300+ REAL TIME Electrical Measurements Objective Questions & Answers 2023

250+ TOP MCQs on Friction and Overload Compensation and Answers

Electrical Measurements & Measuring Instruments Multiple Choice Questions on “Friction and Overload Compensation”.

1. Energy meter reads correctly when the ___________
a) torque is small
b) torque is large
c) torque is medium
d) torque is zero
Answer: a
Clarification: The reading in an energy meter is obtained correctly when the torque value is small at low loads. It is independent of the load on the meter an acts in the same direction as the driving torque.

2. Small torque for energy meter is provided __________
a) by a supply
b) by a shading loop
c) by unshaded loop
d) by a transformer
Answer: b
Clarification: Friction can be compensated in an energy meter by making use of a shading loop. It is placed between the central limb of the shunt magnet and a disc. Figure below illustrates the same.

3. Driving torque is small and is adjustable.
a) True
b) False
Answer: a
Clarification: Interaction between the parts of the shaded and unshaded fluxes, obtained through a shading loop leads to a small driving torque. The value of the torque can be adjusted through lateral movement of the loop.

4. Friction torque is eliminated by _________
a) using lubricating oil
b) by suspending the components in air
c) by adjusting the position of limb
d) by using steel alloy components
Answer: c
Clarification: We can eliminate the friction torque completely by adjusting the position of the shading loop. This enables in providing compensation for the frictional torque.

5. Frictional errors are dominant in an energy meter.
a) True
b) False
Answer: a
Clarification: Frictional errors exist in an energy meter at the top as well as bottom surfaces even at low value of loads. Even when the disc is rotating slowly errors due to friction exist in an energy meter.

6. At full load, disc __________
a) partially revolves and then stops
b) continuously revolves
c) does not revolve at all
d) revolves in an alternating fashion
Answer: b
Clarification: When an energy meter is operated in full load condition, disc revolves continuously due to the field of the series magnet. As a result, an e.m.f is induced in the disc.

7. Self braking torque is _________
a) proportional to cube of load current
b) proportional to load current
c) proportional to square of load current
d) proportional to reciprocal of load current
Answer: c
Clarification: In an energy meter, the self braking torque is dependent on the square of the load current. As a result the disc rotates at a slightly slower speed at high value of loads.

8. Self braking action is minimised by _________
a) maintaining high speed for disc
b) maintaining medium speed for disc
c) keeping the disc at rest
d) maintaining low speed for disc
Answer: d
Clarification: In an energy meter we can minimise or eliminate the self braking action by keeping the disc speed as low as possible at full load condition. This is achieved by maintaining the flux due to the current coil smaller than that due to the shunt coil. The figure below shows the compensating device.

9. Overload compensating devices is _________
a) in the form of a magnetic shunt
b) in the form of a series magnet
c) in the form of a transformer
d) in the form of a supply
Answer: a
Clarification: Magnetic shunt reaches saturation at overloads. As a result, its permeability reduces. Hence the overload compensating device takes the form of a magnetic shunt.

250+ TOP MCQs on Potentiometric Integrating Type DVM and Answers

Electrical Measurements & Measuring Instruments Multiple Choice Questions on “Potentiometric Integrating Type DVM”.

1. In a potentiometric DVM ________
a) voltage is compared
b) current is compared
c) resistance is compared
d) power is compared
Answer: a
Clarification: Potentiometric integrating type DVM employs voltage comparison. The unknown voltage is compared with a reference. The reference value is set by a calibrated potentiometer.

2. How to obtain balance?
a) by using a detector
b) by changing the pot setting
c) by supplying voltage
d) by using a transformer
Answer: b
Clarification: Balance is obtained in a potentiometric integrating type DVM by adjusting the settings of the potentiometer. The dial setting of the potentiometer gives the value of the unknown voltage.

3. In a potentiometric DVM, balance is obtained manually.
a) True
b) False
Answer: b
Clarification: Balance is obtained in a potentiometric integrating type DVM by adjusting the settings of the potentiometer. Hence in a potentiometric integrating type DVM, the balance is obtained automatically.

4. Unknown voltage is __________
a) converted to current
b) boosted
c) filtered
d) measured using a voltmeter
Answer: c
Clarification: In a potentiometric integrated type DVM, the unknown voltage is filtered. It is also attenuated to a suitable magnitude. This forms the input for the comparator. A chopper is usually used as an error detector.

