedance and can be much higher.

Why does a tiny impedance create such a huge current?
Current follows the inverse relationship above. A smaller positive impedance gives a larger current at the same voltage. Zero impedance is outside this calculator’s model.

Can this calculator be used for breaker sizing?
It can help estimate available current, but breaker interrupting ratings and protection studies should account for the full electrical system, including the appropriate fault type, source, conductors, motor contribution and applicable requirements.

What impedance should be entered?
For single-phase, use the complete fault-loop impedance including its return path. For a balanced three-phase fault, use equivalent impedance per phase referred to the entered voltage. Include resistance and reactance as applicable, rather than ordinary load resistance.