What power factor correction actually changes
A load with inductance draws two components of current. One is in phase with the voltage and does real work — that is the kilowatts. The other lags the voltage by 90°, builds and collapses the magnetic fields in motor windings and transformer cores, and returns to the source every half cycle. It does no net work, but it is real current in real conductors, and it heats them exactly as working current does.
A capacitor draws current that leads the voltage by 90° — directly opposite to the inductive component. Connect one in parallel with the load and the two reactive currents circulate between the capacitor and the load instead of coming from the supply. The kilowatts are untouched; the supply simply stops carrying the reactive part. That is the whole mechanism, and it explains what correction does and does not do.
What it does: reduces line current, reduces I²R loss in the conductors and transformer upstream of the capacitors, releases capacity in cables and transformers, raises the voltage slightly at the load, and removes any utility charge based on kVA demand or on a power factor penalty. What it does not do: reduce the kilowatt-hours the load consumes. The motor still does the same work and still has the same losses, and any billing saving comes from the demand or power-factor terms of the tariff, not from the energy term.
The relationship between the three powers is the power triangle: real power kW along the base, reactive power kVAR vertical, apparent power kVA as the hypotenuse. Power factor is the cosine of the angle between kW and kVA. Correction shortens the vertical leg and therefore the hypotenuse while leaving the base alone — see the RLC impedance calculator for the same triangle expressed in ohms.
From power factor to kVAR to microfarads
Start with the angle. Power factor is cos φ, so φ = arccos(PF), and the reactive power at that angle is kVAR = kW × tan φ. The capacitor has to supply the difference between the reactive power the load draws now and the reactive power you are willing to leave on the supply:
kVAR = kW × (tan(arccos PF₁) − tan(arccos PF₂))
Nothing in that expression depends on voltage or frequency, which is why capacitor banks are sold by kVAR rather than by microfarads: the kVAR figure travels with the calculation while the capacitance depends on where you install it.
Converting to capacitance uses the reactive power a capacitor produces, Q = V²/X_C = 2πfCV², rearranged to C = Q/(2πfV²). For a single-phase load that is simply the whole kVAR at the supply voltage. For three-phase, divide the total kVAR by three to get the per-phase share, then apply the voltage each capacitor actually sees: the full line-to-line voltage in a delta bank, or line-to-neutral — line voltage divided by √3 — in a wye bank. Because capacitance goes as the inverse square of voltage, a wye bank needs exactly three times the capacitance of a delta bank for the same kVAR.
The current reduction has a pleasingly simple closed form. Line current is proportional to kVA, and kVA is kW/PF, so the ratio of the two currents is just PF₁/PF₂ and the reduction is 1 − PF₁/PF₂. Going from 0.80 to 0.95 always reduces the current by 1 − 0.80/0.95 = 15.79%, regardless of the load size or the voltage. Conductor loss follows the square, so the same change cuts I²R loss by 1 − (0.80/0.95)² = 29.09%.
Worked example: 100 kW at 0.75 corrected to 0.95
A 480 V, 60 Hz three-phase plant load draws 100 kW at a power factor of 0.75 lagging. The utility tariff penalises anything below 0.95. Size the capacitor bank.
- Angles. φ₁ = arccos(0.75) = 41.4096° and φ₂ = arccos(0.95) = 18.1949°.
- Tangents. tan 41.4096° = 0.881917 and tan 18.1949° = 0.328684.
- Reactive power before and after. kVAR₁ = 100 × 0.881917 = 88.1917 kVAR; kVAR₂ = 100 × 0.328684 = 32.8684 kVAR.
- Capacitor rating. 88.1917 − 32.8684 = 55.3233 kVAR. Specify the next standard bank size at or above that, typically 60 kVAR in steps.
- Apparent power. Before: 100/0.75 = 133.333 kVA. After: 100/0.95 = 105.263 kVA.
- Line current. Before: 133,333/(√3 × 480) = 133,333/831.384 = 160.38 A. After: 105,263/831.384 = 126.61 A.
- Reduction. (160.38 − 126.61)/160.38 = 21.05%, which matches 1 − 0.75/0.95 = 0.210526 exactly. Feeder I²R loss falls by 1 − (0.75/0.95)² = 37.67%.
- Capacitance, delta bank. Per phase Q = 55,323.3/3 = 18,441.1 VAR. ω = 2π × 60 = 376.9911 rad/s, and V² = 480² = 230,400. C = 18,441.1/(376.9911 × 230,400) = 18,441.1/86,858,750 = 2.1231 × 10⁻⁴ F = 212.31 µF per phase.
- Capacitance, wye bank. Each capacitor sees 480/√3 = 277.13 V, so V² = 76,800 and C = 18,441.1/(376.9911 × 76,800) = 636.92 µF per phase — exactly three times the delta figure, as the √3 in the voltage squares to 3.
Now weigh the last step. Going from 0.75 to 0.90 needs 100 × (0.881917 − 0.484322) = 39.76 kVAR; going the rest of the way to 0.95 needs another 15.56 kVAR, and going from 0.95 to unity needs a further 32.87 kVAR. The cost per point of power factor rises steeply as you approach unity, which is why tariff thresholds sit where they do and why almost nobody corrects to unity.
Choosing the target and placing the bank
Pick the target from your tariff, then add a small margin. Utilities commonly set a threshold and bill on kVA demand or apply a surcharge below it, but the number and the mechanism vary by supplier and region — read your own tariff document rather than assuming a figure. Correcting well beyond the threshold buys little: the kVAR needed per additional point rises sharply, and the current reduction per point falls.