5. How is the reference voltage obtained?
a) from a fixed current source
b) from a variable voltage source
c) from a variable current source
d) from a fixed voltage source
Answer: d
Clarification: Reference voltage is applied to the potentiometer. It is obtained from a source of fixed voltage. The position of the slider on the contact surface determines the value of the feedback voltage.

6. Feedback voltage is applied to the ________
a) comparator
b) error amplifier
c) potentiometer
d) sliding contact
Answer: a
Clarification: Comparator accepts the feedback voltage as an input. Comparator compares the values of the unknown voltage and the feedback voltage. Comparator then provides the difference between the feedback voltage and the unknown voltage as its output.

7. Output of the comparator is known as ________
a) amplified signal
b) error signal
c) feedback signal
d) attenuated signal
Answer: b
Clarification: Comparator then provides the difference between the feedback voltage and the unknown voltage as its output. This is also known as the error signal.

8. Slider movement depends on ________
a) current magnitude
b) resistance magnitude
c) voltage magnitude
d) power magnitude
Answer: c
Clarification: In a potentiometric integrating type DVM, the slider moves based on the magnitude of the feedback voltage with respect to the input voltage. Contact is pushed back to the place where the unknown voltage equals the feedback voltage.

9. Accuracy of a potentiometric DVM is ________
a) zero
b) medium
c) low
d) high
Answer: d
Clarification: In a potentiometric integrating type DVM, the accuracy is usually high. It generally depends on the reference of the digital to analog converter. Accuracy of the voltage to frequency converter is less important compared to that of the digital to analog converter.

250+ TOP MCQs on Advanced Miscellaneous Problems on Measurement of Resistance and Answers

Electrical Measurements Question Paper on “Advanced Miscellaneous Problems on Measurement of Resistance”.

1. R_a and R_d are the opposite arms of a Wheatstone bridge as are R_c and R_b. The source voltage is applied across R_a and R_c. Then when the bridge is balanced which one of the following is true?
a) R_a = R_c R_d/R_b
b) R_a = R_b R_c/R_d
c) R_a = R_b Rd/R_c
d) R_a = R_b + R_c + R_d
Answer: b
Clarification: At balance condition, Potential at B = Potential at D
∴ V_A – I_a R_a = V_A – I_b R_b
Or, ( frac{I_a}{I_b} = frac{R_b}{R_a})
Similarly, ( frac{I_a}{I_b} = frac{R_d}{R_c})
∴ R_a = ( frac{R_b R_c}{R_d}).

2. A setup is used to measure resistance R. The ammeter and voltmeter resistance are 0.01 Ω and 2000 Ω respectively. Their readings are 2 A and 180 V respectively, giving a measured resistance of 90 Ω. The percentage error in the measurement is?
a) 2.25 %
b) 2.35 %
c) 4.5 %
d) 4.71 %
Answer: d
Clarification: Current through the voltmeter I_v = (frac{180}{2000})
Current through R, I_R = 2 – 9/100 = 1.91 A
Since, 1.91 R = 180
∴ R = 94.24
∴ Percentage error = (frac{94.24-90}{90}) × 100 = 4.71 %.

3. A 35 V DC supply is connected across a resistance in series with an unknown resistance R. a voltmeter having a resistance of 1.2 kΩ is connected across 600 Ω resistance and reads 5 V. The value of the known resistance is 600 Ω. The value of resistance R will be?
a) 120 Ω
b) 400 Ω
c) 1.8 kΩ
d) 2.4 kΩ
Answer: d
Clarification: Voltage across R₁, V₁ = 35 – 5 = 30 V
Current in the circuit, I = (displaystylefrac{5}{frac{600 ×1200}{600+1200}} = frac{5}{400}) A
∴ R = (frac{30 × 400}{5}) = 2.4 kΩ.

4. A DC ammeter is rated for 15 A, 250 V. The meter constant is 14.4 A-s/rev. The meter constant at rated voltage may be expressed as __________
a) 3750 rev/kW-h
b) 3600 rev/kW-h
c) 1000 rev/kW-h
d) 960 rev/kW-h
Answer: c
Clarification: Meter constant is 14.1 A-s/rev
(frac{14.1}{3600}) A-h/rev = (frac{14.4 × 250}{3600})
So, w = 1 W-h/rev
Hence, 1 rev/W-h = 1000 rev/kW-h.

