Electrical Measurements - 5 - PDF Flipbook

Electrical Measurements - 5

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GATE
EEE

Electrical
Measurements

Test-05Solutions


ELECTRICAL MEASUREMENTS
1. In a multimeter circuit, for a.c, voltage measurement, what is the

function of diode D2?

a) To provide half-wave rectification
b) To short circuit over-range voltages
c) To by-pass reverse leakage current of D1 during the negative

half cycle of the input
d) To make the diode D1 perform full-wave rectification.
Answer: (c)
Solution:
In the absence of D2, leakage current would have flowed in D1.
This reduces the average value for complete cycle than it would
have been in half wave rectification.
2. A wattmeter is being tested under phantom loading condition. If
the wattmeter reading is 60 W, the actual power consumed from
the supply, is
a) much higher than 60 W
b) 60 W
c) much less than 60 W
d) 30 W

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Answer: (c)
Solution:
In phantom loading condition the power consumed by load is
very less. In phantom loading actual power is less than
measured power so, actual power consumed from the supply is
much less than 60 W.
3. The errors in current transformers can be reduced by designing
them with
a) using Primary and secondary winding as close to each other

as possible.
b) high permeability and low loss core materials, avoiding any

joints in the core and also keeping the flux density to a low
value.
c) using large cross section for both primary and secondary
winding conductors.
d) all of the given options.
Answer: (b)
Solution:
High μ, low losses
4. Which one of the following is correct statement?
Chopper stabilized d.c amplifier type electronic voltmeter
overcomes the effect of:
a) Amplifier CMRR
b) Amplifier drift
c) Amplifier sensitivity

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d) Electromagnetic interference
Answer: (b)
Solution:
In direct coupled d.c amplifier, temperature and time induced
drifts are there which are avoided in chopper stabilized d.c
amplifier type voltmeter.
5. Moving coil (PMMC) and moving iron instruments can be
distinguished by observing its:
a) size of terminals
b) pointer
c) range
d) scale
Answer: (d)
Solution:
In PMMC, θ ∝ I (scale is linear)
In MI, θ ∝ I2 (scale is nonlinear)
So, by observing scale (linear or non linear) we can distinguish
these two types of instruments.
6. Why is a MISC meter not recommended for d.c measurement?
a) The meter is calibrated for a.c. and its error I for d.c would be

high
b) The meter does not respond to d.c signals
c) The error is high due to hysteresis effect
d) The error is high due to eddy-current effect
Answer: (c)

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7. A dual trace oscilloscope is set to operate in the alternate. The
control input of the multiplexer used in the y-circuit is fed with a
signal having a frequency equal to
a) the highest frequency that the multiplexer can operate
properly
b) twice the frequency of the time base (sweep) oscillator
c) the frequency of the time base (sweep) oscillator
d) half the frequency of the time base (sweep) oscillator
Answer: (d)
Solution:

∴ Frequency of Mux control signal is f = 1
2T

= 1 fsweep
2

∴ Half the frequency on sweep signal

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8. A load is connected to supply. A current transformer (CT) and a
potential transformer (PT) is used in between load and supply. A
power factor of 0.5 is measured at the secondary side of CT and
PT. If phase angle errors of CT and PT are 0.4 and 0.7., power
factor of the load is
a) cos 60.30
b) cos 58.90
c) cos 59.70
d) cos 61.10
Answer: (d)

9. Hay's bridge is suitable for the measurement of which one of the
following?
a) Inductance with Q > 10
b) Inductance with Q < 10
c) Capacitance with high dissipation factor
d) Capacitance with low dissipation factor
Answer: (a)
Solution:
Hay Bridge: High Q-coil, Q > 10
Maxwell Bridge: Low Q-coil, Q < 10

10. What is harmonic distortion?
a) Linear behavior of circuit elements
b) Non-linear behavior of circuit elements
c) Change in behavior of circuit elements due to change of
temperature

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d) Aging of elements

Answer: (b)

Solution:

Non linearity of circuit element introduces harmonic distortion.

11. Match the items given in List-I and those in List-II

(Temperature coefficient of Resistance). Select your answer

using codes given in the lists:

List-I List-II

(a) Aluminium P. Negligibly small

(b) Manganin Q. Positive

(c) Carbon R. Negative

a) a → R, b → Q, c → P

b) a → Q, b → P, c → R

c) a → P, b → Q, c → R

d) a → R, b → P, c → Q

Answer: (b)

Solution:

Aluminium is conductor so, it has positive temperature

coefficient of resistance.

Carbon is a semiconductor so, it has negative temperature

coefficient of resistance.

Manganin is conductor but its temperature coefficient of

resistance is negligible.

So, option (b) is correct.

