Category: Electrical Measurements Objective Questions

Electrical Measurements Questions and Answers for Aptitude test on “Measurement of Earth Resistance”.

1. What is earthing?
a) connecting electrical machines to earth
b) providing a connection to the ground
c) connecting the electrical machines to source
d) providing a source of current
Answer: a
Clarification: Connecting electrical machines to the general mass of the earth by making use of a conducting material with very low resistance is known as earthing.

2. What is an earth electrode?
a) electrode that is connected to earth
b) material used for earthing
c) electrode connected to the circuit
d) electrode which is connected to the mains
Answer: b
Clarification: Electrode connected to the main is basically a source of e.m.f. Conducting material that is used for connecting electrical machinery to the earth is known as an earth electrode.

3. Earth electrode provides ____________
a) high resistance
b) medium resistance
c) low resistance
d) very high resistance
Answer: c
Clarification: In the case of occurrence of any leakage currents due to poor shielding of the apparatus, the earth electrode is used to provide a very low resistance path from the electrical appliances to the earth.

4. How is the condition of an earth electrode measured?
a) by measuring the voltage
b) by measuring the current
c) by measuring the power
d) by measuring resistance
Answer: d
Clarification: The resistance of the earth electrode is measured in order to check whether it is in a good condition or not.

5. In a three phase system, the neutral is _________
a) earthed
b) connected to low voltage
c) connected to high voltage
d) not connected
Answer: a
Clarification: Earthing can be used to maintain a constant line voltage in a three phase system. This is achieved by earthing the neutral.

6. Earthing is used as return conductor in telephone lines and for traction work.
a) True
b) False
Answer: a
Clarification: The complications involved in laying separate telephone cables and the cast used for traction work is minimised. As a result earthing is used as a return conductor in telephone lines and for traction work.

7. Earthing does not help in protecting the equipment.
a) True
b) False
Answer: b
Clarification: Spike voltages occurring as a result of lightning or any other fault can be dissipated to ground by earthing, thus protecting the equipment.

8. After earthing, the different parts of an electrical machinery are at _________
a) infinite potential
b) intermediate potential
c) zero potential
d) undefined potential
Answer: c
Clarification: After earthing, the various parts of electrical machinery such as casing, armoring of cables, etc are at zero potential.

9. Connection of the various parts of a circuit to earth has a _________
a) medium resistance
b) high resistance
c) very high resistance
d) very low resistance
Answer: d
Clarification: Once an electrical apparatus is grounded, most of its components are at ground potential. When the different parts of electrical machinery are connected to the ground, they possess very low resistance.

10. Specific resistance of soil is _________
a) changes from soil to soil
b) is constant
c) depends on the circuit connected to it
d) depends on the supply voltage
Answer: a
Clarification: Specific resistance depends on the nature and properties of a material. Specific resistance is different for various types of soils such as dry soil, rocky soil, wet soil, etc.

To practice all areas of Electrical Measurements for Aptitude test, .

Electrical Measurements & Measuring Instruments Multiple Choice Questions on “Shunts and Multipliers”.

1. Range of an electrical instrument depends on __________
a) current
b) voltage
c) power
d) resistance
Answer: a
Clarification: The amount of current safely passing through the coil of the instrument and the spiral springs. This acts as the leads of the current to the instrument. As a result, the range of an electrical instrument depends on the current.

2. Moving coil instruments have a current and voltage rating of __________
a) 100 A and 25 V
b) 50 mA and 50 mV
c) 75 nA and 100 μV
d) 25 μA and 75 V
Answer: b
Clarification: Moving coil instruments are designed to function as Ammeters and Voltmeters. They have a maximum current carrying capacity of 50 mA with a voltage rating of 50 mV.

3. A shunt is a __________
a) very high resistance
b) medium resistance
c) very low resistance
d) high resistance
Answer: c
Clarification: Usually shunt is a very low value of resistance. It is connected in parallel with the ammeter coil. Through this we can extend the range of an ammeter.

4. A shunt can be used to measure large currents.
a) True
b) False
Answer: a
Clarification: A shunt is normally a very low value of resistance, connected in parallel with the ammeter coil. By making use of a low range ammeter, large current values can be measured through a shunt.

5. Current terminals have a small current capacity.
a) True
b) False
Answer: b
Clarification: A shunt is normally a very low value of resistance, connected in parallel with the ammeter coil. In a shunt, the current terminals have a large current capacity and are connected in series.

