Without labouring the point
, you are implying the same. ... It is not the N - E pd itself which causes the trip
- it is the fault between N and E - as you, yourself, said
"It's nothing directly to do with potential differences".
I'm becoming more inclined to agree with JohnD (v2) that this is getting 'semantic' - or, at least, linguistic. We're all saying essentially the same thing, but you don't seem to like other people's wording!
There may be a pd between N and E, as there is between L and E, but with no fault the RCD will not trip.
Of course. However, this discussion has been entirely about
N-E faults (actually N-CPC) - i.e. N-E (not L-E) and 'fault' (not 'normal').
However, why can there not be parallel conductors sharing the current with no pd between them?
We're not talking about the pd
between the two 'parallel' conductors (N and CPC), which would obviously fall to a low value when they were connected via a fault (zero pd if the fault were of 'negligible impedance'), but, rather, the potential between the downstream end of the neutral and
earth. Given that both the N and the CPC are ultimately connected to earth (one via the RCD and the other not), no current would flow through either of them unless there was a non-zero N-E potential at the downstream end.
The magnitude of the N-E potential at the downstream end of the neutral (8.6V in my example) is then an indicator of what current will flow through the CPC when an N-CPC fault arises - and the higher that N-E pd, the greater will be that bit of the total 'return' current that flows down the CPC (rather than the N conductor), unbalancing the currents through the RCD.
However, this rather complicated discussion does not tell us anything other than what we already knew - that the higher the current legitimately flowing through the load (and normally all through the neutral), the higher will also be the current that flows down the CPC (reducing the current going through the N-side of the RCD) in the presence of an N-E fault!
Kind Regards, John