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?

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.