Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

1.

Creeping in an energy meter can be found using _________(a) creep adjustment(b) preliminary light load adjustment(c) full load u.p.f adjustment(d) light load adjustmentI got this question during an interview.The above asked question is from Electronic Energy Meter, Adjustments in Single Phase Energy meters topic in chapter Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Correct option is (b) preliminary LIGHT LOAD adjustment

To explain I would SAY: ENERGY meter can be tested for creeping using preliminary light load adjustment. Disc HOLES are so positioned that they aren’t under the poles of a series magnet.

2.

Preliminary light load adjustment involves _________(a) applying rated voltage across current coil(b) making use of a light load(c) applying rated voltage across pressure coil(d) adjusting the light loadThe question was asked by my college director while I was bunking the class.My question is based upon Electronic Energy Meter, Adjustments in Single Phase Energy meters topic in chapter Measurement of Power and Related Parameters of Electrical Measurements

Answer» CORRECT answer is (c) applying rated voltage across pressure coil

Best explanation: Rated voltage is applied across the pressure coil. No current FLOWS through the current coil. Till the disc stops rotating, light LOAD device or equipment is ADJUSTED continuously.
3.

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/\(\sqrt{3}\)The question was posed to me in final exam.My question is from Advanced Miscellaneous Problems on Measurement of Power topic in chapter Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Right ANSWER is (d) 10/\(\SQRT{3}\)

To EXPLAIN I WOULD say: For triangular WAVE- Average value = \(\frac{V_m}{3}\), rms value = \(\frac{V_m}{\sqrt{3}}\)

\(\frac{V_m}{3}\) = 10 V or, Vm = 30 V

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

4.

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) 0.334I got this question in an online interview.My enquiry is from Advanced Miscellaneous Problems on Measurement of Power in chapter Measurement of Power and Related Parameters of Electrical Measurements

Answer» CORRECT choice is (d) 0.334

The EXPLANATION: w1 = 10.5 kW, W2 = -2.5 kW

W = w1 + w2 = 8 kW

Also, tan ∅ = 2.81

∴ ∅ = 70.43° or, cos ∅ = 0.3347.
5.

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) P1 = 35 kW, P2 = 10 kW(b) P1 = 47.25 kW, P2 = -2.25 kW(c) P1 = 39.82 kW, P2 = 5.179 kW(d) P1 = 45 kW, P2 = 0 kWThis question was addressed to me in an interview.I'm obligated to ask this question of Advanced Miscellaneous Problems on Measurement of Power in division Measurement of Power and Related Parameters of Electrical Measurements

Answer» CORRECT option is (c) P1 = 39.82 kW, P2 = 5.179 kW

Easy explanation: P1 + P2 = 45 kW

And, cos ∅ = 0.6

∴ tan ∅ = 1.33

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

∴ P1 – P2 = 34.64 kW

∴ P1 = 39.82 kW

P2 = 5.179 kW.
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 rpmThis question was addressed to me in class test.This interesting question is from Advanced Miscellaneous Problems on Measurement of Power topic in division Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Right OPTION is (c) 12 rpm

For 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 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 % moreI had been asked this question by my school teacher while I was bunking the class.Origin of the question is Advanced Miscellaneous Problems on Measurement of Power in chapter Measurement of Power and Related Parameters of Electrical Measurements

Answer»

The correct CHOICE is (c) 0.2 % more

For EXPLANATION: Power indicated by watt-meter = 400 × 20 + 20^2 × 0.02 = 4008 W

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

8.

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) W1 = 886 W and W2 = 886 W(b) W1 = 500 W and W2 = 500 W(c) W1 = 0 W and W2 = 1000 W(d) W1 = 250 W and W2 = 750 WThis question was posed to me in an internship interview.The question is from Advanced Miscellaneous Problems on Measurement of Power in portion Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Correct option is (c) W1 = 0 W and W2 = 1000 W

To explain I WOULD SAY: W1 = 100 × 20 COS 90° = 0

W2 = \(\frac{100 × 20 × \sqrt{3}}{\sqrt{3}× 2}\) = 1000 W.

