Wattmeter Reactive Power Measurement MCQ || Measurement of 3 Phase Reactive Power Using Wattmeter Questions and Answers

1. Wattmeter cannot be designed on the principle of

  1. Electrostatic instrument
  2. Thermocouple instrument
  3. Moving iron instrument
  4. Electrodynamic instrument

Answer.3. Moving iron instrument

Explanation:

  • We can use electrostatic instruments to measure voltage and power. These types of wattmeters are used for the measurement of a small amount of power, practically when the voltage is high and the power factor is low. It is also used for measurement of dielectric loss of cables on alternating voltage and for calibration of wattmeter and energy meters.
  • Thermocouple instruments are used to measure current, voltage, and power.
  • We can use the moving iron instrument to measure current and voltage only.
  • Electrodynamic instruments are the most preferred instruments for measurement.

 

2. Two wattmeter method is used for measurement of power in a balanced three-phase load supplied from a balanced three-phase system. If one of the wattmeters reads half of the other (both positive), then the power factor of the load is

  1. 0.532
  2. 0.632
  3. 0.707
  4. 0.866

Answer.4. 0.866

Explanation:

Given that,

\({W_2} = \frac{{{W_1}}}{2}\)

We know that,

\(\phi = {\tan ^{ – 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} – {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)

\( = {\tan ^{ – 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} – \frac{{{W_1}}}{2}} \right)}}{{\left( {{W_1} + \frac{{{W_1}}}{2}} \right)}}} \right)\)

\(= {\tan ^{ – 1}}\left( {\frac{{\sqrt 3 \left( {2{W_1} – {W_1}} \right)}}{{2{W_1} + {W_1}}}} \right)\)

\(= {\tan ^{ – 1}}\left( {\frac{{\sqrt 3 {W_1}}}{{3{W_1}}}} \right) = {\tan ^{ – 1}}\left( {\frac{1}{{\sqrt 3 }}} \right) = 30^\circ \)

Power factor, cos ϕ = cos 30° = 0.866

 

3. A single-phase load is connected between R and Y terminal of a 415 V, symmetrical, 3 – phase, 4 – wire system with phase sequence RYB. Load resistance of 100 Ohms is connected in a wattmeter. The power factor of the load is 0.8 lagging. The wattmeter will read.

  1. -795 W
  2. -597 W
  3. +597 W
  4. +795 W

Answer.3. +597 W

Explanation:

Given

VRY = 415 V

Cosφ = 0.8 Lag

φ = 36.96

VRY = 415∠30°

θ = 30° − 36.96

θ = −6.86°

i.e

IR = 415∠30°/100∠36.96°

= 1.45∠−6.86

${W = 4.15 \times \frac{{415}}{{\sqrt 3 }}{\rm{cos}}\left( {120 + 6.86} \right)}$

 

W = −596.46 watt

 

4. A 300 V, 5 A, LPF wattmeter has a full scale of 300 W. The wattmeter can be used for loads supplied by 300 V AC mains with a maximum power factor of ________

  1. 0.2
  2. 0.5
  3. 0.8
  4. 0.1

Answer.1. 0.2

Explanation:

Power measured by 1- ϕ  wattmeter is given by P = VpcIcc cos ϕ

Where

Vpc = Voltage of potential coil

Icc = Current through the current coil

Here, Vpc = 300 V , Icc = 5 A , P = 300 W (Full Scale)

300 = Vpc Icc cos ϕ

300 = 300 x 5 cos ϕ

cos ϕ  = 0.2

 

5. Statement (I): The reactive power of the three-phase balanced load can be measured with the use of a single wattmeter.

Statement (II): For measurement of reactive power the current coil is connected to one of the phases of a 3-ϕ network.

  1. Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I).
  2. Both Statement (I) and Statement (II) are individually true, but Statement (II) is not the correct explanation of Statement (I).
  3. Statement (I) is true, but Statement (II) is false.
  4. Statement (I) is false, but Statement (II) is true.

