Wattmeter MCQ

21. Calculate the percentage error for a wattmeter which is so connected that the current coil is on the load side. The wattmeter has a current coil of 0.03 ohm resistance and a pressure coil of 6000 ohm resistance. It is also known that the load takes 20A at a voltage of 220V and 0.6 power factor.

  1. 0.45 %
  2. 45 %
  3. 5.5 %
  4. 6.5 %
Answer
Answer. a

22. In a single phase dynamometer wattmeter, the instantaneous torque has a component of power which varies as

  1. Twice
  2. Half
  3. Four times
  4. Three times
Answer
Answer. a

23. Laboratory wattmeters are

  1. Induction type
  2. Moving iron type
  3. Electrostatic type
  4. Electro-dynamometer type
Answer
Answer. d

24. For power measurement of three-phase circuit by two wattmeter method, when the value of power factor is less than 0.5 lagging

  1. One of the wattmeter’s will read zero
  2. Both give the same readings
  3. One of the wattmeter connections  will have to be reversed
  4. Pressure coil of the wattmeter will become ineffective
Answer
Answer. c

25. Power consumed by a balanced 3-phase, 3-wire load is measured by two wattmeter method. The first wattmeter reads twice that of the second. Then what will be the load impedance angle in radian?

  1. π/6
  2. π/3
  3. π/2
  4. π/4
Answer
Answer. a

26. A compensated wattmeter has its reading corrected for error due to

  1. Frequency
  2. Friction
  3. Power consumed in current coil
  4. Power consumed in pressure coil
Answer
Answer. d

27. If a dynamometer wattmeter is connected in an AC circuit, the power indicated by the wattmeter will be

  1. Volt-ampere product
  2. Average power
  3. Peak power
  4. Instantaneous power
Answer
Answer. b

28. How many coils are there in a wattmeter?

  1. 2
  2. 3
  3. 4
  4. 1
Answer
Answer. a

29. Which of the following provide deflecting force for voltmeters only?

  1. Electromagnetic effect
  2. Electrodynamic effect
  3. Magnetic effect
  4. Electrostatic effect
Answer
Answer. d

30. In the measurement of 3 phase power by two wattmeter method, for an unbalanced load, the power factor of the load is

  1. $\cos [\tan^{-1}{\frac{\sqrt{3}(W_2-W_1)}{W_2+W_1}}]$
  2. $\cos [\tan^{-1}{\frac{(W_2-W_1)}{W_2+W_1}}]$
  3. $\cos (W_2-W_1)$
  4. None of these
Answer
Answer. a