How voltage drop is calculated
A cable's resistance is R = ρ × L ÷ A, where ρ is the conductor's resistivity, L the length and A the cross-sectional area. Current through that resistance loses voltage by Ohm's law.
In DC and single-phase circuits the current travels out and back, so the drop is Vd = 2 × I × R × L using the one-way length. In a balanced three-phase circuit, Vd = √3 × I × R × L.
| Conductor | Resistivity at 20 °C | Temperature coefficient |
|---|---|---|
| Copper, annealed (100% IACS) | 1.7241 × 10⁻⁸ Ω·m | 0.00393 per °C |
| Aluminum 1350 (61% IACS) | 2.8264 × 10⁻⁸ Ω·m | 0.00403 per °C |
Worked example
A 30 m run of 12 AWG copper (3.31 mm²) at 75 °C carries 16 A on a 230 V single-phase circuit. Resistivity rises to about 2.10 × 10⁻⁸ Ω·m at 75 °C, giving 6.34 mΩ per meter. The drop is 2 × 16 × 0.00634 × 30 = 6.1 V, or 2.6% of 230 V.
Keeping the drop acceptable
Many designers keep branch circuits to about 3% and the total to about 5%. Some codes set firm limits; others treat these as recommendations. If the drop is too high, use a larger conductor, shorten the run, or raise the supply voltage.
This is a resistive estimate. For large conductors and long AC runs, conductor reactance and power factor also matter, and the applicable electrical code's tables should be used for the final design.