This set of Advanced Electrical Measurements Questions and Answers focuses on “Advanced Miscellaneous Problems on Measurement of Power”.

1. A coil (which can be modelled as a series RL circuit) has been designed for high Q performance. The voltage is rated at and a specified frequency. If the frequency of operation is increased 10 times and the coil is operated at the same rated voltage. The new value of Q factor and the active power P will be?

a) P is doubled and Q is halved

b) P is halved and Q is doubled

c) P remains constant and Q is doubled

d) P decreases 100 times and Q is increased 10 times

View Answer

Explanation: ω

_{2}L = 10 ω

_{1}LR will remain constant

∴ Q

_{2}= \(\frac{10 ω_1 L}{R}\) = 10 Q

_{1}

That is Q is increased 10 times.

Now, \(I_1 = \frac{V}{ω_1 L}\)

For a high Q coil, ωL ≫> R,

\(I_2 = \frac{V}{10 ω_1 L} = \frac{I_1}{10}\)

∴ \(P_2 = R (\frac{I_1}{10})^2 = \frac{P_1}{100}\)

Thus, P decreases 100 times and Q is increased 10 times.

2. Two watt-meters, which are connected to measure the total power on a three-phase system supplying a balanced load, read 20.5 kW and -3.5 kW respectively. The total power is?

a) 13.0 kW

b) 17 kW

c) 15 kW

d) 19 kW

View Answer

Explanation: w

_{1}= 20.5 kW, w

_{2}= -3.5 kW

∴ w = w

_{1}+ w

_{2}= 20.5 – 3.5 = 17 kW.

3. A circuit is used to measure the power consumed by the load. The current coil and the voltage coil of the watt-meter have 0.02 Ω and 1000 Ω resistances respectively. The measured power (as compared to the load power) will be?

a) 0.4 % less

b) 0.2 % less

c) 0.2 % more

d) 0.4 % more

View Answer

Explanation: Power indicated by watt-meter = 400 × 20 + 20

^{2}× 0.02 = 4008 W

Percentage increase = \(\frac{4008-4000}{4000}\) × 100 = 0.2 % more.

4. A sampling watt-meter is used to measure the average power of a load. The peak to peak voltage of the square wave is 10 V and the current is a triangular wave of 5A p-p. The reading in watt will be?

a) Zero

b) 25 W

c) 50 W

d) 100 W

View Answer

Explanation: If we consider both the waves, we can see positive power and negative power in each case are equal. So, the net resultant power is zero.

5. The line to line input voltage to the 3-phase, 50 Hz, AC circuit, is 100 V rms. Assuming that the phase sequence is RYB; the watt-meters would read?

a) W_{1} = 886 W and W_{2} = 886 W

b) W_{1} = 500 W and W_{2} = 500 W

c) W_{1} = 0 W and W_{2} = 1000 W

d) W_{1} = 250 W and W_{2} = 750 W

View Answer

Explanation: W

_{1}= 100 × 20 cos 90° = 0

\(W_2 = \frac{100 × 20 × \sqrt{3}}{\sqrt{3} × 2}\) = 1000 W.

6. The meter constant of a single-phase, 100 V induction watt-meter is 600 rev/kW-h. The speed of the meter disc for a current of 15 A at 0.8 power factor lagging will be?

a) 30.3 rpm

b) 16 rpm

c) 12 rpm

d) 33.1 rpm

View Answer

Explanation: Meter constant = \(\frac{Number \,of\, revolution}{Energy}\)

Number of revolution = \(\frac{600 × 100 × 15 × 0.8}{1000}\) = 720

∴ Speed in rpm = \(\frac{720}{60}\) = 12 rpm.

7. A 3-phase, 600 V motor, the load having 0.6 power factor uses two watt-meter to measure the power. If the power measured be 45 kW, then the reading of each instrument will be?

a) P_{1} = 35 kW, P_{2} = 10 kW

b) P_{1} = 47.25 kW, P_{2} = -2.25 kW

c) P_{1} = 39.82 kW, P_{2} = 5.179 kW

d) P_{1} = 45 kW, P_{2} = 0 kW

View Answer

Explanation: P

_{1}+ P

_{2}= 45 kW

And, cos ∅ = 0.6

∴ tan ∅ = 1.33

Or, 1.33 = \(\sqrt{3} \frac{P_1 – P_2}{45} \)

∴ P

_{1}– P

_{2}= 34.64 kW

∴ P

_{1}= 39.82 kW

P

_{2}= 5.179 kW.

8. An average-reading digital Multimeter reads 10 V when fed with a triangular wave, symmetric about the time-axis. If the input is same, the rms reading meter will read?

a) 20/\(\sqrt{3}\)

b) -10/\(\sqrt{3}\)

c) -20/\(\sqrt{3}\)

d) 10/\(\sqrt{3}\)

View Answer

Explanation: For triangular wave- Average value = \(\frac{V_m}{3}\), rms value = \(\frac{V_m}{\sqrt{3}}\)

\(\frac{V_m}{3}\) = 10 V or, V

_{m}= 30 V

So, rms = \(\frac{30}{\sqrt{3}}\) = 10\(\sqrt{3}\).

9. In the Two wattmeter method of measuring power in a balanced three-phase circuit, one wattmeter shows zero and the other positive maximum. The load power factor is?

a) 0

b) 0.5

c) 0.866

d) 1.0

View Answer

Explanation: The load power factor is = 0.5. Since at this power factor one wattmeter shows zero and the other shows a positive maximum value of power.

10. Two watt-meters connected to measure the total power on a three-phase system supplying a balanced load reads 10.5 kW and -2.5 kW respectively. Then the total power factor is?

a) 0.334

b) 0.684

c) 0.52

d) 0.334

View Answer

Explanation: w

_{1}= 10.5 kW, w

_{2}= -2.5 kW

∴ w = w

_{1}+ w

_{2}= 8 kW

Also, tan ∅ = 2.81

∴ ∅ = 70.43° or, cos ∅ = 0.3347.

**Sanfoundry Global Education & Learning Series – Electrical Measurements.**

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