Voltage drop in cables: how to check a cable run

What voltage drop is

A cable has resistance. When load current flows, some voltage is lost along the conductors, so the voltage at the load end is lower than at the source. The loss is usually expressed in volts and as a percentage of the supply voltage. Too much drop can leave equipment under-voltage, reduce motor torque and make protection or control equipment behave differently from its design case.

This guide follows the method in the voltage drop calculator. It is a resistive-only estimate. It is not a cable size recommendation and does not replace the applicable wiring standard, manufacturer data or review by a qualified person.

The calculation step by step

The calculator works in four steps. First it corrects the conductor resistivity for operating temperature, then it finds the one-way conductor resistance, then the drop, then the percentage.

1. Resistivity: ρ(T) = ρ₂₀ × (1 + α × (T − 20)).
2. Resistance: R = ρ(T) × length ÷ area, with length in m and area in mm².
3. Single-phase drop = 2 × I × R. Three-phase drop = √3 × I × R.
4. Drop (%) = drop ÷ supply voltage × 100.

Constants used: copper ρ₂₀ = 0.017241 Ω·mm²/m with α = 0.00393 per °C; aluminium ρ₂₀ = 0.028264 Ω·mm²/m with α = 0.00403 per °C. The length entered is one-way, source to load. The factor 2 (single-phase) accounts for the outgoing and return conductors; the factor √3 gives the line-to-line drop of a balanced three-phase circuit, and the supply voltage entered is line-to-line.

The calculation needs current, not power. If you have a power rating, convert it first. For electrical input power use the kW to amps calculator: three-phase I = kW × 1000 ÷ (√3 × V × PF). For an apparent-power rating use the kVA to current calculator: I = kVA × 1000 ÷ (√3 × V). Motor shaft power must be converted to input power with efficiency before either step. See kW, kVA and amps for the distinction between these quantities.

Worked values below use this method with a stated conductor temperature. Always enter the temperature the conductor will actually reach under the load being assessed; a cable carrying a light current runs cooler than its maximum rating, so 70 °C is a conservative assumption for light loading and a typical one only for heavily loaded cable.

Worked example 1: three-phase copper feeder

A balanced 30 kW load at 400 V line-to-line, power factor 0.85, is fed by an 80 m copper cable of 25 mm² at an assumed conductor temperature of 70 °C.

1. Current: I = 30 × 1000 ÷ (√3 × 400 × 0.85) = 50.94 A.
2. Resistivity: ρ = 0.017241 × (1 + 0.00393 × 50) = 0.020629 Ω·mm²/m.
3. Resistance: R = 0.020629 × 80 ÷ 25 = 0.06601 Ω.
4. Drop = √3 × 50.94 × 0.06601 = 5.82 V.
5. Percentage = 5.82 ÷ 400 × 100 = 1.456 %; load-end voltage = 394.18 V.

These values match the calculator output for the same inputs (resistance 0.0660 Ω, drop 5.8246 V, 1.4562 %).

Worked example 2: single-phase aluminium circuit, and the effect of size

A 32 A single-phase load at 230 V is fed by 45 m of aluminium cable at 60 °C. With 16 mm²: ρ = 0.028264 × (1 + 0.00403 × 40) = 0.032820 Ω·mm²/m, R = 0.032820 × 45 ÷ 16 = 0.09231 Ω, drop = 2 × 32 × 0.09231 = 5.91 V, which is 2.569 % of supply, leaving 224.09 V at the load. With 25 mm² the resistance falls to 0.05908 Ω and the drop to 3.78 V (1.644 %).

The table compares the cases calculated so far.

Doubling the area halves the resistance, so the drop falls in proportion. Assuming 20 °C instead of 70 °C understates the drop in example 1 by about 16 %.

What the estimate leaves out

• Reactance. Only resistance is used; reactance matters more in larger cables.
• Power factor. The result is I × R, not a phasor calculation including load power factor.
• Parallel cables, harmonics, and the voltage dip during motor starting, which is typically much deeper than the running drop.
• Cable derating from installation method, grouping and ambient temperature. These affect the current-carrying capacity, not this calculation.

The allowed percentage depends on the applicable code, the equipment tolerance and the project specification. The Schneider Electric page listed in the sources describes limits taken from an international wiring standard and notes that they apply to steady-state running, not starting.

Common mistakes

• Entering the round-trip length. The calculator expects the one-way length and applies the factor 2 or √3 itself.
• Using line-to-neutral voltage for a three-phase circuit, which overstates the percentage.
• Entering conductor area in the wrong unit, or an AWG number instead of mm².
• Using ambient temperature instead of the conductor operating temperature.
• Treating a pass at running current as proof the circuit is acceptable at start-up.
• Reading the result as a cable size or protection decision.

Checks before you trust the result

• Confirm the length is one-way, measured along the route and not straight-line.
• Check current against the load rating and the actual operating condition, not the cable rating.
• Check the temperature assumption against the loading you expect.
• Compare the result with a manual calculation or the cable manufacturer’s tabulated millivolt-per-amp-per-metre figures.
• Have the cable size, protection and starting performance reviewed against the applicable standard.

Frequently asked questions

Does the calculator tell me which cable size to use?

No. It evaluates one conductor size you enter. Try sizes until the drop is acceptable, then verify current capacity and protection separately.

Why is the drop higher when the cable is warm?

Resistivity rises with temperature, so the same cable at 70 °C has higher resistance than at 20 °C.

Is a larger percentage drop acceptable for a motor?

That depends on the motor, its starting duty and the specification. Check starting voltage separately.

Record the data behind the check

Keep the inputs with the result: current and its source, one-way length, conductor material and area, temperature assumption, supply voltage and the limit used. Record the date and the measured voltage at the load if a test was done later, so future changes in load or route can be compared with the original assumptions.