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    Calculates locally in your browser · IEC 60909 radial fault level
    Power systems / Fault level

    Short-circuit calculator

    The fault level at the end of a cable run, built from the transformer impedance and the cable impedance, with the peak current from the X/R ratio and the adiabatic cable size for the clearing time. Change the cable length and watch the fault level fall.

    Transformer1000 kVAImpedance6.0%Min. cable50 mm²
    Fault current
    18.28 kA
    Three-phase, initial symmetrical
    Peak current
    32.34 kA
    κ = 1.251
    Loop impedance
    13.27 mΩ
    Transformer plus cable
    Fault level
    12.7 MVA
    At 400 V
    Impedance build-up
    Referred to the secondary side, in milliohms.
    Impedance build-up
    ElementR (mΩ)X (mΩ)Z (mΩ)
    Transformer0.969.559.60
    Cable4.802.405.37
    Total5.7611.9513.27

    Cable impedance grows with length; a long run can halve the fault level and change the breaker rating.

    Inspect the model
    Choose a quick result view, a scaled comparison, or the complete diagnostic set.
    Fault current against cable length
    Impedance eats the fault level; this is why the point of fault matters.
    Fault current against cable length0.0005.6611.3216.9722.6328.290 m0 m10 m10 m25 m25 m50 m50 m100 m100 m200 m200 m400 m400 mI″k (kA)
    Canonical equations
    The exact relationships implemented by this engine.

    I″k = c·Un / (√3·Zk) · IEC 60909, radial single source.

    Zt = uk · Un² / Sn · uk as a fraction, Un in kV, Sn in MVA.

    ip = κ·√2·I″k · κ from the X/R ratio at the fault point.

    engineering methodology

    A clear path from inputs to results.

    The prospective fault current decides the breaker rating, the cable damage check and the arc flash study. It is set by the transformer impedance and the cable run, and it falls quickly with distance — which is why the point of fault matters.

    Step-by-step methodWorked exampleAssumptions explained

    How to use

    1. Enter the transformer rating, impedance and X/R ratio.
    2. Set the cable length and its resistance and reactance per kilometre.
    3. Read the fault current, the peak current and the minimum adiabatic cable size.

    The formula

    Ik′′=cUn3Zk,ip=κ2Ik′′I''_k = \frac{c U_n}{\sqrt{3} Z_k}, \qquad i_p = \kappa \sqrt{2} I''_k

    Reference: IEC 60909-0 radial network; adiabatic check from IEC 60364-5-54.

    Worked example

    1 MVA transformer, 400 V, 30 m of cable

    Inputs
    • uk 6%, X/R 10, cable 0.16 + j0.08 Ω/km.
    Calculation path
    1. 1

      Zt = 0.06 × 0.4²/1 = 9.6 mΩ. Cable adds 4.8 + j2.4 mΩ. Zk = 14.4 mΩ. I″k = 1.05 × 400/(√3 × 0.0144) = 16.8 kA.

    Result

    16.8 kA at the far end, 25.3 kA at the terminals.

    Interpreting the result

    A long run can halve the fault level, which is good for the breaker rating and bad for the disconnection time — the same cable that protects the equipment may not trip the breaker at the far end.

    Assumptions & limitations

    • Radial, single source; no motor contribution.
    • Utility impedance is neglected, which raises the fault level.
    • Cable resistance at operating temperature.
    • Arc resistance is not included.

    Frequently asked questions

    Why does c = 1.05?

    IEC 60909 uses a voltage factor to cover tolerance: 1.05 for low voltage, 1.10 for high voltage.

    What is the adiabatic check?

    It sizes the conductor so the fault energy I²t does not take it above its short-time temperature limit.

    Should I use the maximum or minimum fault?

    Maximum for equipment rating, minimum for protection disconnection. This is the maximum.

    Does the X/R ratio matter?

    It sets the peak, which is what a busbar and a breaker’s making capacity see.

    Methodology last reviewed 2026-10-22. Results are informational and not financial or engineering advice — see our disclaimer.