# Network Theory Questions and Answers – Power Measurement in Three-Phase Circuits

This set of Network Theory Multiple Choice Questions & Answers focuses on “Power Measurement in Three-Phase Circuits”.

1. The wattmeter method is used to measure power in a three-phase load. The wattmeter readings are 400W and -35W. Calculate the total active power.
a) 360
b) 365
c) 370
d) 375

Explanation: Wattmeters are generally used to measure power in the circuits. Total active power = W1 + W2 = 400 + (-35) =365W.

2. The wattmeter method is used to measure power in a three-phase load. The wattmeter readings are 400W and -35W. Find the power factor.
a) 0.43
b) 0.53
c) 0.63
d) 0.73

Explanation: We know tanØ = √3((WR – WY)/(WR + WY)) => tanØ = √3 (400-(-35))/(400+(-35))=2.064 => Ø = 64.15⁰. Power factor = 0.43.

3. The wattmeter method is used to measure power in a three-phase load. The wattmeter readings are 400W and -35W. Find the reactive power.
a) 751.44
b) 752.44
c) 753.44
d) 754.44

Explanation: Reactive power = √3VLILsinØ. We know that WR – WY = 400-(-35)) = 435 = VLILsinØ. Reactive power = √3 x 435 = 753.44 VAR.

4. The input power to a three-phase load is 10kW at 0.8 Pf. Two watt meters are connected to measure the power. Find the reading of higher reading wattmeter.
a) 7.165
b) 6.165
c) 6.165
d) 4.165

Explanation: WR + WY = 10kW. Ø = cos-10.8=36.8o => tanØ = 0.75 = √3 (WR-WY)/(WR+WY)=(WR-WY)/10. WR-WY=4.33kW. WR+WY=10kW. WR=7.165kW.

5. The input power to a three-phase load is 10kW at 0.8 Pf. Two watt meters are connected to measure the power. Find the reading of lower reading wattmeter.
a) 1.835
b) 2.835
c) 3.835
d) 4.835

Explanation: WR + WY = 10kW. Ø = cos-10.8=36.8o => tanØ = 0.75 = √3 (WR-WY)/(WR+WY)=(WR-WY)/10. WR-WY=4.33kW. WR+WY=10kW. WY=2.835kW.
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6. The readings of the two watt meters used to measure power in a capacitive load are -3000W and 8000W respectively. Calculate the input power. Assume RYB sequence.
a) 5
b) 50
c) 500
d) 5000

Explanation: Toatal power is the sum of the power in R and power in Y. So total power = WR+WY = -3000+8000 = 5000W

7. The readings of the two watt meters used to measure power in a capacitive load are -3000W and 8000W respectively. Find the power factor.
a) 0.25
b) 0.5
c) 0.75
d) 1

Explanation: As the load is capacitive, the wattmeter connected in the leading phase gives less value. WR=-3000. WY=8000. tanØ = √3 (8000-(-3000))/5000=3.81 => Ø = 75.29⁰ => cosØ = 0.25.

8. The wattmeter reading while measuring the reactive power with wattmeter is?
a) VLILsecØ
b) VLILsinØ
c) VLILtanØ
d) VLILcosØ

Explanation: The wattmeter reading while measuring the reactive power with wattmeter is wattmeter reading = VLILsinØ VAR.

9. The total reactive power in the load while measuring the reactive power with wattmeter is?
a) √3VLILcosØ
b) √3VLILtanØ
c) √3VLILsinØ
d) √3 VLILsecØ

Explanation: To obtain the reactive power, wattmeter reading is to be multiplied by √3. Total reactive power = √3VLILsinØ.

10. A single wattmeter is connected to measure the reactive power of a three-phase, three-wire balanced load. The line current is 17A and the line voltage is 440V. Calculate the power factor of the load if the reading of the wattmeter is 4488 VAR.
a) 0.6
b) 0.8
c) 1
d) 1.2

Explanation: Wattmeter reading = VLILsinØ => 4488 = 440 x 17sinØ => sinØ = 0.6. Power factor = cosØ = 0.8.

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