The formula
R_cable = ρ × (2 × L) / A ; V_drop = I × R_cable
- L
- one-way run length (metres) — doubled for the return path
- A
- conductor cross-section (mm², converted to m²)
- ρ
- resistivity: copper 1.68 × 10⁻⁸, aluminium 2.82 × 10⁻⁸ Ω·m
- I
- load current (amperes)
Worked example
A 20 m run of 2.5 mm² copper carrying 16 A on a 230 V circuit.
- Total conductor length: 2 × 20 = 40 m
- R = 1.68×10⁻⁸ × 40 / 2.5×10⁻⁶ = 0.269 Ω
- V_drop = 16 × 0.269 = 4.30 V
- As a percentage: 4.30 / 230 = 1.87 %
4.3 V lost, or 1.87 % — inside the 3 % usually allowed for a final circuit.
Where you'll use it
Wiring installations, long DC runs to solar panels or LED strips, and diagnosing why a motor at the end of a long feeder never quite reaches full torque. Low-voltage DC is worst affected: 4.3 V is trivial on 230 V and catastrophic on 12 V.