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Hi.

You need the mV/A/m value for the particular cable from the tables in appendix 4 of BS 7671, the (Ib) (design current) of the circuit (not just the fuse/MCB rating), and the length of the circuit. Multiply these 3 figures together and divide the sum by 1000.

Maximum permitted volt drop is 3% (6.9V) for lighting circuits, and 5% (11.5V) for all other circuits.
 
Just a thought - If you use the formula that I gave, you need to correct your R1 + Rn value for temperature.

This is because volt drop calcs are based on conductor operating temps (eg 70 degrees) whereas your R1 or Rn measurement will be taken at around 20 degrees.

So, the actual formula would be:

Volt Drop = (R1 + Rn) x 1.2 x design current :)

This gives a pretty accurate answer.
 
Hi.

You need the mV/A/m value for the particular cable from the tables in appendix 4 of BS 7671, the (Ib) (design current) of the circuit (not just the fuse/MCB rating), and the length of the circuit. Multiply these 3 figures together and divide the sum by 1000.

Maximum permitted volt drop is 3% (6.9V) for lighting circuits, and 5% (11.5V) for all other circuits.

hi, why do we use the sum of 1000 to divide by?
thanks
 
hi, why do we use the sum of 1000 to divide by?
thanks

Hi mate,

It's because the figure used is mV/A/M (milli-volts per Amp per Metre) so you need to divide by 1000 to convert the milli-volts to volts, otherwise your volt-drop answer would be in mV instead of V.

Hope that helps:)
 
Last edited by a moderator:
If you have the book, BS 7671: Requirements for Electrical Installations (etc.....) Study Notes E1 (it's another big red book but with a spiral binding), it all starts on P8/1. There's loads of examples in there too.
 

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