Electrical Measurements - 3 - PDF Flipbook
Electrical Measurements - 3
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GATE
EEE
Electrical
Measurements
Test-03Solutions
ELECTRICAL MEASUREMENTS
1. In dc potentiometer measurements, a second reading is often
taken after reversing the polarities of the dc supply and
unknown voltage and the average of the two readings is taken.
This is with a view to eliminate the effects of
a) ripples in the dc supply
b) stray magnetic field
c) stray thermal emf
d) erroneous standardization
Answer: (c)
2. Match List-I (Parameters to be Measured) with List-II
(Device to be used) and select the correct answer using the code
given below the lists:
List-I
A. High alternating current
B. High alternating voltage
C. High direct current
D. High direct voltage
List-II
1. PT
2. CT
3. Potential divider
4. Hall Effect sensor
1
Codes:
A B CD
a) 4 1 2 3
b) 2 3 4 1
c) 4 3 2 1
d) 2 1 4 3
Answer: (d)
Solution:
High a.c. current - CT
High a.c. voltage - PT
High d.c. - Hall effect
High direct voltage - Potential divider
3. An attenuator probe as shown, is connected to an amplifier of
input capacitance 0 1 μF. Value of C that must be connected
across 100 K to make the overall gain independent of frequency,
is
a) 0.01 μF
b) 0.1 μF
c) 1 μF
d) 10 μF
Answer: (a)
2
Solution:
To make gain independent of frequency time constants of both
RC networks should be equal.
100 K × C = 10 K × 0.1 μF
C = 0.01 μF
4. Which one of the following statements is correct?
Temperature coefficient of a Thermistor is
a) High and negative
b) High and positive
c) Low and negative
d) Low and positive
Answer: (c)
Solution:
Thermistor have negative temperature coefficient (NTC).
5. A d.c voltmeter has a sensitivity of 1000 Ω/Volt? When it
measures half full scale in 100 V range, the current through the
voltmeter will be:
a) 50 mA
b) 100 mA
c) 1 mA
d) 0.5 mA
Answer: (d)
3
Solution:
Sensitivity (Sv) = 1000 Ω/V
Full scale voltage = 100 V
Resistance of the voltmeter = 1000 Ω/V × 100 V
= 100 kΩ
Full load current through voltmeter = 100 = 1 mA
100kΩ
∴ For full load, current through the voltmeter = 1mA = 0.5 mA
2
6. Which one of the following is employed in an optical pyrometer
for measuring temperature?
a) Photocell principle
b) Thermocouple effect
c) Comparing the brightness of the source with the brightness of
a standard source
d) Photo conductivity principle
Answer: (c)
Solution:
The device compare the brightness produced by the radiation of
the object whose temperature is to be measured with that of
reference temperature.
7. If current through the operating coil of a moving iron instrument
is doubled, the operating force becomes
a) one and a half times
b) 2 times
c) 3 times
4
d) 4 times
Answer: (d)
Solution:
The operating torque in a M.I instrument is
= 1 2 (for M.I)
2
Now, Td = force × distance
So, force ∝ I2
So, by increasing current value 2 times, force will increase by 4
times.
8. Two strain gauges are employed for the measurement of strain
in a cantilever. One gauge is mounted on top of the cantilever
and the other is placed at the bottom. The two strain gauges
form two arms of a voltage sensitive Wheatstone bridge. What
is this bridge configuration?
a) Full Bridge
b) Half Bridge
c) Quarter Bridge
d) Null Bridge
Answer: (b)
Solution:
As there are two active strain gauges in 04 arm bridge, it is
called half bridge.
5
9. In phantom loading arrangement, energy consumption in the
calibration test of wattmeter is reduced because of
a) the separate application of low voltage supplies across current
coil.
b) no common point between the two coils.
c) the reduced loss in current coil and pressure coil.
d) the absence of load in the test set.
Answer: (d)
Solution:
In phantom loading arrangement, energy consumption in the
calibration test of wattmeter is reduced because of the absence
of load in the test set, which is why it is called as phantom
loading.
10. Which one of the following is the correct statement?
In an LVDT, the two secondary voltages
a) Are independent of the core positron.
b) Vary unequally depending on the core position.
c) Vary equally depending on the core position.
d) Are always in phase quadrature.
Answer: (c)
Solution:
6
Depending upon core movement, flux linkages with secondary
winding S1, S2 will be different and so is the output,
E0 = ES1 – ES2
11. The output of a piezoelectric crystal has
a) low amplitude and low impedance
b) high amplitude and high impedance
c) low amplitude and high impedance
d) high amplitude and low impedance
Answer: (c)
12. Which one of the following causes the disc in an induction
type of energy meter to rotate in the opposite direction?
a) The breaking magnet is faulty.
b) Both current coil and voltage coil are wrongly connected.
c) Either current coil or voltage coil is wrongly connected.
d) The load is highly reactive.
