What the Power Factor Correction Calculator does
This calculator sizes the capacitor bank needed to move a load from its present power factor to a target one. It gives the required kVAr, the capacitance per phase in both delta and star, the current and apparent power before and after, the transformer capacity the correction releases, and a suggested step arrangement for an automatic bank.
The arithmetic is short - kVAr equals kW times the difference of two tangents - but the decisions around it are not, so the result also flags the two things that spoil real installations: over-correction at light load, and capacitors resonating with harmonics. Everything runs in your browser.
How to use it
- Enter the load's real power in kW. This is the figure that does not change: correction reduces the current and the kVA, never the kW.
- Enter the present power factor. Take it from a power-quality logger or from the utility bill, averaged over a working period - an instantaneous reading at one moment will size the bank wrong.
- Enter the target. 0.95 is the threshold many tariffs use; 0.98 and above starts to risk leading power factor at light load.
- Set the voltage, the frequency and whether the supply is single or three phase. The capacitance depends on all three, although the kVAr does not.
- Choose how many steps the bank should have. A single fixed step suits a constant load; a varying load needs an automatic controller switching three to six steps.
- Read the capacitance per phase for the connection you plan to use, and check the findings before ordering anything.
Reading the results
The kVAr figure is the capacitive reactive power the bank must supply. It is what a capacitor is sold by, and it is specified at a stated voltage and frequency - a 50 kVAr 400 V capacitor produces far less than 50 kVAr at 380 V, because kVAr scales with the square of the voltage.
Released capacity is the kVA the correction frees up in every cable and transformer between the supply and the load. It is often the real justification: it can defer a transformer upgrade.
The loss reduction is the drop in I-squared-R heating in the supply cables, because the current falls in the ratio of the two power factors. It is a genuine energy saving, but a small one compared with the demand-charge saving.
The step table shows the power factor reached as each step switches in, which is what a controller's settings need to reflect.
Worked example: 100 kW at 0.75 power factor, corrected to 0.95
The tangent of arccos(0.75) is 0.8819 and the tangent of arccos(0.95) is 0.3287, so the bank must supply 100 x (0.8819 - 0.3287) = 55.32 kVAr.
Apparent power falls from 100 / 0.75 = 133.33 kVA to 100 / 0.95 = 105.26 kVA, releasing 28.07 kVA. On a 400 V three-phase supply the line current falls from 192.45 A to 151.93 A - a 21% reduction, which is exactly 1 - 0.75/0.95.
Cable losses fall with the square of the current, so they drop by 1 - (0.75/0.95) squared = 37.7%. If those cables were dissipating 3 kW, that is a little over 1.1 kW saved continuously.
For a delta-connected bank at 400 V and 50 Hz, the capacitance per phase is 55,323 / (3 x 314.16 x 400 squared) = 366.9 microfarads. A star bank would need three times as much - 1,100.6 microfarads per phase - at 231 V, which is why delta is the usual choice at low voltage.
If that plant idles overnight at 20 kW, a fixed 55 kVAr bank would over-correct badly: 20 kW with 55 kVAr of capacitance leading is a power factor of about 0.34 leading, which many tariffs penalise as hard as a poor lagging factor. Four steps of roughly 14 kVAr, switched by a controller, avoid that.
Formulas and scoring rules
- Required capacitive kVAr
Q_c = kW x (tan(arccos pf1) - tan(arccos pf2))The standard sizing equation. pf1 is the present power factor, pf2 the target.- Apparent power
kVA = kW / pfBefore and after; the kW is the same in both.- Capacitance, delta connection
C = Q_c / (3 x 2 pi f x V_line^2)Each capacitor sits across the full line voltage.- Capacitance, star connection
C = Q_c / (2 pi f x V_line^2)Three times the delta value, at 1/root-three of the voltage.- Single phase
C = Q_c / (2 pi f x V^2)- Current ratio
I2 / I1 = pf1 / pf2Because real power and voltage are unchanged.- Loss reduction
1 - (pf1 / pf2)^2Applies to the I-squared-R loss in the conductors carrying the corrected current, not to the load itself.
Fixed, automatic, or at the motor
There are three places to put correction and they solve different problems. A fixed capacitor at an individual motor corrects that motor and unloads the cable feeding it, and it switches with the motor so it can never over-correct - but it must be sized below the motor's own magnetising current or the motor can self-excite when it is switched off while still turning.
A fixed bank at the board is cheap and suits a load that never falls far. An automatic bank with a controller and contactor-switched steps suits everything else, and is what most commercial installations end up with. Contactors for capacitor duty have pre-charging resistors, because switching a discharged capacitor onto a live bus draws an enormous transient.