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12. Instrument transformers are known to introduce magnitude and
phase errors in measurements. These are primarily due to
a) improper connections on the primary side
b) measurement errors inherent in the meter connected to the
transfer secondary.
c) open and short circuit parameters of the instrument
transformers.
d) none
Answer: (c)

13. A static combination of control coil and compensating coil is
used in Megger to minimize the effect of
a) stray capacitance
b) surface leakage
c) stray magnetic field
d) aging of magnet
Answer: (c)
Solution:
A static combination of control coil and compensating coil is
used in Megger to minimize the effect of stray magnetic field.

14. The preferred damping condition for instruments is
a) critically damped
b) a damping coefficient of 0.8 to 1
c) over damped
d) underdamped
Answer: (b)

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15. Consider the following statements:
Measurement of currents of more than 10 A requires
1. CT
2. PT
3. Attenuator
4. Hall effect probe
Select the correct answer using the codes given below:
a) 1 only
b) 1 or 4
c) 2 or 3
d) 3 or 4
Answer: (a)
Solution:
Hall Effect probe is useful for current < 10A.

16. An ammeter is obtained by shunting a 30 Ω galvanometer with
30 Ω resistance. What additional shunt should be connected
across it to double the range?
a) 15 Ω
b) 10 Ω
c) 5 Ω
d) 30 Ω
Answer: (a)
Solution:
Given, Rm = Rsh = 30 Ω

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From basic concepts, if we want to make the range m times then

we connect a Rsh as Rsh = Rm across the meter.
m−1

Say the meter is made as shown below.

If the galvanometer rating was I1 then the ammeter rating will be

m times given as, I = mIm and the shunt resistance given by

Rsh = Rm
m−1

⇒ = Rm + 1 = 30 + 1
Rsh 30

= 2 =


Or, I = 2Im

Now, we want this ammeter to have twice the rating again, i.e.,

four times of rating of galvanometer. Let the required shunt

resistance be R2Ω as shown in below.

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To have the four times extension, m = 4.

i.e., I’ = 4Im or m’ = ′ = 4


Now, (30∥R2) =
′−1

= 30
4−1

⇒ 30× 2 = 10
30+ 2

⇒ 30R2 = 300 +10R2

⇒ 20R2 = 300

⇒ R2 = 15 Ω

17. A wattmeter reads 400 W when its current coil is connected in

the R phase and its pressure coil is connected between this phase

and the neutral of a symmetrical 3- phase system supplying a

balanced star connected 0.8 p.f. inductive load. The phase

sequence is RYB. What will be the reading of this wattmeter if

its pressure coil alone is reconnected between the B and Y

phases, all other connections remaining as before?

a) 400.0

b) 519.6

c) 300.0

d) 692.8

Answer: (b)

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Solution:

W = 400 watt

W = Vph Iph cosϕ

Vph Iph = 400/0.8

This type of connection gives reactive power

W = √3VpIp sin ϕ

= √3 400 × 0.6 = 519.6
0.8

18. A 1000 ohms/V meter is used to measure a resistance on 150 V

scale. The meter resistance is

a) 150 k Ω

b) 1 k

c) 6.67 ohms

d) 0.001 ohms

Answer: (a)

Solution:

Meter resistance: 150 x 10000 = 150 kΩ

11


19. The scale of moving iron (M.I) instrument is

a) Uniform

b) cramped

c) linear

d) All the above

Answer: (b)

Solution:

Since the torque equation of MI instrument is

Td = I2 dL or Td ∝ I2


In a MI instrument, to get a linear scale at lower value of , very

high value of dL/d is required, which is not possible

practically due to which the scale is cramped at the lower end of

the scale. Similarly, at higher value of , dL/d is minimum

due to which deflecting torque is minimum. Hence, the scale is

cramped to read this minimum torque at high scale range.

20. The bridge circuit shown in the fig below is used for the

measurement of an unknown element Zx. The bridge circuit is

best suited when Zx is a

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a) Low resistance

b) High resistance

c) Low Q Inductor

d) Lossy Capacitor

Answer: (c)

Solution:

The above bridge is Maxwell Inductance - Capacitance bridge,

It is suitable for the measurement of Low 'Q' Inductor.

