If a three-phase motor with the same line voltage and current has its power factor reduced from 0.8 to 0.5, the input real power will be approximately:

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Multiple Choice

If a three-phase motor with the same line voltage and current has its power factor reduced from 0.8 to 0.5, the input real power will be approximately:

Explanation:
The input real power in a three-phase system with fixed line voltage and current is P = √3 V_L I_L cosφ. If V_L and I_L don’t change, P is directly proportional to the power factor cosφ. Dropping from 0.8 to 0.5 scales the real power by 0.5/0.8 = 0.625. So the new input real power is 0.625 times the original. If the original real power at PF 0.8 is about 7.98 kW, the new power is 7.98 × 0.625 ≈ 4.99 kW.

The input real power in a three-phase system with fixed line voltage and current is P = √3 V_L I_L cosφ. If V_L and I_L don’t change, P is directly proportional to the power factor cosφ. Dropping from 0.8 to 0.5 scales the real power by 0.5/0.8 = 0.625. So the new input real power is 0.625 times the original. If the original real power at PF 0.8 is about 7.98 kW, the new power is 7.98 × 0.625 ≈ 4.99 kW.

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