Do not correct to unity on a varying load. A bank sized for unity at full load overcorrects when the load falls, producing a leading power factor. Leading power factor is penalised on many tariffs in the same way as lagging, it raises voltage at light load, and on a system with generation it can cause self-excitation problems. Fixed banks are best sized against the plant's minimum load, with automatic switched stages handling the rest.
Where you place the bank decides what benefit you get. At the service entrance, correction fixes the utility billing and nothing else — every conductor inside the building still carries the reactive current. At the motor control centre, it unloads the feeders upstream of that point. At the motor terminals, it unloads everything including the branch circuit, and it switches with the motor so it can never overcorrect. The trade is that a capacitor connected directly to a motor must be sized below the motor's magnetising current, or the motor can self-excite when it is switched off while still turning, generating dangerous voltages and causing damaging transients on reclosure.
Beware harmonics. A capacitor bank and the supply transformer's leakage inductance form a parallel resonant circuit, and if that resonance lands near a harmonic order the plant produces — the fifth and seventh from six-pulse drives are the usual suspects — the circulating current can be many times the fundamental and will destroy the capacitors. Where variable-frequency drives, rectifiers or arc loads are present, specify detuned banks with series reactors and confirm the resonant point. The resonant frequency calculator shows where an LC pair resonates.
kVAR required per kilowatt of load
| Present PF | to 0.85 | to 0.90 | to 0.95 | to 1.00 |
|---|---|---|---|---|
| 0.60 | 0.714 | 0.849 | 1.005 | 1.333 |
| 0.65 | 0.549 | 0.685 | 0.840 | 1.169 |
| 0.70 | 0.400 | 0.536 | 0.692 | 1.020 |
| 0.75 | 0.262 | 0.398 | 0.553 | 0.882 |
| 0.80 | 0.130 | 0.266 | 0.421 | 0.750 |
| 0.85 | 0.000 | 0.135 | 0.291 | 0.620 |
| 0.90 | — | 0.000 | 0.156 | 0.484 |
| 0.95 | — | — | 0.000 | 0.329 |
A dash marks a target below the present power factor, which needs no capacitors. Read across a row to see how steeply the requirement rises as the target approaches unity: from 0.80, reaching 0.95 costs 0.421 kVAR per kW and the final step to unity costs another 0.329.
What correction does not fix, and what it can break
- It does not reduce kilowatt-hours. The load does the same work with the same losses. Savings come from demand charges, power factor penalties and reduced conductor loss, not from the energy meter.
- It does not fix an oversized motor. A motor at 25% load has poor power factor because its magnetising current dominates. Capacitors mask the electrical symptom; the motor is still inefficient and still oversized.
- Correcting to unity backfires on a varying load. A fixed bank sized at full load overcorrects at light load, producing a leading power factor that many tariffs penalise and that raises system voltage.
- Capacitors and transformer inductance can resonate. Where harmonic sources are present, an undetuned bank can find itself at a resonant point and draw destructive circulating current. Confirm the resonance and specify reactors.
- kVAR scales with the square of voltage. A capacitor rated at 480 V installed on a 460 V system delivers (460/480)² = 91.8% of its nameplate kVAR. Size from the delivered figure, not the label.
- A capacitor switched with a motor must be smaller than the magnetising current. Otherwise the motor can self-excite while coasting, generating high voltages and causing severe transients if it is reconnected out of phase.
- Capacitor circuits have their own code rules. NEC 460.8(A) requires conductors rated at least 135% of the capacitor's rated current, 460.8(B) covers overcurrent protection, and 460.6 requires a discharge means — a charged capacitor bank is a serious hazard after isolation.
- Displacement power factor is not total power factor. With substantial harmonic distortion, the true power factor includes a distortion component that capacitors cannot correct and may worsen.
Key terms
- Reactive power (kVAR)
- The component of power that circulates between the source and the reactive elements of the load without doing net work. It is what capacitor banks supply locally instead of importing from the supply.
- Apparent power (kVA)
- The product of RMS voltage and RMS current, the hypotenuse of the power triangle. Transformers and conductors are rated in kVA because they are limited by current, not by useful work.
- Displacement power factor
- The cosine of the angle between the fundamental voltage and fundamental current. It is what capacitors correct.
- Detuned bank
- A capacitor bank with a series reactor chosen to move the system's parallel resonance below the lowest significant harmonic order, so the bank cannot resonate with the supply inductance at a frequency the plant produces.
Fixed banks, automatic banks and the alternatives
A fixed bank is the simplest and cheapest arrangement: one capacitor group permanently connected, sized against the plant's minimum reactive demand so it can never overcorrect. It suits a stable base load and needs no controls. An automatic bank switches stages in and out under the command of a power factor controller, holding the target across a varying load, and is the right answer where the reactive demand swings widely across a shift.
Individual correction at the motor is the most thorough option, because the reactive current never leaves the motor terminals and the capacitor is switched by the motor's own starter. Its constraint is self-excitation: the capacitor must be small enough that it cannot sustain the motor's magnetic field once the supply is removed. Manufacturers publish maximum capacitor kVAR by motor rating and speed for exactly this reason, and the limit falls as the number of poles rises.
Where harmonics are the dominant problem rather than displacement, capacitors are the wrong tool. A passive harmonic filter, a multi-pulse rectifier arrangement or an active harmonic filter addresses the distortion directly, and an active front-end drive avoids generating it. Adding plain capacitors to a plant heavy with six-pulse drives is the classic way to turn a distortion problem into a resonance failure.
Finally, keep the economics honest. The saving is the reduction in demand or penalty charges plus the value of the reduced conductor and transformer loss, and both should be computed from your own tariff and your own measured load rather than from a rule of thumb. The motor efficiency calculator measures where the reactive demand is coming from, the three-phase power calculator converts between the three powers, and the electricity bill calculator handles the energy side of the account.