Answer.2.

Explanation:

The power measured by the wattmeter is given by

W = IRVYBcos(IRVYB)

The reactive power of the three-phase balanced load can be measured with the use of a single wattmeter.

From the phasor diagram, we can observe that the phase angle between IR and VYB is 90 – ϕ.​

⇒ W = VYBIR cos (90 – ϕ)

⇒ W = VLIL sin ϕ

⇒ W = √3 VphIph sin ϕ 

But we know that three-phase reactive power Q = 3 VphIph sin ϕ

So, so by multiplying the wattmeter reading by √3 we can get three-phase reactive power.

Q (total) = √3 W = 3-ϕ reactive power

  •  For the measurement of reactive power, the current coil is connected to one of the phases of a 3-ϕ network.
  • And the potential coil is to be connected between the remaining two phases.

 

6. In case of power measurements by two wattmeter method in a balanced 3-phase system with pure inductive load

  1. Both wattmeters will indicate the same value but with opposite signs
  2. Both wattmeters will indicate zero
  3. Both the wattmeters will indicate the same value same sign
  4. One wattmeter will indicate zero and the other will indicate same non-zero value.

Answer.1. Both wattmeters will indicate the same value but with opposite signs

Explanation:

Given that load is pure inductive – Hence it is zero power factor (lag)

Cos ϕ = 0

⇒ ϕ = 90°

The reading of two wattmeters can be expressed as

W1 =  VLILcos(30 + φ)
W2 =  VLILcos(30 − φ)

When P.F reads zero (φ = 90°)

Such a case occurs when the load consist of pure inductance or pure capacitance

W1 =  VLILcos(30 + 90°) = – VLILsin30°
W2 =  VLILcos(30 − 90°) = VLILsin30°

In this condition, the two wattmeter reads equal and opposite i.e W1 + W2 = 0

W1 = -W2

 

7. A 3 phase star-connected load is supplied with 440 V and 50 Hz supply, two wattmeter method is used to measure the power in the load. The readings of the wattmeter are 2000 W and – 1000 W. What is the value of the capacitor in μF to be connected in series with each phase of load such that the power is measured by one wattmeter.

  1. 140.58 μF
  2. 132.79 μF
  3. 138.58 μF
  4. 122.85 μF

Answer.2. 132.79 μF

Explanation:

Given,

Supply voltage V = 440 V

Supply frequency f = 50 Hz

Wattmeter reading W1 = 2000 W

W2 = – 1000 W

Total active power P = W1 + W2 = 1000 W

Power factor angle ϕ1

${\phi _1} = {\tan ^{ – 1}}\left( {\sqrt 3 \frac{{{W_1} – {W_2}}}{{{W_1} + {W_2}}}} \right)$

 

${\phi _1} = {\tan ^{ – 1}}\left( {\sqrt 3 \frac{{2000 + 1000}}{{2000 – 1000}}} \right)$

 

ϕ1 = 79.1∘

Total Real power P = √3 VL IL cos ϕ1

Line current IL = (P / √3 VL) = (1000 / √3 × 440) = 6.94 A

Zph = Vph/Iph (In star connection Vph = VL/√3 and IL = Iph)

Zph = 440 / (√3 × 6.94) = 36.61 Ω

Resistance Rph = Zph cos ϕ1 = 36.61 cos 79.1° = 6.922 Ω

Reactance Xph = Zph sin ϕ1 = 36.61 sin 79.1° = 35.95 Ω

After capacitance is added to each phase of the load, also total real power is measured by only one wattmeter then

Power factor angle ϕ2

${\phi _2} = {\tan ^{ – 1}}\left( {\sqrt 3 \frac{{2000 + 0}}{{2000 – 0}}} \right)$

Reactance after adding capacitance X’ph

X’ph = Rph tan ϕ2 = 6.922 × tan 60∘ = 11.98 Ω 

Capacitive reactance XC = Xph – X’ph = 35.95 – 11.98 = 23.97 Ω

Capacitive reactance XC = 1 / 2πfC 

23.97 = 1 / 2π(50)C

C = 132.79 μF

 

