Power Factor and PF Correction Capacitor Calculator

Q: How do I calculate the kVAR needed to correct power factor?

Qc = kW × (tan φ₁ − tan φ₂), where φ₁ = arccos(present PF) and φ₂ = arccos(target PF). For 10 kW from 0.8 to 0.95 that is 4.21 kVAR.

Q: What size capacitor in microfarads do I need?

C = Qc ÷ (ω V²) with Q in VAR, ω = 2πf and V the voltage across the capacitor. For a delta bank each capacitor sees the line voltage, so it needs a third of the capacitance of a star bank — but a higher voltage rating. The example above works out at 25.96 µF per capacitor in delta.

Your present power factor from kW and kVA — or from kW, volts and amps — then the kVAR and the microfarads of capacitance that lift it to the target, for a single-phase load or a star or delta bank at 50 or 60 Hz.

Last updated 22 Sep 2026

Power factor and correction

kW + kVA → PF, kVAR and µF

What do you know?

Real power (kW)From the meter, the nameplate or the bill: the power that does work.

Apparent power (kVA)Volts × amps ÷ 1000 (× √3 for three-phase). It can never be less than the kW.

System

Voltage (V)Line-to-line for three-phase (400 V, 415 V), line-to-neutral for single-phase (230 V, 120 V).

Line current (A)

Target power factor0.95 is a common utility target; going much past 0.98 costs a lot of kVAR for very little gain and risks over-correction at light load.

Supply frequency50 Hz in India, the UAE and most of the Gulf; 60 Hz in Saudi Arabia and North America. The same kVAR needs a smaller capacitor at 60 Hz.

Capacitor bank connectionDelta is the usual arrangement for LV banks: each capacitor sees the full line voltage and needs a third of the capacitance of a star bank. With a single-phase load this box is ignored: one capacitor goes across the supply.

The capacitor sits across the load. The motor still draws its own current, but the capacitor now supplies the reactive part locally, so the line upstream carries less. The dots show RMS sizes: the capacitor current leads the voltage by 90° and the motor current lags it, which is why they partly cancel rather than add. For a three-phase bank the drawing shows one phase of three; the whole bank is the three capacitors named below.

0.800Example

a 10 kW three-phase load drawing 12.5 kVA at 415 V, 50 Hz, corrected to 0.95 with a delta bank

Power factor and the correction capacitor

PF = P ÷ S = cos φ; Q_c = P (tan φ₁ − tan φ₂); C_{1-phase, star} = Q_c ÷ (ω V²), C_delta = Q_c ÷ (3 ω V_LL²), ω = 2π f

P, S: real power (kW) and apparent power (kVA)
φ₁, φ₂: phase angle before and after correction: φ = arccos(PF)
Q_c: reactive power the capacitors must supply, in kVAR
V: voltage across one capacitor: V_LL in delta, V_LL/√3 in star

Worked example

a 10 kW three-phase load drawing 12.5 kVA at 415 V, 50 Hz, corrected to 0.95 with a delta bank
PF now = 10 ÷ 12.5 = 0.800; φ₁ = 36.9°, Q₁ = √(12.5² − 10²) = 7.50 kVAR
tan φ₁ = 0.7500, tan φ₂ = 0.3287 at PF 0.95
Q_c = 10 × (0.7500 − 0.3287) = 4.21 kVAR
Delta bank: C = 4.21 kVAR ÷ (3 × 2π × 50 × 415²) = 25.96 µF per capacitor, three of them, each carrying 3.38 A
Line current falls from 17.39 A to 14.64 A, a 15.8% saving; a star bank would need 77.87 µF per capacitor instead

kVAR needed per kW of load (multiply by your kW)

tan φ₁ − tan φ₂, computed by this page’s own formula — the same quantity as the coefficient K in ABB’s Table 7.1. Multiply by the load in kW to get the kVAR of capacitors.
Power factor now	To 0.90	To 0.95	To 0.98	To 1.00
0.60	0.849	1.005	1.130	1.333
0.65	0.685	0.840	0.966	1.169
0.70	0.536	0.692	0.817	1.020
0.75	0.398	0.553	0.679	0.882
0.80	0.266	0.421	0.547	0.750
0.85	0.135	0.291	0.417	0.620
0.90	0.000	0.156	0.281	0.484

What power factor is, and how a capacitor fixes it

An induction motor, a transformer and a fluorescent ballast all need a magnetising current that flows out of the supply for a quarter of each cycle and back into it for the next quarter. It does no work, but it is real current: it heats the cable, it loads the transformer and the generator, and it drags the voltage down at the end of a long run. The power factor is the fraction of the current that is actually doing something — kW ÷ kVA.

