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1. (c)
2. (d)
3. (a)
4. (b)
5. (c)
6. (b)
7. (d)
8. (c)
9. (c)
10. (b)
11. (d)
12. (c)
13. (b)
14. (a)
15. (a)
16. (a)
17. (c)
18. (a)
19. (b)
20. (b)
21. (a)
22. (b)
23. (b)
24. (d)
25. (d)
26. (c)
27. (b)
28. (c)
29. (b)
30. (a)
31. (c)
32. (d)
33. (c)
34. (d)
35. (d)
36. (d)
37. (a)
38. (a)
39. (d)
40. (b)
41. (b)
42. (d)
43. (c)
44. (b)
45. (b)
46. (a)
47. (b)
48. (c)
49. (a)
50. (c)
51. (c)
52. (b)
53. (a)
54. (c)
55. (a)
56. (d)
57. (b)
58. (d)
59. (c)
60. (b)
61. (b)
62. (c)
63. (c)
64. (b)
65. (c)
66. (c)
67. (b)
68. (d)
69. (d)
70. (a)
71. (a)
72. (d)
73. (d)
74. (c)
75. (b)
76. (c)
77. (a)
78. (b)
79. (c)
80. (a)
81. (c)
82. (d)
83. (c)
84. (c)
85. (b)
86. (d)
87. (b)
88. (b)
89. (c)
90. (c)
91. (c)
92. (d)
93. (c)
94. (b)
95. (a)
96. (b)
97. (d)
98. (a)
99. (c)
100. (a)
ESE-2019 PRELIMS TEST SERIESDate: 30th December, 2018
ANSWERS
101. (a)
102. (a)
103. (b)
104. (a)
105. (b)
106. (a)
107. (c)
108. (a)
109. (c)
110. (c)
111. (d)
112. (c)
113. (c)
114. (b)
115. (d)
116. (b)
117. (c)
118. (c)
119. (d)
120. (a)
121. (a)
122. (a)
123. (d)
124. (b)
125. (d)
126. (c)
127. (d)
128. (d)
129. (a)
130. (a)
131. (b)
132. (c)
133. (a)
134. (a)
135. (d)
136. (d)
137. (c)
138. (c)
139. (a)
140. (b)
141. (a)
142. (d)
143. (d)
144. (a)
145. (b)
146. (b)
147. (b)
148. (a)
149. (d)
150. (a)
IES M
ASTER
(2) Full Syllabus
1. (c)In a multistage amplifier,
(i) Common-emitter configuration is used forrealizing bulk of the gain of the multistageamplifier.
(ii) Common-collector configuration is required forproper impedance matching. This configurationis not used in the intermediate stages ratherthey are used in the last stage of the multistageamplifier.
2. (d)The circuit diagram of Wien bridge oscillator is shownbelow:
–+ V0
R1
R2
R
R
CC
The above circuit uses negative feedback throughR1 and positive feedback through a series RC anda parallel RC network.
3. (a)
7809 +– +9V
7805
4V
+– 5V–+
V =1Vout
7805 781212V
4V
+– 5V
–+ V =9Vout
+––+4V
V =8Vout
+
–
+
–
+
–
4. (b)The given circuit compares the input voltage withthe reference voltage of 4V and input is applied atthe non-inverting terminal. Thus it is a non-invertingOp-Amp comparator.
8
T/2T12
+Vsat
–Vsat
v0
5T12
vi
t
t
4
0
Diodes D1 and D2 are provided in the circuit toprotect the op-amp against damage due to excessiveinput voltage. Because of these diodes, the differentialinput voltage, Vd, is clamped to either +0.7V or –0.7V, hence these diodes are called clamped diodes.
R1 is used to limit the current through the protectiondiodes D1 and D2 while resistance R is used toreduce the offset problem.
Duty cycle of output voltage
=ON
ON OFF
T100%
T T
=
5T T12 12 100%
T
= 33%
5. (c)Differential gain,
Ad,dB = 60 dB
20 log AAd = 60 dB
Ad = 103
Similarly, CMRRdB=80dB i.e. CMRR = 104
Vd = V1–V2 = 12 mV – 8mV = 4mV
VCM = 1 2V V2
= 10 mV
Output voltage,Vo =CM
d dd
V1A V 1CMRR V
=3
41 10 mV10 4m 1
4 mV10
= 4 + 0.001 m
= 4.00100 V
6. (b)
u1/s x2 x1 5 y
–3 –4x3
1/s
–22
1/s
2x
3x
1x
IES M
ASTER
[EC], ESE Prelims Test Series |FLT - 05| 30th December 2018 (3)
1x = 1 24x x
2x = 23x u
3x = 32x u
1
2
3
x Ax Bu
x 4 1 0 0x 0 3 0 X 1 Ux 0 0 2 1
A =
4 1 00 3 00 0 2
7. (d)Non linearity may be introduced by any of thefollowing given.
