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Active, reactive and apparent power from current

Power of a three-phase or single-phase load from the voltage, current and power factor — for example, from clamp meter and panel voltmeter readings.

Formula
3~: S = √3 · U · I1~: S = U · IP = S · cos φQ = S · sin φ = P · tg φ

The load is balanced, with sinusoidal current and voltage. For an unbalanced load, compute each phase and add them up. At the input of a variable frequency drive the current is distorted by harmonics: use the true power factor λ instead of cos φ, from the drive data or a power quality analyser.

Source: Circuit theory: power of a balanced three-phase circuit

Inputs

You can change a field's unit: the value is converted to the formula's units automatically.

m
U

Three-phase supply: line-to-line voltage, 380 V, 6 or 10 kV; single-phase: phase voltage, 220 V.

I
cos φ

Induction motors at rated load 0.80–0.90, lightly loaded 0.3–0.6; heaters 1.

Unit converter for this formulaVoltage · Current · Power · Apparent power · Reactive power
  • V1
  • kV0.001
All units
Result
PActive power
Fill in all fields
  • SApparent power
  • QReactive power
  • tg φReactive power factor tan φ

More in Power and current

Results are engineering estimates from standard formulas; for design decisions check them against the codes, project documents and specialists' calculations. The formulas carried over from the original set are unchanged, and their errors are described in the notes.