Busch 2006, Abschnitt 5.6 Drehstromleistung

Three-phase reactive power

As with real power, the √3 factor applies to line quantities; the sine of the phase angle enters instead of cos φ.

MINTSI
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Inputs

Inductive reactive power of the balanced three-phase load, in the same power unit as P, conventionally labelled kvar.

RMS voltage between two line conductors of the balanced system.

RMS current in each of the three equally loaded lines.

Displacement factor of the balanced load; reactive power vanishes at cos φ=1.

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Result

Select a target and calculate.

Calculation

Q = √3 · U · I · sinφ with sinφ = √(1−cos²φ)

As with real power, the √3 factor applies to line quantities; the sine of the phase angle enters instead of cos φ.

Understand the inputs
  • Reactive power QInductive reactive power of the balanced three-phase load, in the same power unit as P, conventionally labelled kvar.
  • Line voltage URMS voltage between two line conductors of the balanced system.
  • Line current IRMS current in each of the three equally loaded lines.
  • Power factor cos φDisplacement factor of the balanced load; reactive power vanishes at cos φ=1.
Example

400 V, 10 A and cos φ=0.8 give sinφ=0.6 and Q≈2.771 kvar (compare P≈5.543 kW).

Assumptions and limits

Balanced sinusoidal 50 Hz system without harmonics; only inductive (lagging) reactive power is reported, capacitive correction and imbalance are excluded.

Technical article

Understand Three-phase reactive power

This calculator complements the well-known three-phase real power with the reactive power of a balanced three-phase load, indicating how strongly an inductive load burdens the supply with reactive current before planning correction.

What does this quantity describe?

For a balanced three-phase system, reactive power follows Q=√3·U·I·sinφ, analogous to real power P=√3·U·I·cosφ. U and I are always line quantities. sinφ follows from the entered cos φ via sinφ=√(1−cos²φ) and is positive for 0≤φ<90°.

Formula and variables

Q = √3 · U · I · sinφ with sinφ = √(1−cos²φ)

  • Q = √3·U·I·sinφ
  • sinφ = √(1−cos²φ)
  • S = √3·U·I
Symbol / inputMeaning
Reactive power QInductive reactive power of the balanced three-phase load, in the same power unit as P, conventionally labelled kvar.
Line voltage URMS voltage between two line conductors of the balanced system.
Line current IRMS current in each of the three equally loaded lines.
Power factor cos φDisplacement factor of the balanced load; reactive power vanishes at cos φ=1.

Choose the inputs correctly

U is the RMS voltage between two line conductors, not the phase voltage. I is the RMS current in each of the three equally loaded lines. cos φ is the load's displacement factor from a nameplate or measurement.

How to use the calculator

Record line voltage and line current at the same operating point as the power factor. If cos φ is unknown, determine it first with a power meter or the power-factor calculator from measured real power.

Worked example

400 V, 10 A and cos φ=0.8 give sinφ=0.6 and Q≈2.771 kvar; the same load gives P≈5.543 kW and apparent power S=√3·U·I≈6.928 kVA.

Understand the result and units

The closer cos φ is to 1, the smaller Q becomes relative to P. High reactive power increases line current for the same real power, raising losses upstream in the network.

U is a voltage, I a current and cos φ dimensionless. Q is shown in the same power unit as real power, conventionally labelled kvar in practice.

Useful next calculation

The matching real power comes from three-phase AC power; for single-phase correction see power-factor correction capacitor.

Typical applications

Estimating network loading from motors, transformers and other inductive three-phase loads, and preparing reactive-power correction.

Assumptions, limits and common mistakes

Balanced sinusoidal 50 Hz systems only, without harmonics. Only inductive (lagging) reactive power is reported; capacitive loads, imbalance and distortion reactive power are excluded.

Common mistake: Do not substitute phase voltage for line voltage. Do not confuse cos φ with the total power factor under harmonics, which also includes a distortion component.

Frequently asked questions

What is “Three-phase reactive power” used for?

Estimating network loading from motors, transformers and other inductive three-phase loads, and preparing reactive-power correction.

Where do the input values come from?

U is the RMS voltage between two line conductors, not the phase voltage. I is the RMS current in each of the three equally loaded lines. cos φ is the load's displacement factor from a nameplate or measurement.

What does the result not cover?

Balanced sinusoidal 50 Hz systems only, without harmonics. Only inductive (lagging) reactive power is reported; capacitive loads, imbalance and distortion reactive power are excluded.

Sources, method and review

  • Rudolf Busch, Elektrotechnik und Elektronik für Maschinenbauer und Verfahrenstechniker, 4th ed. 2006, Abschnitt 5.6, Leistung im Drehstromsystem (local chapter PDF)

Our method, source hierarchy and automated checks are documented on the methodology page. Read the methodology

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Last updated
2026-09-16