Electrical

Power Factor Correction

Why capacitors cut your electric bill

Power factor correction adds capacitors in parallel with inductive loads to cancel the lagging reactive current that motors and transformers draw, pulling line current back into phase with voltage. The real power doing work is unchanged — but the apparent power the utility must carry, and bill you for, shrinks dramatically.

  • What it cancelsLagging reactive power (kVAR)
  • Typical before / afterPF 0.75 → 0.95–0.98
  • Sizing lawQc = P (tan φ₁ − tan φ₂)
  • Bill impact20–30 % off the kVA demand charge
  • Phase relationCap current leads V by 90°; inductor lags by 90°
  • Watch outResonance with drive harmonics → detuned banks

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What power factor actually measures

Drive an electric heater from the mains and life is simple: the current rises and falls in perfect lock-step with the voltage, every watt the meter counts turns into heat, and there is nothing to correct. Connect an induction motor instead and something subtler happens. A motor has to build a rotating magnetic field before it can produce any torque, and building that field means pushing current into a winding's inductance. An inductor stores energy in its magnetic field for a quarter of each cycle and then hands it straight back to the supply during the next quarter — so the current it draws lags the voltage by up to a quarter-cycle, ninety degrees.

That lag is the whole story. The instantaneous power — voltage multiplied by current at every moment — now spends part of each cycle negative, meaning energy is flowing back from the load into the supply. Averaged over a cycle, that sloshing energy nets to zero useful work, yet the current carrying it is absolutely real and absolutely present in every cable. Power factor is the ratio of the average real power that does work to the apparent power the cables actually carry. For a clean sinusoidal supply it equals the cosine of the phase angle φ between voltage and current:

power factor  =  P (real, kW)  /  S (apparent, kVA)  =  cos φ

A purely resistive load has φ = 0 and a power factor of 1.0 — perfect. A lightly loaded induction motor can sit at φ = 50° and a power factor of 0.64, meaning that for every kilowatt of mechanical work, the supply is shuttling well over a kilovar of reactive power back and forth that never leaves the wires as anything but heat.

The power triangle

The three quantities — real, reactive, and apparent power — form a right triangle that is the single most useful picture in this whole subject. Real power P (kilowatts) lies along the horizontal. Reactive power Q (kilovolt-amperes reactive, kVAR) lies vertically. Apparent power S (kilovolt-amperes, kVA) is the hypotenuse, and the angle between P and S is the phase angle φ:

S² = P² + Q²            (apparent² = real² + reactive²)
cos φ = P / S           (power factor)
tan φ = Q / P           (the key to sizing correction)

Correction works by attacking the vertical side. A capacitor's current leads the voltage by ninety degrees — the exact opposite of an inductor — so a capacitor generates reactive power that points down the triangle, subtracting from the inductive load's upward Q. Add just enough capacitive kVAR and the vertical side shrinks. As Q falls, the hypotenuse S rotates down toward the horizontal real-power side, φ collapses, and cos φ climbs toward one. The horizontal side P — the actual work — never moves. You have not made the motor do less; you have stopped forcing the supply to carry reactive current it does not need to carry.

Sizing a capacitor bank — the worked numbers

The governing equation falls straight out of the power triangle. To move a load of real power P from an existing power factor cos φ₁ to a target cos φ₂, the capacitive reactive power required is:

Qc = P × (tan φ₁ − tan φ₂)        [kVAR]

Take a concrete factory feeder: 200 kW of mixed motor load measured at a power factor of 0.75 lagging, which we want corrected to 0.98.

Existing:   cos φ₁ = 0.75   →  φ₁ = 41.4°   →  tan φ₁ = 0.882
Target:     cos φ₂ = 0.98   →  φ₂ = 11.5°   →  tan φ₂ = 0.203

Reactive now:        Q₁ = 200 × 0.882 = 176 kVAR
Reactive wanted:     Q₂ = 200 × 0.203 =  41 kVAR
Capacitor must supply:  Qc = 176 − 41 = 135 kVAR

Apparent power before:  S₁ = 200 / 0.75 = 267 kVA
Apparent power after:   S₂ = 200 / 0.98 = 204 kVA
Reduction in kVA demand:   63 kVA  (24 % less current in every cable)

From the kVAR you back out the physical capacitance. Reactive power in a capacitor is Qc = V² ω C, so on a 400 V, 50 Hz three-phase bus the per-phase capacitance for 135 kVAR works out near 900 µF in delta — a stack of dry-film or oil-filled power capacitors rated for continuous duty and the inrush of switching. The crucial engineering judgement is the target: you do not chase exactly unity. A fixed bank sized to hit 1.0 at full load becomes over-corrected (leading) the moment load drops, so the convention is to aim for 0.95–0.98 lagging and leave a small inductive margin.

