main 'Trip' switch opening on change to night rate

[JohnW2]

Johnw2, true but firstly the RCD can operate at anything over half that, and depends what else is leaking on the same RCD.

True. I agree that an N-E leak is definitely one of the possibilities, my only real point being that a significant N-E a leak (even if just a 'last straw') due to stray capacitance is next-to-impossible, assuming that the cables are not literally 'miles' long!

Another thing which often goes unmentioned in these discussions about N-E leaks is that many (most?) circuits have at least some degree of L-E leak (due to filter capacitors in equipment etc.), and that will effectively subtract from any N-E leak, hence meaning that one needs more N-E leak than one might have expected in order to trip and RCD. In the worst case, with nearly 30mA of L-E leak, in simplistic terms (e.g. assuming that both leaks were resistive) it could take an N-E leak of anything up to 60mA (and certainly at least 45mA) to trip a 30mA RCD.

Kind Regards, John

 

[JohnW2]

Yes, I know it's all related, but it's the shared current between N and E which causes the RCD to operated; not the pd between them.

I don't really understand what you're saying. It's the N-E current (through a fault) that causes the imbalance which trips an RCD. You are well familiar with Ohm's law, which tells you that that leakage current will be equal to the N-E pd divided by the resistance/impedance of the fault path. I'm therefore not sure what you are saying.

Kind Regards, John

 

[EFLImpudence]

In the worst case, with nearly 30mA of L-E leak, in simplistic terms (e.g. assuming that both leaks were resistive) it could take an N-E leak of anything up to 60mA (and certainly at least 45mA) to trip a 30mA RCD.

No.

It does not work like that.
The 30mA leak from L will still be missing from the N return.

 

[EFLImpudence]

I don't really understand what you're saying. It's the N-E current (through a fault) that causes the imbalance which trips an RCD. You are well familiar with Ohm's law, which tells you that that leakage current will be equal to the N-E pd divided by the resistance/impedance of the fault path. I'm therefore not sure what you are saying.

I don't think so. It is just the shared current through parallel paths which causes the RCD to trip even if there is NO pd between N and E.

 

[John D v2.0]

If there's no pd between n and e then there wouldn't be a current imbalance due to an n-e fault though. It's the conditional on high load bit we're trying to explain.

However my feeling is we're all violently in agreement here and just debating the terminology

 

[JohnW2]

No. It does not work like that.

On reflection, this is more complicated than I suggested. I need to think - watch this space!

Kind Regards, John

 

[JohnW2]

I don't think so. It is just the shared current through parallel paths which causes the RCD to trip even if there is NO pd between N and E.

If there were no pd between N and E (at the location of the N-E fault), then, as Mr Ohm tells you, there could be no N-E current, hence no L/N imbalance and no RCD trip.

Kind Regards, John

 

[EFLImpudence]

If there's no pd between n and e then there wouldn't be a current imbalance due to an n-e fault though. It's the conditional on high load bit we're trying to explain.

There would be a current imbalance in the RCD - but solely due to the parallel path; it is not dependent on there being a pd between N and E as there may be none.

However my feeling is we're all violently in agreement here and just debating the terminology

Mmmm. It was just your statement that there had to be a pd between N and E which I thought was misleading.

 

[EFLImpudence]

If there were no pd between N and E (at the location of the N-E fault), then, as Mr Ohm tells you, there could be no N-E current, hence no L/N imbalance and no RCD trip.

Connect a wire from RCD 'N in' to 'N out' and see what happens. Is there a pd?

 

[JohnW2]

Connect a wire from RCD 'N in' to 'N out' and see what happens. Is there a pd?

That's not a 'parallel fault path' (to 'earth', aka the neutral side of the tranny at the substation) - it's bypassing one side of the RCD!

I am talking specifically about leaks through N-E fault paths (thereby providing a partial alternative return path, which doesn't go through the RCD). As I said, there can can be no N-E leakage current through the fault if there is no (N-E) pd across that fault. If you don't believe Mr Ohm, Mr Kirchoff will tell you the same

I do not count bypassing one side of the RCD as a 'fault'. Can you suggest any real fault through which current can flow despite their being no pd across the fault? If the fault is very low impedance, then the pd across it could be very low - but, per Messrs Ohm and Kirchoff, if there were NO pd across it, no current could be flowing through it.

FWIW, even with your bypassing of one side of the RCD, there would still have to be a pd (albeit very small) across your 'bypass wire' - otherwise no current would be flowing through it (yet again, I refer you to Mr Ohm).

Kind Regards, John

 

[bernardgreen]

Connect a wire from RCD 'N in' to 'N out' and see what happens. Is there a pd?

If there is a load creating a current in the neutral then yes there is a PD btween RCD "N in" and "N out"

It is an extremely small PD but as the impedance is also extremely small the current can be quite high.

0.01 Volts and 0.001 Ohm is 10 amps

 

[EFLImpudence]

If so, that is still not what is causing the RCD to trip.

 

[bernardgreen]

The RCD is tripping because the current flowing through the Neutral terminal of the load has two parallel routes back to the Neutral terminal at the substation which is connected to the Ground.

One is the intended route along the Neutral conductor and through the RCD sense coil and the other route that bypasses the RCD sense coil is from the Neutral terminal of the load onto the Neutral conductor and then


(a) via a Neutral to Gound fault and through the ground to the substation.
(b) via a Neutral to the CPC ( Earth wire ) and along the CPC to the Main Earth Terminal which is connected to the Incoming Neutral at the cut out ( or to a TT Ground rod )

 

[JohnW2]

If so, that is still not what is causing the RCD to trip.

What is causing the RCD to trip is the fact that you have provided a path back to the neutral of the transformer/substation which almost completely bypasses the N side of the RCD. It's nothing directly to do with potential differences (although, a bernard and I have both said, there still has to be some {very small} pd across your 'bypass wire', else no return current (or any other current) could be flowing through it.

Kind Regards, John

 

[EFLImpudence]

It's nothing directly to do with potential differences

That's what I have been saying.