9.

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 WI got this question during an interview.My doubt stems from Advanced Miscellaneous Problems on Measurement of Power in chapter Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Correct option is (a) Zero

Easiest 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.

10.

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 kWThis question was addressed to me in unit test.The question is from Advanced Miscellaneous Problems on Measurement of Power in chapter Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Right option is (d) 19 KW

Easiest explanation: w1 = 20.5 kW, W2 = -3.5 kW

W = w1 + w2 = 20.5 – 3.5 = 17 kW.

11.

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 timesI had been asked this question in homework.Question is taken from Advanced Miscellaneous Problems on Measurement of Power in section Measurement of Power and Related Parameters of Electrical Measurements

Answer» RIGHT CHOICE is (d) P decreases 100 TIMES and Q is increased 10 times

Explanation: ω2 L = 10 ω1 LR will remain constant

∴ Q2 = \(\frac{10 ω_1L}{R}\) = 10 Q1

That is Q is increased 10 times.

Now, I1 = \(\frac{V}{ω_1L}\)

For a high Q coil, ωL ≫> R,

I2 = \(\frac{V}{10 ω_1L} = \frac{I_1}{10}\)

∴ P2 = \(R (\frac{I_1}{10})^2 = \frac{P_1}{100}\)

Thus, P decreases 100 times and Q is increased 10 times.
12.

The figure shows a three-phase, delta connected load supplied from a 220 V, 50 Hz, 3-phase balanced source. The pressure Coil (PC) and Current Coil (CC) of a watt-meter are connected to the load as shown, with the coil polarities suitably selected to ensure a positive deflection. The watt-meter reading will be?(a) Zero(b) 1600 W(c) 242 W(d) 400 WThis question was posed to me in final exam.I'd like to ask this question from Advanced Problems on Measurement of Power in Polyphase Circuit topic in portion Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Correct CHOICE is (c) 242 W

To elaborate: Watt-meter reading = Current through CC × Voltage ACROSS PC × cos (PHASE angle).

IBR = ICC = \(\frac{220∠120°}{100°}\) = 2.2∠120°

VYB = VPC = 220∠-120°

w = 2.2∠120°× 220∠-120° × cos 240° = 242 W.

13.

Choose the correct statement regarding two watt-meter method for power measurements in 3-phase circuit.(a) When power factor is unity, one of the wattmeters reads zero(b) When the readings of the two watt-meters are equal but opposite sign, the power factor is zero(c) Power can be measured using two watt-meter method only for star connected 3-phase circuit(d) When two watt-meters show identical readings, the power factor is 0.5The question was asked by my school principal while I was bunking the class.I would like to ask this question from Advanced Problems on Measurement of Power in Polyphase Circuit topic in section Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Right option is (b) When the readings of the two watt-meters are EQUAL but opposite sign, the POWER factor is ZERO

The explanation is: When power factor is 0, we have ∅ = 90°

P1 = (3)^0.5 VIcos (30° – ∅) = (3)^0.5 VIcos (30° – 90°) = \(\frac{(3)^{0.5}}{2}\) × VI

P2 = (3)^0.5 VIcos (30° + ∅) = (3)^0.5 VIcos (30° + 90°) = \(– \frac{(3)^{0.5}}{2}\) × VI

∴Total powerP = 0

So, with zero power factor the readings of the two watt-meters are equal but of opposite sign.

14.

In the measurement of 3-phase power by 2 watt-meter method, if the 2 watt-meter readings are equal, the power factor of the circuit is?(a) 0.8 lagging(b) 0.8 leading(c) Zero(d) UnityThe question was posed to me in an internship interview.I'm obligated to ask this question of Advanced Problems on Measurement of Power in Polyphase Circuit in portion Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Right ANSWER is (d) Unity

Best EXPLANATION: Power factor, cosθ = cos arc TAN (3)^0.5 \(\frac{W_1-W_2}{W_1+W_2}\)

W1 = W2 = cos 0° = 1.

