When we moved in there was a switched FCU feeding a 13A skt in the conservatory. I thought I'd also use it to supply a 20A DP switch/neon to feed the shed. I wired it all up and it works fine, then I switched off the FCU, the shed was still live so I removed the fuse, shed still live :D Bloody cowboys.
It is add to have a FCU feeding just one socket. Are you sure it not for the conservatory lights?
Sounds like the cable to the shed got connected to the input side of the FCU instead of the load side.
Yes, I discovered that it does indeed control the lights but not the spur socket next to it.
I simply (good word) assumed that the FCU controlled the 13A socket next to it, so wired the shed switch to the socket.
It is quite normal to do such things. It gives you a 5A supply for the lights from the 32A ring circuit
I suspect that you need a new FCU for the shed.
In that case I'm prolly screwed. AIUI I can't have 2 FCUs from the back of one socket.
Since one is probably fused at 3A for the lights, there is no harm in having a second fused spur from the same point on the ring. The difficulty may be getting adequate terminal space.
Replace the socket with a FCU and job done. (slap another socket in beside it if you need the socket!)
Cheers John, all this info. Lemme explain what I've got - A socket in the kitchen (on the ring) feeds through the wall into the conservatory to another socket and also to the FCU thence to the light. Is that acceptable so far? So can I add another FCU (from the back of the socket in the cons.) to the shed switch? Should all spurs (13A skt) be fused via an FCU? And another? If you had say 2.5mm and 1mm which both needed to connect together say at a skt, is it allowed and how would you do it?
Is that a spur off a spur (ie one cable from the kitchen socket to the conservatory socket and then from the coservatory soket to the FCU)?
Only if you a taking a spur from a spur. Then all the sockets should be protected by a FCU.
No, don't do that. Is that how your conservatory lights are wired up at the moment?
The kitchen socket feeds the FCU and the cons. skt. The cons. skt isn't fed via the FCU
No but I can envisage situations where you might need to do it.
Not ideal. An unfused spur from a ring can feed a single or double socket, more than that (say several sockets) should require the spur to be fused. Now in reality the danger of an overload posed by the addition of the lighting FCU to the socket is minimal. However its not a good place to start to take a power feed to a shed.
Personally I wouldn't - a high load in the shed and on the socket could overload the spur cable, and also place a large point load on the ring circuit (which may or may not be a problem depending on the actual layout and other loading on the circuit).
Other than the specific cases sited, then yes generally.
I would suggest in this case take the spur directly to a FCU rather than the socket, and from there feed the conservatory socket, the shed feed, and the lighting (which you can leave fed via its additional FCU). This will limit the total combined load of shed, conservatory socket, and lighting to 13A - but they may well be adequate unless you have elaborate plans for the shed.
If you want a more beefy supply to the shed, then look at one of the alternative arrangements described here:
The difficulty with connecting 1.0mm^2 directly to a circuit protected at 32A, is whether it will have sufficient fault and overload protection. Now the overload bit you could delegate to a FCU near the lights - or even the fact that a single bayonet lamp fitting will self impose a maximum load well below the cable capacity. The fault protection however you would have demonstrate by measurement and/or calculation[1].
| __|__ 1 _ 2 _ |__R__|======|a|========|L| | |
Say R is your socket on the ring, 1 is a bit of 2.5mm^2 T&E, a is a FCU, and 2 is 1.0mm^2 T&E - then that is fine. The bit of cable at 1 is overload protected by a, and fault protection is provided by the ring circuit protective device. If however 1 is only 1.0mm^2 T&E, then it becomes more complex to prove that its adequately fault protected in the case of someone nailing through the cable between R and A
[1] You would need to first work out the loop impedance at the location of the fault. This will then let you computer the prospective short circuit current. That you can use to check the clearance time for the given protective device. When you have the time, you can use the adiabatic equation to check the survivability of the cable at the PSCC, for the fault clearance duration.
Many thanks John, that should do just fine. The only thing wrong then would be that I have a RCCB CU in the shed with 6A & 16A mcbs :) Can't win 'em all.
*ahem* can I have chips with that? Cheers muchly John, I'll leave the last few paragraphs ;)
No, you will end up safe if not optimal. (a 20A dedicated submain from the main CU would probably be ideal here but possibly more hassle than its worth).
Well its not as bad as it sounds. If doing it all by calculation, then it comes down to working out the length of cable from the CU to the worst case fault position. Looking up the round trip resistance per metre for the size of cable and multiplying by the length. Then add on something for the supply loop impedance (typically 0.35 ohms for TN-C-S or 0.8 for TN-S if you don't have actual measured values available).
You now know R. You also know V (i.e. 230V), so you can work out I using V = IR. Check the response curves for the MCB in question:
Last step use the adiabatic equation: s = sqrt( I^2 x t ) / k to work out "s" the minimum CSA of copper you need to withstand the fault. (k is a constant of 115 for PVC clad T&E)
See:
s = sqrt( 180^2 x 0.1 ) / 115 = 0.5
Which would mean you need at least 0.5mm^2 of conductor to clear the fault without failing - so 1.0mm^2 T&E would be adequately fault protected by a B32 MCB in this circumstance.
I actually almost understand that ;)
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