Three-Phase Power Calculator

Choose whether to enter the current or the power. With the line voltage, the power factor and star or delta, you get the active power in kW, the reactive power in kvar, the apparent power in kVA, the phase values and the single-phase current for comparison.

Free Runs in your browser Checked Tuesday, November 10, 2026

Calculate

Power from the current you measure, or the line current needed for a power you know.

Connection

Star (wye) or delta. For a balanced load the line power is the same; only the phase values differ.

The voltage between two lines, for example 400 V in a European three-phase supply.

The current in one of the three lines, in amperes.

cos φ, between 0 and 1. Motors are often 0.8 to 0.9; resistive heaters are 1.

Result

Active power P — The real power, in kW.
Reactive power Q — The power that swings between the source and the load, in kvar.
Apparent power S — The total power the supply must provide, in kVA.
Line current —
Phase voltage; phase current —
Single-phase current for the same power (230 V) — The current the same active power would need on a single-phase 230 V supply.

Working, step by step

This tool runs in your browser. What you enter is not sent to us.

How to use it

  1. Choose Calculate: Power from current if you measure the current, or Current from power if you know the kilowatts.
  2. Type the Line voltage (V), for example 400, and the Power factor.
  3. Type the Line current (A) or the Active power (kW), depending on your choice.
  4. Read the Active power P, Reactive power Q and Apparent power S. Open Working, step by step for each formula.

The formulas

A balanced three-phase load has three equal phases. The apparent power is the square root of three times the line voltage times the line current. The active power is that times the power factor, and the reactive power is the rest, with the sine of the phase angle. The same line formulas apply to star and delta connections.

S = √3 × VL × IL          apparent power (VA)
P = S × PF = √3 × VL × IL × cos φ   active power (W)
Q = S × √(1 − PF²)     reactive power (var)

Star:  VPH = VL ÷ √3,  IPH = IL
Delta: VPH = VL,       IPH = IL ÷ √3
  • The line-to-line voltage is √3 times the line-to-neutral voltage in a star connection (Wikipedia, Three-phase electric power).
  • The single-phase comparison uses 230 V, the nominal single-phase voltage; the same active power needs a much larger current on a single-phase supply than on a three-phase one.
  • Current from power: IL = P ÷ (√3 × VL × PF).

What this calculator does not cover

  • Unbalanced loads, where the three phases carry different currents and neutral current flows.
  • Harmonics from electronic loads, which change the power factor measured by a meter.
  • Motor starting currents and the protective devices chosen for a circuit: those need an electrician.

Worked example: 400 V, 16 A, power factor 0.9

The apparent power is S = √3 × 400 V × 16 A = 11,085 VA, which is 11.09 kVA.

The active power is P = 11,085 VA × 0.9 = 9,977 W, or 9.977 kW, and the reactive power is Q = 11,085 VA × √(1 − 0.9²) = 4,832 var, or 4.832 kvar.

On a star connection each phase has 230.9 V and 16 A. The same 9.977 kW on single-phase 230 V would need 48.2 A.

Frequently asked questions

Does star or delta change the power?

For a balanced load the total power is the same, because the line voltage and line current give the same √3 relation in both. Only the phase voltage and phase current change.

What is the difference between kW, kVA and kvar?

kW is the power that does work, kVA is the total power the supply must provide and kvar is the power that swings back and forth. kW is kVA times the power factor.

What power factor should I use?

Use the figure on the motor's nameplate or the appliance's datasheet. Resistive heaters have a power factor of 1; many motors and drives are lower.

Why does a three-phase supply carry more power for the same current?

Three phases deliver power continuously with the √3 factor on the line voltage, so for the same line current the power is √3 times the single-phase power at the same voltage.

Is the single-phase comparison a real option?

It shows how much current the same power would need on a 230 V single-phase supply. Use it to see why larger loads use three phases; the supply you actually have decides which applies.

Sources

Figures and text last checked on Tuesday, November 10, 2026.

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