15.

The meter constant of a single-phase, 230 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) 25.02 rpm(c) 27.6 rpm(d) 33.1 rpmThe question was asked in an interview.I need to ask this question from Advanced Problems on Measurement of Power in Polyphase Circuit in portion Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Right choice is (c) 27.6 rpm

Best explanation: Meter CONSTANT = \(\frac{NUMBER \,of\, revolution}{Energy} = \frac{600 × 230 × 15 × 0.8}{1000}\) = 1656

∴ Speed in rpm = \(\frac{1656}{60}\) = 27.6 rpm.

16.

The two-wattmeter method is used to measure active power. The system is a 3-phase, 3- wire system. Then, the power reading is?(a) Affected by both negative sequence and zero sequence voltages(b) Affected by negative sequence voltages but not by zero sequence voltages(c) Affected by zero sequence voltages but not by negative sequence voltages(d) Not affected by negative or zero sequence voltagesI have been asked this question in unit test.My question is based upon Advanced Problems on Measurement of Power in Polyphase Circuit topic in division Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Correct option is (d) Not affected by NEGATIVE or zero sequence voltages

For explanation I would say: If the PHASE voltage is unbalanced, then the power READING is not affected by both negative as well as zero sequence voltages. This is a characteristic property of the 2 watt-METER meter since it measures the active power on a three phase three wire system.

17.

The ratio of the reading of 2 watt-meters connected to measure power in a balanced 3-phase load is 2: 1 and the load is inductive. The power factor of load is?(a) 0.866 lag(b) 0.615 lead(c) 0.866 lead(d) 0.625 lagThe question was asked by my college professor while I was bunking the class.I'm obligated to ask this question of Advanced Problems on Measurement of Power in Polyphase Circuit topic in chapter Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Correct choice is (a) 0.866 LAG

Easy explanation: Power factor, cosθ = COS arc tan (3)^0.5 \(\FRAC{W_1-W_2}{W_1+W_2}\)

Since, \(\frac{W_1}{W_2} = \frac{2}{1}\)

So, power factor = cos arc tan (3)^0.5 \(\frac{2-1}{2+1}\)

= cos 30° = 0.866 lag.

18.

A single-phase load is connected between R and Y terminals of a 220 V, symmetrical, 3-phase, 4 wire systems with phase sequence RYB. A watt-meter is connected in the system. The watt-meter will read (pf = 0.8 lagging)?(a) – 795 W(b) – 168 W(c) + 597 W(d) + 795 WI have been asked this question in homework.I want to ask this question from Advanced Problems on Measurement of Power in Polyphase Circuit in chapter Measurement of Power and Related Parameters of Electrical Measurements

Answer»

The CORRECT CHOICE is (b) – 168 W

Easiest EXPLANATION: VRY = 220∠30°

IRY = \(\frac{220∠30°}{100∠36.86°}\) = 2.2∠-6.68° A

Power = VBN IRY = COS∅ = \(\frac{220}{(3)^{0.5}}\)× 2.2 cos (126.86°)

= -167.6 W ≈ – 168 W.

19.

Power flowing in 3-phase, 3-wire system is measured by two wattmeter whose readings are 7000 W and -2500 W. if the voltage of the circuit is 400 V, then what will be the value of capacitance introduced in each phase to make one wattmeter reads zero? The frequency is 50 Hz.(a) 500 μF(b) 668.6 μF(c) 1000 μF(d) 748.5 μFThe question was asked by my school teacher while I was bunking the class.This question is from Advanced Problems on Measurement of Power in Polyphase Circuit topic in section Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Right CHOICE is (b) 668.6 μF

Explanation: P = P1 + P2 = 7000 – 2500 = 4500 W

∴ Power in each PHASE = 1500 W

Voltage of each phase = 231 V

So, cos∅ = 0.264

∴ Current in each phase = 24.6 A

∴ Z of each phase = 9.39 Ω

∴ R of each phase = 2.48 Ω

Reactance of each phase = 9.056 Ω

For one watt-meter to read zero, power factor should be 0.5

Now, tan∅ = \(\frac{X}{R}\) or, X = 4.29 Ω

∴ Capacitance reactance required = 9.056 – 4.29 = 4.76 Ω

∴ Capacitance C = \(\frac{1}{2π × 50 × 4.76}\) F = 668.6 μF.