8. (c)
State transition matrix –1 –1t L (sI – A)
| sI – A | =
s 0 0 2 s –2sI A –
0 s –2 0 2 s
|(sI – A)| =2s –2
s 42 s
(sI – A)–1 = 2
s 2–2 sadj(sI – A)
sI – A s 4
(sI – A)–1 =2 2
2 2
s 2s 4 s 4
–2 ss 4 s 4
therefore
L–1 [(sI – A)–1]=cos2t sin2t– sin2t cos2t
9. (c)
10. (b)Corner frequency w = 1 rad/sec, 20 rad/sec, 40rad/sec
• w1 = 1 rad.sec, change in slope is 0 – (–20) =20 db/dec so one zero exist at w1 = 1 rad/dec.
• Initially slope is –20 db/dec so one pole exist atorigin.
• At w2 = 20 rad/sec, change in slope is (20 – 0)= 20 db/dec, so one zero exist at w2 = 20.
• At w3 = 40 rad/sec, change in slope is (0 – 20)= –20 db/dec, so one pole exist at w3 = 40.
G(s) = 1 2
3
s sk 1 1w w
ss 1w
= sk 1 s 1
20ss 1
40
G(s) =
2k s 1 s 20
s s 40
for calculation of k
line with slope = –20 db/dec
y = M log w + 20 log k
at w = 1, y = –9 and M = – 20 db/dec
–9 = –20 log (1) + 20 log k
k = 0.35
0.7(s 1) (s 20)G(s)s(s 40)
11.(d)
G(s) H(s) =K (1 s)(1 – s)
given K < 0 – |K|
G(s) H(s) =–|K| (1+ s)
(1 – s)
G(s) H(s) =(1+ s) |K|
(s – 1)
Put s j and H(s) = 1
G( j ) H( j ) =(1 j ) | K |
( j – 1)
2
2
1 KGj( )1
By putting different values of , we get
IES M
ASTER
(4) Full Syllabus
w = 0 1 180w = 0.01 1 –178.85w = 0.1 1 –168.57w = 1 1 –90w = 10 1 –11.521w = 100 1 –1.145w = 1000 1 –0.1145w = 1 0
G (j ) G(j )
w = 0 w = U
jVGH plane
12.(c)Given Open-loop transfer function
So characteristic equation is1 + G(s) H(s) = 0
G(s) = 2K (s – 2)(s 1)
given H(s) = 1
2
k (s – 2)1(s 1)
= 0
(s + 1)2 + k(s – 2) = 0
s2 + 2s + 1 + ks – 2k = 0
s2 + s(k + 2) + 1 – 2k = 0
The routh’s tabulation is
1 1–2kk+2 0
1–2k
s2
s1
s0
For stability K + 2 > 0
k > –2 ...(1)
1 – 2k > 0 1k2
...(2)
so from (1) and (2)
For –2 < k < 12
system is stable
13. (b)Derivative control will increase the damping ofthe the system and thereby transientperformance is improved.
Intergral control changes the system order asit adds a pole to the system at origin andthereby reduction in steady state error. Henceit improves the system performance of steadystate.It changes the order of the system from secondorder, to third order as a pole at origin is addedHence statement 1, 3 is true and 2 is false.
14. (a)
15. (a)The effect of stray magnetic field in AC bridgescan be reduced by shielding the bridges with theuse of high permeable material around the bridge.High permeable material provide low reluctancefor stray magnetic field and hence screened themagnetic field to affect the bridge.Wagner earthing device is used in the ac bridgesto eliminate the effect of stray capacitance.
16. (a)
XCXL
R
R/Xe/XL
Z
30 4020Hz r
0
60Hz
17. (c)V = IR R
= 200 0.02 30
= 4 30
L LV 4 30I I
j L 200 0.5 j
=4 30100 j
= 0.04 60
IES M
ASTER
[EC], ESE Prelims Test Series |FLT - 05| 30th December 2018 (5)
C C 6
V 4 30I I1 10j C 200 50j
= 0.04 120 I = IR + IC + IL
= 0.02 30 0.04 60 0.04 120
I = 0.0230°
18. (a)
60 V 2
i (t) 2
i (t)L
At t = , inductor behaves as short circuitReq = 1
i(t) =601
= 60 AA
Hence IL(t) =60 2 30 A4
For finding time constant, we short circuit the voltagesource. Hence circuit becomes like
2 H2
2
=L 2 2secR 1
19. (b)
BW =fr 150 3KHzQ 50
fUpper = BW 3fr 1502 2
= 151.5 KHz
fLower = BW 3fr 1502 2
= 148.5 KHz
20. (b)
Given bandwidth–length product BL = 400 MHzkm
L = 8 km
B = 400 MHz 50MHz
8
Total pulse broadening 1
2B
= 6
12 50 10 = 10 ns
21. (a)
System of propagation in waveguide is inaccordance of field theory.
