Voltage drop - help working it out.

[MrS]

Volt drop is calculated using ohms law

U = I x R
U is the cable volt drop
I is the circuit current (A)
R is the circuit resistance W(ohms)

Volt drop is allowed as 4% of 230V which is 9.2V

For the everyday standard standard multicore pvc insulated cables we can use this table:
http://www.tlc-direct.co.uk/Figures/Tab4.7.htm

for 2.5 mm cable volt drop is 18 mV/A/m (millivolts per ampere per metre of cable.

So multiply that value by the current the cable is actually carrying and then by the length. Divide your answer by 1000 and you get the volt drop value. This value should be less than 9.2v.

I get this upto now.

Is the current the cable is carrying the design current?

Im obviously wrong here, but if I take the current the cable is carrying as 32A, multiply it 18, then a length of say 50m, then divide it by 1000 I get 28.8V drop. This obviously isn’t right so my figure for design current must be wrong?

 

[Adam_151]

The calculation is correct, such a cable doesn't comply with volt drop requirements over that length, and thats putting aside the fact that the design current is more than the cable can safely carry

even with 6mm you get a volt drop of somewhere around 11v/12v ,so if you wanted to carry 32A over 50m then you'd have to use 10mm² cable, which gives you a volt drop figure of 7.04v

 

[andy]

forget the ohms law for now.

lets say you need to carry 14A a length of 40metres using T&E

voltage drop for 2.5 is 18.5mv

so we have 18X14X40. this gives a drop of 10.08V


again, but using 6mm T&E, voltage drop of 7.3

7.3X14X40 is 4.088V

once you have current, legth in metres, and voltage drop, you times then together (divide by 1000 to give answer in amps, not mA) to get the voltage drop

understand this better?

 

[MrS]

yes i get it thanks

ok so for 10mm cable, it will carry 63A clipped direct, the volt drop is 4.4

if you have 10 mm cable and a 10KW shower running 15m in length. the current is 10000/230 which is 43.5A
(4.4 x 43.5 x15) / 1000 = 2.9V so it is within the limit of 9.2V

ok all clear now, thanks for the help. i just need to get my head round the current on a socket circuit where you dont know what is being plugged in.