Voltage Drop Calculator

Each wire has a small resistance that grows with its length and shrinks with its cross-section. The current has to go out to the load and come back, so on DC and single phase it passes through twice the one-way length. Multiply that resistance by the current and you have the volts lost on the way.

The formula

resistivity at T = resistivity at 20 °C × (1 + α × (T − 20))
resistance per metre (R) = resistivity ÷ cross-section area
DC and single phase: drop = 2 × one-way length × current × R
three phase: drop = √3 × one-way length × current × R
with power factor and reactance: use (R × PF + X × sin φ) in place of R
drop % = drop ÷ supply voltage × 100
voltage at the load = supply voltage − drop
power lost = current² × R × length × number of live wires (2, or 3 on three phase)
current from watts = watts ÷ (volts × PF), or ÷ (1.732 × volts × PF) on three phase

Copper: resistivity 1.724 × 10⁻⁸ Ω·m at 20 °C, α = 0.00393 per °C. Aluminium: 2.82 × 10⁻⁸ Ω·m, α = 0.00403 per °C. 1 mm² = 10⁻⁶ m².

Worked example

A 20 A load on 230 V single phase, 30 m from the board, on 2.5 mm² copper at 20 °C, with a 3% limit.

• Resistance per metre: 1.724 × 10⁻⁸ ÷ 2.5 × 10⁻⁶ = 0.006896 Ω/m (6.896 Ω/km).
• Wire the current runs: 2 × 30 = 60 m, so the loop is 60 × 0.006896 = 0.41376 Ω.
• Voltage drop: 60 × 0.006896 × 20 = 8.2752 V, about 8.28 V.
• As a share: 8.2752 ÷ 230 × 100 = 3.598%, over the 3% limit.
• Voltage at the load: 230 − 8.2752 = 221.72 V.
• Power lost: 20² × 0.006896 × 30 × 2 = 8.2752 × 20 = 165.50 W, which is 3.6% of the 230 × 20 = 4,600 W load.
• Next size up, 4 mm²: 1.724 × 10⁻⁸ ÷ 4 × 10⁻⁶ = 0.00431 Ω/m; 60 × 0.00431 × 20 = 5.172 V = 5.172 ÷ 230 = 2.25%, so 4 mm² is the smallest size within 3%.
• Longest run on 2.5 mm² for 3%: 3% of 230 V = 6.9 V; 6.9 ÷ (2 × 20 × 0.006896 = 0.27584) = 25.01 m.

At a conductor temperature of 75 °C the resistance is 1 + 0.00393 × 55 = 1.21615 times higher, so the drop becomes 8.2752 × 1.21615 = 10.06 V (4.38%). The same run in aluminium 2.5 mm² at 20 °C: 2.82 × 10⁻⁸ ÷ 2.5 × 10⁻⁶ = 0.01128 Ω/m, so 60 × 0.01128 × 20 = 13.536 V (5.89%).

Three phase example: 32 A at 400 V over 50 m of 6 mm² copper: 1.732 × 50 × 32 × 0.0028733 = 7.96 V, which is 1.99% of 400 V.

Resistance of common metric sizes

Worked out from resistivity for solid, plain conductors. Real stranded cables are a little different (standard tables such as IEC 60228 give a maximum, for example 7.41 Ω/km for 2.5 mm² copper), so use the datasheet figure in Advanced when you have it.

AWG to mm²

American Wire Gauge diameter: d = 0.127 mm × 92^((36 − AWG) ÷ 39), and area = π ÷ 4 × d². For 1/0, 2/0, 3/0 and 4/0 use 0, −1, −2 and −3. Bigger sizes are given in kcmil (thousands of circular mils): 1 kcmil = 0.5067 mm².

How much drop is OK?

There is no single world rule. Many national wiring standards and guides use limits of around 3% to 5% from the supply to the furthest point, sometimes split between the main feeder and the final circuit, and some equipment (motors starting, LED drivers, long low-voltage DC runs for solar or boats) needs less. Use the limit your local wiring rules give for your type of circuit.

What this doesn't check

• Current rating (ampacity): a cable must carry the current without overheating. That depends on the insulation, how it's run (in a wall, conduit, buried, in free air), how many cables are bunched together and the room temperature. A size that passes for voltage drop can still be too small to be safe.
• Protection: the breaker or fuse must suit the cable and the load, and the circuit must disconnect fast enough in a fault.
• Starting current: motors draw several times their running current for a moment, so the drop while starting is much bigger.

This calculator uses physics, not any one country's wiring tables. Electrical work is regulated in most places: follow your local rules and have a qualified electrician design and check the installation.