Question

A 3-phase, 600 V motor, the load having 0.6 power factor uses two watt-meter to measure the power. If the power measured be 45 kW, then the reading of each instrument will be?

(a) P1 = 35 kW, P2 = 10 kW

(b) P1 = 47.25 kW, P2 = -2.25 kW

(c) P1 = 39.82 kW, P2 = 5.179 kW

(d) P1 = 45 kW, P2 = 0 kW

This question was addressed to me in an interview.

I'm obligated to ask this question of Advanced Miscellaneous Problems on Measurement of Power in division Measurement of Power and Related Parameters of Electrical Measurements

Answer 1

Correct option is (c) P1 = 39.82 kW, P2 = 5.179 kW

Easy explanation: P1 + P2 = 45 kW

And, cos ∅ = 0.6

∴ tan ∅ = 1.33

Or, 1.33 = \(\sqrt{3} \frac{P_1 – P_2}{45} \)

∴ P1 – P2 = 34.64 kW

∴ P1 = 39.82 kW

P2 = 5.179 kW.