Allied Diesel


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Short Circuit Maximum Current

Short circuit is defined when the impedance of a circuit is suddenly brought close to 0 when the two wires "short" the circuit and it is the job of the circuit breaker to open the ciruit limiting the damage to the alternator which cannot withstand large currents for any significant amount of time. Were there no impedance at all, Ohm's law shows that the current will be infinite and no circuit breaker can "break" such a current. Fortunately there remains always some impedance in the circuit. If the short occurs close enough to the alternator, the only impedance in the circuit is the internal impedance of the rotor coil, which consists of a (negligible) resistance and a reactance which is called the subtransient reactance: the subtransient term refers to the fact that this reactance appears during the very first few cycles after the short circuit before the occurence of the transient and the steady state status. Although under noraml operating conditions this reactance a nuisance in term of losses and wave ditortion and phase shift, it saves the alternator when a short occurs, provided your circuit breaker is correctly spec'ed

The maximum current that could go through the circuit breaker depends also on the exact time the short circuit occurs with respect to the phase of the voltage which could create an aperiodic component which can greatly increase the maximum current that the circuit breaker has to break. From the alternator spec sheet please enter the subtransient reactance (usually indicated by X''d and expreseed in per units) also insert the nominal voltage Phase to neutral of your alternator and press SUBMIT button to get the the breaking capacity required by the circuit breaker and expressed in kA.

Subtransient Reactance % per unit

Phase to Neutral Voltage Volts

Genset Apparent Power kVA


The answer will appear here!

For those interested in the details of the calculations here is an example
From the alternator spec sheet get the relevant data. Assume the X''d is 8.5% pu
Therefore at a nominal voltage of 1 pu the current would have a value of 1/8.5% = 11.76 pu, ie 11.76 times the nominal current
For a 150 kVA genset at 230 Volts LN the nominal current is 150/230/3 = 0.217 kA so the maximum current will be 11.76 times, that is 2.55 kA
However this is an rms value and the maximum for a sinusoidal wave would be 2.55 * sqrt(2) = 3.61 kA
In the worst case scenario the aperiodic component of the reactance will shift the wave upwards (or downwards) completely and this will make the maximum current twice as much. Allowing at least for a 10% safety margin in the evaluation of the reactance and/or the response of the circuit breaker and/or the standby rating of the genset we get 3.61 x 2 x 110% = 7.94 kA
Therefore the circuit breaker should be chosen with a breaking capacity higher than 7.94 kA. A ciruit breaker with a smaller breaking capacity might not be able to trip on a short circuit which was precisely why you bought and installed the breaker