Question

Power flowing in 3-phase, 3-wire system is measured by two wattmeter whose readings are 7000 W and -2500 W. if the voltage of the circuit is 400 V, then what will be the value of capacitance introduced in each phase to make one wattmeter reads zero? The frequency is 50 Hz.

(a) 500 μF

(b) 668.6 μF

(c) 1000 μF

(d) 748.5 μF

The question was asked by my school teacher while I was bunking the class.

This question is from Advanced Problems on Measurement of Power in Polyphase Circuit topic in section Measurement of Power and Related Parameters of Electrical Measurements

Answer 1

Right choice is (b) 668.6 μF

Explanation: P = P1 + P2 = 7000 – 2500 = 4500 W

∴ Power in each phase = 1500 W

Voltage of each phase = 231 V

So, cos∅ = 0.264

∴ Current in each phase = 24.6 A

∴ Z of each phase = 9.39 Ω

∴ R of each phase = 2.48 Ω

Reactance of each phase = 9.056 Ω

For one watt-meter to read zero, power factor should be 0.5

Now, tan∅ = \(\frac{X}{R}\) or, X = 4.29 Ω

∴ Capacitance reactance required = 9.056 – 4.29 = 4.76 Ω

∴ Capacitance C = \(\frac{1}{2π × 50 × 4.76}\) F = 668.6 μF.