22. (b)
Total noise T = Trec+Tant = 70+30 = 100 K
Figure of merit = Gain
T(Figure of merit) dB
= (Gain)dB – 10 log10 (100)
= 40 – 20 = 20 dB
23. (b)
For the positive half-cycle of the input, diode D1is ON and D2 is OFF. Capacitor C1 is charged toVm.
v = Vi m sin( t)
–
+
V0
~ C1
C2
D1
D2
Vm
Vm
+ –
+ –
For negative half-cycle of the input, diode D1 isOFF and D2 is ON. Capacitor C2 is charged toVm with polarity as shown in figure above.
Output voltage, V0 = c1 c2V V
= m mV V
= 2 mV
24. (d)Ic = collector current
IES M
ASTER
(6) Full Syllabus
Ie = emitter current
Ib = base current
Apply KVL in the emitter-base loop, we get
1 e e EB b bV I R V I R 0
1 EB c b e b bV V I I R I R 0
c e b e b 1 EBI R I R R V V
Differentiating above equation w.r.t. Ib, we get
ce e b
b
IR R R 0
I
b e
c e b
I RI R R
Stability factor, S c
bco
c
I 1II 1I
S =e
e b
1R1
R R
25. (d)
26. (c)The point where asymptotes of a root locus meetis called centroid hence statement-I is false
Root locus always starts from open loop pole (k=0)and end at open loop zero (k = ). Hencestatement-II is true.
27. (b)
28. (c)
29. (b)
Hall effect occurs when a transverse magneticfield is applied to a conductor carrying current.Due to the current, a longitudinal electric fieldnormally present in the direction of flow of current.
30. (a)• Superconductors are used for generating very
strong magnetic field.If a superconductor caries a current I, thendeveloped magnetic field
H =I
2 rwhere r is the radius of superconductor wire.• Magnetic bubble memories are built using
ferrites.
31. (c)
ABCABC
ABC
ABC
ABC
ABC
ABC
A B
C
Let shaded area = Y
Y = ABC ABC ABC ABC= m 3,5,6,7
1
11
ABC
A
BC BC BC BC
A 1
0 1 3 2
4 5 7 6
Y = AB BC CA
32. (d)
A
B
C
Y
AB
BC
x
X = AB BC
= AB.BC AB.BC ABC
Then, Y = ABC A C
= A C1 BC
= A C
33. (c)
• When switch is open, corresponding input willbe high.
• When switch is closed, corresponding inputwill be low.
For option (c)
1 2S S A
1 1 0
IES M
ASTER
[EC], ESE Prelims Test Series |FLT - 05| 30th December 2018 (7)
Then,S1S2 = 1.1 = 1
S2.A = 1.0 = 0
Y = 1 2 2 1 0 1 0S S S A
34. (d)
35. (d)
For multiplexer,
Y = 0 1B.I BI
= B B.A1
Hence, X = 0B AB
= .0B AB
= .1B.AB
= B. A B
= BA36. (d)
The 1st section of the shown Wide Bandpassfilter is a High-pass filter and its low cut-offfrequency can be found as fL,
fL =1 1
1 0.032 kHz2 R C
1 1R 5 k , C 1 F
The 2nd section is a Low-pass filter and its highcut-off frequency can be found as fH,
fH =2 2
1 3.979 kHz2 R C
2 2R 10 k , C 0.004 F
Bandwidth of the wide bandpass filter
= fH – fL
= (3.979 – 0.032)kHz
= 3.947 kHz 3.95 kHz
37. (a)
The above circuit represents a Pulse generatorand it is also called as Monostable or One-shotmultivibrator. (Statements ‘1’ and ‘2’ are true)
The pulse width of the output wave is given by
tpw =D
f out
V1R C ln V
1
= f1R C ln
1
, for D outV V
(Statement ‘3’ is false)
38. (a)
Basic causes for non-sinusoidal nature of no loadcurrent of transformer are:
1.Hysteresis of core material
2.Saturation / non-linearity of core material
39. (d)
In a moving iron instrument 2I
1
2
=2
1
2
II
= 22mA
1mA
= 4
2 = 14
=200V
4 = 50V
40. (b)Electric field due to a point charge
E = 20
Q4 r
Electric potential
VAB =B
B AA
V V E.dr [Here A = (0, 6, –8) and B = (–3, 2, 6)
=B
20A
Q dr4 r
=B
A0
Q 14 r
=B A0
1 1Qr r4
A x y z
B x y z
ˆ ˆ ˆr (0 0)a (6 0)a ( 8 0)aˆ ˆ ˆr ( 3 0)a (2 0)a (6 0)a
= 9 95 10 9 10
2 2 2
1
0 0 6 0 8 0
IES M
ASTER
(8) Full Syllabus
2 2 2
1
0 3 2 0 6 0
=
1 14536 64 9 4 36
=
1 14510 7
VAB =345 1.93 V
70
B 3,2,6V = A1.93 V
= 1.93 2 3.93V
41. (b)Given : m = 10 gm; l = 60 cm; g = 10 m/s2
F = mg = Bil
i =m gB l
=310 10 10
0.4 0.6
= 0.41A
42. (d)In free space,
0 00, 1, 1
P = 0 e 0 rE E1
= 0 [ r 1 , Free space]
J = E 0
43. (c)
v =81 3 10
9
E = 8y10 cos 6 10 t x a
E = yA cos t x a
= 86 10 rad m
=3
86 10 9
v 3 10
= 18 rad / m
44. (b)
Directive Gain dG ( , ) of an antenna is ameasure of the concentration of radiated powerin a particular direction ( , ) . It is the abilityof the antenna to direct power in a givendirection.
