I want to power three or four outside security lights off a ring main in the attic. I intend to spur one 1.5mm cable from a switched FCU. I shall have to insert some junction boxes also in this spurred radial circuit to branch off to each light. What are the limits for the (total) lengths of such a spurred radial circuit. And does the (total) load have to be taken into this account?
Spur cable length
- Thread starter biblio
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Unusual to have a ring main in the attic, but there you go...
Yes, the total load very much has to be taken into account, as the maximum length of the cable is governed by voltage drop, and that is directly proportional to the load.
You can find out all about it here: http://www.tlc-direct.co.uk/Book/4.3.11.htm
Yes, the total load very much has to be taken into account, as the maximum length of the cable is governed by voltage drop, and that is directly proportional to the load.
You can find out all about it here: http://www.tlc-direct.co.uk/Book/4.3.11.htm
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Just to be sure bib, do you mean ring main or do you mean lighting circuit?
Sorry to ask, but better safe than sorry.
Sorry to ask, but better safe than sorry.
Thanks for the useful link, ban (the attic in my 1908 town-house forms a third storey, and is below the loft space). securespark: yes, the switched FCU is to go in a 2.5mm ring, not in a lighting circuit, with a 10 amp fuse eg. this means that there will be two sets of 2.5mm core connected together with one set of 1.5mm in the FCU (but instead of in a nearby junction box though). The length of the ring will be close to the 60m too. Taking security lights off the power allows for a choice of higher load, than using a lighting circuit, but the cable run in this instance would be longer than spurring off from a lighting circuit closer to hand. Would the resistance in such a spurred radial circuit add to the resistance in the ring eg? Or would one calculate the length of the radial to include the shorter part of the ring feeding it up to the FCU?
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A spur does not add to the resistance of the ring, but you should include the voltage drop of the ring when calculating the total drop to the end of the spur.
The worst case voltage drop in a ring is equivalent to the drop of a run of 2.5mm² cable ¼ the total length of the ring.
The worst case voltage drop in a ring is equivalent to the drop of a run of 2.5mm² cable ¼ the total length of the ring.
Thanks, ban. Could I just add the voltage drop for 15metres of 2.5mm cable, onto the complete 1.5mm length when working out the calculation? I assume you mean the voltage-drop factor, not the actual drop in voltage in the ring?
No - you can't do that, as the volt drop in the ring will relate to the total load on it, not just that of your spur. You need to decide on the design current for the ring, including the load of your spur - worst case I guess would be the full 32A. The work out what the volt-drop is for that current over 15m of 2.5mm².
You add that volt-drop to the one for your 1.5mm² spur, again, worst case would be 13A, to get the total drop that you will see at the end of your spur.
You add that volt-drop to the one for your 1.5mm² spur, again, worst case would be 13A, to get the total drop that you will see at the end of your spur.
Thanks indeed, ban
ban-all-sheds said:
Is the voltage drop in the cable length of a ring circuit not calculated in the same way as in the single cable length of a radial circuit? When I add the total length of a cable in a ring it either turns out to be too long in my calculations, or the circuit will only support around 10amps, and always the drop is close to the 9volts too; and this is not allowing for other cable-derating factors either.
Could the actual voltage drop in a ring be ascertained (if tested by an Electricity Board eg) by applying the current rating of the protective device in that circuit (the fuse/breaker rating), rather than applying some assumed design current? Satisfying fuse rating would be the worst case scenario anyway wouldn't it?
Do the maths. Consider a ring of length L. Every point on the ring is supplied by 2 cables, one of length L1, the other L2, where L1 + L2 = L.
The current flowing in each leg is proportional to it's length, as is the voltage drop, and the worst case (mid point of the ring, where L1 = L2), the voltage drop is what you'd see at the end of a radial of length L/4.
The current flowing in each leg is proportional to it's length, as is the voltage drop, and the worst case (mid point of the ring, where L1 = L2), the voltage drop is what you'd see at the end of a radial of length L/4.
Thanks, ban-, plugwash.
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