Electrical Measurements - 1 - PDF Flipbook

Electrical Measurements - 1

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EEE

Electrical
Measurements

Test-01Solutions


ELECTRICAL MEASUREMENTS
1. Flux meter is a special type of ballistic galvanometer provided

with which one of the following?
a) Heavy electromagnetic damping and very small controlling

torque.
b) Heavy electromagnetic damping and very large control ling

torque.
c) Small electromagnetic damping and small controlling torque.
d) Large controlling torque and small electromagnetic damping.
Answer: (a)
Solution:
Flux meter is a special type of ballistic galvanometer in which
controlling torque is very small and electromagnetic damping is
heavy. There are no controlling springs as controlling torque is
very small. The high electromagnetic damping is obtaining by
having a low resistance of the circuit comprising the flux meter
and search coil.
2. The simplified block diagram of a 10 bit A/D converter of dual
slope integrator type is shown in fig. The 10-bit counter at the
output is clocked by a 1 MHz clock. Assuming negligible timing
overhead for the control logic, the maximum frequency of the
analog signal that can be converted using this A/D converter is
approximately.

1


a) 2 kHz
b) 1 kHz
c) 500 kHz
d) 250 kHz
Answer: (b)
Solution:
The dual slope integrated type voltmeter output waveform

The maximum frequency can be obtained at 1 − 2 = 0

Therefore, 1 = 2

Where, is clock period

= 1
1

= 1/(2 ) =
2

= 1 ≈ 1
210

2


3. Consider the following statements.
1. Amplifier gain and phase shift
2. Filter transfer functions
3. Two port network parameters
4. Power gain in a two port circuit
Which of the above quantities can be measured using a vector
voltmeter?
a) 1 and 3 only
b) 1, 2 and 4
c) 1, 2 and 3
d) 3 and 4
Answer: (c)

4. A resistance is measured by the voltmeter - ammeter method.
The voltmeter reading is 50 V on 100 V scale and ammeter
reading is 50 mA on 100 mA scale. If both the meters are
guaranteed for accuracy within 2%.of full scale, what is the limit
within which resistance can be measured?
a) 10 Ω
b) 20 Ω
c) 40 Ω
d) 80 Ω
Answer: (d)

3


Solution:

=


= 50 = 103
50×10−3

= 50 ± (2% 100 )

= 50 × 10−3 ± (2% 100 )
= 103 ± 4%

= 1000 ± 40

= 960 1040

Limit = 80 Ω

5. Which of the following statements are correct in case of a power

factor meter?

1. The deflection is proportional to the phase angle between

field coil and crossed coil.

2. The restoring torque is provided by a controlling torque.

3. It consists of two coils mounted at right angles to each other.

Select the correct answer using the code given below:

a) 1 and 2

b) 2 and 3

c) 1 and 3

d) 1, 2 and 3

Answer: (c)

Solution:

No controlling torque in p.f meter, hence point 2 is incorrect.

4


6. A metal strain gauge has gauge factor of two, its nominal

resistance is 120 Ohms. It is undergoes a strain of 10-5, the value

of change of resistance in response to the strain is

a) 240 Ohms

b) 2 × 10-5 Ohms

c) 2.5 × 10-3 Ohms

d) 1.2 × 10-3 Ohms

Answer: (c)

Solution:

= 2 = 120
= 10−5 ∆ =?

= ∆ = 10−5




=




∆ = × ×

= 120 × 10−5 × 2 ⇒ 2.4 × 10−3

7. A capacitor is connected across a portion of resistance of the

multiplier in order to make the pressure coil circuit of the

wattmeter non-inductive. The value of this resistance is r while

the total resistance and inductance of the pressure circuit are

respectively RP and L. The value of the capacitance C is

a)
2

b) 0.41
2

c)
2

5


d) 0.41
2

Answer: (c)

Solution:

= + ( − ) + + ‖ 1


= + ( − ) + +
1+

= + ( − ) + + (1− )
1+ 2 2 2

Neglecting the term 2 2 2 ≪ 1

= + ( − ) + + (1 − )
= � + � + ( − 2)

For non-inductive pressure coil circuit,

= 2

=
2

Hence, option (c) is correct.