45. (b)
L oin o
o L
Z jZ tanZ ZZ jZ tan
ll
The impedance of the shorted line at the junction
ZSC =20
L
ZZ
As , tan4
2tan4
tan2
l = l
But as LZ 0
So, SCZ
The impedance of the open-circuited line at thejunction.
ZOC =20
L
ZZ lAs
4
But as LZ
So, OCZ 0
Now, the effective impedance at the junction.Zef = ZSC || ZOC = 0 || Zeff = 0So, current drawn from the voltage source
I =0 eff 0
V VZ Z Z
46. (a)h – parameter equations :V1 = h11I1 + h12V2I2 = h21I1 + h22V2h-parameter model :
1I 2I
2V1 21I h12 2h V
11h
1V
+
–
+
–
22
1h
Compare the given circuit to h-paramer model,we get
11h = b e 12r r , h 1
h21 = cb 22e d
1, hr r
47. (b)Output voltage
V0 = m Sg V V600
IES M
ASTER
[EC], ESE Prelims Test Series |FLT - 05| 30th December 2018 (9)
Now, apply voltage division rule in input loop,
V = in1000 V
1000 500
V = s2 V3
From (1)
V0 = m2g 6003
V0 = m s400g V
or, 0
S
VV = –400 gm
or, 0
S
VV = –400 × 0.1
= – 40
48. (c)For stability of negative feedback closed loop controlsystem output linearly varies with input, whenany non-linearity should not be introduced by thecomponent of control system.
49. (a)
× × G1G2
H
R(s)++–
+N(s)
C(s)
In a feedback control system, if subjected to noiseN(s), then R(s) = 0
× G1G2
–H
++
N(s)
C(s)
Transfer function
C sN s =
1
1 2
G1 G G H
50. (c)
1010 C1R1 × 10R2 –+
C2
1
1
C 10 10 100R
2
2
C 10 10R 1 10 11
If the forward path gain is reduced by 10% ineach of the system then
90 C1R1 × 9R2–
+C2
1
1
C ´ 90R
2
2
C 9 9´R 1 9 10
Variation in C1 Variation in C2
1 1
1 1
1
1
C CR RVariation 100C
R
90 100 100 10%100
2 2
2 2
2
2
C CR RVariation 100C
R
9 1010 11 100 1%10
11
51. (c)To measure the higher voltage, resistance to beconnected in series with the moving ironinstrument.
750V
IfsI =50mAfs R
R =10m
750V = fs m fsI R I R
750 = 3 350 10 10 50 10 R
R = 3
750 0.5 15k50 10
52. (b)
Emf equation of a 1 transformer is
E1 = max 12 f N
so,Emfturn = 1
1
EN
= max2 f
i.e, max =1
1
EN2 f
=3.142 50
=2100 10
2 50
IES M
ASTER
(10) Full Syllabus
= 22 10
If, Ac = cross sectional area of core
BmaxAc = 22 10
Bmax = 22 10
0.01= 2
= 1.414 T
53. (a)
• Increase in height to width ratio, meansincrease in height or decrease in width ofthe window, which brings the primary andsecondary coils close to each other. Hence, itwill reduce the leakage flux and hence leakagereactance.
• Reduction in leakage reactance of tranformerwill decrease the voltage regulation asregulation = Rpucos ± puX sin
54. (c)For building up of voltage at the terminal of dcshunt motor the field circuit resistance should beless than the critical resistance.
55. (a)
56. (d)
57. (b)
58. (d)
59. (c)
60. (b)
61. (b)62. (c)
Frequency of Rotor emf = Slip × SupplyFrequency
s
120f 120 50N 1500P 4
s o
s
N N 1500 1470S 0.02N 1500
Frequency of Rotor emf = 0.02 × 50 1Hz.
63. (c) I(s) = 2s
(s 2) = 2s 2 2(s 2)
= 2 2(s 2) 2(s 2) (s 2)
= 21 2
(s 2) (s 2)
Taking Inverse Laplace transform, we get,
i(t) = e–2t – 2t.e–2t
= 2te (1 2t)
64. (b)
For the above circuit
i = Rt/Lm mI sin t I sin e ;
where, 1 LtanR
When, 0 or, ie. when voltage applied ispasses through its maximum, there will be notransient.
65. (c)Off the total 70V, 10V is across R2, so 60V isacross R1. 30V is across 60 resistor so, current
= 3060 = 0.5A
R = VI = 60
0.5 = 120
66. (c)A capacitor does not resist any abrupt change inthe current flow. It resists abrupt change involtage.
