Power-factor correction calculator

Capacitor kVAR and µF per phase to raise a lagging power factor, and the line current before and after.

Load
kW
System
V

Result

Capacitor bank
276.6kVAR
Per phase, delta bank
1,062µF
Per phase, wye bank
3,185µF
Apparent power, before
666.7kVA
Apparent power, after
526.3kVA
Line current, before
801.9A
Line current, after
633.1A
Phasor diagram: line current of 801.9 amperes lagging the voltage by 41.41 degrees before correction, and 633.1 amperes lagging by 18.19 degrees afterV = 962.25∠0.0°. I₁ = 801.88∠−41.4°. I₂ = 633.06∠−18.2°.φ₁φ₂VI₁I₂

What it calculates

The capacitor bank that raises a lagging power factor to a target. The real power stays the same; the bank supplies part of the reactive power, so the apparent power and the line current fall. The reactive power the bank must supply is the difference between the two reactive loads, found from the tangent of each power-factor angle:

Qc=P (tan⁡φ1−tan⁡φ2),Q_c = P\,(\tan\varphi_1 - \tan\varphi_2),φ=arccos⁡(pf)\varphi = \arccos(\text{pf})

Dividing that reactive power among the three phases and applying the capacitor equation gives the capacitance to install in each phase, which depends on how the bank is connected:

CΔ=Qc3 ω VLL2C_\Delta = \dfrac{Q_c}{3\,\omega\,V_\text{LL}^{2}}CY=Qcω VLL2C_\text{Y} = \dfrac{Q_c}{\omega\,V_\text{LL}^{2}}ω=2πf\omega = 2\pi f

Delta or wye bank

A delta-connected capacitor sees the full line-to-line voltage. A wye-connected capacitor sees only the line-to-neutral voltage, which is lower by a factor of √3. Reactive power goes with voltage squared, so a wye capacitor must be three times larger to supply the same kVAR. Both results are shown; use the one that matches how the bank is wired.

Worked example

A 480 V plant draws 500 kW at a power factor of 0.75 lagging, and the goal is 0.95. The bank size and the capacitance per phase of a delta bank at 60 Hz are:

Qc=500 (0.88192−0.32868)=276.6 kVARQ_c = 500\,(0.88192 - 0.32868) = 276.6\ \text{kVAR}CΔ=276,6203×376.99×4802=1,062 μFC_\Delta = \dfrac{276{,}620}{3 \times 376.99 \times 480^{2}} = 1{,}062\ \mu\text{F}

A wye bank would need 3,185 µF per phase. Apparent power falls from 666.7 kVA to 526.3 kVA, and the line current from 801.9 A to 633.1 A, which is the saving that lets the same feeder and transformer carry more load.

Assumptions

  • The load is lagging, and the target power factor is higher than the present one and at most 1.
  • The load is balanced, and power factor is the displacement power factor at the fundamental frequency. Harmonics are not included.
  • The capacitors are rated at the system voltage. A capacitor run at a different voltage supplies kVAR in proportion to the square of that voltage.

Questions

Why are the wye microfarads three times the delta microfarads?
A delta capacitor sees the full line-to-line voltage; a wye capacitor sees only the line-to-neutral voltage, which is lower by √3. Reactive power goes with voltage squared, so a wye capacitor needs three times the capacitance to supply the same kVAR.
Why not correct all the way to unity power factor?
The last few points cost the most kVAR for the least current saved, and a bank sized for unity at full load can push the power factor leading at light load. Many utilities set their power factor threshold between 0.90 and 0.95, so 0.95 is a common target.
Does this account for harmonics?
No. It uses displacement power factor at the fundamental frequency. With large non-linear loads, a capacitor bank can resonate with the system inductance near a harmonic, and a detuned or filter bank may be needed.

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