Category: Electrical Measurements Objective Questions

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. Ra and Rd are the opposite arms of a Wheatstone bridge as are Rc and Rb. The source voltage is applied across Ra and Rc. Then when the bridge is balanced which one of the following is true?
a) Ra = Rc Rd/Rb
b) Ra = Rb Rc/Rd
c) Ra = Rb Rd/Rc
d) Ra = Rb + Rc + Rd
Answer: b
Clarification: At balance condition, Potential at B = Potential at D
∴ VA – Ia Ra = VA – Ib Rb
Or, ( frac{I_a}{I_b} = frac{R_b}{R_a})
Similarly, ( frac{I_a}{I_b} = frac{R_d}{R_c})
∴ Ra = ( 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 Iv = (frac{180}{2000})
Current through R, IR = 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: Rsh = ( frac{R_m}{m-1})
Where, m is the multiplication factor = 500/100 = 5
∴ Rsh = 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: Ish Rsh = Im Rm
Ish = I – Im 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}})
∴ Rsh = 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).

To practice all questions papers on Electrical Measurements, .

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 = Ip
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
Ip is the actual primary winding current
Is 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 = Vp
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
Vp is the actual primary winding voltage
Vs 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) Kn = (frac{I_{p(rated)}}{I_{s(rated)}})
b) Kn = (frac{I_{s(rated)}}{I_{p(rated)}})
c) Kn = (frac{1}{I_{s(rated)}})
d) Kn = Ip(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.
Kn = (frac{I_{p(rated)}}{I_{s(rated)}})
where, R is the transformation ratio
Ip(rated) is the rated primary winding current
Is(rated) is the rated secondary winding current.

6. For a P.T. the nominal ratio is given by which of the following relation?
a) Kn = (frac{V_{s(rated)}}{V_{p(rated)}})
b) Kn = (frac{V_{p(rated)}}{V_{s(rated)}})
c) Kn = (frac{1}{V_{s(rated)}})
d) Kn = Vp(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.
electrical-measurements-questions-answers-ratios-instrument-transformers-q6a
where, R is the transformation ratio
Vp(rated) is the rated primary winding voltage
Vs(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 ⁄ Kn
where,
R.C.F is the ratio correction factor
R is the transformation ratio
Kn is the nominal ratio.

8. For a C.T. the turns ratio is defined as the _________
a) n = Np ⁄ Ns
b) n = 1 ⁄ Np
c) n = Ns
d) n = Ns ⁄ Np
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 = Ns ⁄ Np
where, n is the turns ratio
Ns is the secondary turns
Np is the primary turns.

9. For a P.T. the turns ratio is defined as the _________
a) n = Np ⁄ Ns
b) n = 1 ⁄ Np
c) n = Ns
d) n = Ns ⁄ Np
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 = Np ⁄ Ns
where, n is the turns ratio
Np is the primary turns
Ns 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.
electrical-measurements-questions-answers-dual-slope-integrating-type-dvm-q3
where, Vin 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,
Vin = Vreft2 ⁄ 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.

.

250+ TOP MCQs on Introduction to AC Bridges and Answers

Electrical Measurements & Measuring Instruments Multiple Choice Questions on “Introduction to AC Bridges”.

1. In the simplest form, an AC bridge consists of ____________
a) arms, source and a detector
b) arms and source
c) source and detector
d) arms and detector

Answer: a
Clarification: In its simplest form, an AC bridge consists of four arms, a source for excitation and a null detector. The source is connected across a pair of arms while the detector is connected to the pair of opposite arms.

2. Source is ________
a) dc supply
b) ac supply
c) mixed mode supply
d) high voltage supply

Answer: b
Clarification: For an AC bridge we require an AC supply as the source of voltage. It supplies AC voltage at the required frequency.

3. At high frequency, source consists of ________
a) amplifiers
b) regulators
c) oscillators
d) op amps

Answer: c
Clarification: Op amps are basically differential amplifiers. Amplifiers are used in analog circuits for increasing the strength of the signal. Electronic oscillators form sources at high frequencies.

4. Commonly used balance detectors for AC bridges are headphones, tuned amplifiers and vibration galvanometers.
a) True
b) False

Answer: a
Clarification: Headphones, tuned amplifier circuits and vibration galvanometers are used for detecting the balance condition in AC bridges.

5. What is the frequency range for a headphone as a detector?
a) 20 Hz to 20 kHz
b) 10 kHz to 1 MHz
c) 10 MHz to 1 GHz
d) 250 Hz to 4 kHz

Answer: d
Clarification: Headphones can be used as detectors in AC bridges in the low audio frequency range. Low audio frequency range varies from 250 Hz to 4 KHz.

