An industrial facility uses a power factor correction capacitor bank. Before installation, the load was 500 kVA at 0.75 lagging power factor. After installing the capacitors, the power factor improved to 0.95 lagging. What was the approximate reactive power (in kVAR) supplied by the capacitor bank?

Question: An industrial facility uses a power factor correction capacitor bank. Before installation, the load was 500 kVA at 0.75 lagging power factor. After installing the capacitors, the power factor improved to 0.95 lagging. What was the approximate reactive power (in kVAR) supplied by the capacitor bank?

Answer options:

• 150 kVAR ✅ 216 kVAR
• 185 kVAR
• 250 kVAR

Correct answer: 216 kVAR

Explanation: First, calculate the real power (P) and initial reactive power (Q1). P = S1 * PF1 = 500 kVA * 0.75 = 375 kW. Q1 = sqrt(S1^2 - P^2) = sqrt(500^2 - 375^2) = sqrt(250000 - 140625) = sqrt(109375) = 330.7 kVAR. Now, calculate the new reactive power (Q2) with the improved power factor. Since real power (P) does not change, S2 = P / PF2 = 375 kW / 0.95 = 394.7 kVA. Q2 = sqrt(S2^2 - P^2) = sqrt(394.7^2 - 375^2) = sqrt(155788 - 140625) = sqrt(15163) = 123.1 kVAR. The reactive power supplied by the capacitor bank (Qc) = Q1 - Q2 = 330.7 kVAR - 123.1 kVAR = 207.6 kVAR. The closest option is 216 kVAR. (A slight discrepancy exists if using tan(theta) method: Q1 = P * tan(acos(PF1)) = 375 * tan(acos(0.75)) = 375 * tan(41.41) = 375 * 0.8819 = 330.7 kVAR. Q2 = P * tan(acos(PF2)) = 375 * tan(acos(0.95)) = 375 * tan(18.19) = 375 * 0.3287 = 123.2 kVAR. Qc = 330.7 - 123.2 = 207.5 kVAR.) Rounding results in 208 kVAR, making 216 kVAR the nearest practical answer.