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Power factor correction: capacitor kVAR needed

Calculate the capacitor bank in kVAR needed to raise a load's power factor, and the drop in apparent power and current.

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Results

Capacitor bank needed

55.3 kVAR

Size the bank to the next standard step

Reduction in current

21.1%

Apparent power before
133.3 kVA
Apparent power after
105.3 kVA
Current before
160.4 A
Current after
126.6 A
Reduction in cable losses (I²R)
37.7%

Paying for current that does no work

An AC motor needs a magnetic field to run, and building that field draws reactive power, measured in kVAR. It flows back and forth between the grid and the motor each cycle without doing any useful work, but it still travels through the cables and transformers. The power factor measures how much of the total, the apparent power in kVA, is real power in kW:

PF = kW ÷ kVA = cos φ

A plant with a power factor of 0.75 draws a third more current than it would at 1.0 for the same work.

How capacitors fix it

Capacitors draw reactive power in the opposite direction to motors, so they supply the motors’ magnetizing current locally instead of pulling it from the grid. The capacitor bank needed is:

kVAR = kW × (tan φ1 − tan φ2)

where φ1 and φ2 are the angles matching the present and target power factors.

Worked example

A workshop draws 100 kW at a power factor of 0.75 on a 480 V three-phase supply, and the goal is 0.95.

  • Reactive power before: 100 × tan(acos 0.75) = 88.2 kVAR.
  • Reactive power after: 100 × tan(acos 0.95) = 32.9 kVAR.
  • Capacitor bank needed: 55.3 kVAR, so a 60 kVAR standard bank.
  • Apparent power falls from 133.3 kVA to 105.3 kVA.
  • Line current falls from 160.4 A to 126.6 A, a 21.1% reduction.
  • Cable losses, which scale with the square of the current, fall by 37.7%.

Why it pays off

  • Utility charges: many US utilities add a power factor penalty for commercial accounts below about 0.85 to 0.90, or bill demand in kVA so that poor power factor raises the bill directly.
  • Capacity: the same transformer and cables can serve more load once the current drops.
  • Voltage: less current means less voltage drop at the end of long feeders.

Practical points

  • Correct close to large motors, or use an automatic bank at the main switchboard that adds steps as the load changes.
  • Do not overcorrect. A leading power factor at night, when motors are off, can raise voltage.
  • Variable-frequency drives and LED drivers create harmonics. Capacitors can resonate with them, so detuned banks with reactors are used in these installations.
  • Electrical work on capacitor banks must be done by a qualified electrician: capacitors stay charged after being switched off.

Frequently asked questions

What is power factor?

The ratio of real power (kW), which does useful work, to apparent power (kVA), which the wiring must carry. Motors, transformers and fluorescent ballasts draw reactive power that raises the current without doing work, which lowers the power factor.

How is the capacitor size calculated?

kVAR = kW × (tan φ1 − tan φ2), where φ1 and φ2 are the angles whose cosines are the current and target power factors. For 100 kW raised from 0.75 to 0.95, that is 100 × (0.882 − 0.329) = 55.3 kVAR.

Why correct the power factor?

Many utilities charge commercial customers a penalty when the power factor falls below about 0.90, or bill demand in kVA rather than kW. Correction also lowers the current, which frees transformer and cable capacity and cuts I²R losses.

Should I aim for a power factor of 1.0?

No. Most installations target 0.95 to 0.98. Overcorrecting to a leading power factor at light load can raise voltage and cause resonance with harmonics. Automatic capacitor banks switch steps in and out as the load changes.

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Results are estimates provided for information purposes. They do not replace professional advice.

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