20.

Two watt meters, which are connected to measure the total power on a three-phase system supplying a balanced load, read 10.5 kW and – 2.5 kW respectively. What is the total power?(a) 13.0 kW(b) 13.0 kW(c) 8.0 kW(d) 8.0 kWI have been asked this question during an interview for a job.I want to ask this question from Advanced Problems on Measurement of Power in Polyphase Circuit topic in chapter Measurement of Power and Related Parameters of Electrical Measurements

Answer»
21.

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) 1.0This question was posed to me in an interview for job.My question comes from Advanced Problems on Measurement of Power in Polyphase Circuit in chapter Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Correct answer is (b) 0.5

To EXPLAIN: 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.

22.

Disc rotation is determined by ________(a) the supply voltage(b) an arrow(c) the turns ratio(d) the load currentI have been asked this question in exam.This question is from Phase Sequence Indicator topic in division Measurement of Power and Related Parameters of Electrical Measurements

Answer» CORRECT choice is (B) an arrow

Easy explanation: Direction of rotation of the disc is determined by making USE of an arrow that is MARKED on the disc. The phase sequence of the supply voltage is same as that mentioned on the TERMINALS of the meter provided disc rotates in the same direction as the arrow head.
23.

Eddy e.m.f produces a torque.(a) True(b) FalseI got this question in unit test.The origin of the question is Phase Sequence Indicator in portion Measurement of Power and Related Parameters of Electrical Measurements

Answer»

The correct answer is (a) True

The best I can explain: A rotating MAGNETIC field is PRODUCED when the three coils are excited by a supply of three phase. As a result, eddy e.m.f circulates in the DISC. A torque is produced as a result of the interaction between the eddy currents and the rotating magnetic field.

24.

The three coils are ________(a) star connected(b) delta connected(c) not connected(d) shortedThis question was posed to me in an internship interview.This key question is from Phase Sequence Indicator in section Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Right choice is (a) star connected

Explanation: In a rotating type phase sequence indicator, the THREE COILS are connected in the FORM of a star and EXCITED by a three-phase SUPPLY.

25.

Excitation of the three coils produces ________(a) a static magnetic field(b) a rotating magnetic field(c) a static electric field(d) a rotating electric fieldThis question was posed to me by my school principal while I was bunking the class.My question comes from Phase Sequence Indicator in portion Measurement of Power and Related Parameters of Electrical Measurements

Answer»
26.

A rotating phase sequence indicator consists of ________(a) 1 coil(b) 2 coils(c) 5 coils(d) 3 coilsI have been asked this question in examination.I'm obligated to ask this question of Phase Sequence Indicator topic in portion Measurement of Power and Related Parameters of Electrical Measurements

Answer»

The correct choice is (d) 3 coils

The best I can explain: A rotating type of phase sequence indicator basically consists of three coils. These are mounted at 120º to each other is space. The ENDS of the coils are CONNECTED to the terminals R, Y, and B RESPECTIVELY.

27.

How many types of phase sequence indicators are there?(a) 1(b) 5(c) 2(d) 10I had been asked this question in class test.This is a very interesting question from Phase Sequence Indicator topic in portion Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Right choice is (c) 2

To explain: Generally, two types of phase sequence indicators are used. The FIRST ONE is the ROTATING type while the second type is the STATIC type. Usually the rotating type of phase sequence INDICATOR is used.

28.

Synchronization of A.C. supply means __________(a) different phase sequence(b) same phase sequence(c) zero phase(d) using a transformerThis question was posed to me in exam.The doubt is from Phase Sequence Indicator in division Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Right choice is (b) same phase sequence

The EXPLANATION is: During synchronization, the phase sequences of any two GIVEN SUPPLIES are maintained equal. This is ENSURED by MAKING use of a phase sequence indicator.

