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.
Explanation: Wattmeters are generally used to measure power in the circuits. Total active power = W1 + W2 = 400 + (-35) =365W.
2. In the question 2 find the power factor.
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. Find the reactive power in the question 2.
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.
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. Find the reading of higher reading wattmeter in the question 2.
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.
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.
Explanation: Toatal power is the sum of the power in R and power in Y. So total power = WR+WY = -3000+8000 = 5000W
7. Find the power factor in the question 6.
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?
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?
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 reactive power of a three-phase, three-wire balanced load. The line current is 17A and line voltage is 440V. Calculate the power factor of the load if the reading of the wattmeter is 4488 VAR.
Explanation: Wattmeter reading = VLILsinØ => 4488 = 440 x 17sinØ => sinØ = 0.6. Power factor = cosØ = 0.8.
Sanfoundry Global Education & Learning Series – Network Theory.
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