Volt drop query

Ninja1000

Hi guys

Does the regs 4% volt drop rule apply to the size of the MCB on the circuit or the appliances on the circuit?

For example when I calculate the below formula and get the resistance of the cable do I divide it by the size of the mcb (20amps) or do divide it by the overall load the circuit is pulling?



The following is an example of a basic voltage drop calculation:
Refer to Figure 36. Assume a load of 20 Amps is to be supplied at a distance of 25 metres using 2.5 mm2 conductors.




The resistance of 50 metres ( phase and neutral in series ) of 2.5 mm2 must first be established.

Resistance of conductor

l
R =
a
 for copper is 17.2 mm ( 17.2 x 10-6mm )
l is 50 metres ( 50 x 103 mm )
a is 2.5 mm2 ( 2.5 mm2 )

17.2 x 10-6 x 50 x 103
R =
2.5

17.2 x 10-3 x 50
R =
2.5

R = 17.2 x 10-3 x 20

R = 344 x 10-3

R = 0.344 Ohms


The voltage drop across the cable can now be found as follows:

Voltage Drop across Cable VD = IR

VD = 20 x 0.344
VD = 6.88 Volts
This figure is less than the maximum of 9.2 Volts and so the cable is suitable for the load

Risteard81

Use the design current.

However for a circuit supplying socket-outlets this will be the rating of the protective device.

Ninja1000

Risteard81 wrote: »

Use the design current.

However for a circuit supplying socket-outlets this will be the rating of the protective device.

Okay so if its a socket circuit does that mean you use 20amps in the formula? because socket circuits are protected by a 20amp MCB