Formula & Calculator

Short-Circuit Current

Calculates the maximum current that flows during a short-circuit fault, based on source voltage and fault impedance.

Power SystemsFault Analysis

Short-Circuit Current Calculator Isc = V / Zsc

Isc = V / Zsc
Isc = short-circuit current (A)  ·  V = voltage (V)  ·  Zsc = short-circuit impedance (Ω)
⟹ Solve Isc, V, Zsc
V
Ω
A
Please fix the errors above.
Solve for:
Presets:
Short-Circuit Current
V: Zsc: Isc:
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Short-Circuit Current Meter
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Isc = V / Zsc  ·  short-circuit current (A) = voltage (V) / short-circuit impedance (Ω)

Interpretation

Short‑circuit current I_sc = V / Z_sc is the current that flows when a fault (short circuit) occurs.
It is used for protection coordination and device rating.
Example: V=400V, Z_sc=0.2Ω → I_sc = 400/0.2 = 2000 A.

I_sc = V / Z_sc
Short-Circuit Current

Variables

SymbolQuantityUnit
I_scShort-circuit current (RMS symmetrical)A
VLine-to-line RMS voltage at the fault pointV
Z_scTotal equivalent impedance (source + line + transformer etc.)Ω

What it means

The short‑circuit current I_sc is the current that flows during a fault (short circuit) when the impedance is minimal. It is given by I_sc = V / Z_sc, where V is the voltage and Z_sc is the total impedance to the fault point. The short‑circuit current is typically many times the normal current and must be interrupted by protective devices. It is used to calculate the required breaking capacity of circuit breakers and to design the system for fault levels. Example: In a 400V system with a fault impedance of 0.2Ω, the short‑circuit current is 400 / 0.2 = 2000A. Circuit breakers in that circuit must be rated to handle at least this current. Accurate fault current calculation is essential for safety and equipment selection.

Worked example

Short‑Circuit Current – Practical Example

Real‑World
Scenario: A 400 V system has a short‑circuit impedance of 0.05 Ω. Find the short‑circuit current.
ParameterValue
V400 V
Zsc0.05 Ω
FormulaIsc = V / Zsc
1Isc = 400 / 0.05 = 8000 A = 8 kA
Final Design Isc = 8 kA ✓ High fault current
Why: Short‑circuit current is determined by the system voltage and fault impedance – used for breaker sizing.

Common mistakes

Watch for unit mismatches (W vs kW, single- vs three-phase) and remember to include power factor or efficiency where the formula requires it.

Applications

Short‑circuit current I_sc = V/Z_sc is the current that flows during a fault. This is used for equipment rating, protection coordination, and system design. Engineers calculate fault currents to ensure safety and reliability. This formula is essential for power system analysis.

  • Fault current calculation for protection studies
  • Equipment withstand and rating verification
  • Arc flash and incident energy analysis
  • System grounding and protection design
  • Educational understanding of fault currents

Frequently Asked Questions

Q01What is the formula for short‑circuit current?
A01

The short‑circuit current is I_sc = V / Z_sc, where V is the source voltage and Z_sc is the impedance of the short‑circuit path. For three‑phase faults, I_sc = V_L / (√3 Z_sc) for line quantities.

Q02What is the difference between a bolted fault and an arcing fault?
A02

A bolted fault is a direct short with negligible impedance. An arcing fault has an arc resistance, reducing the current.

Q03How do you calculate the symmetrical short‑circuit current?
A03

Use the subtransient reactance of generators and motors, and the impedances of transformers and lines. The total impedance is used in the formula.

Q04What are common mistakes when calculating short‑circuit current?
A04

Common errors: 1) using the wrong voltage, 2) ignoring motor contribution, 3) using the wrong impedance values, 4) not considering the system configuration.

Q05What are practical applications?
A05

Sizing protective devices, determining interrupting capacity, and setting relay coordination.

Q06How does the short‑circuit current vary with the distance from the source?
A06

The current decreases with distance due to the impedance of the line.

Q07What is the effect of fault current on equipment?
A07

High fault currents can cause thermal and mechanical stress on equipment, requiring robust design.

Q08What is the difference between symmetrical and asymmetrical fault current?
A08

Symmetrical current is the steady‑state AC component. Asymmetrical includes a DC offset, which can be higher during the first few cycles.