New home wiring

[AdrianUK]

It sounds as if you might have some interconnections between circuits, so you would have to pull out more than one fuse to remove power from a circuit. If that is the case, then it is dangerous and should be properly investigated straight away.

The chap who owned our house before us clearly believed he had more electrical knowledge than he did...

He decided to fit a second, smaller, consumer unit next to the original wylex board. He fed this using 4mm2 singles eek from the incoming terminals of the original wylex board (protected by the 100A DNO fuse). He then decided that the largest of the 3 ring finals in the house needed to be 'upgraded' so .... one leg of the ring was fed from a 40A () MCB in one consumer unit & returned to a 32A MCB in the second consumer unit! Hey presto a 72A ring final and neither board isolated by turning off just one main switch! Took me a while to sus that one out!

 

[Taylortwocities]

Hey presto a 72A ring final

No it isn't.

the circuit current would still be limited by the 32A MCB. Think about it. Horrendous to wire it like that, though.

 

[EFLImpudence]

Think about it.

Or 40A

 

[Taylortwocities]

Or 40A

Nope

The ring's current will pass through both MCBs. The 32A is the lower current-limiting factor (let's ignore discrimination, shall we?)

 

[AdrianUK]

Or 40A

Nope

The ring's current will pass through both MCBs. The 32A is the lower current-limiting factor (let's ignore discrimination, shall we?)

I'm not convinced. The 'ring' would behave as sorta radial circuit fed from both ends .... so a load exactly in the middle would draw equally from both 'feeders' thus a 64A load at the mid point would take 32A from the 32A MCB end & 32A from the 40A MCB end.

The only way the current would pass thru both MCBs is if the main switch of one consumer unit were OFF so that the second consumer unit is being supplied from the first via the 'ring final'

 

[EFLImpudence]

The ring's current will pass through both MCBs.

Won't it be shared between them, then?
(let's ignore position on ring, shall we?)

 

[JohnW2]

Think about it.

Or 40A

I would say not even that. If one end of the ring is supplied via a 32A MCB, adding another MCB at the other end makes no difference to the distribution of currents. With or without that second MCB, the current through the first (32A) one can theoretically approach 100% of the total load on the circuit if that load is all applied very close to that MCB. I presume that one therefore theoretically has to regard it as a "32A circuit", regardless of what MCB might be on t'other end (40A, 50A, 63A or whatever!).

In any event, having different MCBs on the two ends of a ring clearly ain't safe, and more than two 'cross-connected' rings are (talk of which, I met one of those in my daughter's house!)!

Kind Regards, John

 

[Taylortwocities]

My brain hurts…..

 

[JohnW2]

I'm not convinced. The 'ring' would behave as sorta radial circuit fed from both ends .... so a load exactly in the middle would draw equally from both 'feeders' thus a 64A load at the mid point would take 32A from the 32A MCB end & 32A from the 40A MCB end.

Indeed, but that's close to be the 'best case' scenario. If a 64A load were drawn from a point very close to the 32A MCB, most of that current would flow through the 32A MCB. The closer the load got to the 64A MCB, the closer would the current flowing through that MCB get to 64A.

Kind Regards, John

 

[EFLImpudence]

If one end of the ring is supplied via a 32A MCB, adding another MCB at the other end makes no difference to the distribution of currents.

How does the load know from which end to draw the current?

 

[AdrianUK]

If one end of the ring is supplied via a 32A MCB, adding another MCB at the other end makes no difference to the distribution of currents.

How does the load know from which end to draw the current?

The load doesn't. This is a 'load flow' problem... the division of the current ratios will be the inverse of the impedances of the two 'feeders'

 

[JohnW2]

If one end of the ring is supplied via a 32A MCB, adding another MCB at the other end makes no difference to the distribution of currents.

How does the load know from which end to draw the current?

From the ratio of the impedances of the two paths back to the CU.

Consider an extreme example of a 50 metres (total length) ring, with a 64A load applied 1 metre from one end. The impedances of the two paths from load to CU will be in the ratio 49:1, hence the currents through those two paths will also be in the ratio 1:49. With 64A load, that means that about 62.72A will flow through the nearest MCB and about 1.28A through the one at t'other end - totally regardless of what OPDs might be fitted to the two ends.

Kind Regards, John

 

[EFLImpudence]

Yes, but the discussion is: would the circuit load be limited to 32A or could it rise to 72A?

 

[stillp]

Neither. Imagine ramping up the current. At some point the 32A will trip, and current will continue to flow via the 40A breaker. Eventually that will trip.

 

[JohnW2]

Neither. Imagine ramping up the current. At some point the 32A will trip, and current will continue to flow via the 40A breaker. Eventually that will trip.

That's not quite right. One could arrange a 72A load around the ring in such a manner that exactly 32A flowed through the 32A MCB and exactly 40A flowed through the 40A one - but (assuming it was 2.5mm² cable), much (or all) of the cable would be 'well overloaded'.

Kind Regards, John