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

View Answer

Explanation: Wattmeters are generally used to measure power in the circuits. Total active power = W

_{1}+ W

_{2}= 400 + (-35) =365W.

2. In the question 2 find the power factor.

a) 0.43

b) 0.53

c) 0.63

d) 0.73

View Answer

Explanation: We know tanØ = √3((W

_{R}– W

_{Y})/(W

_{R}+ W

_{Y})) => tanØ = √3 (400-(-35))/(400+(-35) )=2.064 => Ø = 64.15⁰. Power factor = 0.43.

3. Find the reactive power in the question 2.

a) 751.44

b) 752.44

c) 753.44

d) 754.44

View Answer

Explanation: Reactive power = √3V

_{L}I

_{L}sinØ. We know that W

_{R}– W

_{Y}= 400-(-35)) = 435 = V

_{L}I

_{L}sinØ. 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

View Answer

Explanation: W

_{R}+ W

_{Y}= 10kW. Ø = cos

^{-1}0.8=36.8

^{o}=> tanØ = 0.75 = √3 (W

_{R}-W

_{Y})/(W

_{R}+W

_{Y})=(W

_{R}-W

_{Y})/10. W

_{R}-W

_{Y}=4.33kW. W

_{R}+W

_{Y}=10kW. W

_{R}=7.165kW.

5. Find the reading of higher reading wattmeter in the question 2.

a) 1.835

b) 2.835

c) 3.835

d) 4.835

View Answer

Explanation: W

_{R}+ W

_{Y}= 10kW. Ø = cos

^{-1}0.8=36.8

^{o}=> tanØ = 0.75 = √3 (W

_{R}-W

_{Y})/(W

_{R}+W

_{Y})=(W

_{R}-W

_{Y})/10. W

_{R}-W

_{Y}=4.33kW. W

_{R}+W

_{Y}=10kW. W

_{Y}=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.

a) 5

b) 50

c) 500

d) 5000

View Answer

Explanation: Toatal power is the sum of the power in R and power in Y. So total power = W

_{R}+W

_{Y}= -3000+8000 = 5000W

7. Find the power factor in the question 6.

a) 0.25

b) 0.5

c) 0.75

d) 1

View Answer

Explanation: As the load is capacitive, the wattmeter connected in the leading phase gives less value. W

_{R}=-3000. W

_{Y}=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) V_{L}I_{L}secØ

b) V_{L}I_{L}sinØ

c) V_{L}I_{L}tanØ

d) V_{L}I_{L}cosØ

View Answer

Explanation: The wattmeter reading while measuring the reactive power with wattmeter is wattmeter reading = V

_{L}I

_{L}sinØ VAR.

9. The total reactive power in the load while measuring the reactive power with wattmeter is?

a) √3V_{L}I_{L}cosØ

b) √3V_{L}I_{L}tanØ

c) √3V_{L}I_{L}sinØ

d)√3 V_{L}I_{L}secØ

View Answer

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

_{L}I

_{L}sinØ.

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.

a) 0.6

b) 0.8

c) 1

d) 1.2

View Answer

Explanation: Wattmeter reading = V

_{L}I

_{L}sinØ => 4488 = 440 x 17sinØ => sinØ = 0.6. Power factor = cosØ = 0.8.

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