Three-Phase Voltage Drop Calculator

Calculate balanced three-phase voltage drop from current or electrical input kW, with power factor, cable resistance, reactance and conductor temperature. The result is an engineering estimate under the assumptions below.

Calculate three-phase voltage drop

Use a decimal point or comma. A single separator is treated as decimal: write 1000 for one thousand.

Use line-to-line voltage for balanced three-phase AC.
Source to load; do not double the distance.
The entered length is interpreted in the selected unit.
Current in one line conductor.
Total three-phase electrical input, not motor shaft output.
Greater than 0 and no greater than 1.
Select an area or enter a custom value. This does not select ampacity.
Per conductor, for the selected material. No extra material multiplier is applied.
Replace example reactance with manufacturer data.
Zero is allowed for a resistance-only approximation.
Conductor temperature, not ambient temperature. Supported calculation range: −20 to 120°C.
Your comparison target; this is not a universal code requirement.

Upload a datasheet or diagram photo to suggest conductor and installation values.

Verify suggested values against your source document before relying on a calculation.

Compare cable scenarios

Calculate an option, save it, then change the inputs and save another. Compare up to three snapshots with their assumptions.

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Cable planning: choose your next check

Use ampacity and voltage drop as separate checks. Choose the tool for the information you have; review each model before using its result.

Three-phase voltage drop formula

For a balanced sinusoidal load with lagging power factor, the approximate line-to-line voltage drop is ΔV = √3 × I × L × (R × PF + X × √(1 − PF²)). Use line current I in A, one-way length L in km, and per-conductor R and X in Ω/km. R must describe the operating temperature. The ELEK voltage-drop guide explains this longitudinal AC approximation.

Percentage drop = 100 × ΔV / VLL. Enter 400 V for a 230/400 V three-phase system, rather than the 230 V line-to-neutral value. Do not multiply line current by three or double the route length.

Current input versus kW input

Current mode uses the line current directly. Power mode estimates current from I = 1000 × P / (√3 × VLL × PF), where P is total three-phase electrical input power in kW. Power factor remains editable in both modes.

A motor nameplate may specify shaft output power. For an illustrative 30 kW output and efficiency 0.90, electrical input is 30 / 0.90 = 33.333 kW. At 400 V and PF 0.85, that corresponds to approximately 56.60 A. Use actual nameplate or measured line current when available. The kW mode uses the entered voltage to estimate current; it does not iteratively solve a constant-power load after voltage drop.

Using cable resistance and reactance

Select a reference cross-section or choose “Custom R20”. Custom resistance must be the resistance of one conductor of the selected material at 20°C, in Ω/km. It is used directly: selecting aluminum does not multiply your supplied R20 by a copper-to-aluminum ratio.

Temperature correction is R(T) = R20 × [1 + α × (T − 20)]. This calculator uses approximate α values of 0.00393/°C for copper and 0.00403/°C for aluminum. If your datasheet gives resistance only at 75°C or 90°C, do not enter that number as R20 and apply a second correction. Obtain the 20°C value or use a method appropriate to the supplied data.

Copper presets are reference R20 values; aluminum presets use an ideal estimate of 28.26 / area in mm². Neither substitutes for the chosen cable’s datasheet. The reactance presets 0.06, 0.08 and 0.12 Ω/km are example inputs. Choose Custom to supply an actual value, including zero for a resistance-only model. Reactance depends on cable construction, spacing and frequency.

Worked example: balanced feeder

Assume 400 V line-to-line, 100 A, a 100 m one-way route, PF 0.8 lagging, R = 0.524 Ω/km at 20°C, and X = 0.08 Ω/km. These are stated calculation inputs, not a cable-sizing approval.

  1. Convert length: L = 100 / 1000 = 0.1 km.
  2. Find sinφ: √(1 − 0.8²) = 0.6.
  3. Combine components: 0.524 × 0.8 + 0.08 × 0.6 = 0.4672 Ω/km.
  4. Calculate ΔV = √3 × 100 × 0.1 × 0.4672 = 8.09214 V.
  5. Percentage = 100 × 8.09214 / 400 = 2.02304%; approximate load voltage = 391.91 V.

Three-phase conductor losses

The total resistive loss for all three identical phase conductors is Ploss = 3 × I² × R × L. In the example, 3 × 100² × 0.524 × 0.1 = 1,572 W. A result of 524 W would represent one phase conductor. This figure excludes connections, neutral or sheath losses and upstream equipment.

Variant of the exampleDropTotal phase-conductor loss
100 m at 20°C8.09 V (2.02%)1,572.00 W
50 m at 20°C4.05 V (1.01%)786.00 W
100 m at 75°C; same R20 and X9.66 V (2.42%)1,911.79 W

Frequently asked questions

What does the voltage-drop target mean?

It compares the calculated percentage with your project setting. The default 3% does not certify NEC or IEC compliance. Ampacity, protective devices, terminal ratings and equipment operating voltage require separate checks.

Does this model cover motor starting or unbalanced loads?

The displayed calculation is a balanced steady-load estimate. Starting current, starting power factor, supply impedance, harmonics and unbalance can require a different circuit analysis. A large calculated drop also calls for a more complete model.

Can I calculate single-phase or DC voltage drop here?

Use the DC and single-phase voltage drop calculator for those circuits. The √3 factor on this page applies to balanced three-phase line-to-line results.

Related: three-phase current calculator · kW to amps.