Voltage Drop on Long Cable Runs: How to Size the Wire

Everyday September 2, 2026

Why long cable runs lose voltage, and how to pick a conductor size that stays inside the 3 per cent limit.

Quick answer: Voltage drop is the voltage lost along a cable because the conductor has resistance. It rises with current and length, and falls as the conductor gets thicker. The common limit is 3 per cent for a final circuit. A 100 ft run at 15 A on 12 AWG loses about 4.8 V, which is 4 per cent of 120 V and too much.

Cable is not a perfect wire. Push current down 100 feet of it and some of the voltage arrives as heat in the copper instead of at the socket. On a long run the motor gets hot, the lights dim when the compressor starts, and the fault stays invisible unless you go looking.

Why long runs lose voltage

Drop equals current multiplied by the resistance of the conductor, and that resistance is proportional to length and inversely proportional to cross sectional area. Double the run and you double the loss. Double the current and you double it again.

The detail people miss is that the current goes out and comes back, so the length in the sum is twice the physical distance. A 100 ft run is 200 ft of copper. If the underlying arithmetic feels shaky, our explainer on Ohm's law covers the relationship everything here sits on.

What counts as too much

The working figures are 3 per cent on a final circuit and 5 per cent across the installation. Those are percentages of the system voltage, not fixed volt figures, which is the thing to grasp here. Three per cent of 230 V is 6.9 V. Three per cent of 12 V is 0.36 V.

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Two worked examples

A 120 V garage circuit

Say a 15 A circuit running 100 ft to a detached garage on 12 AWG copper. Twelve gauge is about 1.59 ohms per 1,000 ft, so 200 ft of it is 0.318 ohms. At 15 A the drop is 4.8 V. Against 120 V that is 4 per cent, over the guideline, and the tool at the far end sees 115 V.

Step up to 10 AWG, about 1.0 ohm per 1,000 ft. Round trip resistance becomes 0.20 ohms, the drop is 3.0 V, and you are at 2.5 per cent. One size up solved it, which is the normal answer to a voltage drop problem.

A 12 V run to a shed or a boat

Here is where it bites. Take 20 m of 2.5 sq mm cable carrying 10 A. Using the tabulated figure of roughly 18 mV per amp per metre, the drop is 18 times 10 times 20, divided by 1,000, which is 3.6 V. On a 230 V circuit that is 1.6 per cent and perfectly fine. On a 12 V system it is 30 per cent, and the load at the other end is trying to work on 8.4 V.

This is why low voltage runs use cable that looks absurdly thick for the current. It is not about heat capacity but about keeping the drop small against a small starting voltage.

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Using the voltage drop calculator

System voltage is nominal, so 120 V, 230 V or 12 V rather than what a meter reads today. Current should be the actual design load, not the breaker rating, though sizing for the breaker is the cautious choice if the load may grow.

Length is the one way route the cable physically takes, including vertical drops and the detour round the doorway, not the straight line on the plan. Measure the route, then add ten per cent. The calculator doubles it internally.

Conductor size is AWG in the US and square millimetres in the UK, and the two do not line up neatly. If you are working from a mixed source, our unit converter guide keeps feet, metres and areas straight faster than doing it in your head.

Pick copper or aluminium honestly, too. Aluminium carries around 60 per cent of copper's conductivity, so it needs roughly two sizes larger for the same drop.

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Common questions

Does voltage drop waste electricity? Yes, though rarely enough to notice on a bill. The lost voltage becomes heat in the cable. That 4.8 V drop at 15 A is 72 watts warming your wall whenever the circuit is loaded.

Why is my drop worse than the calculator says? Usually connections. A loose terminal or a corroded joint adds resistance no cable sum accounts for. Published tables also assume the conductor is running warm, so a hot loft reads worse than a cold bench test.

Can I fix it without replacing the cable? Sometimes. Splitting the load across two circuits halves the current in each and halves the drop. Moving the supply point closer shortens the run. Otherwise it is heavier cable, and pulling it once is cheaper than twice.

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