How to Calculate Voltage Drop (UK Guide)

There are two ways to calculate volt drop, and the one taught in physics class isn't the one used on site.

The textbook method is Ohm's law. Work out the conductor resistance from its length and cross-sectional area, then multiply by the current. It's correct and it's slow, because you need resistivity figures and a temperature correction before you start.

The method that gets used in the UK is the tabulated one. BS 7671 publishes a millivolt drop per amp per metre figure for every cable size, in Appendix 4. For twin and earth that's Table 4D5. The figure already includes the return conductor and it's already corrected to the cable's operating temperature, so the sum collapses to one line:

Volt drop (V) = (mV/A/m x design current x route length in metres) / 1000

Route length means the actual path the cable takes, over joists, round door frames, down to the board. Not the straight-line distance between the two ends. On a long run this is the input people get wrong most often, and it's usually wrong in the direction that flatters the result.

BS 7671 Appendix 4 gives 3% for lighting circuits and 5% for everything else, measured from the origin of the installation to the point of use. On a 230V supply that's 6.9V and 11.5V respectively.

Lighting gets the tighter limit because filament and some LED drivers visibly change output as voltage sags, and flicker is a comfort and safety issue rather than just an efficiency one.

There's a nuance about whether those percentages are a regulation or guidance, and what Regulation 525.202 actually requires instead. That's covered in our guide to sizing a cable to BS 7671, because it belongs with the sizing method. For the purposes of calculating volt drop, treat 3% and 5% as the numbers you have to hit.

Takeaway: 6.9V for lighting, 11.5V for everything else, on 230V. Measured from the origin, not from the nearest board.

A converted barn in Derbyshire. External and outbuilding lighting runs off a dedicated 6A circuit at the house consumer unit. The route to the furthest fitting, measured properly along the actual cable path, is 60 metres. Total connected load is around 690W, so 3A.

The installer has priced 1.0 mm² twin and earth, which carries 16A clipped direct and is therefore massively over-rated for a 3A load. On current alone it's not just adequate, it's generous.

Volt drop: (44 x 3 x 60) / 1000 = 7.92V

As a percentage of 230V: 3.44%

The lighting limit is 3%. It fails, on a cable rated for more than five times the current it's carrying.

Step up to 1.5 mm²: (29 x 3 x 60) / 1000 = 5.22V, which is 2.27%. That passes with room to spare, and the extra copper over 60 metres costs a few pounds at CEF or Screwfix.

Takeaway: on lighting circuits, current-carrying capacity almost never decides the cable. Distance does, and the 3% limit arrives far sooner than people expect.

This is the one worth reading twice, because it's where compliant-looking work goes wrong.

A detached workshop at the end of a garden in Sheffield. There's a submain from the house consumer unit to a small board in the workshop, then final circuits inside the workshop off that board.

Submain: 16 mm² two-core SWA, 30 metres, design current 40A. Volt drop: (2.8 x 40 x 30) / 1000 = 3.36V, which is 1.46% of 230V. Comfortable.

Final circuit: a 2.5 mm² radial socket circuit inside the workshop, 25 metres from the workshop board to the furthest socket, design current 20A. Volt drop: (18 x 20 x 25) / 1000 = 9.0V, which is 3.91% of 230V. Also inside the limit.

Both legs pass. Two separate calculations, two passes, and if you check them one at a time you'll sign it off.

Add them up, which is what the regulation actually asks for, because the measurement runs from the origin of the installation to the point of use:

3.36V + 9.0V = 12.36V, which is 5.37% of 230V. That fails.

Now the interesting part, which is how you fix it. The instinct is to uprate the submain, because it's the big cable carrying the big current and it feels like the important one. Go to 25 mm² SWA and the submain drop falls to (1.75 x 40 x 30) / 1000 = 2.1V. Total becomes 11.1V, or 4.83%. It scrapes a pass, and you've paid for 30 metres of 25 mm² armoured cable.

Uprate the final circuit instead. Go from 2.5 mm² to 4 mm² and its drop falls to (11 x 20 x 25) / 1000 = 5.5V. Total becomes 8.86V, or 3.85%. That's a comfortable pass, from 25 metres of the cheapest cable on the job.

The smaller cable had the higher millivolt figure, so it was contributing three times the drop of the submain despite carrying half the current over a shorter distance. That's where the fix was.

