1. (c) Induction principle is more generally used for Watt-hour meters than for ammeters and voltmeters owing to their comparatively high cost, and inaccuracy of induction instruments of the latter types.
2. (b)
6. (a) Average current = 117.02 117.08 117.11 117.03 4
9. (d) Refer potentiometric type digital voltmeter 10. (b)
11. (b) LVDT & strain gauge measure linear displacement.
12. (d) Td = qC , Td =KI2
15. (c) Work done by the electric field = eV Kinetic energy = 1mv2
17. (a) 18. (c)
19. (c) 20. (d)
21. (a) 22. (c)
23. (a)
24. (a) The makers specify the maximum value as 100 mA. This means that the current drawn from the cell should be less than 100 mA and this current should flow momentarily.
25. (d) 26. (a)
27. (a) 28. (b)
29. (b) 30. (c)
31. (a) 32. (c)
33. (c) Both spectrum analyzer and distortion factor meter are signal analysers. A spectrum analyser sweeps the signal frequency band and displays a plot of amplitude versus frequency. Distortion factor meter tunes out the fundamental signal and gives an indication of the harmonics.
34. (c) 35. (d)
36. (b) 37. (b)
38. (d) There will be two deflecting torques, one acting on coil A and the other on coil B. The coil windings are so arranged that the torques due to the two coils are opposite in direction. So, there is no controlling torque.
39. (c) The error caused by pressure coil inductance is VI sin f tan b. With low power factor, the value of f is large and therefore, the error is correspondingly large.
40. (d) Braking torque TD ¥ R2 where R is the radial position of braking magnet with respect to the centre of the disc.
41. (a) Bigger size mirror results in increased inertia constant J.
m
as J increases both fn & A decrease.
42. (b) Actual loading arrangements would involve a considerable waste of power. In order to avoid this
‘Phantom’ or ‘Fictitious’ loading is done.
43. (a) Kelvin bridge overcomes the difficulties that arise in a Wheatstone bridge on account of the resistance of the leads and the contact resistances while measuring low valued resistors. The Kelvin double bridge incorporates the idea of a second set of ratio arms, hence the name double bridge and the use of four terminal resistors for the low resistance arms. The second set of ratio arms eliminate the effect of connecting lead resistance.
44. (d) Self-capacitance Cd = C1 4C2 3
-45. (a) The expansion (deflection) is proportional to the heating effect of the current and hence to the square of the rms value of the current. Therefore, the meter may be calibrated to read the rms value of the current.
46. (c)
47. (c) Simple a.c. amplifiers may be used to amplify a dc input through use of an additional circuit component known as chopper. The approach is used to build a dc amplifier
where in the dc is first converted to an equivalent a.c.
signal that is amplified by a standard a.c. amplifier. The a.c. signal is finally converted back to a dc signal.
Chopper stabilized amplifier eliminates the effects of dc offset currents & the drift of other dc parameters by using an a.c. coupled amplifier for necessary gain.
48. (b) In some meters a slow but continuous rotation is observed even when there is no current flowing through the current coil and only pressure coil is energized.
This is called creeping. The major cause for creeping is over-compensation for friction.
49. (a) A three digit display on a digital voltmeter for 0-1 V range will be able to indicate values from zero to 999 mV, with smallest increment or resolution of 1 mV. In practice a fourth digit usually capable of indicating either 0 or 1 only is placed to the left of active digits.
This permits going above 999 to 1999 to give overlap between ranges for convenience. This is called over ranging. This type of display is known as a 3½ digit display.
50. (c) Refer Piezoelectric type accelerometer.
51. (c) 53. (b) The wattmeters read
One µ cos (30 + f) The other µ cos (30 – f)
Thus whenever load p.f. angle f becomes 60° one of the wattmeters would read zero.
\ For the given condition, the load p.f. is cos 60° = 0.5
54. (d) M should be able to detect phase reversal if not strictly phase sensitive. Phase sensitive detector can well serve the purpose.
55. (d) Strictly speaking, an Ayrton shunt is used to protect the galvanometer during initial stages of balancing by reducing the galvanometer sensitivity. However, it can also be used to convert a galvanometer into a multistage ammeter
57. (b)
Voltmeter resistance = 100 × 10 = 1000 k-ohm = 1 M-ohm Photovoltaic cell internal resistance = 1 M-ohm The voltmeter reads 5 V. Therefore, the internal voltage is 10 V.
58. (b) Differential transformer transducer is used for measurement of displacement.