6. For single frequency value, the most sensitive detector is ________
a) tuned detector
b) vibration galvanometer
c) headphone
d) oscillator

Answer: a
Clarification: Vibration galvanometer is used for detecting the balance condition. Oscillator is used as a source of supply voltage. Tuned detector is the most sensitive detector for a single frequency value.

7. Tuned detectors are used in the frequency range of ________
a) 1 Hz to 100 Hz
b) 10 Hz to 100 Hz
c) 1 kHz to 100 kHz
d) 1 MHz to 100 MHz

Answer: b
Clarification: Tuned amplifier circuits are used as detectors in the low frequency range. Low frequency range usually ranges from 10 Hz to 100 Hz in AC bridges.

8. Vibration galvanometers are used for ________
a) very high frequency
b) very low frequency
c) low audio frequency
d) high audio frequency

Answer: c
Clarification: Vibration galvanometers are used as detectors in AC bridges for low audio frequency. Low audio frequency ranges from 5 Hz to 1000 Hz.

9. AC bridge is an outcome of ________
a) Kelvin bridge
b) Megger
c) De Sauty bridge
d) Wheatstone bridge

Answer: d
Clarification: Wheatstone bridge is the simplest form of bridge for the measurement of resistance and forms the basis for an AC bridge. Kelvin bridge is used for the measurement of low resistance and a megger is used for the measurement of high resistances.

250+ TOP MCQs on Errors in Current Transformers and Answers

Electrical Measurements & Measuring Instruments Multiple Choice Questions on “Errors in Current Transformers”.

1. Errors are introduced in Current Transformers.
a) True
b) False
Answer: a
Clarification: A current transformer is used for the measurement of very high currents. The use of a C.T. leads to the introduction of two errors in power measurement.

2. Ratio error is defined as ___________
a) Ratio error = Kn⁄R
b) Ratio error = Kn – R⁄R
c) Ratio error = Kn – R
d) Ratio error = 1⁄R
Answer: b
Clarification: Ratio error of a C.T. is defined as the ratio of the magnitude of the difference between the nominal and actual ratio with respect to the actual ratio.
electrical-measurements-questions-answers-errors-current-transformers-q2

3. Phase angle in a C.T. is defined as ____________
a) (frac{180}{π} [frac{I_m cosδ}{nI_s}]) degrees
b) (frac{180}{π} [frac{I_c sinδ}{nI_s}]) degrees
c) (frac{180}{π} [frac{I_m cosδ – I_c sinδ}{nI_s}]) degrees
d) (frac{180}{π} [frac{I_m sinδ – I_c cosδ}{nI_s}]) degrees
Answer: c
Clarification: During power measurement, there exists phase angle error in a C.T. The phase angle is defined as
Phase angle = (frac{180}{π} [frac{I_m cosδ – I_c sinδ}{nI_s}]) degrees.
where. Im is the magnetising component of the excitation current
Ic is the core loss component of the excitation current
Is is the secondary winding current.

4. Phase angle error is given by ____________
a) (frac{180}{π} [frac{1}{nI_s}]) degrees
b) (frac{180}{π} [frac{I_m}{I_s}]) degrees
c) (frac{180}{π} [frac{I_m}{n}])degrees
d) (frac{180}{π} [frac{I_m}{nI_s}]) degrees
Answer: d
Clarification: The power measurement in a C.T. leads to phase angle error. Phase angle error is given by the relation
θ = (frac{180}{π} [frac{I_m}{nI_s}]) degrees
where, θ is the phase angle error
Im is the magnetising component of the excitation current
Is is the secondary winding current.

5. Ratio error is due to _________
a) iron loss
b) C.T.
c) magnetising component
d) supply voltage
Answer: a
Clarification: We know that the ratio error in a C.T. is given by the relation
Ratio error = Kn – R⁄R = n + Ie⁄Is
where, Ie is the iron loss component of the excitation current
n is the turns ratio.

6. Phase angle error is due to _________
a) C.T.
b) magnetising component
c) iron loss
d) supply voltage
Answer: b
Clarification: We know that the phase angle error in a C.T. is given by the relation
θ = (frac{180}{π} [frac{I_m}{nI_s}]) degrees
where, θ is the phase angle error
Im is the magnetising component of the excitation current
Is is the secondary winding current
It is observed from the equation for the phase angle error that it depends on the magnetising component of the excitation current.

7. In power measurements 180° phase shift is required.
a) True
b) False
Answer: a
Clarification: For eliminating errors in power measurement, there must be a phase difference of 180° between the primary and the secondary currents.

8. Errors in a C.T. can be minimised by _________
a) making use of laminations
b) having low reactance
c) increasing the secondary winding turns
d) decreasing the primary winding turns
Answer: b
Clarification: The excitation current Io can be minimised thus eliminating the errors in a C.T. by minimising the iron loss. The core must have a low iron loss and a minimum value of leakage reactance.

.