8. The apparent power drawn by an AC system is 50 kVA, reading of wattmeter is 20 kW. Reactive power consumed will be

  1. 45.82 kVAR
  2. 50 kVAR
  3. 70 kVAR
  4. 20 kVAR

Answer.1. 45.82 kVAR

Explanation:

Given that,

Apparent power (S) = 50 kVA

Reading of wattmeter = Active power (P) = 20 kW

Apparent Power

$S = \sqrt {{P^2} + {Q^2}}$

$ \Rightarrow 50 = \sqrt {{{20}^2} + {Q^2}}$

Q = 45.82 kVAR

 

9. In a particular test, the readings of two wattmeters are 5 kW and 1 kW. If the second-meter connections are reversed, the total power and power factor are

  1. 5 kW, 0.6933 lag
  2. 4 kW, 0.3592 lag
  3. 3 kW, 0.327 lag
  4. 5 kW, 0.327 lag

Answer.2. 4 kW, 0.3592 lag

Explanation:

P1 = 5 kW, P2 = 1 kW

Pt = P1 – P2

= 5 – 1 = 4 kW

p.f = cos ϕ

${\phi _1} = {\tan ^{ – 1}}\left( {\sqrt 3 \frac{{{W_1} – {W_2}}}{{{W_1} + {W_2}}}} \right)$

 

${\tan ^{ – 1}}\left( {\sqrt 3 \left( {\frac{6}{4}} \right)} \right)$

= 68.94°

Hence, p.f. = cos ϕ = cos 68.94° = 0.3592

Therefore the pf is 0.3592 lag

 

10. Two wattmeters are connected to measure the power consumed by a 3-ϕ load with a power factor of 0.4. The total power consumed by the load as indicated by the two wattmeters is 30 kW. The difference between the two wattmeter readings is _____ kW

  1. 40.44 kW
  2. 35.59 kW
  3. 39.69 kW
  4. 42.25 kW

Answer.3. 39.69 kW

Explanation:

P = 30 W, cos ϕ = 0.4

If P1 and P2 are two individual wattmeter readings then according to the problem,

P = P1 + P2 = 30 kW

Power factor angle ϕ = cos-1 (0.4) = 66.4°

$\phi = {\tan ^{ – 1}}\left( {\sqrt 3 \frac{{\left( {{P_1} – {P_2}} \right)}}{{\left( {{P_1} + {P_2}} \right)}}} \right)$

 

${66.4^ \circ } = {\tan ^{ – 1}}\left( {\frac{{\sqrt 3 \left( {{P_1} – {P_2}} \right)}}{{30}}} \right)$

 

(2.288 × 30)/√3 = P1 − P2

P1 – P= 39.69 kW

 

11. A 3 phase 500 V motor load has a power factor of 0.5. Two wattmeter are connected to measure the input. They show the input to be 36 kW. The readings of both the instruments are.

  1. W1 = 36 kW, W2 = 0 kW
  2. W1 = 9 kW, W2 = 27 kW
  3. W1 = W2 = 18 kW
  4. W1 = 45 kW, W2 = -9 kW

Answer.1. W1 = 36 kW, W2 = 0 kW

Explanation:

Given that, line voltage (VL) = 500 V

Cos ϕ = 0.5

Cos ϕ = 0.5 ⇒ ϕ = 60°

W1 + W2 = 36 kW

${\phi _1} = {\tan ^{ – 1}}\left( {\sqrt 3 \frac{{{W_1} – {W_2}}}{{{W_1} + {W_2}}}} \right)$

 

$\Rightarrow \tan 60 = \frac{{\sqrt 3 \left( {{W_1} – {W_2}} \right)}}{{\left( {{W_1} + {W_2}} \right)}}$

⇒ W1 – W2 = 36

⇒ W1 = 36 kW, W2 = 0 kW

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