8. The simultaneous application of signals x(t) and y(t) to the

horizontal and vertical plates, respectively, of an oscilloscope,

produces a vertical figure-of-8 display. 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)

6


c) Q sin (8t + 60)
d) Q sin (4t + 30)
Answer: (b)
Solution:

( ) = sin 4 + 30

( ) =?

= = 1
2

= 1 ( )
2

The frequency of “Q sin(2t + 15)” is half of x(t) therefore

answer is option (b).

9. Assertion (A): Electrodynamometer wattmeter is not suitable

for low power factor power measurement.

Reason (R): Many wattmeters are compensated for errors

caused by inductance of voltage coil by means of a capacitor

connected in parallel with a portion of multiplier series

resistance.

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 the correct explanation

of A.

7


c) A is true but R is false.
d) A is false but R is true.
Answer: (b)
Solution:
Electrodynamometer wattmeter is not suitable for low power
factor power measurement due to

i) Low value of deflecting torque
ii) Increase errors due to pressure coil inductance.
iii) Power loss in the pressure coil current'
Hence, option (b) is correct.
10. The current and potential coils of a wattmeter were
accidentally interchanged while connecting. After energizing the
circuit, it was observed that the wattmeter did not show the
reading. This would be due to
a) damage done to the potential coil
b) damage done to the current coil
c) damage done to both potential and current coils
d) loose contact
Answer: (b)
Solution:
Current coil has very low resistance, if it is connected in
parallel, then due to high current passing through it, it will get
damage.

8


11. Which one of the following statements is not correct?
a) Correctness is measurement requires both accuracy and
precision.
b) Reproducibility and consistency are expressions that best
describe precision in measurements.
c) It is not possible to have precise measurements which are
not accurate.
d) An instrument with 2% accuracy is better than another with
5% accuracy.
Answer: (c)
Solution:
Accuracy and precision are two separate aspect of correctness
in measurement.
Accuracy- Degree of conformity compared to true value.
Precision - It is state of strict exactness. How consistently
something is strictly exact.
e.g. three reading of current (true value = 10 A)
By meter: 12.5, 12.5, and 12.5
The meter is precise but not accurate

12. Two resistors with nominal resistance values Rl and R2 have
additive uncertainties ΔR1 and ΔR2, respectively. When these
resistances are connected in parallel, the standard deviation of
the error in the equivalent resistance R is
a) ±�� 1 ∆ 1�2 + � 2 ∆ 2�2

9


b) ±�� 2 ∆ 1�2 + � 1 ∆ 2�2
c) ±�� 1 �2 ∆ 2 + � 2 �2 ∆ 1
d) ±�� 1 �2 ∆ 1 + � 2 �2 ∆ 2
Answer: (a)
13. By mistake, an ammeter is used as a voltmeter.
In all probabilities, it will
a) give much higher reading.
b) give extremely low reading.
c) indicate no reading at all.
d) get damaged.
Answer: (d)
Solution:
An ammeter has low internal resistance and is connected in
series to the circuit, if we use it as a voltmeter then the high
current will flow through the ammeter and then it will get
damaged.
14. An energy meter connected to an immersion heater (resistive)
operating on an AC 230 V, 50 Hz, AC single phase source reads
2.3 units (kWh) in t hour. The heater is removed from the supply
and now connected to a 400 V peak to peak square wave source
of 150 Hz. The power in kW dissipated by the heater will be
a) 3.478
b) 1.739

10


c) 1.540

d) 0.870

Answer: (b)

Solution:

Given, 2.3 units for one hour at 230 V and energy consumed =

kWH

2.3 = 2 ×


2.3 = 2302 × 1 ℎ ⇒ = 23


Now heater connected to peak to peak of 400V square wave.