60. (a) It is an elementary piece of knowledge. For low resistance measurements, if circuit II is used, the ammeter resistance, which may be comparable with the low resistance being measured, is added to the unknown and thereby causing larger error.
61. (c) The setting of the potentiometer is at 1.0138 instead of 1.0813. Therefore, the working current in the slide wise is more than what it would have been if the setting was at 1.0813, i.e. the correct value for applied voltage of 1.0813 V,
c
I 1.0813 I =1.0138
If the setting is at 1.0138 and the applied voltage is also 1.0138, the standardisation is then correct.
(Statement of the problem appears ambiguous) 62. (c) If D1 were not there in the negative half cycle of the
applied voltage, there would be some small reverse current through D2 (in the reverse direction) and the meter, thereby giving erroneous mean value and hence of the applied voltage. With D1 as in circuit, the current in the negative half cycle flows through Rsh and D1 and there is no voltage across D2 and hence no current through D2.
63. (c) When major area of the Lissajous figure lies in the 1st and 3rd quadrant, then the angle is within 0 and ± 90°.
If the loop lies in 2nd and 4th quadrant, then the angle is 180° – q where q is within 0 and ± 90°. In the present case the angle is given by q = sin–1 (Y1/Y2) = sin–1 0.5
= 30°, therefore, the phase difference is 30° or – 30°, i.e., 30° or 330°. Y1 is the deflection on the y-plates when x-plates deflection is zero, i.e., when Vx(t) = 0 and Vy(t) = Vmy sinf and Y2 is the deflection due to Vmy. 64. (c) The total length observed will be proportional to
2Vm = 2 2 × 30. 30 V dc causes 1 cm deflection from the centre point. Therefore, the length observed is 2 2 cm » 3 cm.
65. (b) Form factor = RMS Value
Mean Valueof the wave M.I. Instrument will show RMS value. Rectifier voltmeter is calibrated to read rms value of sinusoidal voltage, i.e., with form factor of 1.11. Therefore, mean value of the applied voltage is V2
The current coil and voltage coil can be connected in two ways. The wattmeter reads high by an amount equal to power loss in current coil in one case and in voltage coil in the other. The two possible power losses are
The two percentage errors are (i) 4 100 2
67. (b) In the induction type of instrument there is no moving coil. Therefore, the wattmeter is question is of dynamometer type. The fixed coil is the current coil and moving coil the pressure or voltage coil.
68. (b) It is simple resistive circuit calculation,
( )
From the given data,
G G
69. (a)
70. (d) Only 1 and 3 statements are correct. Measurement range is not wide.
71. (a) Applying the usual balance condition relation Z1Z4 = Z2Z3, we have
72. (c) Q = Lw / R. With Q-meter resistance considered, the measured or indicated Q is 1/1.11 times the actual Q.
Thus, statement 1 is not correct. Statements 2 and 3 are correct.
73. (d)
74. (c) For a half radiated wave
peak peak
76. (d) Assuming that gate open when the ramp is 8V and closes when the ramp reached zero, then the number of pulses counted by counter would be :
400 × 103 × 20 × 10–3 = 8000 77. (b) f = 1 × 106 / 1 × 104 = 100
\ T=1001 =10 ms
78. (d) All the three errors may occur in bridge method of measurement.
79. (d) For 10 V to produce 1 mA requires a 10 k ohm resistor.
Thus, the meter will appear as:
M
10 V + –
10 kW
When 5 kilo-ohms is placed across the terminals, the current will be 10 V / (10 + 5) k.
Thus, 10 V/15 K = 2/3 mA and thus 2/3 deflection.
80. (a)
81. (d) The meter reading will either decrease or increase depending upon the direction of flux.
F Fm
Fs
As shown in fig. let Fm is the flux due to meter coil and Fs is due to stray magnetic field, then resultant will be F. Similar but opposite will happen when Fs is in opposite direction. Thus, the reading will either increase or decrease.
82. (d) Ammeter error, ID = ±4%
Voltmeter error, VD = ±24%
Resistance to be measured, R = V/I
In case, error in resistance measurement is ± D , thenR the maximum possible error will be
R+ D =R V+ DV / I- DI or RI I R R I R+ D - D D = + DV V
Neglecting DR being small and substituting V = RI we have,
Therefore, maximum percentage error
= 2.4% + 1% = 3.4%
83. (c) Error in the potentiometer is due to thermo electric emf’s set up at junctions of dissimilar metals and also by the heat from the operator’s hand during adjustment of the working parts of the potentiometer. Reversing of polarity and taking reading will give error in the negative direction. Suppose, the reading with error due to one direction is D + DR and in the other direction is D – DR, then the mean value is (D + DR + D – DR) / 2 = D; which is the actual value.