Takeaway: add every leg from the origin before you check compliance, and when it fails, attack the cable with the highest mV/A/m rather than the one that looks most important.

Every volt drop calculation in this article uses 230V, because that's the declared nominal supply voltage in the UK. It is not what's necessarily arriving at your meter.

Under the Electricity Safety, Quality and Continuity Regulations 2002, the declared voltage is 230V with a tolerance of plus 10% and minus 6%. That's a legal range from 216.2V to 253V, and distribution network operators work inside it.

So take that workshop after it's been fixed, sitting at a compliant 3.85% drop. Calculated on 230V that's 8.86V, leaving 221V at the socket. Now suppose the incoming supply is running at 220V, which is entirely lawful and not unusual at the end of a rural feeder. The same 8.86V of drop now leaves 211V at the socket, and the calculation never knew.

For most kit that's fine. Modern switch-mode power supplies are relaxed about it. For motors, older induction loads and some inverters, a supply already low in tolerance plus a drop already at the limit is exactly the combination that produces intermittent faults nobody can reproduce.

Takeaway: design to the limits, but if you know the supply runs low, treat 5% as a ceiling to stay well under rather than a target to reach.

One technical point that separates a rough calculation from a defensible one.

Below 16 mm², conductor reactance is small enough to ignore, so Appendix 4 gives a single millivolt figure and that's all you need. At 16 mm² and above the tables split into three columns: resistance (r), reactance (x) and impedance (z). On AC circuits at those sizes you should be working from z, not r.

Use the resistance column alone on a long run of large cable and you'll understate the drop. It's usually a modest error, but on a 70 metre run of 35 mm² SWA feeding a workshop with motor loads, modest errors are how a design lands the wrong side of a limit.

Our voltage drop calculator is resistance-based and states this in its own notes. For anything at 16 mm² and above on a long AC run, cross-check against the impedance figure from the manufacturer's data. Doncaster Cables and Prysmian both publish full tables for their own ranges, and manufacturer data for the specific cable you're installing always beats a generic table.

Takeaway: at 16 mm² and above on AC, use the impedance column and check the manufacturer's figures for long runs.

Four options, roughly cheapest first.

Uprate the cable. The default answer, and usually right. Work out which leg contributes most drop and fix that one rather than uprating everything.

Shorten the route. Sometimes a cable takes a long way round for reasons that made sense to whoever ran it and nobody since. Ten metres saved is ten metres of drop removed.

Move the distribution point. On a big installation, a submain to a board nearer the loads, with short final circuits off it, will often beat long final circuits run from a single central board. This is the fix that most changes the outcome on large or sprawling sites.

Reduce the load, or split it. Two circuits at half the current each will each drop half as much as one circuit carrying the lot.

What you should not do is quietly recalculate on a shorter length than the cable actually runs, or drop the power factor to a flattering number to make the figure work. Both happen. Both produce a design that passes on paper and sags in service.

Takeaway: fix the leg with the biggest contribution, and if you find yourself adjusting inputs rather than the design, stop.

I asked an electrician in Sheffield, who's been testing and inspecting for around fifteen years, what he sees most often on volt drop. His answer wasn't about the maths.

He said the recurring problem is people calculating from the wrong starting point. Someone runs a circuit from a garage board, calculates the drop from that board to the load, gets a comfortable 2% and moves on, without ever adding the drop on the submain that feeds the garage board in the first place. He described it as the most common thing he picks up on periodic inspections of extended installations, and said it's almost always outbuildings, because that's where submains live.

He also made a point about the difference between a calculation and a measurement. A calculation tells you what should happen at design current. Measuring at the far end under actual load, with everything running, is the only thing that tells you what does happen. On anything long or important, he tests rather than trusts the sum.

Takeaway: find out where the origin of the installation really is before you calculate, and on long runs measure at the far end under load.

The arithmetic is one line, but it's one line repeated for every leg of the run, and that's where slips happen. Our voltage drop calculator handles single-phase, three-phase and DC, applies power factor, corrects for ambient temperature and will auto-size the conductor to hit a target percentage.

If you need the full sizing method rather than just the drop, including current-carrying capacity and correction factors, use the cable sizing calculator instead.

Neither replaces a design by a competent person. Electrical work in dwellings in England and Wales is notifiable under Part P of the Building Regulations, and every figure here should be confirmed against the current edition of BS 7671 and the IET On-Site Guide before you install anything.