6. Potential terminals have a __________
a) high current capacity
b) low voltage capacity
c) low current capacity
d) high voltage capacity
Answer: c
Clarification: A shunt is normally a very low value of resistance, connected in parallel with the ammeter coil. In a shunt, the potential terminals have a low current carrying capacity. As a result, a low range ammeter is used to measure the large current.

7. In case of AC ammeters, shunts consist of __________
a) impedance
b) capacitance
c) resistance
d) inductance
Answer: d
Clarification: AC ammeter shunts comprise of the inductances of the ammeter as well as the shunt. In order to extend the range of an AC ammeter, inductances of the ammeter and the shunt are taken into account.

8. What is the effect of the ammeter range on the shunt resistance?
a) no effect
b) varies by a factor of multiplying factor
c) varies by a factor of the resistance
d) varies by a factor of unity
Answer: b
Clarification: We know that
N = 1 + ^R_a⁄_Rs
where, N is the multiplying factor
R_a is the ammeter resistance
R_s is the shunt resistance
It is clear from the above equation that in order to increase the ammeter range by N times, the shunt resistance is equivalent to ¹⁄_N-1.

9. A multiplier is __________
a) non-inductive
b) resistive
c) capacitive
d) non-capacitive
Answer: a
Clarification: A multiplier is basically a non-inductive, high resistance that is used to extend the range of a D.C. voltmeter. Multiplier consists of a low range D.C. voltmeter connected in series with it.

10. What is the condition for using a multiplier in A.C. voltmeters?
a) by using ac supply
b) by maintaining a uniform impedance
c) by maintaining a uniform frequency
d) by using a galvanometer
Answer: c
Clarification: A multiplier can be used for A.C. voltmeters. The condition to be satisfied is that the total impedance of the voltmeter and the multiplier circuit must be constant for a wide range of frequencies.

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

1. Ramp type DVM uses ___________
a) a linear ramp technique
b) a non-linear ramp technique
c) an exponential ramp technique
d) an asymptotic ramp technique
Answer: a
Clarification: A ramp type DVM makes use of a staircase ramp technique or a linear ramp technique. Compared to the linear ramp technique, the staircase ramp technique is much simpler.

2. Linear ramp technique is based on __________
a) voltage measurement
b) time measurement
c) current measurement
d) resistance measurement
Answer: b
Clarification: Linear ramp technique works on the principle of measurement of time required by a linear ramp to rise from 0 V to the input voltage. It can also be the time required by the input voltage to fall to 0 V.

3. Time is measured using __________
a) clock
b) logic gates
c) counter
d) flip-flops
Answer: c
Clarification: An electronic counter is used for the measurement of time in the linear ramp technique. A digital display is used to show the numerical value of time.

4. Which is the main device used in the linear ramp technique?
a) exponential ramp
b) asymptotic ramp
c) non-linear ramp
d) linear ramp
Answer: d
Clarification: The linear ramp technique essentially consists of a linear ramp. It is either positive or negative going. Range of the linear ramp varies from –12 V to +12 V. Base range varies from –10 V to +10 V.

5. Resolution depends on __________
a) frequency
b) resistance
c) voltage
d) current
Answer: a
Clarification: In the linear ramp technique, resolution is dependent on the frequency of the local oscillator. By adjusting the frequency of the local oscillator the resolution of the linear ramp can be made higher.

6. How is input voltage measured?
a) by using a voltmeter
b) by counting the pulses
c) by using a multimeter
d) by using a transformer
Answer: b
Clarification: The electronic counter used in the linear ramp technique counts the definite number of pulses during the start and end pulse. This count is a measure of the input voltage signal.

7. Which determines the rate of measurement cycles?
a) oscillator
b) amplifier
c) mutivibrator
d) oscilloscope
Answer: c
Clarification: Initiation of the measurement cycles is taken care of by the sample rate multivibrator. This vibrator oscillates at the rate of 1000 cycles per second. It is adjusted by a front panel control.

8. What is the typical value of the multivibrator?
a) 10 cycles/second
b) 0.2 cycles/second
c) 50 cycles/second
d) 5 cycles/second
Answer: d
Clarification: A multivibrator has a typical value of 5 cycles/second. It has an accuracy of the order of ±0.005 %. The ramp generator starts the next ramp voltage based on the sample rate.

9. Swing of the ramp is __________
a) ±12 V
b) ±10 V
c) ±8 V
d) ±5 V
Answer: a
Clarification: In the linear ramp technique, a swing of ±12 V is produced by the ramp. This limits the effective voltage available to ±10 V considering the voltage drop across the components in the circuit.

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Electrical Measurements Objective Questions & Answers on “Advanced Problems on Measurement of Low Medium and High Resistance”.