5. A DC ammeter has a resistance of 0.1 Ω and its current range is 0-100 A. If the range is to be extended to 0-500 A, the meter requires which of the following shunt resistance?
a) 0.010 Ω
b) 0.011 Ω
c) 0.025 Ω
d) 1.0 Ω
Answer: c
Clarification: R_sh = ( frac{R_m}{m-1})
Where, m is the multiplication factor = 500/100 = 5
∴ R_sh = 0.1/4 = 0.025 Ω.

6. A 100 μA ammeter has an internal resistance of 100 Ω. The range is to be extended to 500 μA. The shunt required is of resistance __________
a) 20.0 Ω
b) 22.22 Ω
c) 25.0 Ω
d) 50.0 Ω
Answer: c
Clarification: I_sh R_sh = I_m R_m
I_sh = I – I_m or, (frac{I}{I_m} – 1 = frac{R_m}{R_{sh}})
Now, m = (frac{I}{I_m})
Or, m – 1 = (frac{R_m}{R_{sh}})
∴ R_sh = 25 Ω.

7. Resistance is measured by the voltmeter-ammeter method employing DC excitation and a voltmeter is connected directly across the unknown resistance. If the voltmeter and ammeter readings are subject to maximum possible errors of ±2.4 % and ±1% respectively, then the magnitude of the maximum possible percentage error in the value of resistance deduced from the measurement is?
a) 1.4 %
b) 1.7 %
c) 2.4 %
d) 3.4 %
Answer: d
Clarification: Ammeter error ∆I = ± 1%
Voltmeter error ∆V = ± 2.4%
We know that (frac{∆R}{R} = frac{∆V}{V} + frac{∆I}{I})
∴ Maximum percentage error = 2.4% + 1% = 3.4%.

8. A galvanometer with a full-scale current of 10 mA has a resistance of 1000 Ω. The multiplying power of a 100 Ω shunt with this galvanometer is?
a) 110
b) 100
c) 11
d) 10
Answer: c
Clarification: Multiplying factor = m = (frac{I}{I_1} )
Now, (frac{I_1}{I_2} = frac{100}{1000} )
∴ (frac{I_1}{100} = frac{I_2}{1000} = frac{I}{1000} )
∴ (frac{I}{I_1}) = 11.

9. A slide wire potentiometer has 10 wires of 2 m each. With the help of a standard voltage source of 1.045 V, it is standardized by keeping the jockey at 104.5 cm. If the resistance of potentiometer wires is 2000 Ω, then the value of working current is?
a) 1 mA
b) 10 mA
c) 0.1 mA
d) 0.5 mA
Answer: b
Clarification: Total length of the slide wire = 10 × 200
Total resistance of slide wire = 2000 Ω
∴ Resistance per cm = 1 Ω
Resistance of 104.5 cm = 104.5 Ω
So, the current = (frac{1.045}{104.5}) = 10 mA.

10. The simultaneous applications of signals x (t) and y (t) to the horizontal and vertical plates respectively, of an oscilloscope, produce a vertical figure of 8 displays. If P and Q are constants and x(t) = P sin (4t + 30), then y(t) is equal to _________
a) Q sin (4t -30)
b) Q sin (2t +15)
c) Q sin (8t +60)
d) Q sin (4t +30)
Answer: b
Clarification: (frac{f_y}{f_x} = frac{x-peak}{y-peak})
Here, x-peak = 1 and y-peak = 2
∴ y(t) = Q sin (2t + 15).

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250+ TOP MCQs on Ratios of Instrument Transformers and Answers

Electrical Measurements & Measuring Instruments Multiple Choice Questions on “Ratios of Instrument Transformers”.

1. Transformation ratio of an instrument is defined as ____________
a) ratio of primary to secondary phasor
b) ratio of secondary to primary phasor
c) reciprocal of the primary phasor
d) reciprocal of the secondary phasor
Answer: a
Clarification: For an instrument transformer, the transformation ratio is defined as the ratio of the magnitude of the actual primary phasor to the magnitude of the secondary phasor.

