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) to improve the power factor seen from the source to 0.96 (lagging). Which of the following is closest to the reactive power the capacitor must supply?

Answer: B. About 3.7 kvar

Method 1 (difference of tan φ): the reactive power before correction is Q1 = P tan φ1 = 8 × 0.75 = 6 kvar. After correction, cos φ2 = 0.96 gives sin φ2 = 0.28 and tan φ2 = 0.28/0.96 ≈ 0.292, so Q2 = 8 × 0.292 ≈ 2.33 kvar. The capacitor's reactive power is Qc = Q1 − Q2 ≈ 6 − 2.33 ≈ 3.7 kvar. Method 2 (power triangle): the apparent power after correction is 8/0.96 ≈ 8.33 kVA, and Q2 = √(8.33² − 8²) ≈ 2.33 kvar, the same result. Raising the power factor all the way to 1 would need 6 kvar, but for 0.96 a capacitor of about 60% of that size is enough.

• A. About 2.3 kvar is the reactive power Q2 that remains after correction, not the amount the capacitor supplies.
• B. Correct. Qc = 6 − 2.33 ≈ 3.7 kvar.
• C. 6.0 kvar is the value for raising the power factor to 1.