∴ = 2 = 2002 = 1.739
23

15. Electronic voltmeters which use rectifiers employ negative

feedback. This is done

a) to increase the overall gain

b) to improve the stability

c) to overcome the non-linearity of diodes

d) to increase the bandwidth

Answer: (c)

Solution:

Electronic voltmeters which use rectifiers uses a high open loop

gain and large negative feedback to overcome the non-linearity

of the rectifier diodes.

11


16. In the circuit shown below, the ammeter reads 0.1 A and the
voltmeter reads 10 V. The internal resistance of the ammeter is 1
Ω and that of the voltmeter is 500 Ω. What is the value of R?

a) 100 Ω
b) 125 Ω
c) 90 Ω
d) 120 Ω
Answer: (b)
Solution:

= 500
500+

⇒ 10 = 500 × 0.1
500+

⇒ 5000 + 10 = 50

⇒ 40 = 5000 ⇒ = 125

17. Assertion (A): High resistance standards have separate

potential and current terminals.

Reason (R): Having separate terminals ensures accurate

measurement of voltage drop.

12


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: (d)
Solution:
Low resistance standard have separate potential and current
terminals.
18. Modern electronic mustimeters measure resistance by
a) taking advantage of an electronic bridge compensator for

nulling.
b) forcing a constant current and measuring the voltage across

unknown resistance.
c) using a bridge circuit
d) applying a constant voltage across the unknown resistance

and measuring the current through it
Answer: (b)
Solution:
In electronic mustimeters, when the function switch is placed in
the OHM position, a resistor is connected in series with an
internal battery and the meter simply measures the voltage drop
across the unknown resistor.

13


19. The figure shows a three phase delta connected load supplied
from a 400V, 50Hz, 3ϕ balanced source. The pressure coil and
current coil of a wattmeter are connected to the load as shown.
With the coil polarities suitably selected to ensure a positive
deflection. The wattmeter reading will be.

a) 0

b) 1600 Watt

c) 800 Watt

d) 400 Watt

Answer: (c)

Solution:

= = ( )
2

= 400∠1200 ⇒ 4∠1200
100

Voltage across pressure coil = = 400∠ − 1200

14


Wattmeter reading = × cos(∠ & )
= 4 × 400 cos(2400)
= 1600 �− 12� = −800

20. Match List-I (Device/network) with List-II (Application)
and select the correct answer,
List-I
A. Twin-T network
B. Shielded decade capacitance box selective
C. Wagner earthing arrangement
D. Inter bridge transformer
List-II
1. For use in accurate a.c. bridge
2. For realizing frequency amplifier
3. To match impedance and block d.c noise in a.c bridge
4. For minimizing earth capacitance leakage
Codes:
AB C D
a) 2 1 4 3
b) 3 4 1 2
c) 2 4 1 3
d) 3 1 4 2
Answer: (a)
Solution:
• Twin-T network is used in notch filter for realizing frequency
selection.
15


• Wagner earthing device removes all the earth capacitances
from the bridge network.

21. A Wien-bridge is used to measure the frequency of the input
signal. However, the input signal has 10% third harmonic
distortion. Specifically the signal is 2 sin400πt + 0.2 sin 1200πt
(with t in sec). With this input the balance will
a) Lead to a null indication and setting will correspond to a
frequency of 200 Hz
b) Lead to a null indication and setting will correspond to 260Hz
c) Lead to a null indication and setting will correspond lo 400Hz
d) Not lead to null indication
Answer: (d)

22. A moving iron instrument has 10 Ω resistance and gives a full
scale deflection when carrying 50 mA. It can be used measure
750 V by using what resistance?
a) 0.005002 Ω in series
b) 0.005002 Ω in parallel
c) 14990 Ω in series
d) 14990 Ω in parallel
Answer: (c)
Solution:

