When a fault occurs in an electrical system, the current that flows is not the normal load current -- it is a fault current that can be tens or hundreds of times larger, limited only by the impedances of the source, transformer, and conductors between the supply and the fault point. Protective devices such as circuit breakers, fuses, and busbars must be rated to withstand and interrupt this fault current without failing. If they are undersized, the consequence is not a slow overload -- it is an explosive arc flash event that can destroy equipment and injure or kill personnel.
This guide covers the IEC 60909 short circuit calculation method, the physical difference between symmetrical and asymmetrical fault current, the kappa factor, and a full worked example for a low voltage distribution system -- with an interactive calculator.
Symmetrical fault current is the steady state AC component only. Asymmetrical adds the decaying DC offset in the first few cycles.
Breaker interrupting ratings use symmetrical RMS. Mechanical withstand ratings use peak asymmetrical.
Symmetrical vs Asymmetrical: What Is the Difference
At the instant a fault occurs, the current waveform is not a clean sine wave. Two components appear simultaneously: an AC symmetrical component that continues indefinitely, and a DC offset component that decays exponentially.
The DC offset arises as energy stored in the system inductance dissipates through resistance.
The symmetrical component is the theoretical worst-case steady state AC value -- what the system would produce at zero DC offset. This is the value published as a breaker's interrupting rating.
The asymmetrical component adds the transient DC offset on top of the AC waveform, making the total current higher in the first few cycles. Mechanical parts of switchgear must withstand this higher peak.
Symmetrical Fault Current (I''k)
Pure AC component only. Waveform is symmetric about the zero axis. Used for: breaker interrupting rating, fuse selection, relay current settings.
Formula: I''k = c × V_n / (√3 × Z_total)
Steady state, no DC offsetPeak Asymmetrical Current (ip)
AC component plus decaying DC offset. Peaks in the first half cycle. Used for: busbar mechanical bracing, switchgear frame, cable support spacing.
Formula: ip = κ × √2 × I''k
First half cycle maximumThree Phase Fault (Highest Severity)
All three phases shorted together. Produces the highest symmetrical fault current because all three phase voltages drive current through the lowest impedance path. Always the worst case for symmetrical current.
Used for breaker sizingSingle Phase to Earth Fault
One phase shorted to earth. Can produce higher or lower fault current than a three phase fault depending on transformer earthing and zero-sequence impedance. Often governs earth fault relay settings and earth conductor sizing.
Governs earth protectionThe Calculation Formulas: IEC 60909 Method
IEC 60909 is the international standard for short circuit current calculation. It uses a voltage factor c to account for pre fault voltage levels above nominal and the fact that the worst-case fault current does not occur at nominal voltage.
V_n = nominal system voltage (V, line to line)
Z_k = total short circuit impedance at the fault point (Ω)
Z_k = Z_source + Z_transformer + Z_cable + Z_motor (sum all upstream impedances)
I''k = initial symmetrical short circuit current (A RMS)
κ = kappa factor = 1.02 + 0.98 × e−3R/X
R/X = resistance to reactance ratio of the total impedance path
Range of κ: 1.02 (purely resistive, R/X very high) to 2.0 (purely inductive, X/R very high)
√2 = 1.414 converts RMS symmetrical to peak AC value
Building the Total Impedance: What to Include
The total fault impedance at any point is the sum of all series impedances between the grid connection and the fault location.
For a typical LV industrial system, this includes the grid source impedance, the main transformer, the LV busbar, the cable run, and any running motors. The transformer guide covers how to read Vz% from a nameplate.
| Impedance Source | How to Calculate | Typical Value |
|---|---|---|
| Utility grid source (Zs) | From the utility short circuit MVA at the point of common coupling: Zs = V² / Ssc | Varies -- utilities publish Ssc at the metering point |
| Transformer (Zt) | Zt = (Vz% / 100) × (V² / Sn) where Vz% is transformer percentage impedance and Sn is rated kVA | Vz% = 4% to 6% for distribution transformers |
| Cable (Zc) | Zc = R + jX from cable datasheet. At LV, resistance usually dominates. R from IEC 60228 at operating temperature. | 0.2 to 15 Ω/km depending on size |
| LV busbar (Zb) | Calculated from busbar dimensions and material, or from manufacturer data. Often small enough to neglect for preliminary calculations. | Often below 0.001 Ω for main busbars |
| Motor contribution | Running motors act as generators during a fault -- similar to the motor contribution discussed in the electrical energy guide. Contribution per motor ≈ Isc_motor = 6 × In. Add to fault current for first few cycles only. | Typically adds 10 to 30% to fault current near large motors |
Worked Example: 630 kVA Transformer LV Panel
A 630 kVA, 415V transformer with 6% impedance feeds a main LV panel via 35 mm² copper cable, 30 m long (R = 0.554 Ω/km at 70°C). Grid source impedance is negligible (strong grid assumption).
Calculate the symmetrical fault current, peak asymmetrical current, and required circuit breaker rating.
Zt = (6/100) × (415² / 630,000) = 0.06 × 0.2734 = 0.01640 ΩZc = 0.554 × 30/1000 = 0.01662 ΩZ_total = 0.01640 + 0.01662 = 0.03302 ΩI''k = (1.05 × 415) / (1.732 × 0.03302) = 435.75 / 0.05719 = 7,620 A = 7.62 kAκ = 1.02 + 0.98 × e^(−3 × 0.2) = 1.02 + 0.98 × 0.549 = 1.558ip = 1.558 × 1.414 × 7,620 = 16,800 A = 16.8 kAShort Circuit Current Calculator
Applying the Results to Equipment Selection
| Equipment | Use Symmetrical (I''k) | Use Peak (ip) | Standard Reference |
|---|---|---|---|
| Circuit breaker interrupting rating | Yes -- breaker must interrupt I''k | No | IEC 60947-2 |
| Circuit breaker momentary/close and latch rating | No | Yes -- must withstand ip | IEC 60947-2 |
| Fuse breaking capacity | Yes | No | IEC 60269 |
| Busbar mechanical bracing | No | Yes -- magnetic force ∝ ip² | IEC 61439 |
| Cable support spacing | No | Yes -- separation force during fault | IEC 61200-53 |
| Earth conductor sizing (adiabatic) | Yes -- uses I²t at I''k | No | IEC 60364-5-54 |
| Current transformer (CT) accuracy | Yes -- rated on symmetrical current | No | IEC 61869-2 |
Watch: Short Circuit Calculation Explained
Short Circuit Calculation Questions
External References
- IEC 60909-0: Short Circuit Currents in Three Phase AC Systems
- Short Circuit and Fault Current Calculation Guide -- Electrical World
What We Learn Today
- Symmetrical I''k = (c × V_n) / (√3 × Z_total) -- use this for breaker interrupting rating
- Peak ip = κ × √2 × I''k -- use this for busbar bracing and mechanical withstand
- Kappa (κ) = 1.02 + 0.98 × e^(−3R/X) -- depends on system X/R ratio
- Total impedance = transformer + cable + source -- sum all series paths to the fault point
- Running motors add approximately 6 × In each to the fault current for the first few cycles
- Always verify both symmetrical interrupting rating and peak momentary withstand on the breaker datasheet