1. Circuit shows the method of Measurement of low resistance by Ammeter-Voltmeter method. The measured resistance Rm for the given Circuit is _________

a) R_x + R_v
b) (frac{R_x^2}{R_x + R_v})
c) (frac{R_v^2}{R_x + R_v})
d) (frac{R_x R_v}{R_x + R_v})
Answer: d
Clarification: Measured resistance R_m = (frac{V_x}{I_A} = frac{V_x}{I_v + I_R})
(frac{I_v}{V_x} = frac{1}{R_v}) And (frac{I_R}{V_x} = frac{1}{R_X})
So, R_m = (frac{R_x R_v}{R_x+R_v}).

2. Circuit shows the method of Measurement of low resistance by Ammeter-Voltmeter method. What is the percentage error?

a) Zero
b) (frac{R_x}{R_x + R_v}) × 100
c) (– frac{R_x}{R_x + R_v}) × 100
d) (– frac{R_v}{R_x + R_v}) × 100
Answer: c
Clarification: Percentage Error = (frac{R_m – R_x}{R_x}) × 100
(= frac{R_x R_v – R_x(R_x-R_v)}{R_x (R_x+R_v)}) × 100
∴ Percentage Error = (– frac{R_x}{R_x + R_v}) × 100.

3. The readings of polar type potentiometer are I = 12.4∠27.5°, V = 31.5∠38.4°. Then, reactance of the coil will be ________
a) 2.51 Ω
b) 2.56 Ω
c) 2.54 Ω
d) 2.59 Ω
Answer: c
Clarification: Here, V = 31.5∠38.4°
I = 12.4∠27.5°
Z = (frac{31.5∠38.4°}{12.4∠27.5°}) = 2.54∠10.9°
But Z = R + jX = 2.49 + j0.48
∴ Reactance X= 2.54 Ω.

4. The voltage drop across a standard resistor of 0.2 Ω is balanced at 83 cm. Find the magnitude of the current, if the standard cell emf of 1.53 V is balanced at 42 m.
a) 13.04 A
b) 10 A
c) 14.95 A
d) 12.56 A
Answer: c
Clarification: Voltage drop per unit length = (frac{1.53}{42}) = 0.036 V/cm
Voltage drop across 83 cm length = 0.036 × 83 = 2.99 V
∴ Current through resistor, I = (frac{2.99}{0.2}) = 14.95 A.

5. A resistance R is measured using the connection shown in the below figure.

The current measured is 10 A on ranges 100A and the voltage measured is 125 V on 150 V range. The scales of the ammeter and voltmeter are uniform. The total number of scale divisions of the ammeter is 100 and that of the voltmeter is 150. The scale division can be distinguished. The constructional error of the ammeter is ± 0.3% and that of voltmeter±0.4%. The resistance of the ammeter is 0.25 Ω.
The value of R is?
a) 12.75 Ω
b) 12.0 Ω
c) 12.25 Ω
d) 12.5 Ω
Answer: c
Clarification: Percentage error in ammeter = (± frac{1}{10×100} × 100) = ± 0.1%
Percentage error in voltmeter= (± frac{1}{10×150} × 100) = ± 0.067%
So, δI = ± 0.3 ± 0.1 = ± 0.4%
δV = ± 0.4 ± 0.067 = ± 0.467%
R = (frac{V}{I})
So, error = ± δV ± δI = ± 0.867
Measured value of resistance = (R_m = frac{125}{10}) = 12.5
∴ True value = (R_m(1-frac{Ra}{R_m})) = 12.25 Ω.

6. A resistance R is measured using the connection shown in the below figure.

The current measured is 10 A on ranges 100A and the voltage measured is 125 V on 150 V range. The scales of the ammeter and voltmeter are uniform. The total number of scale divisions of the ammeter is 100 and that of the voltmeter is 150. The scale division can be distinguished. The constructional error of the ammeter is ± 0.3% and that of voltmeter±0.4%. The resistance of the ammeter is 0.25 Ω.
The possible error in the measurement of R is?
a) ±0.11 Ω
b) ±0.15 Ω
c) ±0.867 Ω
d) ±0.625 Ω
Answer: a
Clarification: Possible error is ± 0.867, so,
12.25 ± 0.867%
Or, 12.25 ± 0.11 Ω.

7. Low resistance is measured by ___________
a) De-Sauty’s bridge
b) Maxwell’s bridge
c) Kelvin double bridge
d) Wein’s bridge
Answer: c
Clarification: De-Sauty’s bridge is used for measurement of Capacitance; Maxwell’s bridge is used for measurement of Inductance and Wein Bridge for Frequency. Kelvin double bridge is used for measurement of Low resistance.

