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NeonSparks

just moved an mcb to rcd side of board and now a lighting radial on NON rcd side trips rcd...any ideas? I'lm sure i have moved correct wires.
 
Agree both would have to be on same mcb as you have stated.

Why's this? If both circuits share the same neutral, by putting both MCB's on the same neutral bar (and same RCD) then surely it's the same thing?

To have upstairs on 1 mcb and downstairs on a different MCB (albiet on the same RCD), would that not be better than powering 2 circuits through the same MCB?
 
Why's this? If both circuits share the same neutral, by putting both MCB's on the same neutral bar (and same RCD) then surely it's the same thing?

To have upstairs on 1 mcb and downstairs on a different MCB (albiet on the same RCD), would that not be better than powering 2 circuits through the same MCB?

As i have already stated in an early post the reason for puting the "2" circuits on the one mcb is because a borrowed neutral is dangerous. If there is a borrowed neutral then the are not truely 2 separate circuits.
 
No because if one mcb is isolated to work on a fitting then the borrowed neutral will become live if the other circuit is not also disconnected.

I know your right, but I can't get it straight in my head.

Lets say there was no borrowed neutral and both circuits wired correctly, both MCB's connected to the same RCD and same neutral bar, are the neutrals still not connected together, only this time at the board rather than at the light fitting?
 
I know your right, but I can't get it straight in my head.

Lets say there was no borrowed neutral and both circuits wired correctly, both MCB's connected to the same RCD and same neutral bar, are the neutrals still not connected together, only this time at the board rather than at the light fitting?
The difference there is that if you isolate one circuit to work on it and the other circuit is still on the circuit that is still on has a neutral connected to the board and therefore a return path for te current. In the case of a borrowed neutral when you split the circuit at the borrowed neutral you are breaking the return path to the board therefore that neutral will be live.
 
The difference there is that if you isolate one circuit to work on it and the other circuit is still on the circuit that is still on has a neutral connected to the board and therefore a return path for te current. In the case of a borrowed neutral when you split the circuit at the borrowed neutral you are breaking the return path to the board therefore that neutral will be live.

Thank you, that's a very good explanation. I do understand what your saying, but I'm struggling to picture it in my mind...must be getting tired :confused:
 

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