29.

Phase sequence indicator gives the maximum value of phase voltages.(a) True(b) FalseThis question was addressed to me in exam.This interesting question is from Phase Sequence Indicator in section Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Right option is (a) True

The EXPLANATION is: The phase sequence for a three phase supply indicates the order in which MAXIMUM values of phase voltages ER, EY, and EB occur. A three phase supply can be in EITHER RYB or RBY configuration.

30.

Full load u.p.f adjustment involves _________(a) adjusting the loads at unity power factor(b) applying rated voltage to pressure coil and a p.f. of unity for current coil(c) only applying rated voltage to pressure coil(d) only a p.f. of unity for the current coilThe question was posed to me in an online quiz.My question comes from Electronic Energy Meter, Adjustments in Single Phase Energy meters in division Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Right option is (c) only applying rated voltage to PRESSURE coil

To explain: Rated voltage is applied to the pressure coil. The current coil is provided with a current at UNITY p.f. ERRORS are kept minimum and the POSITION of the BRAKE magnet is so adjusted that disc rotates at the correct speed.

31.

Low power factor adjustment involves _________(a) adjusting the power factor at lower loads(b) applying rated voltage to pressure coil and a p.f. of 0.5 for current coil(c) only applying rated voltage to pressure coil(d) only a p.f. of 0.5 for the current coilThe question was posed to me during an internship interview.My doubt stems from Electronic Energy Meter, Adjustments in Single Phase Energy meters in portion Measurement of Power and Related Parameters of Electrical Measurements

Answer»

The correct CHOICE is (b) applying rated voltage to pressure coil and a p.f. of 0.5 for current coil

Best EXPLANATION: Rated voltage is applied to the pressure coil. The current coil is provided with a current at 0.5 p.f. lagging. TILL the disc rotates at correct speed, LAG device is ADJUSTED.

32.

Light load adjustment involves _________(a) applying rated voltage across the pressure coil(b) adjusting a light load(c) applying rated current across the transformer(d) applying rated voltage across the current coilThis question was addressed to me in quiz.This is a very interesting question from Electronic Energy Meter, Adjustments in Single Phase Energy meters in chapter Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Right option is (a) applying RATED VOLTAGE across the pressure coil

Explanation: Disc rotation is adjusted in such a WAY that CORRECT speed is maintained. The pressure coil is supplied with the rated voltage and the current coil is PROVIDED with only about 5 % of the full load at u.p.f.

33.

Creep adjustment involves _________(a) adjusting the creep(b) exciting the current coil(c) adjusting the turns ratio(d) exciting the pressure coilThe question was posed to me in an online interview.My enquiry is from Electronic Energy Meter, Adjustments in Single Phase Energy meters topic in portion Measurement of Power and Related Parameters of Electrical Measurements

Answer»
34.

What is the role of a multiplier?(a) it multiplies the voltage and current(b) divides the alternating voltage and current(c) supplies instantaneous voltage and current(d) multiplies alternating voltage and currentThis question was addressed to me by my college director while I was bunking the class.This interesting question is from Electronic Energy Meter, Adjustments in Single Phase Energy meters in section Measurement of Power and Related Parameters of Electrical Measurements

Answer»

The correct choice is (d) multiplies alternating VOLTAGE and CURRENT

Explanation: A multiplier basically performs the multiplication of the alternating voltage and the current. Multiplier also PROVIDES the current in the form of INSTANTANEOUS power to a voltage CONTROLLED oscillator.

35.

Energy meter can be directly used in measurement.(a) True(b) FalseThe question was posed to me in an online quiz.The doubt is from Electronic Energy Meter, Adjustments in Single Phase Energy meters topic in chapter Measurement of Power and Related Parameters of Electrical Measurements

Answer» RIGHT choice is (b) False

The explanation is: Adjustments need to be made in an energy meter before it is used for the measurement of energy. This is done in order to keep the ERRORS DUE to measurement WITHIN allowable limits of ±5 %.
36.

