Short Circuit FAQ

Symmetrical fault current (Ik″) is the RMS value of the AC component of fault current at the instant of fault inception. Peak fault current (ip) includes the DC offset component and can reach 1.8 to 2.55 times Ik″ depending on the X/R ratio of the fault circuit. Peak current determines mechanical forces on busbars and equipment, while symmetrical current is used for thermal withstand calculations.

Lower transformer impedance allows higher fault currents at the secondary terminals. A 1000 kVA transformer with 6% impedance has a maximum secondary fault level of approximately 24 kA (Isc = In × 100/Zk%). If the impedance were 4%, the fault level would rise to approximately 36 kA — requiring all downstream protective devices and cables to be rated accordingly.

The voltage factor c accounts for the difference between actual system voltage and nominal voltage, and variations in generator, motor, and load voltages. For maximum fault current calculations (equipment rating), cmax = 1.05 for low voltage (≤ 1 kV) and 1.10 for medium voltage (1–35 kV). For minimum fault current calculations (protection operation verification), cmin = 0.95. The factor is multiplied into the fault current formula.

Yes. Induction motors act as generators momentarily during a fault, contributing additional current for approximately 50–100 milliseconds. IEC 60909-0 models motor contribution based on the motor’s locked-rotor current ratio and rated power. For large motor groups, this contribution can increase the total fault level by 10–30% above the utility contribution alone. Motor contribution must be considered in protection coordination.

Add the cable impedance to the total upstream impedance. Cable impedance equals the per-metre resistance and reactance values multiplied by cable length in metres. Use conductor resistance at maximum operating temperature for worst-case thermal calculation, or at 20°C for maximum fault current. The fault current at the cable end is Ik = cUn / (√3 × Ztotal).

Minimum fault current is the lowest fault current that could occur at a given point — typically a phase-to-earth fault at the end of the longest cable at maximum conductor temperature. This value must be high enough to operate the protective device within the required disconnection time to prevent electric shock. If minimum fault current is too low, the device may not trip, leaving dangerous voltages on exposed metalwork.

Calculate the maximum prospective fault current at the circuit breaker’s installation point using IEC 60909 or by measurement. The circuit breaker’s rated ultimate breaking capacity (Icu) must equal or exceed this value. For main switchboard incomers, obtain the utility’s fault level contribution and add internal source contributions. Common MCB ratings are 6 kA or 10 kA; MCCBs are available from 25 kA to over 150 kA.

The adiabatic equation k²S² ≥ I²t verifies that a cable can thermally withstand a short circuit. k depends on conductor and insulation material (115 for copper/PVC, 143 for copper/XLPE). S is the conductor cross-sectional area in mm². I is the fault current in amperes, and t is the protective device total clearing time in seconds. If k²S² is less than I²t, a larger cable is needed.

Yes, and it is recommended for complex installations. Software like ECalPro, ETAP, and DIgSILENT PowerFactory model complete network impedances, automatically apply IEC 60909 methodology, and check equipment ratings against calculated fault levels. Software is especially valuable for installations with multiple sources, ring networks, or motor contributions that make manual calculation error-prone.