Short-Circuit Current Calculator
Model a bolted short-circuit fault end to end — solve for fault current, source voltage, or short-circuit impedance, then see the fault's apparent power, X/R ratio, and peak asymmetrical current.
Holds source voltage (and, for the peak curve, resistance R) at your current inputs and sweeps Zsc — showing the steady rms fault current alongside the much larger first-cycle peak (asymmetrical) current, and how fast both fall as impedance rises.
Left: the fault path as a single-line diagram — source voltage driving current through the resistive (R) and reactive (X) parts of Zsc into a bolted fault. Right: the impedance triangle, Zsc² = R² + X², with the power-factor angle φ where cos φ = R/Zsc.
| Isc | Typical read |
|---|---|
| < 1 kA | Low — light branch circuit or long feeder |
| 1 – 10 kA | Moderate — typical LV distribution board |
| 10 – 50 kA | High — main LV switchboard / MV feeder |
| > 50 kA | Severe — utility-scale infeed, needs high breaking-capacity gear |
| X/R | κ (peak factor) |
|---|---|
| 1 | ≈ 1.07 |
| 5 | ≈ 1.56 |
| 10 | ≈ 1.75 |
| 20 | ≈ 1.86 |
| 30 | ≈ 1.91 |
Source Voltage ↑
Fault current scales linearly with voltage — a stiffer, higher-voltage source drives more current through the same impedance.
SC Impedance ↓
Isc is inversely proportional to Zsc — a stiffer source (low source impedance) or shorter, heavier-gauge cable means a higher fault current.
X/R ratio ↑
A more reactive (less resistive) circuit has a higher X/R ratio, which pushes the asymmetry factor κ toward its maximum of 2.0 — a much larger first-cycle peak.
Peak vs. rms current
Breakers and bus bars must be rated for the peak asymmetrical current (κ·√2·Isc), which can be nearly twice the steady rms value right after the fault starts.
What Is Short-Circuit Current?
Short-circuit current is the current that flows when a fault — an accidental, low-impedance connection — bridges two points that shouldn't be connected, typically a live conductor and ground or neutral. With the normal load removed from the circuit, only the short-circuit impedance between the source and the fault point limits how much current flows, which is why fault currents are usually far larger than normal operating currents.
The Short-Circuit Current Formula
This is simply Ohm's law applied to a bolted fault: the source voltage V drives current through whatever impedance Zsc sits between the source and the short circuit — the source's own internal impedance plus any transformer, cable and busbar impedance in the path. Because Zsc is normally small, Isc comes out large: this is the basis for sizing breakers, fuses and busbar bracing.
How to Calculate Short-Circuit Current
Choose the unknown above — Fault Current, Source Voltage, or SC Impedance — enter the other two, and the calculator rearranges the formula automatically.
| Solve for | Rearranged formula |
|---|---|
| Isc | Isc = V / Zsc |
| V | V = Isc × Zsc |
| Zsc | Zsc = V / Isc |
Short-Circuit Current Formula Variables
| Symbol | Meaning |
|---|---|
| Isc | Short-circuit (fault) current — steady-state rms symmetrical value |
| V | Source voltage at the fault point |
| Zsc | Short-circuit impedance magnitude between source and fault |
| R, X | Resistive and reactive components of Zsc, where Zsc = √(R² + X²) |
| κ | Asymmetry (peak) factor, set by the X/R ratio of the fault path |
| Ip | Peak asymmetrical fault current, Ip = κ·√2·Isc (derived, not entered) |
Short-Circuit Impedance: Resistance vs. Reactance
Zsc isn't a single physical component — it's the combined impedance of everything between the source and the fault: the source or transformer's internal impedance, cable or busbar impedance, and any other series elements. It has both a resistive part R (which dissipates energy) and a reactive part X (from inductance in windings, cables and busbars), related by Zsc = √(R² + X²). Larger equipment and higher-voltage systems tend to have a higher X/R ratio, since conductor reactance doesn't shrink as fast as resistance does with larger conductor sizes.
Why Peak (Asymmetrical) Fault Current Matters
The Isc formula gives the steady-state rms current once any transient has settled. But the instant a fault occurs, the current can start asymmetrically offset from zero, and the first current peak can be substantially higher than √2·Isc. That extra margin is captured by the asymmetry factor κ, which depends only on the X/R ratio: κ = 1.02 + 0.98·e^(−3R/X). Equipment — breakers, busbars, switchgear — must be braced for this peak value, Ip = κ·√2·Isc, not just the steady rms current.
Short-Circuit Current Calculation Example
Isc = 400 / 0.2 = 2000 A = 2 kA
X = √(0.2² − 0.02²) ≈ 0.199 Ω → X/R ≈ 9.95 → κ ≈ 1.75 → Ip ≈ κ·√2·Isc ≈ 4.94 kA peak
Frequently Asked Questions
Normal load current is limited by the load's own impedance, which is typically much larger than the wiring and source impedance. A short circuit removes the load entirely, so the only thing left limiting current is the small Zsc of the source and feeder path — which is why Isc can be tens or hundreds of times the rated load current.
Isc is the steady-state rms current once the fault has settled into a symmetrical sine wave. Ip is the much higher first-cycle peak, inflated by a DC offset that decays over a few cycles — equipment must be mechanically and electrically rated to survive that peak, not just the steady Isc.
It's the ratio of reactance to resistance in the fault path, and it controls how asymmetrical the first current cycle is. A low X/R (very resistive path) decays quickly and stays close to symmetrical; a high X/R (very reactive path, common near large transformers) produces a much higher, slower-decaying peak.
No — this calculator models a single-impedance bolted fault with the basic Isc = V/Zsc relationship plus a standard asymmetry approximation. A full study (per IEC 60909 or ANSI/IEEE methods) accounts for multiple contributing sources, motor contribution, and network topology in detail.
In terms of raw current magnitude, yes — lower Zsc means higher Isc and a higher required equipment breaking capacity. But a very high Zsc isn't necessarily "safer" either, since it can mean fault currents too low to reliably trip protective devices quickly, which is its own hazard.
Variables
| Symbol | Quantity | Unit |
|---|---|---|
| I_sc | Short-circuit current (RMS symmetrical) | A |
| V | Line-to-line RMS voltage at the fault point | V |
| Z_sc | Total 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| Parameter | Value |
|---|---|
| V | 400 V |
| Zsc | 0.05 Ω |
| Formula | Isc = V / Zsc |
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
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