67. (b)
68. (d)
69.(d) Quality factor,
Q =resonant frequency
Bandwidth
=0ff
=
6
320 10
100 10
= 200
IES M
ASTER
[EC], ESE Prelims Test Series |FLT - 05| 30th December 2018 (11)
70. (a)When in series, we have to add the Z-parametersto combine the two, two network.
71. (a)L-C network has only energy storage element.Power dissipation elements ‘R’ is not there, so thepoles and zeros lie on the imaginary axis. Networkfunction should be determined by zero initialcondition.
72. (d)
The correct sequence of steps involved for Analog-to-Digital conversion isFiltering Sampling Quantization Encoding
Filtering: An anti-aliasing filter is used to lowpass filter the input signal to half the samplingrate, so as to avoid aliasing in the sampledsignal
Sampling: Sampling converts the continuous-time continuous-amplitude analog signal intodiscrete-time continuous amplitude signal.
Quantization: It is the process of discretizingthe sampled signal (that is continuous inamplitude) into discrete-amplitude signal.
Encoding: It is the process of assigning digitalvalue to the quantized signal.
73. (d)
For n-bit Analog to Digital conversion,
1. Successive Approximation Register (SAR) ADCrequires ‘n’ clock cycles.
2. Dual slope integrating ADC requires
approximately n 12 clock cycles.
3. Counter method ADC requires (2n – 1) clockcycles.
4. Simultaneous ADC requires only 1 clock cycle.
Note: Counter type ADC is also known as Ramp
type ADC
Dual slope integrating type ADC is the slowestamong all ADCs. It is the most accurate ADCand thus, is used in Digital voltmeters.
The number of clock cycles (given above) forvarious ADCs is the maximum conversion timerequired.
74. (c)
Slip is s
s
N NN
.
So for the slip to be negative sNN and in thesame direction. In the opposite direction slip willbe greater than one.
75. (b)
Slip of the induction motor is given as ratio ofrotor copper loss to rotor input power.
76. (c)Small-signal equivalent of the above circuit is
gm2VSG2
v0
rd2
rd1vi
S2
+
– gm1VGS1
1
S1
G1, G2
Here VGS1 = v i and VSG2 = –v i
Apply KCL at node ‘1’
gm1VGS1 + 0m2 SG2
d1 d2
v v g Vr r
m1 i 0 m2 id1 d2
1 1g v v g vr r
0
i
vv
=
d1 d2m1 m2
d1 d2
r rg g
r r
= m1 m2 d1 d2g g r r 77. (a)
V(s) = Z s .I s
=
2 s 1.s s 2 s 3
si t 2u t ; I 2 / s Initial Value :
t 0LimV t =
sLims.V s
=
s
2 s 1Lims. 0s s 2 s 3
IES M
ASTER
(12) Full Syllabus
Steady State Value :
tLimV t =
s 0Lims.V s
=
s 0
2 s 1 2 1Lims.s s 2 s 3 6 3
78. (b)
A potentiometer is used as a error detection device.Its input is the position error and potentiometerconvert it to corrersponding voltage.
79. (c)
t s 0Limy LimsGt s
and the input is given at instant t = 2
= 2ss 0LimsG es
= 2s
2s 0
1Lims es s 2s 2
=12 = 0.5
80. (a)XRA A instruction execution resets contents ofaccumulator and hence sets zero flag.
81. (c)
x1
e f g
a x2 b x3 c x4 d x5
L1 L2 L3
Applying Mason’s gain formula
5
1
xx =
n
K KK 1
P
Forwards path :
P1 = abcd, 1 1
Loops : L1 = be, L2 = cf, L3 = dg
Non-touching Loop :
L1L3 = bedg
5
1
xx =
1 1
1 31 2 3
P1 L LL L L
5
1
xx =
abcd1 bedgbc cf dg
82. (d)
The wave is propagating in +z direction. Theexpression of electric field for TM mode is givenby :
i zZ 0
m x n yE E sin sin ea b
comparing with
i zZE 30sin 50 x sin 60 y e
Now, m 50a
2m 50a 50 8 10 4
n 60b
2n 60 5 10 3
Mode is 43TM
83. (c)
A parasitic element shorter than the driven one orradiator tends to increase radiation in its owndirection and behaves as convergent convex lensewhich is called a Director.
Radiator
Director
Reflector
A parasitic element longer than the driven oneand which reduces signal strength in its owndirection and increases in opposite direction iscalled a reflector.
84. (c)
The main purpose of the buried layer is to minimizethe series resistance of the collector.
