Electrical Measurements - 2 - PDF Flipbook
Electrical Measurements - 2
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GATE
EEE
Electrical
Measurements
Test-02Solutions
ELECTRICAL MEASUREMENTS
1. What is the correct sequence of blocks A, B, C and D in the
block diagram of a data acquisition system?
a) A/D converter, Signal converter, Multiplexer and Digital
recorder
b) Signal converter, Multiplexer, Digital recorder and A/D
converter.
c) Digital recorder, Signal converter, Multiplexer and A/D
converter.
d) Signal converter, Digital recorder, Multiplexer and A/D
converter
Answer: (a)
Solution:
Digital recorder block cannot be prior to A/D converter
2. If 3-phase power is measured with the help of two-wattmeter
method in a balanced load with the application of 3-phase
balanced voltage, variation in readings of wattmeters will
depend on
a) load only
b) power factor only
c) load and power factor
d) neither load nor power factor
Answer: (c)
1
Solution:
1 = √3 cos(300 + )
2 = √3 cos(300 − )
Hence readings of wattmeters depend on both loading as well as
power factor.
3. Which of the following units are present in a spectrum analyzer?
1. Mixer
2. Saw-tooth generator
3. Local oscillator
Select the correct answer using the code given below:
a) 1 only
b) 1 and 3 only
c) 2 and 3 only
d) 1, 2 and 3 only
Answer: (d)
Solution:
2
4. A phase sequence indicator is used to determine the phase
sequence of any energized
a) 3-phase line
b) single phase line
c) dc line
d) telephone line
Answer: (a)
5. For a dual ADC type 3½ digit DVM, the reference voltage is
100mV and the first integration time is set to 300ms. For some
input voltage, the "deintegration" period is 370.2ms. The DVM
will indicate
a) 123.4
b) 199.9
c) 100.0
d) 1.414
Answer: (a)
Solution:
We know: 1 = 2
⇒ = × 2
1
= 100 × 370.2
300
= 123.4
3
6. One single-phase energy meter operating on 230 V and 5 A for
5 hours makes 1940 revolutions. Meter constant is 400 rev/kWh.
The power factor of the load is
a) 1.0
b) 0.8
c) 0.7
d) 0.6
Answer: (b)
Solution:
Energy consumed = �1490400� = 4.85 ℎ
Power = �4.585� ℎ = 970
cos ∅ = 970
230 × 5 cos ∅ = 970
cos ∅ = �23907×0 5�
= 0.843
7. Match List-I with List-II and select the correct answer:
List-I List-II
A. Wien Bridge 1. Power
B. 3-Voltmeter method 2. Frequency
C. Tri-vector meter 3. Phase
D. Watt-hour meter 4. Maximum demand
5. Energy
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Codes:
AB C D
a) 2 1 4 5
b) 4 5 3 1
c) 2 5 4 1
d) 4 1 3 5
Answer: (a)
Solution:
3-voltmeter method - Power measurement
8. The main advantage of a strip-chart recorder is that it provides a
a) record you can see at a glance
b) record for long unattended periods
c) daily record on a single sheet
d) record of voltage and current, using instrument
Answer: (b)
5
9. A Hall Effect pick-up can he used for the measurement of
a) Temperature change
b) Pressure
c) Magnetic flux
d) Relative humidity
Answer: (c)
Solution:
Flux to be measured drive current and produce hall voltage
according to Hall Effect.
10. In a 3 ½ digit voltmeter, the largest number that can be read is:
a) 9999
b) 0999
c) 1999
d) 5999
Answer: (c)
Solution:
For 1-bit, range is 0 – 9
For 2-bit, range is 00 – 99
For 4-bit, range is 0000 – 9999
For ½ bit MSB will be either 0 or 1
So, for 312 digit, largest number is 1999
11. Which one of the following statements is correct?
In distortion factor meter, a filter is used to suppress
a) d.c component
b) Odd harmonics
6
c) Even harmonics
d) Fundamental frequency component
Answer: (d)
Solution:
Fundamental frequency is removed by filter to calculate THD
(Total harmonic distortion).
12. In electrodynamometer ammeter, the deflection of the pointer
is proportional to
a) mean of currents in fixed coil a moving coil.
b) square of the current in moving coil.
c) Rms value of currents in fixed coil.
d) mean-square of currents in fixed coil and moving coil.
Answer: (d)
Solution:
In electrodynamometer ammeter the deflection pointer torque:
= 1 2 cos ∅
Also, =
At balance condition,
=
Or, = 1 2 cos ∅
Or, ∝ 1 2
Where I1, I2 are rms values of current in fixed/current coil and
pressure coil/moving coil respectively.
7
13. Air friction damping should not be used where the deflecting
torque in the instrument is produced due to
a) magnetic field
b) electrostatic field
c) thermo-electric emf
d) none of these
Answer: (d)
14. When a multiplier is added to an existing voltmeter for
extending its range, its electromagnetic damping
a) remains unaffected
b) increases
c) decreases
d) changes by an amount depending on the controlling torque
Answer: (c)
Solution:
When a multiplier is added to an existing voltmeter for
extending its range its electromagnetic damping decreases.
