Voltage Drop, Three Phase

Also known as three phase VD

Vd=3ρLIAV_{d} = \frac{\sqrt{3} \, \rho L I}{A}

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Learning zone

On a balanced three-phase circuit the three currents sum to zero, so no return conductor carries them home. Each line drops ρLI/A line-to-neutral, and converting that to a line-to-line figure multiplies by √3 — which is why three-phase drop is only 86.6% of the single-phase drop for the same current, length and cable. Running 50 A 100 m on 25 mm² copper gives 1.732 × 1.72e−8 × 100 × 50 / 25e−6 ≈ 5.96 V, about 1.2% on a 480 V system.

This resistive form is what most codes accept for typical building circuits, but it ignores reactance. On large conductors, long runs, or poor power factor, cable inductance adds its own drop and the true answer needs Vd = √3 I (R cos φ + X sin φ). For 4/0 and larger, or anything over a few hundred feet, use the impedance tables — the resistive answer can be optimistic by a third.

Voltage Drop, Three Phase
Vd=3ρLIAV_{d} = \frac{\sqrt{3} \, \rho L I}{A}
Where
  • VdV_{d}= Line-to-line voltage drop
  • ρ\rho= Conductor resistivity
  • LL= One-way run length
  • II= Line current
  • AA= Conductor area