Electrician Red Seal · Question
An electrician is tasked with improving the power factor of a 100 kW, 480 V, three-phase load that currently has a power factor of 0.70 lagging. The goal is to raise the power factor to 0.95 lagging. What is the approximate reactive power (kVAR) that needs to be compensated by capacitors?
First, calculate the initial reactive power (Q1) for PF1 = 0.70. cos(θ1) = 0.70 → θ1 = 45.57°. Q1 = P * tan(θ1) = 100 kW * tan(45.57°) ≈ 100 kW * 1.020 = 102.0
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Question: An electrician is tasked with improving the power factor of a 100 kW, 480 V, three-phase load that currently has a power factor of 0.70 lagging. The goal is to raise the power factor to 0.95 lagging. What is the approximate reactive power (kVAR) that needs to be compensated by capacitors?
Answer options:
- 36 kVAR
- 52 kVAR ✅ 71 kVAR
- 94 kVAR
Correct answer: 71 kVAR
Explanation: First, calculate the initial reactive power (Q1) for PF1 = 0.70. cos(θ1) = 0.70 → θ1 = 45.57°. Q1 = P * tan(θ1) = 100 kW * tan(45.57°) ≈ 100 kW * 1.020 = 102.0 kVAR. Next, calculate the final reactive power (Q2) for PF2 = 0.95. cos(θ2) = 0.95 → θ2 = 18.19°. Q2 = P * tan(θ2) = 100 kW * tan(18.19°) ≈ 100 kW * 0.328 = 32.8 kVAR. The required compensation is Q_c = Q1 - Q2 = 102.0 kVAR - 32.8 kVAR = 69.2 kVAR. The closest option is 71 kVAR.
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