Why correcting the phase angle saves real money

Every conductor on the site, plus the supply transformer and the utility's distribution network, has to be rated for the apparent power S, because the full vector current — real plus reactive — flows through all of it. The reactive current does no work but it still obeys ohmic heating: copper losses scale with the square of current, so reducing line current 24 % cuts those I²R losses by about 42 %. Three things follow directly:

  • Recovered capacity. A 250 kVA transformer feeding the 200 kW load at 0.75 was 107 % loaded — overloaded. Correct to 0.98 and the same load draws 204 kVA, leaving 46 kVA of headroom for new machines with no new transformer.
  • Lower losses and higher voltage. Less current means smaller voltage drop along the feeders, so motors at the end of a long run see a healthier terminal voltage and run cooler.
  • A smaller bill. Industrial tariffs charge a maximum-demand component in kVA (or apply an explicit power-factor penalty, or meter reactive energy in kVARh). All three reward a higher power factor. Cutting the billed demand from 267 to 204 kVA on a tariff of, say, 12 currency units per kVA per month removes about 750 units a month — paying back a capacitor bank in well under two years.

Methods of correction, from a single capacitor to a STATCOM

  • Fixed correction. A single capacitor bolted to a motor's terminals, sized to cancel that motor's magnetising current, switched by the motor's own contactor. Cannot over-correct because it is matched to one known load and is gone whenever the motor is. The default for large fixed-speed pumps and fans.
  • Automatic (bulk) correction. A switched capacitor bank at the main board, split into steps — say 6 × 25 kVAR — energised and shed by a power-factor controller through contactors as the measured plant power factor drifts. Tracks a varying load, but in coarse discrete jumps and with finite switching life.
  • Detuned banks. Each capacitor step has a small series reactor that pulls the bank's resonant frequency below the 5th harmonic (to roughly 189 Hz on a 50 Hz system). Standard wherever variable-speed drives or rectifiers are present, because it makes harmonic resonance physically impossible.
  • Synchronous condenser. An unloaded synchronous machine run over-excited so it generates reactive power smoothly and continuously, and provides inertia. Once the workhorse of large substations; making a comeback on grids with lots of inverter-based renewables.
  • Active power factor correction (PFC). Power electronics that shape the current. A boost-converter PFC stage in a switch-mode supply forces input current to track the voltage sinusoidally — mandated by IEC 61000-3-2 for most equipment over 75 W. At grid scale a STATCOM or active harmonic filter injects a continuously variable correction current within a few milliseconds and corrects distortion, not just phase.

Capacitor banks versus active correction

PropertySwitched capacitor bankDetuned capacitor bankActive filter / STATCOM
Corrects displacement PFYesYesYes
Corrects harmonics (distortion PF)No — can worsen themMitigates resonance onlyYes — cancels them actively
Response speedSeconds (contactor steps)SecondsMilliseconds, stepless
ResolutionCoarse (discrete kVAR steps)Coarse stepsContinuous
Can over-correct (go leading)Yes, at light loadYes, at light loadNo — closed-loop controlled
Capital cost per kVARLowModerateHigh (3–5× capacitors)
Typical useStable motor plantDrives + motors mixedWelding, EAF, drive-heavy, weak grid

Failure modes and what goes wrong

  • Harmonic resonance. A capacitor bank and the supply transformer's leakage inductance form a parallel LC circuit with a natural frequency. If that frequency lands near a harmonic generated by drives or rectifiers — usually the 5th (250/300 Hz) or 7th — the bank amplifies that harmonic current several-fold, overheating and bursting capacitor cans. The cure is the detuned reactor that shifts resonance safely below the 5th harmonic.
  • Over-correction at light load. A fixed bank sized for full load keeps pumping its full kVAR into a near-idle system at night. The surplus leading current pushes bus voltage above nominal, can trip over-voltage protection, and earns a leading-power-factor penalty from many utilities.
  • Switching transients. Energising a capacitor into a live bus draws a large inrush spike and a high-frequency transient that can stress the capacitor, weld contactor tips, and disturb sensitive electronics. Banks use inrush-limiting reactors or zero-crossing (point-on-wave) switching contactors.
  • Capacitor ageing and dry-out. Dielectric loss heats the film; over years capacitance drifts down and internal pressure rises. Self-healing metallised-film units clear small faults automatically but lose capacitance each time; a bank quietly under-corrects long before it fails outright. Periodic kVAR measurement catches it.
  • Treating distortion as displacement. Throwing capacitors at a site whose poor power factor is mostly harmonic distortion (rectifier and LED-driver dominated) does little for true power factor and can trigger resonance. The fix is filters or active correction, not more capacitance.

