How it works
To improve PF from cos φ1 to cos φ2 for a load of P kW, the reactive power that a capacitor bank must supply is Qc = P(tan φ1 − tan φ2) kVAr. For P = 100 kW, PF from 0.7 to 0.9: tan φ1 = 1.020, tan φ2 = 0.484, Qc = 100(1.020−0.484) = 53.6 kVAr. Shunt capacitors are preferred over series because they improve PF at the load bus without risking ferroresonance. Synchronous condensers (overexcited synchronous motors on no load) provide continuously variable VAr and can both absorb and supply reactive power. STATCOMs based on VSC technology respond in less than one cycle and inject smooth reactive current without bulky capacitors.
Key points to remember
The BEE (Bureau of Energy Efficiency) mandates a minimum power factor of 0.85 for HT consumers in India; failure attracts a monthly surcharge of 1% of the energy bill per 0.01 drop below 0.85. Fixed capacitor banks are switched manually or by a timer; automatic power factor correction (APFC) panels use a PF controller IC (e.g., BCR series) to switch capacitor stages as load varies, maintaining PF above 0.95 continuously. Series capacitors are used on long transmission lines to reduce inductive reactance and improve stability, but they can cause sub-synchronous resonance (SSR) with generator shaft torsional modes — an important distinction from shunt correction. Leading PF caused by lightly loaded cables must also be controlled to prevent voltage rise.
Exam tip
The examiner always asks you to calculate the capacitor bank size in kVAr required to correct PF from a given lagging value to 0.9 — draw the power triangle showing S1, P, Q1 and S2, P, Q2, then state Qc = Q1−Q2 = P(tanφ1−tanφ2) before substituting numbers.