An electronic energy meter is advantageous compared to conventional ones.(a) True(b) FalseThe question was posed to me in a national level competition.This is a very interesting question from Electronic Energy Meter, Adjustments in Single Phase Energy meters topic in portion Measurement of Power and Related Parameters of Electrical Measurements

Answer»

The correct option is (a) True

To explain: An ELECTRONIC energy meter does not have frictional losses, creeping is not NEEDED irrespective of the nature of the load such as LOW load, full load POWER factor, etc and the accuracy in the reading is of the ORDER of ±1%.

37.

Frequency of oscillation in an electronic energy meter depends on __________(a) output current of multiplier(b) output voltage of multiplier(c) output power of multiplier(d) input resistance of multiplierThis question was posed to me in an international level competition.This is a very interesting question from Electronic Energy Meter, Adjustments in Single Phase Energy meters in section Measurement of Power and Related Parameters of Electrical Measurements

Answer»
38.

Analog signal is converted _________(a) into oscillations(b) into digital(c) into pulses(d) into currentThis question was posed to me in an online quiz.This question is from Electronic Energy Meter, Adjustments in Single Phase Energy meters in portion Measurement of Power and Related Parameters of Electrical Measurements

Answer»

The correct ANSWER is (b) into DIGITAL

The explanation is: The analog signal OBTAINED in an electronic energy meter is converted into digital by MAKING use of a digital circuit. By making use of a seven-segment display, energy is MENTIONED in watt-hours.

39.

Average power is _________(a) product of voltage and current(b) product of average current and voltage(c) product of instantaneous voltage and current(d) product of absolute voltage and currentI had been asked this question in semester exam.Question is taken from Electronic Energy Meter, Adjustments in Single Phase Energy meters topic in portion Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Right answer is (c) PRODUCT of instantaneous voltage and current

For explanation: The AVERAGE power is COMPUTED as the product of the instantaneous voltage ACROSS the load and the instantaneous current flowing through the load. A scaling device is used to bring the supply voltage to a proper LEVEL.

40.

Measurement of energy involves _________(a) inductance and capacitance measurement(b) power consumption and time duration(c) resistance measurement and voltage drop(d) current consumption and voltage dropI have been asked this question in exam.The question is from Electronic Energy Meter, Adjustments in Single Phase Energy meters topic in chapter Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Right answer is (b) power consumption and time duration

Explanation: An ELECTRONIC energy meter is used in two stages. Energy measurement basically involves the measurement of power and the time duration. In the first stage, it is used as a WATTMETER while in the SECOND stage it is used MONITORING the power CONSUMED in a time interval.

41.

An electronic energy meter makes use of ___________(a) IC(b) Transformer(c) CRO(d) MultimeterI got this question in a job interview.I want to ask this question from Electronic Energy Meter, Adjustments in Single Phase Energy meters in section Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Correct option is (a) IC

For EXPLANATION: BASICALLY, integrated circuits ALSO known as IC’s are used for the operation of an ELECTRONIC energy meter.

42.

In some energy meters, creeping can be avoided by __________(a) attaching small gold pieces(b) attaching small aluminium pieces(c) attaching small iron pieces(d) attaching small zinc piecesThis question was posed to me by my school principal while I was bunking the class.Asked question is from Errors in Single Phase Energy Meters in chapter Measurement of Power and Related Parameters of Electrical Measurements

Answer»

The correct ANSWER is (c) ATTACHING small iron pieces

For explanation: By attaching some iron pieces to the edge of the disc, CREEPING can be limited in some energy meters. Force of attraction that is EXPERIENCED by the brake magnet as a result of the iron piece is ENOUGH to eliminate the creeping.

43.