Answer: (c)
Explanation:
Due to change in connection of either voltage of current coil, the
direction of rotation will be reversed as direction of one of the
field will be reversed.
7
13. The time/div and voltage/div axes of an oscilloscope have been
erased. A student connects a 1 kHz, 5V p-p square wave
calibration pulse to channel-1 of the scope and observes the
screen to be as shown in the upper trace of the fig. An unknown
signal is connected to channel-2 (lower trace) of the scope. If the
time/div and V/div on both channels are the same, the amplitude
(p-p) and period of the unknown signal are respectively.
a) 5 V, 1 ms
b) 5 V, 2 ms
c) 7.5 V, 2 ms
d) 10 V, 1 ms
Answer: (c)
Solution:
• Channel – 1 display:
→ Known signal (Vpp = 5 V, f = 1 kHz)
→ Volt = 5V = 2.5 Volt/div
div 2 div
→ Time = 1/1kHz
div 4 div
8
= 1ms = 0.25ms/div
4 div
• Channel – 2 display:
→ Vpp = 3div × 2.5 volt = 7.5V
div
→ T = 8 div × 0.25 = 2
14. Assertion (A): Laboratory Cathode Ray Oscilloscopes employ
electrostatic deflection system.
Reason (R): Electrostatic deflection systems are most suitable
for small screens.
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: (c)
Solution:
• Electrostatic deflection system can operate at high
frequency 200 - 300 MHz
• Electromagnetic deflection system 20 - 25 kHz.
9
15. A potentiometer is used to measure the voltage between two
points of a dc circuit, which is found to be 1.2 V. This is also
measured by a voltmeter, which is found to be 0.9 V. The
resistance of the voltmeter is 60 kΩ. The input resistance
between two points is
a) 60 kΩ
b) 20 kΩ
c) 45 kΩ
d) 80 kΩ
Answer: (b)
Solution:
Method – 1:
Voltage across two points = 1.2 V (Using potentiometer)
Voltage across voltmeter = 0.9 V
Resistance of voltmeter = 60 kΩ
By voltage division,
0.9 → 60 kΩ
0.3 → 60 × 0.3 = 20
0.9
Method – 2:
Input resistance = 1.2−0.9 × 60kΩ
0.9
= 0.3 × 60kΩ
0.9
= 20kΩ
10
16. For the bridge shown Z1 = 200∠200 Ω, Z2 = 150∠300 Ω and Z3
= 300∠-300 Ω. What is the value of Z4 so that the bridge is
balanced?
a) 225∠200 Ω
b) 225∠-200 Ω
c) 100∠800 Ω
d) 100∠-800 Ω
Answer: (b)
Solution:
For balance condition,
1 = 3
2 4
200∠200 = 300∠−300
150∠300 4
4 = 225∠ − 200
11
17. In reference to the figure, the voltage waveform v(t) is
measured by a PMMC, a PMMC combined with bridge rectifier
and a moving iron (MI) instrument. Two lists are prepared there
after
Instrument List List of Instrument reading
a. PMMC i. 5 V
b. PMMC rectifier ii. 2.75 V
c. M.I. iii. 2.5 V
The correct option relating the instruments and their reading is
a) a – i, b – ii, c – iii
b) a – iii, b – ii, c – i
c) a – ii, b – iii, c – i
d) a – iii, b – i, c – ii
Answer: (b)
Solution:
Given waveform is shown below.
12
Average value for above waveform is
= 1 ∫0 ( )
= 10 × 4 = 2.5
16
R.M.S value of given waveform is
= � 1 ∫0 2( ) = 10�146 = 5
PMMC measures only average value, so reading of PMMC =
Vavg = 2.5V.
PMMC half wave rectifier measures = 1.1 × avg value of
PMMC = 1.1 × 2.5 = 2.75 volt
M.I instrument measures rms value of waveform.
Hence reading of M.I instrument = Vrms = 5 Volt
18. When the Wheatstone bridge shown in the figure is used to
find the value of resistor Rx, the galvanometer G indicates zero
current when R1 = 50 Ω, R2 = 65 Ω and R3 = 100 Ω. If R3 is
known with ±5% tolerance on its nominal value of 100 Ω. What
is the range of Rx in Ohms?
a) [123.50, 136.50]
b) [125.89, 134.12]
13
c) [117.00, 143.00]
d) [120.25, 139.75]
Answer: (a)
Solution:
As per given data,
R1 = 50Ω, R2 = 65Ω, R3 = 100Ω
The value of R3 with ±5% tolerance R3 = 100±5%
= 100 + 100 × 5
100
= 105
(or)
= 100 − 100×5 = 95
100
In both conditions, the bridge is balanced, so under balance
condition,
1 = 2
3
⇒ = 2 3
1
i) When 3 = 105
∴ = 105×65
50
= 136.50
14
ii) When 3 = 95
∴ = 95×65
50
= 123.50
19. Electrodynamic type wattmeters have large errors while
measuring power in ac circuits at low power factor conditions,
since the voltage across and the current through the
a) current coil are not in phase
b) current coil are not in quadrature
c) pressure coil are not in phase
d) pressure coil are not in quadrature
Answer: (c)
20. To minimize the errors due to lead and contact resistances, low
resistance used in electrical measurement work are provided
with
a) guard rings
b) four terminals
c) thick insulation
d) metal shields
Answer: (b)
Solution:
For measuring low resistance four terminal method is used
because in this lead resistance does not affect the measuring
quantity.