15. The dynamic characteristic of capacitive transducers are
similar to those of
a) low pass filters
b) high pass filters
c) notch filter
d) band stop filters
Answer: (b)
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16. An unshielded moving iron voltmeter is used to measure the
voltage in an a.c circuit. If a stray d.c magnetic field having a
component along the axis of the meter coil appears, the meter
reading would be
a) unaffected
b) decreased
c) increased
d) either decreased or increased depending on the direction of
the d.c field
Answer: (d)
17. Standardization of potentiometers is done in order that, they
become
a) accurate
b) precise
c) accurate and direct reading
d) accurate and precise
Answer: (c)
18. How would a reading 0.6973V be displayed on a 4 ½ digit
digital voltmeter on its 10 V scale?
a) 0.697
b) 0.6973
c) 0.6900
d) 0.7000
Answer: (a)
9
Solution:
Resolution on 10 V scale = 10
10
→ Number of full bits
Resolution = 10 = 1 = 0.001
104 103
Reading of 0.6973 = 0.697 volts
19. A piece of oil soaked paper has been inserted between the
plates of a parallel plate capacitor. Then the potential difference
between the plates will
a) increase
b) decrease
c) remain unaltered
d) become zero
Answer: (b)
Solution:
For parallel plate capacitor,
= ∝
Where, E = Electric field
and x = distance between plates
here, E ∝ 1 ⇒ ∝ 1
4 0
As increases (due to oil soaked paper), E decreases that
means V also decreases.
10
20. A 200/1 Current transformer (CT) is wound with 200 turns on
the secondary on a Toroidal core. When it carries a current of
160 A on the primary, the ratio and phase errors of the CT are
found to be – 50% and 30 minutes respectively. If the number of
secondary turns is reduced by 1 the new ratio error (%) and
phase error (min) will be respectively.
a) 0.0, 30
b) -0.5, 35
c) -1.0, 30
d) -1.0, 25
Answer: (a)
Solution: % = − × 100
Where
R =Transformation ratio
Kn = nominal ratio
By reducing the secondary winding turns slightly, the actual
transformation ratio is made nearly equal to the nominal ratio.
And phase error doesn’t change.
Therefore, percentage error = 0% and phase error is 30 minutes
only.
21. A moving coil galvanometer is made into a d.c, ammeter by
connecting
a) a low resistance across the meter
b) a high resistance in series with the meter
c) a pure inductance across the meter
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d) a capacitor in series with the meter
Answer: (a)
22. Magneto elastic transducers work on the principle of
a) change of dimension with change of applied stress
b) change of permeability with change in stress
c) change of coercive force with change of stress
d) none of these
Answer: (b)
23. In eddy current damping systems, the disc employed should be
of
a) Conducting and magnetic material
b) Conducting but non-magnetic material
c) Magnetic but non-conducting material
d) Non-conducting and non-magnetic material
Answer: (b)
Solution:
Disc is conducting to produce eddy current for magnetic field. It
must be non-magnetic, i.e., so that it does not affect the
measurement.
24. The full-scale deflection current of a meter is 1 mA and its
internal resistance is 100 ohms. This meter is to have full
deflection when 100V is measured. What is the value of series
resistor to be used?
a) 99.90 k ohms
b) 100 k ohms
12
c) 99.99 ohms
d) 100 ohms
Answer: (a)
Solution:
Total value of resistance = 100 = 100000 = 100
1×10−3
Meter resistance = 0.1 kΩ
∴ Series resistance required = 100 – 0.1
= 99.9 k ohms
25. The figure shows input attenuator of a multi-meter. The meter
reads full-scale with 12V at M with the range switch at position
B obtain full-scale deflection with the range switch position at
D?
a) 1 V
b) 150 V
13
c) 120 V
d) 147 V
Answer: (c)
Solution:
R = 2 MΩ + 6 MΩ + 1.2 MΩ + 600 kΩ + 120 kΩ + 60 kΩ + 20
kΩ = 10 MΩ
Switch at B, VM = 12 V:
= × 2 Ω
10 Ω
= 12 × 0.2
= 2.4
Switch at D, VFS = 2.4 V:
= × 200 Ω
10 Ω
2.4 = × 200 Ω
10 Ω
= 120
26. The advantage of the LVDT over the inductive bridge-type
transducer is that
a) it produces high output voltage for small changes in core
position
b) it can be used for operation at temperatures as low as - 2650C
and as high as 5000C
c) it can measure displacement of a moving object
d) the error due to stray magnetic field is less
Answer: (a)
14
27. Electronic voltmeters which use rectifiers employ negative
feedback. This is done
a) to increase the overall gain
b) to improve stability
c) to overcome non-linearity of diodes
d) none of these
Answer: (c)
28. Assertion (A): In general, moving iron voltmeters are less
accurate than those of the dynamometer type
Reason (R): Moving iron voltmeters often use air friction
damping.
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:
Moving iron voltmeters - 5% -10% error
Dynamometer - 0.5 - 1% of full scale
Both uses air friction damping
29. Magneto-electric transducers work on the principle of
a) change of dimensions with change of applied stress
b) change of permeability with change in stress
c) change of coercive force with change of stress
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d) none of these
Answer: (b)
30. A 3-phase balanced load which has a power factor of 0.707 is
connected to a balanced supply. The power consumed by the
load is 5kW. The power is measured by the two wattmeter
method. The readings of the two wattmeter’s are
a) 3.94 kW and 1.06 kW
b) 2.50 kW and 2.50 kW
c) 5.00 kW and 0.00 kW
d) 2.96 kW and 2.04 kW
Answer: (a)
Solution:
As per given data
3-ϕ balanced load pf = 0.707
ϕ = 450
Power consumed by load = 5 kW
As per two wattmeter method,
1 = cos(30 − )
2 = cos(30 + )
= √3 cos = 5000
= 5000 = 4083.099
√3×0.707
1 = 4083.099 cos(30 − 45)
2 = 4083.099 cos(30 + 45)
1 = 3.94 2 = 1.06
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