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Circuit Analysis: AC, Linear Circuits and Transients

As in the figure, a capacitor C is connected in parallel with an inductive load of 8 kW active power and a power factor of 0.8 (lagging) so that the power factor seen from the source becomes 1. Which of the following is the reactive power the capacitor must supply?

Power-factor correction of an inductive load with a capacitor V C R L Load I IC IL
Power-factor correction of an inductive load with a capacitor · View in diagrams
  1. A6.0 kvar
  2. B8.0 kvar
  3. C10 kvar
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Answer: A. 6.0 kvar

Method 1 (power triangle): the apparent power is S = P/cos φ = 8/0.8 = 10 kVA, and the reactive power is Q = √(S² − P²) = √(10² − 8²) = 6 kvar. To raise the power factor to 1, the capacitor must supply 6.0 kvar of leading reactive power, cancelling all of this lagging reactive power. Method 2 (using tan φ): Q = P tan φ, and for cos φ = 0.8, tan φ = 0.6/0.8 = 0.75, so Q = 8 × 0.75 = 6 kvar. At 200 V and 50 Hz the required capacitance is C = Q/(ωV²) = 6000/(2π × 50 × 200²) ≈ 477 µF. The active power of the load is unchanged, and the source current falls from 10 kVA/200 V = 50 A to 8 kW/200 V = 40 A.

  • A. Correct. It cancels Q = P tan φ = 6 kvar.
  • B. 8.0 is the value of the active power [kW]; the capacitor does not cancel active power.
  • C. 10 is the value of the apparent power [kVA].
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