17th wiring regs revision Question - Electrical Courses

Dazzol2009 · 03-07-2009

I will be doing my 2382-10 in the coming weeks, got stuck on a question about fault current, not sure if I have been doing too much revision and my brain has gone.
Anyway question is:
The fault current due to an earth fault of negligible impedance in a 400v 3 phase, 4 wire circuit, having an earth fault loop impedance of 0.3 ohms.

The choices are:

A 1383 A
B 767 A
C 124.5 A
D 72 A

B is the given answer , do not understand how the answer has been arrived upon, I am obviously missing a formula, can anybody help?

**Mac** · 03-07-2009

You need to find the power factor and power values.I did work it out and it came out with the answer of 767A.
I=P/(1.732 x E x PF)
Electrical & Electronics, Ohm's Law, Formulas & Equations

P= 1333.33
PF= 0.0025091
E= 400V
1.732 x 400 x 0.0025091=1.739. 1333.33 / 1.739 =766.72.Rounded up gives you 767A

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