Relationship ze and pfc

So I’m currently studying Ze and pfc etc kind of understanding it.
main question really is about the relationship between them both I understand that the lower the Ze the higher the fault current but why is this? Is it just simply to do with ohms law? And I’m just over thinking and confusing myself

Any responses would be appreciated
 
So I’m currently studying Ze and pfc etc kind of understanding it.
main question really is about the relationship between them both I understand that the lower the Ze the higher the fault current but why is this? Is it just simply to do with ohms law? And I’m just over thinking and confusing myself

Any responses would be appreciated

A lower loop impedance allows more current to flow, and yes this is Ohms law.

Think of what happens when each variable is changed, either up or down. Example - what would happen to the PFC figure if the supply voltage was higher?
 
For a single phase domestic supply, where Ze/Zs is dominated by the cable resistance, it is just Ohm's law:

PFC = U / Ze

Typically your MFT reports both the PFC and Ze when measured, the PFC is based on the measured voltage U, but Ze is independent of that. You would use Ze along with R1+R2 (or Zs at the end of a circuit) to check that any over-current protection will disconnect fast enough for safety even at the lowest likely supply voltage.

The worst case PFC (of both earth-fault 'PEFC' and supply short-circuit 'PSCC' ) would be used to check that any over-current protection is able to safely interrupt any fault.

For three-phase and/or high current systems it gets more complicated, but the same basic principles apply. The IET's "Electrical Installation Design Guide" has some examples of the slightly more tricky cases on pages 76-81
 
It's a long time since I was at school, but isn't I = V/R the same as V = IR something like transposing one of the divisors to the other side of the resultant make it a multiplicand rather than a divisor?
 
It's a long time since I was at school, but isn't I = V/R the same as V = IR something like transposing one of the divisors to the other side of the resultant make it a multiplicand rather than a divisor?

It is indeed. I think they were joking though.
 
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Kyle_Trainee,
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R-fur,
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