AC Power Measurement MCQ || Measurement of AC Power MCQ Questions and Answers

11. Voltage applied to a load is 100√2 sin500t. Current through the load is 10√2sin(500t + π/3). The power consumed by the load is:

  1. 500 W
  2. 200 W
  3. 1000 W
  4. 2000 W

Answer.1. 500 W

Explanation:

v = 100√2  sin (500t) V

I = 10√2 sin (500t + π/3) A

Power factor angle φ = π/3 − 0 = π/3

Active power consumed by load = Vrms× Irms × cos ϕ

$\frac{{100\sqrt 2 }}{{\surd 2}} \times \frac{{10\sqrt 2 }}{{\surd 2}} \times \cos \frac{\pi }{3}$

P = 500 W

 

12. The 3-phase balanced source in the given figure delivers 1500W at a leading power factor of 0.844, then the values of ZL in Ohm is approximately

The 3-phase balanced source in the given figure delivers 1500W at a leading power factor of 0.844, then the values of ZL in Ohm is approximately

  1. 90 ∠32.44 deg
  2. 80 ∠32.44 deg
  3. 80 ∠-32.44 deg
  4. 90 ∠-32.44 deg

Answer.4. 90 ∠-32.44 deg

Explanation:

Calculation:

P = 1500 W, VL = 400 V,

cosϕ = 0.844

∴ ϕ = cos-1(0.844) = – 32.44° (Negative due to leading power factor)

∴ 1500 = √3 x 400 × IL × 0.844

I= 2.565 A

Given 3 – ϕ balanced source is in star connection

VL = √3 VP

= IP

Impedance ZL (per phase impedance) can be calculated as

|ZL| = VP/IP = 400/√3(2.165)

|ZL| = 90 Ω

ZL =  |ZL|∠φ = 90∠−32.44°

Z= 90∠-32.44 deg

 

13. An iron-cored choke coil has an equivalent resistance of 4 Ω. It draws 10A from a single-phase ac source of voltage 200 V, 50 Hz. Then, the power consumed by the coil and its power factor respectively are:

  1. 200 W, 0.2 lag
  2. 400 W, 0.2 lag
  3. 200 W, 0.2 lead
  4. 400 W, 0.2 lead

Answer.2. 400 W, 0.2 lag

Explanation:

The Power consumed by the coil is given as

P = V × I × Cosφ = I2×R

Where,

V = voltage across the coil,

I = current through the coil,

cos ϕ = power factor angle

R = resistance offered by the choke coil

Power consumed by the coil,

P = I2 R = 102 × 4

400 W

Using the Power formula

P = V × I × cos ϕ

⇒ 400 = 200 × 10 × cos ϕ

⇒ cos ϕ = 0.2

power factor = cos ϕ = 0.2 lag

 

14. A wattmeter has a full-scale range of 2500 Watts. It has an error of 1.2% of true value. What would be a range of reading if the true power is 1250 Watts.

  1. 1235 W to 1265 W
  2. Exactly 1250 W
  3. 1200 W to 1300 W
  4. 1500 W to 2500 W

Answer.1. 1235 W to 1265 W

Explanation:

We have,

Error = 1.2 % of true value

Wattmeter has full scale range of 2500 W

True Power Measured = 1250 W

True Power Measured = 1250

⇒ Error = ±1.2 % of 1250 = 0.012 × 1250 = ±15 W

Hence, range of Wattmeter reading will be,

From (1250 – 15) W to (1250 + 15) W

∴ Reading of wattmeter = 1235 W to 1265 W

 

15. A wattmeter reads 10 kW when its current coil is connected in R phase and the potential coil is connected across R and neutral of a balanced 400 V (RYB sequence) supply. The line current is 54 A. If the potential coil reconnected across B-Y phases with the current coil in R phase, the new reading of the wattmeter will be nearly

  1. 10 kW
  2. 13 kW
  3. 16 kW
  4. 19 kW

Answer.2. 13 kW

Explanation:

Line voltage (V) = 400 V

Line current (I) = 54 A

When the current coil is connected in the R phase and the potential coil is connected across R and neutral.

Wattmeter reading (P1) = VRN IR cos ϕ1

ϕ1­ = angle between VRN and IR

=  10 × 103 = 400/√3 × 54 × Cosφ1

⇒ cos ϕ1 = 0.8

When the current coil is connected to R phase and potential coil between B and Y phase.