15
21. A current of {2 + √2sin (314t + 30) + 2√2cos (952t + 45)} is
measured with a thermocouple type, 5 A full scale meter. What
is the meter reading?
a) 2A
b) 3A
c) 5A
d) (2+3√2) A
Answer: (b)
Solution:
Thermocouple type instruments read rms value consider a
signal,
( ) = 1 + 2 sin( − 1) + 3 sin(2 − 1) + ⋯
= � 1 2 + �√ 22�2 + �√ 32�2 + ⋯
= �(2)2 + �√√22�2 + �2√√22�2
= √4 + 1 + 4 = 3
22. In a digital frequency meter, the Schmidt trigger is used for
a) sinusoidal waveforms into rectangular pulses
b) scaling of sinusoidal waveforms
c) providing time base
d) none of these
Answer: (a)
16
23. The waveform given below is observed on the screen of a
CRO. If the time/div switch is set at 10 μs and the volt/div at
200 mV the frequency and peak to peak voltage are,
respectively
a) 33.33 kHz, 800 mV
b) 25 kHz, 600 mV
c) 33.33 kHz, 600 mV
d) 25 kHz, 400 mV
Answer: (a)
Solution:
• Numbers of voltage division peak to peak = 4 volt/division =
200 mV
VP-P = (4)(200) = 800 mV
• Numbers of time division = 1.5 + 1.5 = 3
Time/div = 10 μs
Time period = (3)(106)sec
= 1 = 1 = 33.33
3×10−6
17
24. An ammeter has a current range of 0 – 5A, and its internal
resistance is 0.2Ω. In order to change the range to 0 – 25 A we
need to add a resistance of
a) 0.8 Ω in series with the meter
b) 0.02 Ω in series with the meter
c) 0.04 Ω in parallel with the meter
d) 0.05 Ω in parallel with the meter
Answer: (d)
Solution:
Multiplying power m = = 25 ⇒ 5
5
∴ ℎ =
−1
= 0.2 ⇒ 0.05
5−1
25. The total power delivered to a three-phase load is equal to
a) algebraic difference of two-wattmeter readings
b) algebraic sum of two-wattmeter readings
c) vectorial difference of two-wattmeter readings
d) vectorial sum of the two-wattmeter readings
Answer: (b)
18
26. The circuit in fig is used to measure the power consumed by
the load. The current coil and the voltage coil of the watt meter
have 0.02 Ω and 1000 Ω resistance respectively. The measured
power compared to the load power will be
a) 0.4% less
b) 0.2% less
c) 0.2% more
d) 0.4% more
Answer: (c)
Solution:
Measured power = True power + error
True power = VI cosϕ
= 200 × 20 ×1 ⇒ 4000
Error is due to only load side connected coil i.e., current coil.
2 = 202 × (0.02) ⇒ 400 × 0.02 ⇒ 8
Measured power, = 4000 + 8 ⇒ 4008
%error = 4008−4000 × 100
4000
= 0.2% more
19
27. A CRO probe has an impedance of 500 kΩ in parallel with a
capacitance of 10pF. The probe is used to measure the voltage
between P and Q as shown in Fig. The measured voltage will be
a) 3.53 V
b) 4.37 V
c) 4.54 V
d) 5.00 V
Answer: (a)
Solution:
C = 10pF
XC = 1� = 1
2 ×100×103×10×10−12
100 ∥ ∥ = 159 ∥ 500 ∥ 100
= 54.676
20
∴ = 10 × 54.676 = 3.53
54.67+100
28. When selecting a measuring instrument, be sure its range is
a) higher than the value being measured
b) below that of the circuit being tested
c) mid-way on the scale
d) on the lower third of the scale
Answer: (a)
29. A 35 V d.c supply is connected across a combined resistance
of 600Ω and an unknown resistance of RΩ in series. If a
voltmeter having a resistance of 1.2kΩ is connected across
600Ω resistor and reads 5 V, then what is the value of the
resistance R?
a) 1.20 Ω
b) 500 Ω
c) 1.7 kΩ
d) 2.4 kΩ
Answer: (d)
21
Solution:
600Ω ∥ 1.2kΩ = (600)(1200) = 400Ω
1200+600
= 5
400
Also, 30 = iR
30 = 5 ⇒ = 2400
400
= 2.4
30. 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)
22
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