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Power factor correction

Motors and other inductive loads draw reactive power, which gives a low power factor and unnecessarily large current. Capacitors connected in parallel supply the reactive power locally. The current in the network then drops, and so do the losses.

QC=P (tan⁡φ1−tan⁡φ2)Q_C = P\,(\tan\varphi_1 - \tan\varphi_2)reactive power the capacitors must supply
cos⁡φ2>cos⁡φ1\cos\varphi_2 > \cos\varphi_1better power factor afterwards

Symbols

QCQ_Creactive power of the capacitor bankkvar
PPactive powerkW
φ1, φ2\varphi_1,\ \varphi_2phase angle before and after°

Example

P=50P = 50 kW from cos⁡φ=0.75\cos\varphi = 0.75 to 0.950.95:

QC=50 (0.882−0.329)≈27.7Q_C = 50\,(0.882 - 0.329) \approx 27.7 kvar.

Correct close to the load. Then the rest of the installation does not have to carry the reactive current.
Practise three-phase and power for free →

← Three-phase power

Part of Electric Circuits: Three-phase and power.