8. The resistance can be measured most accurately by _________
a) Voltmeter-Ammeter method
b) Bridge method
c) Multimeter
d) Megger
Answer: b
Clarification: Bridge method applies the concept of null point or bridge balance condition. Multimeter and Megger are used for measuring very high resistances and Voltmeter-Ammeter method is used for Low resistances. A null type instrument has higher accuracy as compared to a deflection type instrument.

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
This corresponds to a voltage of 1.045 V
∴ Current = (frac{1.045}{104.5}) = 10 mA.

10. Which of the following method is used for the measurement of Medium Resistance?
a) Kelvin’s double bridge method
b) Carey-Foster bridge method
c) Anderson Bridge
d) Direct-Deflection method
Answer: b
Clarification: Kelvin’s double bridge method is used for measurement of Low Resistance, Anderson Bridge is not used for measurement of Resistance and Direct-Deflection method is used for Measurement of High Resistance.

11. In the Wheatstone bridge shown below, if the resistance in each arm is increased by 0.05%, then the value of Vout will be ________

a) 50 mV
b) Zero
c) 5mV
d) 0.1mV
Answer: b
Clarification: In Wheatstone bridge, balance condition is
R₁R₃ = R₂R₄
Here, R₁ = 5, R₂ = 10, R₃ = 16, R₄ = 8
And when the Wheatstone bridge is balanced then, at V_out voltage will be Zero.

To practice all objective questions on Electrical Measurements, .

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

1. What is the current transformer?
a) transformer used with an A.C. ammeter
b) transformer used with an D.C. ammeter
c) transformer used with an A.C. voltmeter
d) transformer used with an D.C. voltmeter
Answer: a
Clarification: A transformer used to extend the range of an A.C. ammeter is known as a current transformer. A current transformer is also abbreviated as C.T.

2. What is the potential transformer?
a) transformer used with an D.C. ammeter
b) transformer used with an A.C. voltmeter
c) transformer used with an D.C. ammeter
d) transformer used with an A.C. voltmeter
Answer: b
Clarification: A transformer used to extend the range of an A.C. voltmeter is known as a potential transformer. A potential transformer is also abbreviated as P.T.

3. C.T. and P.T. are used for _________
a) measuring low current and voltages
b) measuring very low current and voltages
c) measuring high currentsand voltages
d) measuring intermediate currents and voltages
Answer: c
Clarification: C.T. is basically used for the measurement of high currents. A P.T. is usually used for the measurement of high voltages. They are used with A.C. ammeters and voltmeters in order to extend their range.

4. The primary winding of a C.T. has _________
a) a larger number of turns
b) no turns at all
c) intermediate number of turns
d) a few turns
Answer: d
Clarification: The primary winding of a C.T. has a very few number of turns. It is connected in series with the load circuit through which the primary current flows.

5. The secondary winding of a C.T. has _________
a) a large number of turns
b) a few turns
c) no turns at all
d) intermediate number of turns
Answer: a
Clarification: Secondary winding of a C.T. has a large number of turns. It is connected in series to an ammeter through which a small portion of the current flows through.

6. Turns ration for a C.T. is _________
a) n = ^N_p ⁄ _Ns
b) n = ^N_s ⁄ _Np
c) n = ¹ ⁄ _Np
d) n = N_s
Answer: b
Clarification: The turns ratio for a C.T. is defined as the ratio of the number of turns in the secondary to the number of turns in the primary. It is given by the relation
n = ^N_s⁄_Np

7. The primary winding of a P.T. has _________
a) intermediate number of turns
b) no turns at all
c) a larger number of turns
d) a few turns
Answer: c
Clarification: The primary winding of a P.T. has a very large number of turns. It is connected in parallel with the load whose voltage is to be measured.

8. The secondary winding of a P.T. has _________
a) a large number of turns
b) intermediate number of turns
c) no turns at all
d) a few turns
Answer: d
Clarification: Secondary winding of a P.T. has a few number of turns. A low range voltmeter is connected in parallel with the secondary winding.

9. Turns ration for a C.T. is _________
a) n = ^N_p ⁄ _Ns
b) n = ^N_s ⁄ _Np
c) n = ¹ ⁄ _Np
d) n = N_s
Answer: a
Clarification: The turns ratio for a P.T. is defined as the ratio of the number of turns in the primary to the number of turns in the secondary. It is given by the relation
n = ^N_p⁄_Ns.

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