Displacement versus distortion — the true power factor

Everything above the harmonics section assumes a clean sinusoid, where power factor is simply cos φ — the displacement power factor. Modern loads break that assumption. A diode-bridge rectifier in a laptop charger or a variable-frequency drive draws current in sharp non-sinusoidal pulses near the voltage peaks. Even if those pulses are perfectly in phase with the voltage (zero displacement), the waveform is so distorted that a large fraction of the current does no average work. The honest figure is the true power factor:

true PF  =  displacement PF × distortion PF
distortion PF = 1 / √(1 + THD_i²)

where THDi is the total harmonic distortion of the current. An uncorrected diode-bridge supply can have a displacement factor near 1.0 yet a distortion factor of only 0.6 because its current THD exceeds 100 %, leaving a true power factor around 0.6. No capacitor fixes this — the reactive power is not at the fundamental frequency. That is precisely why regulators mandate active boost-PFC front ends in equipment and why heavy industrial sites pair detuned capacitor banks with active harmonic filters: the capacitors handle the fundamental-frequency phase shift, the active filter handles the distortion, and only together do they push the true power factor above 0.95.

Frequently asked questions

What is power factor correction and how does it work?

It is the addition of capacitance in parallel with inductive loads — motors, transformers, ballasts — to cancel the reactive power they draw. An inductor's current lags the voltage by up to 90°; a capacitor's current leads it by 90°, the mirror image. Wire a correctly sized capacitor across the same bus and its leading reactive current supplies the motor's lagging reactive demand locally instead of dragging it from the generator. The reactive component of line current shrinks, the phase angle collapses, and the power factor — the cosine of that angle — climbs from a typical 0.75–0.85 toward 0.95 or higher. The real power doing work is unchanged.

Why does a low power factor cost money?

Cables, transformers, and switchgear must be sized for apparent power (kVA), the vector sum of real power (kW) and reactive power (kVAR), because the full current flows through every conductor. A 100 kW load at 0.7 draws 143 kVA; at 0.95 it draws only 105 kVA. The extra reactive current heats cables (losses scale with current squared), eats transformer headroom, and does no work. Utilities recover that cost through a kVA demand charge, a kVARh reactive-energy charge, or a power-factor penalty. Correcting from 0.75 to 0.98 routinely cuts the demand part of an industrial bill by 20–30 %.

How do you size a power factor correction capacitor?

From the load's real power P and its power factor cos φ₁, pick a target cos φ₂. The capacitive reactive power needed is Qc = P × (tan φ₁ − tan φ₂) in kVAR. Example: 200 kW at 0.75 (tan φ₁ = 0.882) draws 176 kVAR; to reach 0.98 (tan φ₂ = 0.203) it needs 41 kVAR, so the bank supplies 176 − 41 = 135 kVAR. Capacitance follows from Qc = V²ωC. You aim for 0.95–0.98, not exactly unity, because a fixed bank that is correct at full load goes leading at light load.

What is the difference between fixed, automatic, and active correction?

Fixed correction is one capacitor at a motor's terminals, switched with the motor, matched to a single known load so it never over-corrects. Automatic correction is a switched bank at the main board, split into steps that a power-factor relay energises and sheds as load varies. Active correction uses power electronics — a boost-converter PFC stage forces input current to follow the voltage sinusoidally (IEC 61000-3-2), while a STATCOM or active filter injects a continuously adjustable correction current in milliseconds and also fixes distortion that capacitors cannot.

Can power factor be too high or go leading?

Yes. A fixed bank sized for full load keeps supplying full kVAR when most motors are off at night; the surplus capacitive current leads the voltage, raises bus voltage above nominal, and earns a leading-PF penalty. Capacitors also form an LC circuit with the transformer's inductance — if that resonance lands near a drive harmonic (typically the 5th or 7th) the bank amplifies it and the capacitors burst. The fixes are automatic switched banks and detuned (reactor-connected) banks tuned below the lowest harmonic.

Does power factor correction reduce harmonics?

Plain capacitors improve only the displacement power factor at the fundamental frequency; they do nothing for distortion from rectifiers and drives, and can make harmonics worse by resonating with the supply. To address harmonics you need detuned banks (a series reactor tuned below the 5th harmonic), passive tuned filters, or an active harmonic filter that injects equal-and-opposite harmonic current. True power factor is displacement × distortion, so on a drive-heavy site ignoring distortion leaves true PF below 0.9 even with the capacitors in.