Power Factor Correction Calculator

This power factor correction calculator finds capacitor power and capacitance from real power and the present and desired power factor. Example: 30 kW from 0.70 to 0.95 needs about 20.7 kvar, or 3 × 138 µF in delta at 400 V.

Source: Wikipedia – Blindleistungskompensation. Updated: 2 October 2026.

Real power P

Power factor before

Power factor of the installation from a measurement or nameplate, between 0.05 and 1.

Power factor after (target)

Voltage (line to line)

Three-phase: voltage between two lines (e.g. 400 V); single-phase: e.g. 230 V.

Frequency

Capacitor connection

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Result

Result: Capacitor power Qc = 20.75 kvar, capacitance: 3 × 137.6 µF in delta
Before and after: Quantity Before After
Power factor 0.7 0.95
Phase angle φ 45.57° 18.19°
Real power P 30 kW 30 kW
Reactive power Q 30.61 kvar 9.861 kvar
Apparent power S 42.86 kVA 31.58 kVA
Current per line 61.86 A 45.58 A
Steps: Reactive power Q = P·tan φ: before 30·tan(arccos 0.7) = 30.61 kvar, after 30·tan(arccos 0.95) = 9.861 kvar.
Capacitor power Qc = Q₁ − Q₂ = 30.61 − 9.861 = 20.75 kvar.
Delta: per capacitor C = Qc / (3·ω·U²) = 20.75 kvar / (3·2π·50 Hz·(400 V)²) = 137.6 µF.
The current drops from 61.86 A to 45.58 A, which is 26.32 % less.; The calculation assumes sinusoidal currents. With harmonics (frequency converters, LED drivers) detuned capacitor banks may be needed. Selection and connection belong in the hands of a qualified electrician; capacitors store charge.

Quantity	Before	After
Power factor	0.7	0.95
Phase angle φ	45.57°	18.19°
Real power P	30 kW	30 kW
Reactive power Q	30.61 kvar	9.861 kvar
Apparent power S	42.86 kVA	31.58 kVA
Current per line	61.86 A	45.58 A

How it is calculated

Inductive loads such as motors and transformers need reactive power Q in addition to real power P, which burdens the grid; capacitors supply it locally and so lower the current. Correction raises the power factor, ideally to 0.95 to 1.

The calculation

Reactive power is Q = P·tan φ. To go from power factor cos φ₁ to cos φ₂ you need the capacitor power Qc = P·(tan φ₁ − tan φ₂). For 30 kW from 0.70 to 0.95, tan φ₁ = 1.020 and tan φ₂ = 0.329, so Qc ≈ 20.7 kvar. The capacitance follows: single-phase C = Qc / (ω·U²); three-phase in delta per capacitor C = Qc / (3·ω·U²); in star C = Qc / (ω·U²), with U the line-to-line voltage and ω = 2π·f. In the example that is 3 × 138 µF in delta at 400 V and 50 Hz.

Benefit

Apparent power falls from 42.9 kVA to 31.6 kVA and the current from 61.9 A to 45.6 A, about 26 percent less. That relieves cables and transformers and can avoid reactive-energy charges.

Do not over-correct to a power factor of 1: the installation becomes capacitive and resonances can occur.
Delta connection needs one third of the star capacitance per capacitor.
With harmonics choose detuned banks and respect the grid operator’s rules.

Frequently asked questions

How do you calculate the capacitor power for power factor correction?

Qc = P·(tan φ₁ − tan φ₂). For 30 kW from 0.70 to 0.95 that is about 20.7 kvar.

How large must the capacitor be?

Single-phase C = Qc / (ω·U²). In three-phase delta each of the three capacitors needs C = Qc / (3·ω·U²), in star C = Qc / (ω·U²), with U the line-to-line voltage.

Why improve the power factor?

A low power factor means more current for the same real power. That loads cables and transformers and can cause reactive-power charges from the grid operator.

Sources and legal basis

As of: 2 October 2026 · Editorial policy