Electrical

Power factor correction calculator

Work out the capacitor bank (kVAr) needed to raise a load's power factor to a target, the capacitance per phase for delta or star connection, and how much the apparent power and line current fall as a result.

Load and target
Average kW over the period you are correcting for; from the bill (kWh ÷ hours) or a meter.
From the bill (kWh ÷ kVAh) or a power analyser.
0.95 avoids penalties with margin; do not aim for 1.0 with fixed capacitors.
Line-to-line for three-phase, line-to-neutral for single-phase.

Correction needed

Enter your values and press Calculate.

What the calculator does

It takes the load's real power and its present power factor, works out the reactive power the load draws, and reports how many kVAr of capacitors must be added to bring the power factor up to the target. It also gives the capacitance per phase for a three-phase bank in delta or star (or a single-phase capacitor), the apparent power before and after, and the reduction in line current, which is where the savings in cable loss and switchgear loading come from.

Formula

Qc = P × (tan φ1 − tan φ2) φ = arccos(PF)

Three-phase, delta: Cphase = Qc ÷ (3 × 2πf × VLL²)
Three-phase, star: Cphase = Qc ÷ (3 × 2πf × (VLL ÷ √3)²) = Qc ÷ (2πf × VLL²)
Single-phase: C = Qc ÷ (2πf × V²)

S = P ÷ PF I = S ÷ (√3 × VLL) I = S ÷ V

where P is real power in watts, Qc the capacitive reactive power to add in var, f the supply frequency and C the capacitance in farads (the calculator reports microfarads). A star bank needs three times the capacitance of a delta bank for the same kVAr because each capacitor sees only the phase voltage, which is why low-voltage banks are almost always delta.

Worked example

A workshop averages 100 kW at a power factor of 0.75 on a 400 V, 50 Hz three-phase supply and wants 0.95.

  1. tan(arccos 0.75) = 0.8819; tan(arccos 0.95) = 0.3287.
  2. Qc = 100 × (0.8819 − 0.3287) = 55.3 kVAr.
  3. Apparent power falls from 100 ÷ 0.75 = 133.3 kVA to 100 ÷ 0.95 = 105.3 kVA; line current from 192.5 A to 151.9 A, a 21 % reduction.
  4. Delta bank: C = 55 320 ÷ (3 × 2π × 50 × 400²) = 367 µF per phase. In star it would be 1 101 µF per phase.

In practice you would buy a 60 kVAr automatic bank made of 10 and 20 kVAr steps rather than a single fixed 55 kVAr unit, so that the correction follows the load through the day.

Multiplier table: kVAr per kW to reach the target

Present PFTarget 0.90Target 0.95Target 0.98
0.600.8491.0051.130
0.650.6850.8400.966
0.700.5360.6920.817
0.750.3980.5530.679
0.800.2660.4210.547
0.850.1350.2910.417
0.9000.1560.281

tan φ1 − tan φ2 for common combinations; multiply by the load in kW. The calculator uses the exact value for any pair.

Assumptions and limitations

  • Displacement power factor only. Capacitors correct the phase lag of inductive loads. They do nothing for the distortion caused by drives, rectifiers and switch-mode supplies, and can resonate with them; where more than about a fifth of the load is electronic, the bank needs detuning reactors or an active filter, and a power quality survey comes first.
  • Average load. A fixed bank sized for the average will over-correct at light load and push the power factor leading, which utilities may also penalise and which raises the voltage. Use an automatic bank with steps for varying loads, or size fixed capacitors for the minimum load only.
  • Nominal capacitor rating. Capacitor kVAr is rated at nominal voltage; at a lower voltage the output falls with the square of the voltage. Buy banks rated for the actual system voltage (440 V rated units on a 400 V system give only 83 % of their nameplate kVAr).
  • Individual motor correction is not covered. A capacitor connected directly across a motor must be limited to about 90 % of the motor's magnetising kVAr to avoid self-excitation when the motor is switched off; use the motor manufacturer's table.
  • Not a harmonic, resonance or switching-transient study. Those need measured data.

Frequently asked questions

Where do I get the present power factor?

Industrial and commercial bills in Pakistan print kWh and kVAh (or kVArh) for the month; power factor = kWh ÷ kVAh, or cos(arctan(kVArh ÷ kWh)). Otherwise a clamp power analyser at the incomer for a working day gives a more useful profile than a single reading.

Why not correct to 1.0?

The last few points cost the most capacitance for the least benefit, and a fixed bank that reaches 1.0 at full load will be leading at part load. 0.95 to 0.98 is the usual target.

Does correction reduce my kWh?

Only slightly, through lower I²R loss in your own cables and transformers between the bank and the loads. The main savings are the removal of the power-factor penalty, lower demand (kVA) charges where they apply, and capacity freed in transformers, cables and switchgear.

Where should the bank be connected?

For penalty avoidance, at the main incomer downstream of the utility meter. For freeing capacity in your own distribution, closer to the large inductive loads. Many plants do both: a central automatic bank plus fixed capacitors at big motors.

References

  • IEC 60831-1:2014, Shunt power capacitors of the self-healing type for a.c. systems having a rated voltage up to and including 1 000 V — ratings, tolerances and test conditions for LV capacitor units
  • IEEE Std 1459-2010, Definitions for the Measurement of Electric Power Quantities — displacement versus distortion power factor
  • IEEE Std 519-2022, Harmonic Control in Electric Power Systems — harmonic limits relevant when capacitors are added to systems with non-linear loads
  • NEPRA, Terms and conditions of the consumer-end tariff for distribution companies (power factor penalty clause) — the 90 % threshold applied to industrial and commercial consumers in Pakistan; check the current schedule

Last reviewed 2026-09-20.