16


( + 10)50 × 10−3 = 750
⇒ = 14990 is series

23. Assertion (A): In a bridge type of measurement, it is required
that the indicator used to show the balance condition of the
bridge should have very high sensitivity.
Reason (R): The accuracy of the null-indicator does not play
any role in a bridge measurement.
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 the correct explanation
of A
c) A is true but R is false
d) A is false but R is true
Answer: (b)

24. A 50Hz, bar primary CT has a secondary with 500 turns. The
secondary supplies 5A current into a purely resistance burden of
1Ω. The magnetizing ampere-turns is 200. The phase angle
between the primary and secondary current is
a) 4.60
b) 85.40

17


c) 94.60

d) 175.40

Answer: (a)

Solution:

1 = 1 2 = 500 = 5 2 = 1

Magnetizing ampere turns = 200 AT

0 = Magnetizing ampere turns = 200 = 200
Primary turns 1

Transformation ratio, = 2 = 500 = 500
1 1

The phase angle between primary and secondary current is

= 180 � 0 co s ( + )� = 180 �20500c0o×s5(0)�


≈ 4.60 [∵ ( + )is very small]

25. Match List-I with List-II and select the correct answer using

the code given below the lists:

List-I

A. Mutual inductance

B. High-Q inductance

C. Audio frequency

D. Dielectric loss

List-II

1. Wien-bridge

2. Schering bridge

3. Hay bridge

4. Heaviside-Campbell bridge

18


Codes:
A BC D

a) 4 1 3 2
b) 2 3 1 4
c) 4 3 1 2
d) 2 1 3 4
Answer: (c)
Solution:
• Heaviside-Campbell bridge is used for the measurement of

mutual inductance.
• Hay bridge is used for coils of having Q > 10.
• Wien bridge is suitable for frequency measurement (100 Hz

to 100 kHz).
26. In electrodynamometer type wattmeters, the inductance of

pressure coil produces error. The error is
a) constant irrespective of the power factor of the load
b) higher at higher power factor loads
c) higher at lower power factor loads
d) highest at unity power factor loads
Answer: (c)
Solution:
The error caused by pressure coil inductance is sin tan .
With low power factor, the value of ϕ is large and therefore, the
error is correspondingly large.

19


27. Match List-I (Bridge Circuit) with List-II (Measured
Parameter) and select the correct answer:
List-I
A. Hay's bridge
B. Kelvin's double bridge
C. Schering Bridge
D. Wheatstone bridge
List-II
1. Low resistance
2. Medium resistance
3. High Q inductance
4. Capacitance
Codes:
ABCD
a) 4 1 3 2
b) 3 2 4 1
c) 4 2 3 1
d) 3 1 4 2
Answer: (d)
Solution:
• Hay bridge : High Q - coil > 10
• Wheatstone bridge : Medium resistance
• Schering bridge :Capacitance
• Kelvin double wire bridge : Low resistance

20


28. Which one of the following methods decreases the error due to
connections in a dynamometer type Wattmeter?
a) Using bifilar compensating winding in place of current coil
b) Using non-inductive pressure coil circuit
c) Using a capacitor across a part of high resistance of pressure
coil circuit
d) Using a swamping resistance
Answer: (a)
Solution:
For compensation of error due to connection bifilar
compensating winding is used.

29. Fig. shows the electrostatic vertical deflection system of CRT.
Given that VA is the accelerating voltage, the deflection
sensitivity (deflection/volt) is proportional to

a)


b)


c)


21


d)


Answer: (a)

30. Assertion (A): The rotating disc in an energy meter is made up

of a magnetic material.

Reason (R): Braking takes place due to eddy current generated

by the braking magnet.

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 the correct explanation

of A

c) A is true but R is false

d) A is false but R is true

Answer: (d)

Solution:

The rotating disc in an energy meter is made up of aluminium.

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