Blondel’s Theorem MCQ || Blondel’s Theorem Questions and Answers

Electrical Measurement MCQ, Electronics MCQ / March 21, 2022

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

 

12. Two wattmeters are used to measure the power in a 3-phase balanced system. What is the power factor of the load when one wattmeter reads twice the other?

  1. 0
  2. 0.5
  3. 0.866
  4. 1

Answer.3. 0.866

Explanation:

Given that, W1 = 2 W2

Power factor = cos ϕ

$\phi = {\rm{ta}}{{\rm{n}}^{ – 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} – {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)$

 

$\tan \phi = \frac{{\sqrt 3 \left( {2{W_2} – {W_2}} \right)}}{{\left( {2{{\rm{W}}_2} + {W_2}} \right)}} = \frac{1}{{\sqrt 3 }}$

⇒ ϕ = 30°

Power factor = cos 30° = 0.866

 

13. 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

 

14. A wattmeter is connected as shown in the figure. The wattmeter reads

A wattmeter’s current coil is connected in series with the load, whereas its pressure coil is connected across Z2. T

  1. Zero always
  2. Total power consumed by Z1 and Z2
  3. Power consumed by Z1
  4. Power consumed by Z2

Answer.4. Power consumed by Z2

Explanation:

Wattmeter reads the power and it is given by

P = VPC ICC cos ϕ

VPC is the voltage across the pressure coil

ICC is current flows through the current coil

ϕ is the phase angle between VPC and ICC

Application:

The circuit representation of the given question is as shown below.

A wattmeter’s current coil is connected in series with the load, whereas its pressure coil is connected across Z2. T

  • The potential coil is connected across Z2.
  • It reads the voltage across Z2 only.
  • So, Wattmeter reads only power consumed by Z2.

 

15. In a two-wattmeter method of measuring power in a balance 3-phase circuit, the ratio of the two wattmeter reading is 1 : 2. The circuit power factor is

  1. 0.707
  2. 0.5
  3. 0.866
  4. indeterminate

Answer.3. 0.866

Explanation:

Given that, W2 = 2W1

Power factor = cos ϕ

$\phi = {\rm{ta}}{{\rm{n}}^{ – 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} – {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)$

 

$\phi = {\tan ^{ – 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} – 2{W_1}} \right)}}{{\left( {{W_1} + 2{W_2}} \right)}}} \right)$

φ = 30°

Power factor = cos ϕ = 0.866

 

16. In the measurement of power of balanced load by two wattmeter method in a 3-phase circuit, The readings of the wattmeters are 4 kW and 2 kW respectively, the later is being taken after reversing the connections of current coil. the power factor and reactive power of the load is

  1. 0.2 & 6 kVAR
  2. 0.2 & 6 √3 kVAR
  3. 0.32 & 2 kVAR
  4. 0.32 & 2 √3 kVAR

Answer.2. 0.2 & 6 √3 kVAR

Explanation:

Given that,

Reversing the connections

W1 = 4 kW, W2 = – 2 kW

Total reactive power (Q) = √3[4 – (- 2)] = 6 √3 kVAR

$\phi = {\rm{ta}}{{\rm{n}}^{ – 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} – {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)$

 

$\phi = {\rm{ta}}{{\rm{n}}^{ – 1}}\left( {\frac{{\sqrt 3 \left( {4 – ( – 2)} \right)}}{{4 – 2}}} \right)$

 

ϕ = 79.10o

cosϕ = 0.2 (approx)

 

17. In a balanced 3-phase 200 V circuit, the line current is 115.5 A. When the power is measured by two wattmeter method, one of the wattmeter reads 20 kW and the other one reads zero. What is the power factor of the load?

  1. 0.5
  2. 0.6
  3. 0.7
  4. 0.8

Answer.1. 0.5

Explanation:

Given that, one of the wattmeter reads zero.

W2 = 0

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

 

$\tan \phi = \frac{{\sqrt 3 \left( {{W_1} – 0} \right)}}{{\left( {{W_1} + 0} \right)}}$

 

⇒ tan ϕ = √3

⇒ ϕ = tan-1 (√3) = 60°

Power factor = Cosφ = Cos60° = 1/2 = 0.5

 

18. The minimum number of wattmeter(s) required to measure 3-phase, 3-wire balanced or unbalanced power is

  1. 1
  2. 2
  3. 3
  4. 4

Answer.2. 2

Explanation:

According to Blondell’s theorem, the no. of watt meters to be required to measure the total power in n-phase system is either N (or) N–1.

When a separate neutral wire is available in the system the no. of watt meters to be required is N.

When the neutral wire is not available in the system, then the no. of watt meters to be required is (N–1).

Power is measured according to Blondel’s theorem

Phase/conductor

Wattmeter required

n

n-1

n with neutral wire

n

 

One line is acting as a common line for the return path. Hence the minimum number of wattmeters required is 2.

 

19. In a three-phase supply driving a load, what is the minimum number of single-phase watt meters required to measure the power and power factor?

  1. One
  2. Two
  3. Three
  4. Four

Answer.2. Two

Explanation:

According to this theorem for the n-wire system, (n – 1) single-phase wattmeters are used.

Hence for 3-phase power and power factor measurement, the total number of single-phase wattmeters used is (n – 1) = 3 – 1 = 2 wattmeter.

W1 = reading of 1st wattmeter = VLIL cos (30 + ϕ)

W2 = reading of 2nd wattmeter = VLIL cos (30 – ϕ)

Now, P3-ϕ = W1 + W2

 

20. In two wattmeter method of measuring 3 phase power, the power factor is 0.5, then one of the wattmeters will read

  1. W/2
  2. 0
  3. 0.577 W
  4. 1.414 W

Answer.2. 0

Explanation:

he reading of two wattmeters can be expressed as

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

(i) When PF is unity ( φ = 0°)

W1 =  VLILcos30°
W2 =  VLILcos30°

Both wattmeters read equal and positive reading i.e upscale reading

(ii) When PF is 0.5 (φ = 60°)

W1 =  VLILcos90° = 0
W2 =  VLILcos30°

Hence total power is measured by wattmeter W2 alone

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