Voltage drop and cable size calculator
On a long cable run part of the voltage is lost on the way: the cable has resistance and, when current flows, some voltage "drops" across it. If the drop is too large, motors struggle to start and run hot, and electronic equipment misbehaves. This calculator estimates the voltage drop of a run and the minimum cross-section to stay within a limit. It works in mm², the IEC sizes used in Europe, the UK, Australia, India and most of the world.
How it is calculated
With the usual approximation for low-voltage circuits, ignoring the cable’s reactance:
- Three-phase: ΔV = √3 × L × I × cos φ / (γ × S)
- Single-phase: ΔV = 2 × L × I × cos φ / (γ × S) (the 2 is because the current goes out and comes back)
L is the length of the run in metres, I the current in amps, S the cross-section in mm² and γ the conductivity of the conductor in m/(Ω·mm²), which falls as the cable heats up. The minimum cross-section comes from solving for S with the maximum allowed drop.
Example
A 60 m three-phase run at 400 V carrying 16 A, with a power factor of 0.85 and 2.5 mm² PVC-insulated copper cable (γ = 48): ΔV = 1.732 × 60 × 16 × 0.85 / (48 × 2.5) ≈ 11.8 V, that is 2.9 %. With 1.5 mm² it would be almost 20 V, 4.9 %.
Voltage drop limits
- IEC 60364-5-52 (informative Annex G) suggests a maximum of 3 % for lighting and 5 % for other uses in installations supplied from the public low-voltage network. National rules such as BS 7671 follow similar values.
- NEC (United States): its informational notes recommend no more than 3 % on a branch circuit and 5 % for feeder and branch circuit together.
- On machines, also follow what the equipment manufacturer asks for.
Using AWG?
Convert the wire size to mm² first: 14 AWG ≈ 2.08 mm², 12 AWG ≈ 3.31 mm², 10 AWG ≈ 5.26 mm², 8 AWG ≈ 8.37 mm², 6 AWG ≈ 13.3 mm². Lengths go in metres (1 ft = 0.3048 m).
To understand the topic in depth: Practical electrical theory, with theory and a practice test.
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