Harmonics change the answer
A capacitor's impedance falls as frequency rises, so it attracts harmonic current from every non-linear load on the system. Worse, the capacitance and the supply transformer's leakage inductance form a parallel resonant circuit, and if that resonance lands near the fifth or seventh harmonic the resulting current can be several times what anyone intended - enough to blow capacitor fuses repeatedly or destroy the capacitors.
Where there are variable-speed drives, rectifiers, UPS systems or significant LED lighting, the usual answer is a detuned bank: a reactor in series with each step, tuned to around 189 Hz or 210 Hz on a 50 Hz system, so the circuit is inductive at every harmonic frequency. That decision needs a harmonic survey, not a calculator.
Limitations: what the result does not prove
- It sizes capacitance for a displacement power factor. If the poor power factor is caused by harmonic distortion rather than by phase lag, capacitors will not fix it and may make it dramatically worse.
- It assumes a balanced three-phase load. Unbalanced loads need per-phase correction, which is unusual but not unheard of.
- The step arrangement is a starting suggestion, not a design. The steps a controller can actually use depend on the equipment ordered, and a bank is usually built from standard module sizes.
- Nothing here covers the switchgear, the fusing, the discharge resistors, the enclosure, the ventilation or the harmonic survey - and none of it is a compliance statement. A qualified engineer must verify the design against the applicable standard and local regulations.
Privacy: where your data goes
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Standards and sources
- IEC 60364-5-52 - Selection and erection of wiring systems - checked 19 Sep 2026
- IEC 60831-1 - Shunt power capacitors of the self-healing type for AC systems up to 1 kV
- IEC 61921 - Power capacitors: low-voltage power factor correction banks
- IEEE 519 - Harmonic control in electric power systems
Frequently asked questions
How many kVAr do I need to correct from 0.8 to 0.95?
Multiply the load kW by 0.421. That is tan(arccos 0.8) = 0.750 minus tan(arccos 0.95) = 0.329. So 100 kW needs 42.1 kVAr, 250 kW needs 105.3 kVAr, and so on - the requirement is proportional to the kW, not to the kVA.
Will power factor correction reduce my electricity bill?
It removes a kVA demand charge or a low-power-factor penalty where the tariff has one, which is usually the main saving, and it cuts the I-squared-R losses in your own cables by a few per cent of the load. It does not reduce the kWh the equipment consumes, so on a pure kWh tariff the payback is much slower.
What happens if I over-correct?
The power factor goes leading: the current now leads the voltage instead of lagging it. That raises the supply voltage at light load, can make a standby generator unstable or trip on reverse VAr, stresses the capacitors, and is penalised by many tariffs. It is the commonest fault with fixed banks on plant that idles overnight.
Should the capacitors be connected in star or delta?
Delta is usual at low voltage because it needs a third of the capacitance for the same kVAr, which makes the units smaller and cheaper. The capacitors must then be rated for the full line voltage. Star is used at higher voltages, where capacitor voltage rating becomes the limiting cost, and where an unbalanced-current protection scheme needs the star point.
Can I fit capacitors to a motor directly?
Yes, and it is a neat solution because the correction switches with the motor and cannot over-correct the installation. The rule is to keep the capacitor kVAr below about 90% of the motor's no-load magnetising kVAr, otherwise the motor can self-excite when it is switched off while still spinning and generate a damaging over-voltage. Never fit capacitors on the motor side of a variable-speed drive.
Why do capacitors need discharge resistors?
A disconnected capacitor holds its charge. Standards require it to fall to a safe voltage within a set time - commonly 75 V within three minutes for low-voltage equipment - so that someone working on the bank, or a contactor reclosing onto a charged capacitor, does not meet full voltage. Discharge resistors or reactors do that automatically.
Does correction help with voltage drop?
Usually yes, modestly. Voltage drop along a cable depends on both the resistive and the reactive components of the current, and correction removes the reactive part. On a long, heavily loaded feeder supplying a poor power factor, correcting at the load end can recover a useful fraction of a per cent - but it is not a substitute for a properly sized cable.
Is a power factor of 1.0 the ideal target?
Not in practice. Aiming at exactly unity leaves no margin, so any drop in load pushes the installation leading. Most designers target 0.95 to 0.98 lagging, which clears the usual tariff thresholds while keeping the system on the safe side of unity across the whole load range.
Last reviewed by the A2Z.Tools team against the sources listed above.