85.(b)R1 = 500 ± 1% R2 = 1000 ± 1%
R = 1 2R R (Parallel combination equivalentresistance)
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[EC], ESE Prelims Test Series |FLT - 05| 30th December 2018 (13)
Nominal value of R = 500 1000 = 1000
3
Tolerance (or limiting error) for R1 and R2respectively
1
1
R1% 0.01
R
1 1R R 0.01
2
2
R1% 0.01
R
2 2R R 0.01
Hence R 1 = ± 5
R 2 = ±10
Now1R
= 1 2
1 1R R
2
1
R 1R R =
2
2 211 2
R1 1RR R
2
RR
=
1 22 2
2
R RR R
Hence, the limiting error in the resultant
R =
2 2
1 21 2
R RR RR R
= 4 15 109 9
= 2.22 + 1.11 R = 3.33
Tolerance =RR
= 3333 9.99 101000
3
= 0.99% ±1%86. (d)
i. The sacle is uniformly divided
ii. The power consumption is very low (as 25 W
to 200 W )
iii. The torque to weight ratio is high which giveshigh accuracy
iv. Since the operating forces are large on accountof large flux densities which may be as high0.5 Wb/m2 the errors due to stray magneticfields are small.
87. (b)Common base amplifier has excellent high-frequency. Thus, common base amplifiers haveinvariably high cut-off frequency.
88. (b)
Conduction current density, CJ E
Displacement current density, dDJt
D E
dJ
=E j Et
If CJ
= dJ
Then, =
=
f =02 3
=
2
8
1012 3 10
360
f = 60 MHz
89. (c)
The radiations from a heated body at hightemperatures fall within the visible region of theelectromagnetic specturm. Within the visibleregion a given wavelength has a fixed colour andthe energy of radiation is interpreted as intensityor brightness. Therefore, if we measure brighnessof the light of a given colour emitted by a hotsource, we can have an indication of temperature.
90. (c)
The material required for resistance thermometershould possesses :
(i) high temperature resistance coefficient sothat a small change in temperature producesa large change in resistance.
(ii) high resistivity so that minimum volume ofmaterial is used for the construction ofthermometer.
91. (c)
92. (d)
93. (c)
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(14) Full Syllabus
Td = n(IPd)FF
Total propagation delay, Td = 4 × 20 ns
= 80 ns
So, counting time T 80ns
So frequency of counting, 1fT
i.e. counting speed,
f 91 1T 80 10
61000 1080
12.5 MHzSo, maximum counting speed in the given option= 10 MHz.
94. (b)
67 H 0110 0111=81H 1000 0001=
1110 1000
Flag = S Z AC P CY
(because result is positive)
So sign, S = 0
Zero, Z = 0
AC, AC = 0
CY, CY = 0
95. (a)
To reset all the flip-flops
Q1 = 1, Q2 = 0, Q3 = 1
(101)2 = (5)10
The given circuit is a Mod-5 counter becauseafter the counter state is 101, the counter resetsand starts counting again from 000.
96. (b)
Microwave link repeaters are kept typically 50km apart due to earth’s curvature.
97. (d)
Split-phase code is also known as Manchestercode and it suppresses dc component irrespectiveof same probability of occurance of 1’s and 0’s.
98. (a)
Doppler Radar functions on doppler effect andhence displays only moving targets.
99. (c)Power transformers use silicon steel as amagnetic material and alnico is a permanentmagnet.
100. (a)
In paramagnetic materials interaction betweenneighbouring dipoles is negligible.
101. (a)
102. (a)
103. (b)
8 xˆE 4sin a V m10 t 0.8x Standard equation is :
o xˆE E sin( t x)a
Here, 800.8, 10 , nonmagnetic
Hence,
= 0 0 r
= rc
0 0
1c speed of light
So, r = c
=8
80.8 3 10 2.4
10
r = 2 5.762.4
104. (a)
105. (b)
106. (a)
107. (c)Depletion region is created due to diffusion ofmajority charge carriers across junction so itdoes not contain any free charge (electron orhole). It only contains ions, negative ions on p-side and positive ions on n-side.
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[EC], ESE Prelims Test Series |FLT - 05| 30th December 2018 (15)
108. (a)
1mA
5V
CRCV 2V50k
5V
Given,IE = 1 mA
IC = E EI I1
=75 1mA
75 1
IC = 0.987mA
109. (c)From the Fermi-dirac statistics, the probabilityof electron occupation of an energy level equalto the fermi level is given by
f(E) =f
1E E1 exp
KT
For E = Ef,
f(E) = 1
1 exp 0 = 0.5
110. (c)LED will glow only when output from OR willbe ‘0’ and it is possible only when both inputof OR gate will be ‘0’ i.e. output of NOT gateand EXNOR gate should be zero. It is possibleonly when S1 is closed and S2 is open.
0
S = open i.e. 12
S = closed i.e. 01
00
1
111. (d)Number of chips required
=
Required sizeAvailable size
=32K 81024 8
=32 1024 8
1024 8
= 32
112. (c)
CLK Q– 0 Q2 Q1 Q0
0 – 0 1 0
1 1 1 0 1
2 0 0 1 0
So for 2 clock pulses, the values of Q2Q1Q0
resets, so frequency = 18KHz2 = 9KHz
113. (c)For 4-flip flops the count is 24 = 16, So the no.of unused or missed states are
16 – 10 = 6 counts.