The example. A 10 kW load drawing 12.5 kVA has a power factor of 0.80 and 7.5 kVAR of reactive power. To reach 0.95 it needs 4.21 kVAR of capacitors: at 415 V, 50 Hz in delta that is 26.0 µF in each of three capacitors, or 77.9 µF each in star. The line current falls from 17.4 A to 14.6 A — 16% less current for exactly the same work — and the cable, the breaker and the transformer all get that margin back.

Why capacitors. A capacitor’s current leads the voltage by 90°, a motor’s magnetising current lags it by 90°, so putting the two side by side lets the reactive current circulate locally between them instead of travelling back to the substation. Nothing about the motor changes: it still needs its magnetising current, and the kW on the meter is the same. What changes is the current in everything upstream of the capacitor.

Where to be careful. Do not over-correct: a fixed bank sized for full load will push the power factor leading when the motor idles, which raises the voltage and can be worse than the problem. Use stepped, automatically switched banks on a varying load, and never connect a capacitor directly across a motor that is fed through a variable-speed drive. Capacitors must be discharged before handling — they hold a lethal charge after the supply is off, which is why IEC 60831-1 requires every capacitor unit to carry a discharge device (clause 22) — and they need their own protection, rated for the high inrush current at switch-on. Where the load is electronic (rectifiers, LED drivers, computer supplies) the poor power factor is caused by harmonics, not by phase shift; capacitors can resonate with the supply inductance and make things worse, and the answer is a detuned filter chosen by a specialist.

For the kVA and kW behind the ratio, see the kVA to kW calculator; for the three-phase relationships, the three-phase power calculator; for the current a corrected load draws, the watts to amps calculator.

Frequently asked questions

How do I calculate the kVAR needed to correct power factor?

Q_c = kW × (tan φ₁ − tan φ₂), where φ₁ = arccos(present PF) and φ₂ = arccos(target PF). For 10 kW from 0.8 to 0.95 that is 4.21 kVAR.

What size capacitor in microfarads do I need?

C = Q_c ÷ (ω V²) with Q in VAR, ω = 2πf and V the voltage across the capacitor. For a delta bank each capacitor sees the line voltage, so it needs a third of the capacitance of a star bank — but a higher voltage rating. The example above works out at 25.96 µF per capacitor in delta.

Should the bank be star or delta?

Low-voltage banks are usually delta: a third of the capacitance for the same kVAR, and no dependence on a neutral. Star banks appear at higher voltages, where the lower voltage per capacitor matters more than the extra capacitance.

Will power factor correction reduce my bill?

On a domestic kWh tariff, no — the meter counts real power. On commercial and industrial tariffs with a kVA demand charge or a power-factor penalty, yes, and often quickly. It always reduces current, so it frees up cable, breaker and transformer capacity.

Does a capacitor help with LED or computer loads?

Usually not. Their poor power factor comes from a distorted current waveform, not a phase shift, and a capacitor cannot correct distortion; it can even resonate with the supply and amplify harmonics. Those loads need active power-factor correction inside the equipment, or a detuned filter.

Related calculators

References

ABB, Technical Application Papers No. 8: Power factor correction and harmonic filtering in electrical plants (1SDC007107G0202): Q_c = P (tan φ₁ − tan φ₂), and C_Y = 3 × C_Δ for a three-phase bank.
Hughes E, Hiley J, Brown K, Smith I M. Electrical and Electronic Technology, 12th ed. Pearson 2016: three-phase star and delta relationships, the power triangle and power-factor improvement.
IEC 60364-4-43:2008, Low-voltage electrical installations — Protection against overcurrent, clause 433.1: I_B ≤ I_n ≤ I_Z and I₂ ≤ 1.45 × I_Z.
IEC 60831-1, Shunt power capacitors of the self-healing type for a.c. systems having a rated voltage up to and including 1 000 V — Part 1: General, clause 22 (discharge device).