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If a 240V circuit is protected by a 15A simi-enclosed fused and has an electrical earth-fault loop impedance of 1.9 ohms determine the earth fault current in the event of a zero impedance earth fault?
Wiki User
∙ 15y ago
I am assuming that its a 240 Volt AC circuit supplying an inductive load with a fault loop impedance of 1.9 ohms at the time of the short circuit. The power factor is assumed to be 0.8
The instantaneous earth fault current value would be;
Current = (Voltage x Power Factor) / Impedance
(240 x 0.8) / 1.9
192 / 1.9
= 101 Amps.
However this may be a trick question as it doesn't ask for an instantaneous value, the fuse will limit the fault current to 15 amps and should disconnect the circuit within 0.4 seconds.
Wiki User
∙ 15y ago
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