114. (b)
Fading refers to the fluctuations in signal strengthwhen received at the receiver.
115. (d)
In f requency-hopping spread spectrum,processing gain
GP = RF bandwidth N. f N
Message bandwidth f
where f is frequency separation betweenadjacent discrete frequencies.
116. (b)We know
X(z) =n
nx[n] z
=n 3 3
n
15
u[n-3] z–n–3 + 3
3 (n 3)
3 (n 3)
n
1 1z u[n 3] z5 5
n–3 = n'
n'–3
–n'
n'=–
z 1× u[n'] z125 5
Since u[n'] = 0 n 0
X(z) =
3
1
z 1 1, | z |1125 51 z5
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(16) Full Syllabus
So X(z) =n '3
n '
0
z 1 z125 5
= 0 1 23z 1 1 1 ...
125 5z 5z 5z
= 23z 1 11 ...
125 5z 5z
X(z) = 3z 1 1provided 1
1125 5z15z
117. (c)
0s 6 j8 s 6 j8
2 28 0s 6
2 2s 2 6 s 10 0
Comparing with 2 2n ns 2 s
n = 10
= 2 62 10
= 0.6
118. (c)
2a aT I I (series motor aI )
2
1
TT =
2
1
2a2a
I
I
T2 =2
1 220T10
= 20 Nm × 4
= 80 Nm
119. (d)
120. (a)
The reactance voltage is developed due to thepoor commutation which causes the delay inquick reversal of current because of inductiveeffect occuring due to reversal.
121. (a)N = 630 rpm (given)
Ns = 120 50 750r.p.m.8
s =s
s
N NN
s = 750 630 0.16750
s = 2
2
RX
X2 = 2R 0.07 0.44s 0.16
122. (a)Skewing is done to eliminate the spaceharmonics in the synchronous machine. Thiswould give a voltage profile near to sinusoidal.
123. (d)Here, fm = 30 kHz
No. of levels L = 6Now, sampling frequency fs = 2fm = 60kHzAlso, L = 2nor 2n = 6 n must be 3 bit
Bit rate Rb = n.fs = 3×60 = 180 kbps124. (b)
Variable slope integrator is able to producevariable step size according to variation insignal.
125. (d)For 8–PSK, no. of bits per symbol = 3Bit Transmission rate = 3600 bps
Symbol rate = BitRate
No.of bitspersymbol
= 3600 1200 symbol sec.3
126. (c)(i) Minimum error probability of a matched
filter is given by e0
1 EP erfc2 N . It is
obvious from equation that the errorprobability depends only upon the signalenergy E. It does not depend on the shapeof the signal.
Pe = 0
1 Eerfc2 N
(ii) The maximum signal to noise ratio forthe matched filter is found to be,
0max
SN =
0
2EN
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[EC], ESE Prelims Test Series |FLT - 05| 30th December 2018 (17)
(iii) We know that impulse response of thematched filter is given by,
h(t) = 0
2k x T tN
If 2K/N0 = 1 (assumed) then h(t) = x T t . The outpt of the matched filter
is obtained by convolution of x(t) andh(t) is y(t) = x t h t .
(iv) The equation
T
0 0
2ki.e. y T f t x t dtN gives the
output of matched filter. It may beobserved that this equation and equation
T
0
y T f t x t dt (which gives
output of correlator) are identical. In
equation
T
0 0
2ki.e. y T f t x t dtN
the constant 02k N is present whichcan be normalized to 1. The similarity
between equation
T
0
y T f t x t dt
and equation
T
0 0
2ki.e. y T f t x t dtN shows
that the matched filter and correlatorprovides same output.
127. (d)H(x) = 2p x log p x
Since it is a discrete source, let p(x) = p H(x) = 2plog p
nH x = n np x log p x
n n2 2p log p pnlog (P)
Since it is memoryless source. pn = P128. (d)
B.W. of channel = 1MHz = 106HzChannel capacity C=10 Mbps=10×106bps
C = 2SB.log 1N
where N = N0B
C = 20
SBlog 1N B
as B
C 0
S1.44N ...(i)
Given for B = 106, C = 107 bps
107 = 62 6
0
S10 log 1N 10
10 = 2 60
Slog 1N 10
210 = 60
S1N 10
N0 = 6S
1023 10...(ii)
from (i) & (ii)as B
C = 6S1.44 1023 10S
C = 61.44 10 1023 C 1.5 Gbps
129. (a)Fidelity is the ability of the receiver toreproduce all the modulating frequenciesequally. It depends on the bandwidth andlinearity of various components in a receiver.
130. (a)
131. (b)In the dual trace C.R.O., two inputs can bedisplayed simultaneously as there are two seperatevertical input channels and they use seperateattenuator and preamplifier.Aquadag is an aquous solution of graphite used tocollect the secondary electrons produced duringbombarding of electrons on screen.
