additional RCD protection from old Wylex fuse oard

[haptism]

My mate has an old wylex board with the push in rewireable fuses; and has asked about fitting an rcd to one circuit (a ring main) that supplies one outside socket (at the mo he's using a plug in portable rcd when using the mower). He does not want a new consumer unit, and i know that rcbo's are not compatible with these old fuseboards so thats not an option; a rcd spur to the outside socket is not an option either because that means taking out the kitchen sink to access the ring where it loops outside. Therefore he wants to protect the whole circuit with an rcd located near the fuseboard.

Could I take one leg of the ring via the rcd and leave the other leg in the fuseboard, or would both legs of the ring need to go through the rcd ?

Thanks

 

[markc123]

Cheapest option would be to find said socket he's plugging his lawn mower into and replace the socket faceplate with an RCD socket.

Why would you only take one leg off the circuit? presumably you'll be installing the RCD at or near the CU ?

 

[telectrix]

how you do this is to take both the L and N legs of the RFC out of the fuse and N bar. extend with in line connectors to reach RCD in an enclosure adjacent to the CU. then you feed the RCD from the fuse and the N bar with a couple of bits of 6mm stripped out of some 6mm T/E . you're then responsible for the whole circuit. test and then fill in a MWC

 

[haptism]

Cheapest option would be to find said socket he's plugging his lawn mower into and replace the socket faceplate with an RCD socket.

Why would you only take one leg off the circuit? presumably you'll be installing the RCD at or near the CU ?

That is another way to do it, but he might aswell have the entire circuit protected. Im thinking of adding an rcd into the circuit just after it leaves the fuseboard, just not sure if both ends of the ring need to go through the rcd, or just one. Im assuming the rcd would still operate if just one side of the ring went through it.

 

[telectrix]

no..you need both legs and both N's into the RCD or it will trip.

 

[haptism]

how you do this is to take both the L and N legs of the RFC out of the fuse and N bar. extend with in line connectors to reach RCD in an enclosure adjacent to the CU. then you feed the RCD from the fuse and the N bar with a couple of bits of 6mm stripped out of some 6mm T/E . you're then responsible for the whole circuit. test and then fill in a MWC

So you mean feed the supply side of the rcd with a 6mm radial from the fuse and connect the existing two legs of the ring to the load side ? Why couldn't I just divert one leg of the ring via the rcd ? Which im not sure about so this really is why im asking. If only one leg of the ring went via the rcd then would the whole circuit be protected ? or not ? if you see where im coming from !

 

[telectrix]

NO. both L and N legs needs to be diverted through the RCD. this is fairly basic if you know how a RCD works.

 

[Midwest]

So you mean feed the supply side of the rcd with a 6mm radial from the fuse and connect the existing two legs of the ring to the load side ? Why couldn't I just divert one leg of the ring via the rcd ? Which im not sure about so this really is why im asking. If only one leg of the ring went via the rcd then would the whole circuit be protected ? or not ? if you see where im coming from !

Are you a qualified electrician?

 

[haptism]

An rcd measures any imbalance between line & neutral, so how be the principles that it would trip if rcd was connected to just one leg of the ring. Thanks telectrix for noting it will trip which I take your word for it, just dont see how this is immediately.

 

[Bobster]

And what happens if it trips with only one leg of the ring going through it!

I'd recommend your friend to get an electrician in.

 

[haptism]

And what happens if it trips with only one leg of the ring going through it!

I'd recommend your friend to get an electrician in.

There would still be a return path neutral (via the fuseboard leg) if thats where you going

 

[Midwest]

haptism, I don't won't to appear rude, but you post #4 suggest you are not a qualified electrician. At the moment your mate is safely using his mower with a plug in RCD. If you alter the ring final circuit he uses for his mower, and don't get it correct, it an have potentially dangerous consequences for your mate. I note from your profile you have a C&G 2360 and 16th ed?

 

[haptism]

Amazing what you find out when you ask a few questions innit

 

[SJD]

An rcd measures any imbalance between line & neutral, so how be the principles that it would trip if rcd was connected to just one leg of the ring. Thanks telectrix for noting it will trip which I take your word for it, just dont see how this is immediately.

In a ring final circuit, the current drawn at each outlet (socket, FCU, etc) is split between the two legs of the circuit. Since the L and N of one leg is roughly the same resistance (and same for the other leg) then roughly the L and N of each leg have the same current flowing. But not exactly so, small differences such as cable tolerances, poor connections, will cause a slight imbalance. As soon as the imbalance reaches 30mA (actually maybe 20-odd mA) the RCD will trip. With a 30A load on the circuit, an imbalance of just 0.1% will cause tripping.

And once it has tripped, the circuit will still be live via the other leg, and the cable potentially overloaded since a single 2.5mm T&E is not rated to 30A.

 

[haptism]

haptism, I don't won't to appear rude, but you post #4 suggest you are not a qualified electrician. At the moment your mate is safely using his mower with a plug in RCD. If you alter the ring final circuit he uses for his mower, and don't get it correct, it an have potentially dangerous consequences for your mate. I note from your profile you have a C&G 2360 and 16th ed?

Yes thats right I do; instead of just taking someone word for it and doing things parrot fashion thought I would continue and ask why an rcd on just one leg of a ring would trip. My reasoning is there would be proportional current draw through both lines of a ring to a load, and therefore proportional return currents via neutral, so in this instance why should it trip. As Rob said, what would happen if there was a fault and it did trip, answer it would still be energized. So that is an answer in itself. At least im a bit wiser now. Guys lay off the persecution its all good stuff.