Power factor correction: sizing capacitor kVAR
Calculate the reactive compensation needed to move from an existing to a target power factor and see the effect on kVA and current.
Why power factor matters
An inductive load such as an induction motor draws active power (kW), which does useful work, and reactive power (kVAR), which sets up magnetic fields. The supply must carry both, so apparent power is S = kW ÷ PF and current rises as power factor falls. Capacitors supply reactive power locally, so less reactive current has to flow through the cables and transformer.
The power factor correction calculator estimates the capacitor kVAR needed and the apparent power before and after. It is a screening estimate; the equipment selection needs supplier and qualified electrical review.
The formulas
With power factor PF = cos φ, tan φ = √(1 ÷ PF² − 1). The required reactive compensation is:
kVAR = kW × (tan φ1 − tan φ2)
where φ1 is the existing angle and φ2 the target. Apparent power is kW ÷ PF before and after. The calculator rejects a target that is not higher than the existing power factor, and both values must be above zero and no higher than 1.
Because kW is unchanged, the percentage reduction in apparent power is 1 − PF1 ÷ PF2. At constant voltage current falls by the same proportion.
The relationship behind this is the power triangle: S² = P² + Q², where P is kW, Q is kVAR and S is kVA. The existing reactive power is Q1 = kW × tan φ1; the target leaves Q2 = kW × tan φ2; the capacitor supplies the difference. Capacitors on a bus supply leading reactive power, so they cancel part of the lagging reactive power of the load without changing kW.
Correcting an individual motor at its terminals needs extra care in how the capacitor is switched with the motor; confirm the arrangement with the supplier rather than relying on this estimate.
Worked example 1: 250 kW plant load
Average active power is 250 kW at an existing PF of 0.78, with a target of 0.95 on a 400 V three-phase supply.
- tan φ1 = √(1 ÷ 0.78² − 1) = 0.8023.
- tan φ2 = √(1 ÷ 0.95² − 1) = 0.3287.
- kVAR = 250 × (0.8023 − 0.3287) = 118.40 kVAR.
- Apparent power before = 250 ÷ 0.78 = 320.51 kVA; after = 250 ÷ 0.95 = 263.16 kVA.
- Reduction = (1 − 263.16 ÷ 320.51) × 100 = 17.89 %.
These figures match the calculator. For current, use the kW to amps calculator: at 400 V, I = 250 × 1000 ÷ (√3 × 400 × 0.78) = 462.62 A before correction and 379.84 A at 0.95.
Worked example 2: smaller motor group, higher target
A 90 kW average load at PF 0.82, target 0.97, at 400 V. tan φ1 = 0.6980 and tan φ2 = 0.2506, so kVAR = 90 × 0.4474 = 40.26 kVAR. Apparent power falls from 109.76 kVA to 92.78 kVA, a 15.46 % reduction, and line current from 158.42 A to 133.92 A.
| Case | kW | PF before → after | kVAR | kVA before → after |
|---|---|---|---|---|
| Example 1 | 250 | 0.78 → 0.95 | 118.40 | 320.51 → 263.16 |
| Example 2 | 90 | 0.82 → 0.97 | 40.26 | 109.76 → 92.78 |
The second case shows that a higher target does not need a proportionally larger bank when the starting power factor is already fair: the last few points of PF cost progressively more kVAR per point gained.
Energy, cost and limits
Capacitors reduce current and apparent power, but active energy (kWh) is nearly unchanged. For the 250 kW load running 24 hours a day for 30 days, the energy and running cost calculator gives 250 × 24 × 30 = 180,000 kWh whether or not capacitors are installed. Any saving comes from a tariff that charges for kVA demand or reactive energy, from freed transformer and cable capacity, and from slightly lower losses. Check your own tariff terms; the calculator does not model them.
- Use average kW and PF over the period that matters, such as the billing interval, not a single instantaneous reading.
- Capacitor output scales with the square of applied voltage: a unit rated at 400 V on a 415 V supply delivers about (415 ÷ 400)² = 1.076 times its rated kVAR.
- Harmonics, resonance, switching steps and detuning reactors need separate engineering review.
- Loads that vary need stepped or automatic banks; a fixed bank sized for peak can over-correct at light load.
Common mistakes
- Entering kVA in the kW field.
- Using a nameplate or full-load power factor when the real average load is light and the power factor is lower.
- Choosing a target of 1.00, which gives no margin and risks a leading power factor when load drops.
- Entering percentages such as 85 instead of 0.85.
- Ignoring harmonic-producing loads such as drives when selecting capacitors.
Checks before you trust the result
- Take PF and kW from the same metering interval.
- Confirm whether the meter reports displacement power factor or total power factor including harmonics.
- Compare the kVAR result with the existing bank, if any, to avoid double correction.
- Ask the supplier for a harmonic and resonance check before purchase.
Run the calculator once with the lower and upper end of the plausible existing power factor. If the kVAR result changes a lot, the data are not yet good enough to buy a bank. As a cross-check, reactive power can be read directly as Q = √(kVA² − kW²): for example 1, √(320.51² − 250²) = 200.5 kVAR before and √(263.16² − 250²) = 82.2 kVAR after, a difference of 118.4 kVAR, which agrees with the calculator.
Frequently asked questions
Does correction reduce my kWh?
Not materially. It reduces reactive current and apparent power; kWh changes only by the small reduction in losses.
Can I just round the kVAR up?
Pick a standard rating near the calculated value and check the resulting power factor at light load to avoid leading.
Where should the capacitors go?
That is a design decision, at the load, at the motor control centre or at the main board, and is outside this calculator.
Record the data
Record the meter interval, average kW, existing and target PF, supply voltage, capacitor rating chosen, switching steps and the date installed. After commissioning, log the measured PF so you can confirm the result and notice later drift.