How to size a cable: current, length and voltage drop

Two separate checks, not one

Sizing a cable correctly means passing two independent checks, and the larger of the two required sizes is the one that's actually used. The first is current-carrying capacity (sometimes called ampacity): whether the conductor can carry the design current continuously without its insulation overheating, which depends on the installation method – clipped to a surface, run in conduit, buried, or in contact with thermal insulation all give different capacity ratings for the same cable size. The second is voltage drop: whether too much voltage is lost as heat along the length of the cable before it reaches the load.

A short, heavily loaded cable is often limited by current-carrying capacity; a long, lightly loaded cable is often limited by voltage drop instead. Both checks use the same starting inputs – current and cable length – but they're separate calculations with separate rules, and BS 7671, the UK Wiring Regulations (BS 7671:2018+A2:2022), requires both to be satisfied.

The voltage-drop formula

For a single-phase or DC circuit, the voltage drop over a cable run is calculated as: ΔU = (2 × L × I × cos φ) / (γ × A), where L is the one-way cable length in metres, I is the current in amps, cos φ is the power factor (1 for DC or purely resistive loads), γ is the conductor's conductivity (56 m/(Ω·mm²) for copper at normal operating temperature, 35 for aluminium), and A is the cross-sectional area in mm². For a three-phase circuit, the factor of 2 is replaced with √3 (approximately 1.73), since three-phase power delivery loses proportionally less voltage per amp over the same cable.

Rearranged to find the minimum required cross-section directly: A = (2 × L × I × cos φ) / (γ × ΔU_max), where ΔU_max is the maximum allowed voltage drop in volts (the percentage limit converted to volts at the supply voltage). The result is then rounded up to the next standard conductor size – 1.5, 2.5, 4, 6, 10, 16, 25 mm² and so on, per IEC 60228.

BS 7671 voltage drop limits

BS 7671 sets maximum permitted voltage drop from the origin of an installation (the consumer unit or distribution board) to the point of use: 3% for lighting circuits and 5% for other circuits, on a low-voltage supply. At the standard UK domestic supply of 230 V, that's a maximum of 6.9 V for lighting and 11.5 V for other circuits.

Worked example: 32 A circuit, 20 m run

Take a single-phase 230 V circuit carrying 32 A over a 20 m one-way run, copper conductor, power factor 1 (a resistive or near-resistive load), with a 3% drop limit. Applying the formula, the required cross-section works out to 3.31 mm², which rounds up to the next standard size, 2.5 mm² only if it's sufficient – here it isn't quite, so the calculation rounds up to 4 mm². Checking the actual drop on 4 mm²: ΔU = (2 × 20 × 32 × 1) / (56 × 4) = 5.71 V, which is 2.48% of 230 V – within the 3% limit, with roughly 0.5 percentage points of headroom.

The relationship is close to linear: doubling the length from 10 m to 20 m roughly doubles the required cross-section, and the actual drop on the recommended size stays in a similar range (1.99% to 2.48%) because each step rounds up to the next standard size rather than landing exactly on the limit.

Copper vs. aluminium

Aluminium conductors have lower conductivity than copper (35 versus 56 m/(Ω·mm²)), so an aluminium cable needs a larger cross-section than copper for the same current, length and drop limit. On the same 32 A, 20 m, 3% example above, copper needs 4 mm²; aluminium needs 5.30 mm², rounding up to 6 mm² – one standard size larger. Aluminium is lighter and often cheaper for a given current rating on longer runs, which is why it's common in larger supply cables, but it needs a bigger cross-section than copper to do the same job.

Reference method C: typical current-carrying capacities

For context, BS 7671 Table 4D5 gives current-carrying capacities for twin-and-earth copper cable under Reference Method C (clipped direct to a surface, no thermal insulation contact): approximately 20 A for 1.5 mm², 27 A for 2.5 mm², 37 A for 4 mm², 47 A for 6 mm², 64 A for 10 mm² and 85 A for 16 mm². These figures assume the standard 30°C ambient temperature with a single circuit and no grouping or insulation effects – a real installation is often derated below these figures if the cable runs through loft insulation, is grouped with other circuits, or sits in a hotter ambient environment, and the correct derated figure always comes from the full BS 7671 Appendix 4 tables for the actual installation conditions, not from this list.

Where this calculation stops being enough

The voltage-drop method above covers straightforward, single-circuit runs at a known ambient temperature. It doesn't account for grouping factors (multiple cables bundled together, which reduce each cable's safe current), thermal insulation contact, ambient temperature correction, protective device coordination (making sure a fuse or circuit breaker disconnects fast enough under fault conditions), or earth fault loop impedance – all separate checks required under BS 7671 for a compliant installation. This calculation is a useful planning and verification tool, not a substitute for a full design carried out by, or checked against the work of, a qualified electrician working to the current edition of BS 7671.