21. The sensitivity of a voltmeter with a full scale indication of

500 μA having an internal resistance of 250 ohms is

a) 2000 ohm/V

b) 50000 ohm/V

c) 1000 ohm/V

d) 2000 ohm/V

Answer: (d)

Solution:

Full scale voltage = 500 × 10-6 × 250

= 125000 × 10-6

= 0.125 V

Internal resistance = 250 ohms

∴ Sensitivity of the meter = 250
0.125

= 2000 ohm/V

13


22. Assertion (A): In a spectrum analyzer, the local oscillator may
be automatically swept and the spectra indicated on a CRT for a
high frequency instrument.
Reason (R): Spectrum analyzer contains intermediate frequency
(IF) amplifier.
a) Both A and R are true, and R is the correct explanation of A.
b) Both A and R are true, but R is not a correct explanation of
A.
c) A is true, but R is false
d) A is false, but R is true.
Answer: (b)
Solution:
In spectrum analyzer Saw tooth generator makes local oscillator
sweep through its frequency to produce IF.
RF input and IF are mixed and amplified through IF amplifier.

23. Thermocouple instruments can be used for a frequency range
a) upto 100 Hz
b) upto 5000 Hz
c) upto 1 MHz
d) 50 MHz and above
Answer: (b)

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24. The resistance of a circuit is found by measuring the current

flowing and the power fed into the circuit. The limiting error of

power and current measurements are found to be ± 2% and ±

1.25% respectively. Which one of the following is the limiting

error in the measurement of resistance?

a) ± 4.5%

b) ± 0.45%

c) ± 2%

d) ± 0.045%

Answer: (a)

Solution:

P = I2R, ⇒ R =
2

% = ± � + 2 � × 100 ⇒ = 2 ∙ = 1.25
100 100

% = ±(2 + 2 × 1.25) = ±4.5%


25. Sensitivity inaccuracy of a recording instrument means

a) the amount of input required to produce unit pen deflection

b) the smallest signal required to produce detectable output

c) the maximum error in sensitivity displayed by a pen

d) degree to which the instrument is not sensitive enough to

repeat readings

Answer: (c)

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26. In two wattmeter method of measuring 3-phase power, power

factor is 0.5, then one of the wattmeter will read

a)
2

b) Zero

c) √2 W

d)
√2

Answer: (b)

Solution:

To have 0.5 power factor, Cos θ = 0.5 ⇒ θ = 60

We have tan = √3 1− 2
1+ 2

Hence, one of the wattmeter “w2” should be zero to have power

factor of 0.5.

27. Consider the following statements pertaining to strain gauges:

1. In metal strain gauge, the basic effect is a change of

resistance with strain.

2. Gauge factor of metal strain gauges is larger than that of

semiconductor strain gauges.

3. Gauge factor of semiconductor strain gauges varies with

strain.

4. Metal strain gauges are insensitive to temperature.

Which of the statements given above is/are correct?

a) 1 and 2

b) 2 only

c) 1 and 3

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d) 3 and 4

Answer: (c)

Solution:

Gauge factor of semiconductor > Gauge factor of metal. Gauge

factor of semiconductor changes with strain.

28. In three-phase power measurement the power factor of load

will be

a) cos tan−1 √3 (W1−W2)
W1+W2

b) (W1+W2)
W1−W2

c) √3 W1−W2
W1+W2

d) tan−1 W1−W2
�W1+W2

Answer: (a)

29. For the circuit shown in the figure, the voltage and current

expressions are

v(t) = E1 sin(ωt) + E3 sin(3ωt) and
i(t) = I1 sin(ωt − ∅1) + I3 sin(3ωt − ∅3) + I5 sin(5ωt)
The average power measured by the Wattmeter is

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a) 1 E1I1 cos ∅1
2

b) 1 [E1I1 cos ∅1 + E1I3 cos ∅3 + E1I5]
2

c) 1 [E1I1 cos ∅1 + E3I3 cos ∅3]
2

d) 1 [E1I1 cos ∅1 + E3I1 cos ∅1]
2

Answer: (c)

Solution:

Pave = P1 + P2 + P5

P1 ⇒ V1 I1 Cosϕ1 = E1 ∙ I1 cos ϕ1
√2 √2

= E1I1 cos ϕ1
2

P2 ⇒ V2I2Cosϕ2 = E3 ∙ I3 cos ϕ3
√2 √2

P5 = 0 as 5th harmonic does not contribute power as it is not

present in v(t).

∴ Pave = 1 [E1I1 cos ϕ1 + E3I3 cos ϕ3]
2

30. The resistance of thermistor is 5000 Ω at 250 C and its

resistance temperature co-efficient is 0.04/0C. A measurement

with a lead resistance of 10 Ω will cause an error of

a) 0.020C

b) 0.20C

c) 0.10C

d) 0.050C

Answer: (d)

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Solution:

Rθ = Rθ0(1 + ∝θ ∆θ)

Rθ = 5000 + 10 = 5010 Ω

0 = 5000 Ω
∝ = 0.04 Ω/0C

5010 = 5000 (1 + ∝ ∆ )

∝ ∆ = 10
5000

∆ = 10 = 1 = 0.05℃
(5000)(0.04) 20

error = 0.05℃

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