Energy, electricity and mechanics
Voltage drop in a cable
Voltage drop, power lost and minimum cross-section of a cable based on its length, current and material.
How it is calculated
One-way conductor resistance: R = ρ × L ÷ S (ρ in Ω·mm²/m, L in metres, S in mm²). Resistivity at 20 °C: copper ρ ≈ 0.0172 Ω·mm²/m, aluminium ρ ≈ 0.0282 Ω·mm²/m; when hot (conductor at 70-90 °C) it rises by around 20 %.
Voltage drop: for direct current or single-phase AC (there and back) ΔV = 2 × R × I; for balanced three-phase (one way per phase) ΔV = √3 × R × I. Drop as a percentage: 100 × ΔV ÷ V.
Power lost to Joule heating: single-phase/DC P = 2 × I² × R; three-phase P = 3 × I² × R.
Minimum cross-section to stay within the allowable drop: S is solved from the ΔV formula, setting ΔV to the allowable value. The recommended commercial cross-section is the first size in the range 1.5 - 2.5 - 4 - 6 - 10 - 16 - 25 - 35 - 50 - 70 - 95 - 120 mm² equal to or greater than the calculated minimum.
Keep in mind
- A cable's cross-section is also limited by its current-carrying capacity due to heating (which depends on how it is installed: in free air, embedded, in trunking, bundled with other cables). This calculator does not check that limit: always use the larger of the size required by voltage drop and the size required by current-carrying capacity per the REBT table or other applicable regulation.
- Every electrical installation needs circuit-breaker and residual-current protection and, in AC, earthing. Working with voltages above 50 V AC is dangerous: always cut the power before handling the wiring.