Calculating voltage drop…
You won’t find this formula anywhere in BS7671:2018 as this is another example of where G&G / EAL assume some prior electrical theory knowledge, but the formula is…
Voltage Drop = (mV/A/m) x Ib x L / 1000.
(mV/A/m): To find the value for (mV/A/m) we need to look in the current carry capacity and voltage drop tables within Appendix 4, and the fastest way to finding the table you need is by using Table of Tables on page 522.
Ib: is the design current which is usually just watts divided by voltage if you need to work it out.
L: is just the length of the circuit in metres.
We divide it all by 1000 to convert the figure from Millivolts into Volts
Note: By finding the value for volt-drop (mV/A/m) within Appendix 4 we can calculate the voltage drop for every type of cable we are likely to use. It’s worth remembering though that there are a couple of things that you cannot calculate voltage drop for using your copy of BS7671:2018 and that is power track and lighting trunking systems. This is because it doesn’t give any values for voltage drop for those systems anywhere in the book and you would need to get the information from the manufacturer.
Let’s have a look at a question…
A multi-core 70 degrees C thermoplastic non-armoured cable with 2.5 mm conductors supplies a single-phase load of 20 amp at 230 volts over 22m. What is the voltage drop in the cable going to be?
To work this out first need to put these values into the formula. We can see that the load is 20 amp which is your design current (Ib) and the length is 22 m long.
Next, we need to find the value of (mV/A/m) for a multi-core 70 degrees C thermoplastic non-armoured cable with 2.5 mm conductors.
If you go to the Table of tables on page 522 to start.
Within Table of tables, just towards the bottom of page 523 you’ll find current carrying capacities and voltage drop for cables and the table for we are looking for is multi-core 70 degrees C thermoplastic non-armoured cable is which is Table 4D2A with the voltage drop table for this cable being Table 4D2B on page 404
Table 4D2B on page 404…
It’s a 2.5 mm sq conductor and we are looking for a single-phase so have a look in column 3 and we can see that it’s 18 mv/A/m.
And now that we have all of the figures it’s time to work it out…
18 x 20 x 22 = 7920 divided by 1000 gives us 7.92V
Which means that the voltage drop for a multi-core 70 degrees C thermoplastic non-armoured cable with 2.5 mm conductors supplying a single-phase load of 20 amp at 230 volts over 22m is 7.92V
There’s two more common formula question that you may come across in the 18th Edition exam.