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Wire voltage drop calculator

Every wire is a resistor. A long, thin cable takes some of the supply's volts for itself, so the load gets less, and turns them into heat. Type the run's length, the cable size and the current, and this gives the voltage drop, what's left at the load and the cable size that keeps the loss small.

At the battery or power supply end.

Watts ÷ volts, if you only know its power.

In metres. Doubled for you: the current goes out and back.

Sets the cross-section below.

Copper. Printed on the cable, or pick a size above.

Voltage drop, I × R

0.84 V
Share of the supply
7.0%
Voltage at the load
11.16 V
Heat in the cable, I² × R
4.2 W
Cable resistance
168 mΩ
Thinnest for under 3%
2.5 mm²

Over 3% is lost in the cable: the load gets 11.16 V and runs weaker. 2.5 mm² cable keeps it under 3%.

For copper at 20 °C. Hot cable has more resistance, and aluminium about 1.6 times as much.

The formulas

A cable's resistance grows with its length and shrinks with its cross-section. Count the length out and back:

R = ρ × 2L over A

For copper, ρ is about 0.0168 Ω for each metre of 1 mm² wire. Then Ohm's law on the cable gives the drop, and the heat goes as current squared:

Vdrop = I × R and Ploss = I² × R

A worked example

A 12 V pump draws 5 A at the end of a 5 m run of 1 mm² cable, 10 m in all:

R = 0.0168 × 10 over 1 = 0.168 Ω
Vdrop = 5 A × 0.168 Ω = 0.84 V

The pump gets 11.16 V, 7% less than it should, and the cable turns 5² × 0.168 = 4.2 W into heat. For under 3%, the drop must stay below 0.36 V, which needs at least 2.33 mm²: fit 2.5 mm² cable, and the drop falls to 0.34 V.

Send it at a higher voltage

For the same power, P = V × I, twice the voltage needs half the current. The drop halves and the heat, I² × R, falls to a quarter. Sending 20 W through a 2 Ω cable wastes 5.6 W at 12 V, 1.4 W at 24 V and 0.35 W at 48 V. That's why the grid runs at hundreds of thousands of volts, and why big solar and boat systems move up to 24 or 48 V. The load has to be built for the higher voltage, though.

Free lesson

Now build it yourself

Each lesson comes in three short parts: see the idea, try it on a live board, then build it yourself, with quick checks as you go.

Try it yourself: free lesson

“Lost in the wire” is lesson 4 of 6 in Watt Harbour, Power & Energy World 1. Free with an account. Lessons unlock in order, so you’ll do the 3 before it first.

Or start from the very first lesson, no account needed

The ideas behind it

Questions people ask

How do you calculate voltage drop in a wire?

Find the cable's resistance, R = ρ × L ÷ A, with ρ = 0.0168 Ω·mm²/m for copper, L the length out and back, and A the cross-section in mm². The drop is I × R. 10 m of 1 mm² copper is 0.168 Ω, so 5 A drops 0.84 V.

Why is the length doubled?

The current goes out to the load on one wire and comes back on the other, and both drop volts. A lamp 5 m away has 10 m of wire in its loop. The calculator doubles the one-way length for you.

How much voltage drop is acceptable?

A common rule of thumb is 3% for anything important. On 12 V that's only 0.36 V, which is why low-voltage systems need much thicker cable than mains for the same power.

How do I reduce voltage drop?

Use thicker cable, a shorter run, or less current. For the same power, a higher voltage needs less current: twice the volts halves the drop and cuts the heat to a quarter.

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