External garage electricity wrongly connected to neighbours supply

[EFLImpudence]

It is right Part of the N current from a load protected by a different RCD goes 'backwards' through the N side of the RCD protecting the circuit with a N-E fault, thereby creating an L/N imbalance (hence a trip, if enough imbalance) of that RCD ...



No. The full N current from load (as well as the full L current from the load) goes through the RCD protecting the circuit(s) with a load ("RCD1" in diagram above), hence no trip of that/those RCD(s)

Yes, I obviously misread your post. I thought you were saying both RCDs would trip.

This is the situation in Bernard's frequently posted diagram .

 

[JohnW2]

Yes, I obviously misread your post.

Fair enough.

I thought you were saying both RCDs would trip.

As you now realise, I obviously wasn't saying that.

The important point is that many people don't seem to realise that a load protected by a different RCD can result in tripping of an RCD protecting a circuit with an N-E fault - and, even if they do realise that, don't also realise that, if it happens, the RCD protecting the load (but not the circuit with the fault) will not trip.

 

[ericmark]

With anything other than a TN-C-S it is surely inevitable that there will be some N-E potential difference, isn't it? Absence of some N-E pd could only occur if the length of neutral conductor from transformer to installation is zero (impossible), or if no customer served by the transformer is drawing any current (incredibly unlikley)?

I agree highly unlikely, but with a split or three-phase supply, zero current can be flowing through the neutral, even when there is a load.

 

[JohnW2]

I agree highly unlikely, but with a split or three-phase supply, zero current can be flowing through the neutral, even when there is a load.

I would say that, even with a ('loaded') split-phase or 3-phase supply, it would be incredibly unlikely that the neutral current would be exactly zero, wouldn't it?

 

[SUNRAY]

It is right Part of the N current from a load protected by a different RCD goes 'backwards' through the N side of the RCD protecting the circuit with a N-E fault, thereby creating an L/N imbalance (hence a trip, if enough imbalance) of that RCD ...

View attachment 405160

No. The full N current from load (as well as the full L current from the load) goes through the RCD protecting the circuit(s) with a load ("RCD1" in diagram above), hence no trip of that/those RCD(s)

Your diagram correctly describes my earlier comments and demonstrates the most likely reason there is a PD betwix N & E.
The load you show is anywhere on the system, including the properties along the road etc, the fact you have drawn this makes me surprised you (much more than some on here) didn't understand my comment from the start, unless of course you were not aware of the situation at the time.

However; I'll add if there was no N to E PD at the point you applied the short then there must be a low resistance/impedance N to E link somewhere prior but close to RCD2, there will be no or negligable current in the fault and threfore no trip.

However#2; The same would be true if there were no link at all with N floating.

 

[SUNRAY]

I would say that, even with a ('loaded') split-phase or 3-phase supply, it would be incredibly unlikely that the neutral current would be exactly zero, wouldn't it?

I would say that's true unless all of the loads are 3 wire 3ph

 

[JohnW2]

This is the situation in Bernard's frequently posted diagram .

I should perhaps have added ...

My diagram goes further than bernard's oft-posted one. His diagram shows a N-E fault resulting in an RCD tripping when there was a lot on some other circuit protected by the same RCD.

In contrast, mine shows that there can still be a trip even if the only load(s) are on circuit(s) protected by a different RCD (different from the one protecting the circuit with the N-E fault) - which, as I said, I think a good few folk actually realise.

Indeed, as we have discussed in the past, I think one can go even further than that - since I think even load(s) in different installations (i.e. some neighbours' houses) connected to the same neutral can result in a N-E fault causing an RCD protecting its circuit to trip. In other words, the "RCD 1" (and circuits protected by it) in my diagram above could be in a different installation/premises.

 

[JohnW2]

I would say that's true unless all of the loads are 3 wire 3ph

Agreed, but this is a DIY forum and the situation you describe in never going to arise in domestic premises.

In fact, probably pretty rarely even in non-domestic premises since, even if most of the loads are 3-wire 3-phase, there would often/usually be at least some single-phase circuits/loads, wouldn't there?

 

[JohnW2]

Your diagram correctly describes my earlier comments and demonstrates the most likely reason there is a PD betwix N & E.

As I wrote to eric, and as you have now agreed, if there are any single-phase loads being supplied via the transformer, then (other than in a TN-C-S) installation, a N-E potential difference is inevitable.

I'm therefore coming to wonder whether your statement which started this whole side discussion was perhaps simply a 'statement of the obvious' - i.e. that you were talking specifically about trips due to N-E faults (you didn't say that) and that when you appeared to be saying that the presence of a N-E potential difference facilitated such RCD trips, you merely meant that an N-E fault will only cause an RCD trip if there is at least some neutral current 'somewhere' (within the same installation, or maybe even a different one sharing the neutral), manifested by a measurable N-E pd - which, I say, is obvious (and, of course, virtually always the situation which exists).

However; I'll add if there was no N to E PD at the point you applied the short then there must be a low resistance/impedance N to E link somewhere prior but close to RCD2, there will be no or negligable current in the fault and threfore no trip.

We have perhaps not thought or written enough about the meaning of "E" and, particularly, about TN-C-S.

With TN-C-S there is a very low impedance ('approaching zero') 'link' between N and E, upstream of, and usually very close to, the RCDs - and hence usually a very small ('approaching zero') potential difference between N and the 'E' within the installation (i.e. MET/CPCs). In that case, a fault between N and the installation's "E" (MET/CPCs) is presumably unlikely to result in an RCD trip, although a fault between N and 'true E' might well result in a trip (manifested, using your language, as a pd between N and true E but not between N and the installation's 'E' (MET/CPCs) ?