How to use it
- Type any two of Active power P, Reactive power Q, Apparent power S and Power factor cos φ, and leave the other two empty. The calculator fills in the rest.
- Type the Target power factor, for example 0.95, and choose the Supply: single-phase or three-phase.
- For three-phase, choose how the capacitors are wired, then type the Voltage and the Frequency of the network.
- Read Capacitor power to add and the Capacitance per capacitor; the working is in Working, step by step.
The power triangle
A load with a power factor below 1 draws more current than its useful work needs. Its apparent power S is the vector sum of the active power P, which does the work, and the reactive power Q, which the source and the load exchange every cycle.
S = √(P² + Q²)
cos φ = P ÷ S
Q = P × tan φ = S × sin φ
P active power (kW) Q reactive power (kvar) S apparent power (kVA)- An inductive (lagging) load, such as an induction motor, has a positive Q. A capacitor supplies that Q, so the source carries less current.
- The power factor is P ÷ S: at 1 all the current does useful work.
Sizing the capacitor
To raise the power factor from cos φ1 to cos φ2, the bank must supply Qc = P × (tan φ1 − tan φ2). A capacitor of reactance X supplies Q = V² ÷ X, and X = 1 ÷ (2πfC), so the capacitance follows from the reactive power.
Qc = P × (tan φ1 − tan φ2)
Single-phase, or star with V the line voltage: C = Qc ÷ (2πfV²)
Three-phase delta, V the line voltage: C = Qc ÷ (3 × 2πfV²)
C capacitance of one capacitor (F) f frequency (Hz) V voltage across it (V)- In delta each capacitor works across a full line voltage. For the same total kvar it is one third of the capacitance of a star-connected capacitor.
What the calculator does not cover
- Harmonics, and the resonance a capacitor can form with the supply. Some networks need a detuned or filtered bank.
- Leading (capacitive) loads, where Q has a negative sign. The calculator assumes the usual lagging load.
- Voltage rise, switching and protection of the bank. Check them against the rules of your network operator and of the capacitor maker.
Worked example: 80 kW at cos φ 0.7, corrected to 0.95 with a 400 V delta bank
The apparent power is S = 80 ÷ 0.7 = 114.29 kVA, and the reactive power is Q = 80 × tan(arccos 0.7) = 81.62 kvar. The phase angle is 45.57°.
To correct to 0.95, tan φ1 = 1.020 and tan φ2 = 0.329, so Qc = 80 × (1.020 − 0.329) = 55.32 kvar.
For a delta bank at 400 V and 50 Hz, C = 55,320 var ÷ (3 × 2π × 50 × 400²) = 366.9 µF for each of the three capacitors. In star at the same voltage each would need 1,101 µF.
Frequently asked questions
What is a power factor?
The power factor is cos φ, the ratio of active power to apparent power. A value of 1 means all the current does useful work. Inductive loads such as motors have a lower power factor, so they draw more current for the same useful power.
Why does correcting the power factor reduce the current?
The current is the apparent power divided by the voltage. Correction lowers the reactive part of that current while the active power stays the same, so cables and transformers carry less current for the same work.
Why does a delta bank need smaller capacitors?
In delta each capacitor sees the full line voltage, so it supplies more reactive power per microfarad. In star it sees the line voltage divided by √3, so each capacitor has to be three times larger for the same total kvar.
What if my power factor is already above the target?
The calculator then shows zero kvar and no capacitance, because the load already meets the target and needs no capacitor.
Can I enter the powers in watts or volt-amperes?
The boxes take kilowatts, kilovolt-amperes and kilovars. Type 80 for 80 kW, not 80,000 W.
Sources
- Wikipedia: Power factor (S = P + jQ, cos φ = P/S, Q = P tan φ)
- Wikipedia: Reactive power (Q = V²/X for a reactance)
- Wikipedia: Power factor correction (capacitors added to a load)
Figures and text last checked on Tuesday, November 10, 2026.