132. (c)Bifilar winding is most commonly used method ofwinding arrangement to make noninductive coil.This method is used to reduce the inductive effectof coil. In this method, two wires wound side byside carrying current in opposite direction producetwo equal and opposite magnetic fields and sincetwo wires are very close to each other, the netmagnetic field is almost zero. Therefore, the netinductance of the coil is almost negligible.
133. (a)“Guard circuit” is used in the high resistancemeasurement to eliminate the error due to leakagecurrent over high resistance. “Guard Circuit” isconnected in parallel with the measuring ammeter.So, the leakage current passes through the GuardCircuit and hence does not affect the reading ofammeter.
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(18) Full Syllabus
134. (a)
Intramodal dispersion results due to propagationdelay differences between the different spectralcomponents of the transmitted signal.
135. (d)Forbidden energy gap of Si and Ge are given as1.12 eV and 0.72 eV and as we know
2in = gE /KT3
0A T e
ni gE /KTehigher the band gap lower the intrinsicconcentration, which results into low conductivity
= i nn q niTherefore, Si has higher forbidden energy gapbut low conductivity.
136. (d)In case the moving coil instruemnt is used as avoltmeter a large series resistance of negligibletemperature coefficient (made of material likemanganin) is used. This eliminates the errordue to temperature. This is because the coppercoil forms a very small fraction of the totalresistance of the instrument circuit and thusany change in its resistance has negligible effecton the total resistance.
137. (c)The phasor diagram of a wattmeter for a laggingpower factor is given below
V
IP
I
Here,V Voltage applied to pressure coil circuitIP Current in pressure coil circuitI Current in current coil circuit Angle by which IP lags V
Angle by which I lags V(load angle)
As the know,True power
Actual wattmeter reading
=cos
cos cos( )
× Actual wattmeter reading[For lagging loads]
138. (c)
Clipper circuit is used to remove the certainportion of a waveform. A clipper can be made ofresistor and diode e.g. half wave diode can beused to remove one half (positive or negative) ofsinusoidal waveform.
139. (a)
140. (b)Though passive components does not provideelectrical energy for continuous period of time theyare still important part of electrical circuits asheat dissipation can be provided through thereelement or as they can be used in biasing circuitsof a transistor during amplification.
141. (a)
For a good Tunnel diode, the essential requirementsare large (IP/IV) ratio and large voltage swing.
P V
Ge GaAs SiI I 8 15 3.5
Since (IP/IV) ratio of Silicon is less, Tunnel diode isfabricated with Germanium or Gallium Arsenide materialand not with Silicon.
142. (d)Sequential circuits :• It gives outputs depending on the present input
and past input also.• It contains atleast one feedback path.• It contains some memory.• It exhibit cyclic nature i.e. its output sequence
repeats after some clock cycle.
143. (d)With increase in temperature mobility of chargecarriers decreases as vibration of atom’s inmolecule increases but at the same time carrierconcentration increases.
2in = gE /KT3
0A T e
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[EC], ESE Prelims Test Series |FLT - 05| 30th December 2018 (19)
in increases with temperature increase.Now conductivity of intrinsic semiconductor .
= i nn qAs with temperature intrinsic concentration niincreases and mobility decreases, but increasein carrier concentration is much more thandecrease in carrier mobility therefore conductivityeffectively increases. Hence resistivity decreases.
144. (a)
An LED is a light emitting diode that gives outspontaneous emissions, when it is energized.
The recombination of holes and electrons insemiconductor materials gives off energy (equalto the Band gap energy) in the form of light andheat.
145.(b) Shunt enhancement allows the ammeter tomeasure currents higher than the meter current.
I Im
Rsh
I sh
Rm
A
I is measured, which is larger than the full scalemeter currents Im.The voltage drop across the meter for full scaledeflection always remains ImRm.
146. (b)Both statements are individually correct butreason is not the correct explanation ofassertion.
147. (b)Whenever a signal is passed through acommunication channel, distortion will beintroduced. To compensate for the lineardistortion, we can use a network calledequalizer connected in cascade with thechannel or system as shown.
ChannelH (f)C
EqualizerH (f)eq
Input
Delayed Version of Channel
Input
The equalizer is designed in such a waythat inside the frequency band of interestthe overall amplitude and phase responsesof the cascade system shown in figure areapproximately equal to amplitude and phaseresponses for the distortionless transmission.The transfer function of the equalizer is givenby
Heq(F) = 0j2 ft
c
keH f
Equation shows the ideal value of theequalizer transfer function. Practically, theequalizer transfer function should be as closeas possible to the expression given inequation.
148. (a)Linear polarization is most widely used forradio communication applications. Verticalpolarization is often used for mobile radiocommunication.
149. (d)• FM is less susceptible to noise but is not
noise free.• In FM, the message is contained in the
carrier as frequency variation.So, Assertion is false but reason is true.
150. (a)As temperature increases, the band gapenergy decreases because the crystal latticeexpands and interatomic bonds are weekened.Weaker bonds means less energy is neededto break a bond and get an electron in theconduction band. Because of decreasedenergy gap and increased electron energyconductivity of insulator increases withincrease in temperature.
