Power-Factor Correction kvar

Also known as capacitor bank sizing · PF correction

Qc=P(tan⁡φ1−tan⁡φ2)Q_{c} = P \left( \tan\varphi_{1} - \tan\varphi_{2} \right)

Worked example: 100 kW from 0.70 to 0.95 PF → 69.15 kvar — press Try an example to run it live, then adjust anything.

Enter your known values, leave one input blank, and solves for the missing one. Tap a variable’s symbol to see what it means, with a typical value. Try different units for next level excitement!

Here the solver did the work — could you?

Power-factor correction →

UniversityCircuits & Electrical Power

Test your skills in the Exam Room: new numbers every attempt — free lessons for students, no sign-up, just pure learning. Find 1 more lesson on this formula.

See your Report Card
Compete with your friends
share your results
Learning zone

Power-Factor Correction kvar explained

QcPF₁PF₂P

Each power factor corresponds to a phase angle φ = arccos(PF), and the reactive burden of a load is P tan φ. To move from a poor PF₁ to a target PF₂ you must cancel the difference in those vars, which is what a capacitor bank does. A 100 kW plant at 0.70 PF carries 100 × 1.020 = 102 kvar; at 0.95 PF it would carry only 32.9 kvar, so about 69 kvar of capacitors closes the gap.

Two field cautions. Chasing unity is a mistake — the last few points cost the most capacitors, and an overcorrected plant goes leading, which can push voltage up and, on a motor that keeps spinning after the contactor opens, cause damaging self-excitation. And where the harmonic content is high, plain capacitors resonate with the supply inductance and amplify the harmonics; detuned reactors are then part of the package.

Power-Factor Correction kvar formula

Qc=P(tan⁡φ1−tan⁡φ2)Q_{c} = P \left( \tan\varphi_{1} - \tan\varphi_{2} \right)
Where
  • QcQ_{c}= Correction reactive power (var) (W)
  • PP= Load real power (W)
  • PF1\text{PF}_{1}= Existing power factor
  • PF2\text{PF}_{2}= Target power factor