84. (c) In a dual slope integrating type digital voltmeter,
( )
in 2 1
V =Vref t / t
Where, t1 is the first integration time.
t1 = 10 × 1/50 = 0.2 s Vm = 1 V
Vref = 2
t2 = VinT1 / Vref = 1 × 0.2 / 2 = 0.1 sec 85. (b) The lines base setting = 5 msec/div.
Input frequency = 1 3 100050 5 10- =
´ = 200 Hz
Frequency of input signal = 314 = w, or, 2pf = 314 or f = 314 / 2p = 50 Hz Number of cycles of signal displayed
= 50 × 10 / 200 = 2.5
86. (a) The “accuracy” of a measuring instrument is determined by the closeness of the value indicated by it to the correct value of the measurement.
87. (c) (100 50 )Z° d =(300- 90 ) (200 0 )° ° Þ Zd=600 140- °
88. (d) From balance condition
( )
b89. (d) As both the signals are of same frequency, same phase and of same amplitude hence these are applied at the X-Y input of the CRO, hence on the CRO’s screen, these will produce a straight line with 45° with respect to x-axis.
90. (a)
91. (b) Magnitude of the limiting error, ¶ = eA r.As
= 0.01 × 150 = 1.5 V Relative error, r 1.5
75 0.02
e = =
voltage between the limits,
( )
75 1 0.02± =75 1.5V±
% limiting error = 1.5 100 2%
75´ =
92. (b) As voltmeter has high impedance, therefore, current through it is zero or negligible.
10 W
94. (a) The overall uncertainty;
1 2 3
A.Vibration Galvanometer 100 Hz
B. Head phone 1 kHz
C. D 'Arsonval Galvanometer 0 Hz
D. CRO L arg e frequency range
Instruments Frequency
97. (a) L
The resulting Lissajous pattern is a circle as the magnitudes of the two signal are same and phase difference is 90°.
Note: A circle can be formed in Lissajous pattern only when the magnitudes of the two signals are equal and the phase difference between them is 90° or 270°.
99. (c) The damping torque in the disc of an ac energy meter is produced by Eddy current effect (by interaction of the eddy current and permanent magnet’s field).
100. (b) Error as percentage of true value;
= Full scale voltage Accuracy Voltage value ´
102. (c) The resolution of a wire wound potentiometer can be improved by reducing the diameter of the resistance wire.
103. (d) Loading effect error: When the voltage across impedance is measured by a voltmeter, the voltmeter acts as a load and due to its internal impedance an error in reading occurs. This error is called loading effect error.
104. (a) The PMMC is used for measurement of DC. Therefore, the current
i 3 4 2 sin 314t 3A= + = .
105. (d) In a trapezoidal pattern, modulation factor (M) Large line – Smaller line
M=Large line + Smaller line
Smaller line
Larger line
So, in the trapezoidal pattern smaller line = 0.
Hence, Large line – 0
M 1
Large line + 0
= =
106. (b) The shape of the figure seen will be a straight line as both the voltages are equal and in phase with each other (the shape will be a straight line even if the two voltages are out of phase by 180°).
107. (b) CRO has minimum loading effect on the quantity under measurement.
108. (d) Owen’s bridge is used for measurement of inductance in terms of capacitance.
At balance condition; Equating imaginary terms; x a b
c R R C
= C
109. (c) Servo type Potentiometric recorder has better frequency response than galvanometric recorder.
110. (c) The Wheatstone bridge is ideally suited for measuring resistance in the range of 100 W to 10 kW.
111. (c) Electronic voltmeter requires external source for its operation.
Electronic voltmeter consumes low power.
112. (d) Given;
Oscillator frequency = 400 kHz = 400 × 103 Hz.
Time period of oscillation; T 1 3 400 10
= ´
Number of pulses counted by the counter (in 20 ms);
3 3
113. (c) In electronic voltmeter MOSFET is used at the first stage.
114. (c) Percentage change in length = 3%;
Change in resistivity = 0.3%; hence, Gauge factor
Internal resistance of voltmeter (Rv) = 200 kW.