P2 = VBY IR cos ϕ2

A wattmeter reads 10 kW, when its current coil is connected in R phase

Φ2 = angle between VBY and IR

P2 = VBY IR cos (90 + ϕ1)

= 400 × 54 × sin ϕ1

= 400 × 54 × 0.6

= 12.96 kW

 

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

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

Answer.1. 0.45%

Explanation:

The circuit for the above question can be drawn as

Calculate the percentage error for a wattmeter that is so connected

Given,

Rc = 0.03 Ω

Rp = 6000 Ω

Where Rc and Rp are current coil resistance and potential coil resistance respectively.

IL = 20 A

VL = 220 V

cos ϕ = 0.6

Actual Power measured by wattmeter (Pa) is given as,

Pa = VIcos ϕ = 220 × 20 × 0.6 = 2640 W

Therefore error is due to the current coil is given as,

E = I2L×RC = 202 × 0.03 = 12 W

Wattmeter reading

Pm = Pa + e = 2640 W + 12 W = 2652 W

error = (Pm −Pa)/Pa = (2652 − 2640)/2640

e = 12/2640

%error = 0.45

 

17. Which of the following devices are required to measure three-phase balanced power?

  1. One wattmeter
  2. One wattmeter and one voltage transformer of 1 : 1 ratio
  3. One wattmeter and two voltmeters
  4. None of the above

Answer.1. One wattmeter

Explanation:

  • The one-wattmeter method is used for the measurement of three-phase power in case of balanced loads only. It is used for both star and delta connected loads.
  • Two wattmeter methods and three wattmeter methods are used for both balanced and unbalanced loads.
  • Two wattmeter method is best suitable for the three-phase three-wire system and the three wattmeter method is suitable for the three-phase four-wire system.

 

18. A circuit has an impedance of 22 Ω and draws a current of 10 amperes at 0.8 power factor lagging. Which one of the following will be the apparent power?

  1. 2.2 kVA
  2. 22 kVA
  3. 220 kVA
  4. 1.76 kVA

Answer.1. 2.2 kVA

Explanation:

Given-

Z = 22 Ω

Irms = 10 A

cosϕ = 0.8

Apparent Power:

S = (Vrms Irms) = (Irms )2 Z

∴ S = (10)2 x 22

S = 2200 VA

S = 2.2 kVA

 

19. The circuit is used to measure the power consumed by the load. Resistance of the current and potential coils of the wattmeter is 0.02 Ω and 1000 Ω respectively. With this circuit measured power compared to load power will be:

  1. 0.2 percent less
  2. 0.4 percent more
  3. 0.4 percent less
  4. 0.2 percent more

Answer.4. 0.2 percent more

Explanation:

Let’s assume voltage across the load as 200 V and current entering the load or current coil be 20 A.

Also, the actual voltage and currents are V1 and Irespectively.

potential coils of the wattmeter is 0.02 Ω and 1000 Ω respectively.

Let,

I = 20 A

V = 200 V

Rpc = 1000 Ω

Rcc = 0.02 Ω

Using Current division,

$I = {I_1}(\frac{{{R_{pc}}}}{{{R_{pc}} + {R_{cc}}}})$

$20 = {I_1}[\frac{{1000}}{{1000 + 0.02}}]$

I1 = 20.0004 A ≈ 20 A

Applying KVL to the circuit,

V1 = V + I(Rcc)

V1 = 200 + (0.02) (20) = 200.4 V

Measured Power, Pm

= V1 I1 = 200.4 × 20 = 4008 W

True Power, Pt = V I

= 200 x 20 = 4000 W

% Change or % Error is given as (%E),

error = (Pm −Pa)/Pa = (4008 − 4000)/4000

e = 8/4000

%error = 0.2

Since, P> Pt, so, 0.2 % more.

 

20. The voltage across a load is 200 V and the current flowing through it is 50 A at a lagging power factor of 0.8. The wattmeter used to measure the power consumed by the load is connected such that its current is in series order with the coil load and the voltage is at full load in the side of the coil load. The resistances of current and voltage coils are 0.01 Ω and 1000 Ω respectively. What will be the reading of the wattmeter?

  1. 4040 W
  2. 8040 W
  3. 8400 W
  4. 4400 W

Answer.2. 8040 W

Explanation:

%Error = Load in Current Coil/True Power

The voltage across a load is 200 V and the current flowing through it is 50 A at a l

Given that pressure coil resistance RPC = 1000Ω

Current coil RCC = 0.01Ω

Power factor = 0.8

Error in wattmeter = V2/RPC

= (2000)2/1000 = 40 W

Reading = 200 × 50 × 0.8 + 40 = 8040 W

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