Creeping is avoided by __________(a) reversing the polarity of the voltage(b) drilling two diametrically opposite holes(c) holding the disc(d) increasing the frictionI have been asked this question by my school principal while I was bunking the class.Enquiry is from Errors in Single Phase Energy Meters in division Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Right answer is (B) drilling two diametrically opposite holes

Easiest explanation: In an energy meter, creeping causes the disc to rotate even when there is no current FLOWING. By drilling two diametrically opposite holes under the edge of the poles of a shunt magnet, rotation of the disc is limited to a MINIMUM VALUE.

44.

Disc rotates slowly in some energy meters.(a) True(b) FalseI got this question during an interview.This question is from Errors in Single Phase Energy Meters topic in division Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Right option is (a) True

Easiest explanation: Even when there is no current FLOW through the ENERGY meter, disc rotates SLOWLY. This is known as creeping. This is occurs as a result of the over compensation PROVIDED for friction.

45.

How can temperature effect be compensated in an energy meter?(a) through heat sinks(b) by a temperature shunt(c) by using resistance(d) by using a coolantI got this question in examination.I'd like to ask this question from Errors in Single Phase Energy Meters topic in division Measurement of Power and Related Parameters of Electrical Measurements

Answer» RIGHT option is (b) by a temperature shunt

To explain: The resistance of the copper and aluminium parts in an energy METER increase with an increase in the temperature. As a result the disc rotates with a speed that is higher than ACTUAL. Temperature EFFECTS can be compensated by making USE of a temperature shunt on the brake magnet.
46.

Energy meter creeps __________(a) due to change in supply(b) due to reversal in polarity of voltage(c) due to asymmetry in magnetic circuit(d) due to turns ratio of transformerThis question was addressed to me by my school teacher while I was bunking the class.My query is from Errors in Single Phase Energy Meters in section Measurement of Power and Related Parameters of Electrical Measurements

Answer»

Right option is (c) due to ASYMMETRY in MAGNETIC circuit

Explanation: In an ENERGY meter, when the magnetic circuit is ASYMMETRICAL, a driving TORQUE is produced. As a result of this driving torque, the energy meter creeps.

47.

How is the flux of shunt coil related to voltage?(a) flux is proportional to square of voltage(b) directly proportional(c) inversely proportional(d) independent of each otherThis question was posed to me during an interview.I want to ask this question from Errors in Single Phase Energy Meters in section Measurement of Power and Related Parameters of Electrical Measurements

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The correct ANSWER is (a) flux is PROPORTIONAL to SQUARE of voltage

The best I can EXPLAIN: In an energy meter, the supply voltage may fluctuate as a result of unavoidable reasons leading to errors in the reading. Supply voltage causes the shunt flux to induce an e.m.f in the disc. This is results in a self braking TORQUE proportional to square of the voltage.

48.

Supply voltage in an energy meter is __________(a) constant always(b) zero always(c) depends on the load(d) can fluctuateThe question was asked in a national level competition.The origin of the question is Errors in Single Phase Energy Meters in portion Measurement of Power and Related Parameters of Electrical Measurements

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49.

Phase angles in an energy meter cannot be incorrect.(a) True(b) FalseThis question was addressed to me during an internship interview.This key question is from Errors in Single Phase Energy Meters in division Measurement of Power and Related Parameters of Electrical Measurements

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Correct ANSWER is (b) False

The explanation: In an energy meter, phase angle errors occur as a RESULT of improper adjustments of lag condition, abnormal FREQUENCIES etc. Due to temperature, changes in resistance VALUES ALSO lead to error in the phase angle.

50.

Magnitude of flux in an energy meter varies __________(a) due to abnormal currents and voltages(b) due to high resistance and inductance values(c) due to changes in the transformer turns(d) due to the induced e.m.f in the windingsThis question was addressed to me during an interview.This intriguing question originated from Errors in Single Phase Energy Meters topic in chapter Measurement of Power and Related Parameters of Electrical Measurements

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Correct option is (a) DUE to ABNORMAL currents and voltages

For explanation: In the driving system of an ENERGY meter, magnitude of FLUX can be INCORRECT as a result of abnormal values of currents and voltages. This occurs due to a change in the resistance of the pressure coil circuit.