Voltage Drop, Single Phase

Also known as single phase VD · wire size voltage drop

Vd=2ρLIAV_{d} = \frac{2 \rho L I}{A}

Worked example: 20 A, 50 m, 4 mm² copper → 8.6 V drop — press Try an example to run it live, then adjust anything.

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Voltage Drop, Single Phase explained

AρIVdL

Every conductor is a resistor, and Ohm's law does the rest: the current travels out and back, so the drop is 2 × ρL/A × I. Work in SI and no mystery constant is needed — copper is about 1.72 × 10⁻⁸ Ω·m (1.72 μΩ·cm) at 20 °C, aluminium about 2.82 × 10⁻⁸. Feeding a 20 A load 50 m away on 4 mm² copper drops 2 × 1.72e−8 × 50 × 20 / 4e−6 = 8.6 V, unacceptable on a 230 V circuit and a clear call for larger cable.

North American practice hides the same physics in the constant K in Vd = 2KIL/cmil, where K ≈ 12.9 Ω·cmil/ft for copper and 21.2 for aluminium — those numbers are just ρ expressed in circular-mil-feet. Two traps: use the one-way run length (the 2 is already there), and remember K rises with temperature, which is why 12.9 is a 75 °C figure while cold-copper calculations use about 10.4.

Voltage Drop, Single Phase formula

Vd=2ρLIAV_{d} = \frac{2 \rho L I}{A}
Where
  • VdV_{d}= Voltage drop (V)
  • ρ\rho= Conductor resistivity (Ω·m)
  • LL= One-way run length (m)
  • II= Load current (A)
  • AA= Conductor area (m²)