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Electrician Red Seal · Question

A manufacturing plant has a main electrical feeder rated for 800 A at 600 V. Power factor correction has been installed, bringing the average power factor from 0.75 lagging to 0.95 lagging. With the improved power factor, how much more active power (in kW) can the feeder now supply without exceeding its current rating? Assume a three-phase system.

The maximum apparent power (S) the feeder can supply is √3 * V * I = √3 * 600 V * 800 A = 831,384 VA = 831.38 kVA. Initial active power P1 = S * PF1 = 831.38 kV

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Question: A manufacturing plant has a main electrical feeder rated for 800 A at 600 V. Power factor correction has been installed, bringing the average power factor from 0.75 lagging to 0.95 lagging. With the improved power factor, how much more active power (in kW) can the feeder now supply without exceeding its current rating? Assume a three-phase system.

Answer options: ✅ 166.3 kW

  • 83.3 kW
  • 222.1 kW
  • 138.8 kW

Correct answer: 166.3 kW

Explanation: The maximum apparent power (S) the feeder can supply is √3 * V * I = √3 * 600 V * 800 A = 831,384 VA = 831.38 kVA. Initial active power P1 = S * PF1 = 831.38 kVA * 0.75 = 623.53 kW. New active power P2 = S * PF2 = 831.38 kVA * 0.95 = 789.81 kW. The increase in active power is P2 - P1 = 789.81 kW - 623.53 kW = 166.28 kW. Option A (166.3 kW) is the closest.

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