Voltage drop in a ring circuit (calling ban-all-sheds)

[biblio]

ban-all-sheds wrote:
"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."

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?

 

[Damocles]

You may not be interested in worst case scenarios. Design current means what it says, how much current you are expecting it to carry.

Any point on a ring has two paths back to the CU. You are therefore dealing with two resistors in parallel. If you imagine a point exactly half way round the ring, then each half has only half the resistance of the whole length of cable. (of course). And since there are two identical resistances connected in parallel, then the actual resistance will be half again the resistance of either length of cable on its own.

So we have got 1/2 times 1/2 times the resistance of the whole cable length.

The result is different for different places around the ring. But you can prove it will only get lower then this.

 

[biblio]

Got it. Thanks Damocles.