Voltmeter Reading = 250 V Ammeter reading = 10 mA
Internal resistance of Ammeter » 0 Value of unknown resistance (say RU) = ? Now, effective resistance, e 3
R 250 25 k
10 10
-= = W
´ To calculate unknown resistance (RU):
e U v U
220 V produces deflection of 2 cm. So, for 3 cm deflection:
220 3 330V 2
´ =
117. (d) 118.(b)
119. (a) As given VO/Vin is independent of frequency then;
C = 1 × 10/1 = 10 mF.
120. (a)
121. (c) Total length of the slide wire = 1000 cm Total resistance of the slide wire = 1000 ohm
\ Resistance per cm = 1 ohm
Resistance of 101.8 cm segment of the wire = 101.8 ohm This corresponds to a voltage of 1.018 V
\ Current = 1.018 101.8 = 1 mA
122. (d) Parallel sum of 600 ohm and 1200 ohm
= 600 12001800´ = 400 ohm Revolution made = 1380
\ Meter constant = 1380
9.2 = 150 rev/kWh 125. (a) Two wattmeter readings are
W1 = VLIL cos (30 – f)
= VLIL cos (30 – 90°) = VLIL cos 60°
W2 = VLIL cos (30 + f)
= VLIL cos 120° = – VLIL cos 60°
126. (c) Energy consumed as indicate by energy meter
= 180090 = 0.05 k Wh
127. (d) Absolute error = exp ected value measured value exp ected value
-=80 79 100 1.25%
80
- ´ =
Relative Accuracy = measured value 100 exp ected value´ or Relative Accuracy = 100 – % absolute error
= 100 % – 1.25 % = 98.75 % 128. (b) Thermocouple instruments are suitable for very high
frequencies upto 50 Mhz.
129. (d) By nyquist theorem minimum sampling frequency should be twice of highest signal frequency. Here sampling frequency is 2 kHz.
Resolution = n1 0.01 100
2 1= ´
-or n = 14
130. (d) Nominal ratio Knom = 1000 5 =200 Secondary burden Re = 1.1W
Since the burden of secondary winding is purely resistive therefore, secondary winding power factor is unity or d = 0. The power factor of exciting current is 0.45
\ cos (90° – a) = 0.45
or a = 90° – cos–1 0.45 = 26.74°
Since there is no turn compensation, the turns ratio is equal to nominal ratio or Kt = Knom = 200. When the primary winding carries rated current of 1000 A, the secondary winding carries a current of 5 A.
Rated secondary winding current, Is = 5A Actual transformation ratio,
\ Ratio error = Nominal ratio Actual ratio Actual ratio 100
132. (c) The frequency is obtained by following equation f =
133. (c) Pattern observed on the screen is an ellipse. So, phase angle
f = sin–1 3 5 æ öç ÷
è ø = 36.9° or 143.1°
We can see from the figure that ellipse is in second and fourth quadrants so the valid value of phase angle is 143.1°. range, cause hardly any variation in Im and thus the scale is highly cramped for high resistance values.
135. (b) In the following configuration
XC
writing node equation at P
P P
V 10 1 1 j
V 0
100 100 500 159
- + æçè + - ö÷ø=
Voltage developed across voltmeter will be in proportional to their resistance.
141. (b) fs = 400 rad/s
For reconstruction in SH circuit for signal we keep frequency fs/2 to have a linear phase response.
\ Sampling rate of signal = 400
144. (a) x l FS
Þ R = 999 ohm (in series with first resistance)
146. L L
147. When p.f. < 0.5 one of the wattmeter reads negative power;
which is read after reversing the connections to the current coil or pressure coil. Also by formula:
1 1 2
150. Voltage drop per unit length = 1.45/50 = 0.029 V/cm Voltage drop across 70 cm. length = 0.029 × 70 = 2.03 V
\ Current through resistor = 2.031W =2.03A 151. No. of pulses = 5.75 × 100 = 575 pulses.
152. Total energy supplied by the source in one minute.
W = 3 × 1017 × 1.6 × 10–19 × 10000 = 4.8 × 102 J = 480 J
\ Average power supplied to the beam = 480/60 = 8 W 153. 1% accuracy means that a maximum possible error of 300 1100´
= 3 may be present in any reading. Since the deflection is 83 V the present limiting error is 3 100 3.62
83´ =
154. In a dual slope integrating type digital voltmeter.
\ Vin = Vref (t2/t1)
where t1 is the first integration line t1 = 10 × 1/50 = 0.2 sec