Electrician Red Seal · Question
Consider a 400 V, 3-phase system. An industrial client is incurring penalties for a low power factor of 0.70 lagging. They are operating at 500 kW of real power. To avoid penalties, they need to improve the power factor to 0.95 lagging. What is the approximate reduction in line current (I_L) achieved by this power factor correction?
Initial apparent power (S1) = P / PF1 = 500 kW / 0.70 = 714.28 kVA. Initial line current (I_L1) = S1 / (√3 * V_L) = 714280 VA / (√3 * 400 V) = 1031 A. New appar
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Question: Consider a 400 V, 3-phase system. An industrial client is incurring penalties for a low power factor of 0.70 lagging. They are operating at 500 kW of real power. To avoid penalties, they need to improve the power factor to 0.95 lagging. What is the approximate reduction in line current (I_L) achieved by this power factor correction?
Answer options:
- 150 A reduction
- 220 A reduction ✅ 300 A reduction
- 380 A reduction
Correct answer: 300 A reduction
Explanation: Initial apparent power (S1) = P / PF1 = 500 kW / 0.70 = 714.28 kVA. Initial line current (I_L1) = S1 / (√3 * V_L) = 714280 VA / (√3 * 400 V) = 1031 A. New apparent power (S2) = P / PF2 = 500 kW / 0.95 = 526.31 kVA. New line current (I_L2) = S2 / (√3 * V_L) = 526310 VA / (√3 * 400 V) = 759 A. The reduction in line current is I_L1 - I_L2 = 1031 A - 759 A = 272 A, which is approximately 300 A.
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