2. For a C.T. the transformation ratio is given by which of the following relation?
a) R = (frac{I_s}{I_p})
b) R = (frac{I_p}{I_s})
c) R = (frac{1}{I_p})
d) R = I_p
Answer: b
Clarification: In a current transformer, the transformation ratio is given by the ratio of the magnitude of the actual primary current to the magnitude of the actual secondary current.
R = (frac{I_p}{I_s})
where, R is the transformation ratio
I_p is the actual primary winding current
I_s is the actual secondary winding current.

3. For a P.T. the transformation ratio is given by which of the following relation?
a) R = (frac{V_s}{V_p})
b) R = (frac{1}{V_s})
c) R = (frac{V_p}{V_s})
d) R = V_p
Answer: c
Clarification: In a potential transformer, the transformation ratio is given by the ratio of the magnitude of the actual primary current to the magnitude of the actual secondary current.
R = (frac{V_p}{V_s})
where, R is the transformation ratio
V_p is the actual primary winding voltage
V_s is the actual secondary winding voltage.

4. Nominal ratio of an instrument transformer is defined as the __________
a) reciprocal of the rated primary value
b) ratio of rated secondary value to primary value
c) reciprocal of the rated secondary value
d) ratio of rated primary value to secondary value
Answer: d
Clarification: In an instrument transformer, nominal ratio is defined as the ratio of the rated primary current or voltage to the rated secondary winding current or voltage.

5. For a C.T. the nominal ratio is given by which of the following relation?
a) K_n = (frac{I_{p(rated)}}{I_{s(rated)}})
b) K_n = (frac{I_{s(rated)}}{I_{p(rated)}})
c) K_n = (frac{1}{I_{s(rated)}})
d) K_n = I_p(rated)
Answer: a
Clarification: In a current transformer, the transformation ratio is given by the ratio of the magnitude of the actual primary current to the magnitude of the actual secondary current.
K_n = (frac{I_{p(rated)}}{I_{s(rated)}})
where, R is the transformation ratio
I_p(rated) is the rated primary winding current
I_s(rated) is the rated secondary winding current.

6. For a P.T. the nominal ratio is given by which of the following relation?
a) K_n = (frac{V_{s(rated)}}{V_{p(rated)}})
b) K_n = (frac{V_{p(rated)}}{V_{s(rated)}})
c) K_n = (frac{1}{V_{s(rated)}})
d) K_n = V_p(rated)
Answer: b
Clarification: In a potential transformer, the transformation ratio is given by the ratio of the magnitude of the actual primary voltage to the magnitude of the actual secondary voltage.

where, R is the transformation ratio
V_p(rated) is the rated primary winding voltage
V_s(rated) is the rated secondary winding voltage.

7. Ratio correction factor is defined as _________
a) reciprocal of nominal ratio
b) ratio of nominal ratio to transformation ratio
c) ratio of transformation ratio to nominal ratio
d) reciprocal of transformation ratio
Answer: c
Clarification: The ratio correction factor for an instrument transformer is defined as the ratio of the transformation ratio to the nominal ratio.
R.C.F = ^R ⁄ _{K_n}
where,
R.C.F is the ratio correction factor
R is the transformation ratio
K_n is the nominal ratio.

8. For a C.T. the turns ratio is defined as the _________
a) n = ^N_p ⁄ _{N_s}
b) n = ¹ ⁄ _{N_p}
c) n = N_s
d) n = ^N_s ⁄ _{N_p}
Answer: d
Clarification: For a current transformer, the turns ratio is defined as the ratio of the number of turns in the secondary winding to the number of turns in the primary winding.
n = ^N_s ⁄ _{N_p}
where, n is the turns ratio
N_s is the secondary turns
N_p is the primary turns.

9. For a P.T. the turns ratio is defined as the _________
a) n = ^N_p ⁄ _{N_s}
b) n = ¹ ⁄ _{N_p}
c) n = N_s
d) n = ^N_s ⁄ _{N_p}
Answer: a
Clarification: For a potential transformer, the turns ratio is defined as the ratio of the number of turns in the primary winding to the number of turns in the secondary winding.
n = ^N_p ⁄ _{N_s}
where, n is the turns ratio
N_p is the primary turns
N_s is the secondary turns.

250+ TOP MCQs on Errors in Single Phase Energy Meters and Answers

Electrical Measurements & Measuring Instruments Multiple Choice Questions on “Errors in Single Phase Energy Meters”.

1. 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 windings
Answer: a
Clarification: 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.

2. Phase angles in an energy meter cannot be incorrect.
a) True
b) False
Answer: b
Clarification: 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.

3. 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 transformer
Answer: c
Clarification: 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.

4. Supply voltage in an energy meter is __________
a) constant always
b) zero always
c) depends on the load
d) can fluctuate
Answer: d
Clarification: Generally the supply voltage is constant in an energy meter. It can fluctuate as a result of unavoidable reasons leading to errors in the reading of the energy meter.

5. 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 other
Answer: a
Clarification: 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.

6. 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 coolant
Answer: b
Clarification: 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.

7. Disc rotates slowly in some energy meters.
a) True
b) False
Answer: a
Clarification: 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.

8. Creeping is avoided by __________
a) reversing the polarity of the voltage
b) drilling two diametrically opposite holes
c) holding the disc
d) increasing the friction
Answer: b
Clarification: 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.

9. 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 pieces
Answer: c
Clarification: 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.

250+ TOP MCQs on Dual Slope Integrating Type DVM and Answers

Electrical Measurements & Measuring Instruments Multiple Choice Questions on “Dual Slope Integrating Type DVM”.

1. Why is dual slope method preferred over ramp techniques?
a) no noise
b) partial noise
c) average noise
d) maximum noise
Answer: a
Clarification: During the process of integration, noise is canceled out by the positive and negative ramps in the dual slope method. The input signal is integrated only for a fixed interval of time and this is the basis for the dual slope method.

2. What is the significance of the name dual slope method?
a) it has two slopes
b) it integrates the input twice
c) it uses two inputs
d) it has two outputs
Answer: b
Clarification: The input signal is integrated only for a fixed interval of time and this is the basis for the dual slope method. Reference voltage is integrated with a negative slope. Hence the method is known as dual slope integrating type DVM.

3. What is the output voltage in a dual slope integrating type DVM?
a) differential of the input
b) multiple of the input
c) integral of the input
d) zero
Answer: c
Clarification: In a dual slope integrating type DVM, the output voltage is given by the integral of the input voltage.

where, V_in is the input voltage
R₁ is the series resistance
t₁ is the time for which the capacitor is charged.

4. Input voltage depends on ____________
a) resistance
b) capacitance
c) current
d) time-period
Answer: d
Clarification: The input voltage in a dual slope integrating type DVM is given by the relation,
V_in = V_ref^t₂ ⁄ _t₁
From the above equation it is seen that the input voltage in a dual slope integrating type DVM depends on the time periods t₁ for which the capacitor is charged and t₂ during which the capacitor is discharged.

5. Noise rejection is poor.
a) True
b) False
Answer: b
Clarification: In a dual slope integrating type DVM, the noise is cancelled out by the positive and negative ramps during the process of integration. As a result, the noise rejection is excellent.

6. What is the effect of the capacitor on the output?
a) no effect
b) charging effect
c) electrostatic effect
d) magnetic effect
Answer: a
Clarification: In the dual slope integrating type DVM method, the capacitor is connected through means of an electronic switch. As a result the effects due to offset voltage wherein there exists an output voltage without the application of any input are eliminated.

7. What is the effect of clock on the voltage?
a) voltage doubles with clock input
b) voltage halves with clock input
c) no effect
d) voltage becomes zero with clock input
Answer: c
Clarification: In a dual slope integrating type DVM, the value of the unknown voltage is independent of the frequency of the clock. It depends only on the number of counts read by the electronic counter.

8. What is the counter value at the beginning?
a) one
b) ten
c) three
d) zero
Answer: d
Clarification: In a dual slope integrating type DVM, the electronic counter is reset to 0 at the beginning of the measurement of voltage. Flip-flop output is also maintained at zero and is given to control logic.

9. What is the maximum count of the counter?
a) 9999
b) 0
c) 500
d) 1000
Answer: a
Clarification: In a dual slope integrating type DVM, the electronic counter reaches a maximum value of 9999 before resetting. A carry pulse is generating pulling down all the digits to zero. Flip-flop then activates the control logic.