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ESE-2018 PRELIMS TEST SERIESDate: 26th November, 2017
ANSWERS
1. (c)
2. (d)
3. (b)
4. (a)
5. (a)
6. (c)
7. (c)
8. (b)
9. (b)
10. (a)
11. (b)
12. (d)
13. (a)
14. (c)
15. (d)
16. (b)
17. (a)
18. (c)
19. (b)
20. (a)
21. (a)
22. (d)
23. (b)
24. (a)
25. (c)
26. (c)
27. (d)
28. (b)
29. (c)
30. (d)
31. (b)
32. (c)
33. (c)
34. (b)
35. (c)
36. (b)
37. (a)
38. (b)
39. (d)
40. (c)
41. (a)
42. (d)
43. (a)
44. (b)
45. (b)
46. (b)
47. (b)
48. (a)
49. (a)
50. (a)
51. (b)
52. (d)
53. (c)
54. (c)
55. (c)
56. (d)
57. (d)
58. (d)
59. (b)
60. (b)
61. (d)
62. (c)
63. (a)
64. (c)
65. (b)
66. (c)
67. (d)
68. (a)
69. (b)
70. (d)
71. (c)
72. (d)
73. (d)
74. (b)
75. (b)
76. (d)
77. (b)
78. (d)
79. (b)
80. (d)
81. (a)
82. (d)
83. (a)
84. (c)
85. (d)
86. (b)
87. (b)
88. (a)
89. (d)
90. (d)
91. (a)
92. (d)
93. (d)
94. (c)
95. (c)
96. (a)
97. (a)
98. (b)
99. (d)
100. (d)
101. (c)
102. (b)
103. (c)
104. (b)
105. (b)
106. (a)
107. (b)
108. (b)
109. (a)
110. (c)
111. (b)
112. (b)
113. (a)
114. (b)
115. (b)
116. (d)
117. (b)
118. (d)
119. (a)
120. (c)
121. (c)
122. (c)
123. (c)
124. (b)
125. (a)
126. (a)
127. (d)
128. (a)
129. (c)
130. (d)
131. (d)
132. (c)
133. (c)
134. (d)
135. (b)
136. (a)
137. (d)
138. (c)
139. (b)
140. (d)
141. (a)
142. (d)
143. (d)
144. (a)
145. (c)
146. (d)
147. (c)
148. (b)
149. (a)
150. (c)
(Test - 10)-26 November 2017 (2)
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1. cWhen thyristor is ‘ON’ at t = 0Vs = VT + VL + VC ...(1)
Vs = di 10 L i dtdt C
Taking Laplace transform both side
sVs = s
1L I s I sCs
I(s) =
sV1s Ls
Cs =
s s
2V C
s LCs 1
=
s
2
V1L s
LCTaking inverse Laplace transform both sides
I(t) = s0
V LC sin tL where 0
1LC
I(t) = s 0CV sin t iL ...(2)
VL =diLdt [by putting value from
equation (1)]
VL =
s 0Cd V sin tLLdt
VL = 0 s 0CL V cos tL
= s 0V cos t ...(3)and VC = Vs – VL
VC = s 0V 1 cos t ...(4) Thyristor conduct for W0t =
t =0
= LC
so VC = Vs [1 – (–1)] = 2Vs [using equation(4)]since when thyristor stop conducting then nocurrent will flow through inductor then
LdiV Ldt
d 0L 0Vdt
So, using equation (1)Vs = VT + VL + VC
Vs = VT + 0 + 2Vs c s V 2V
VT = –Vs2. d
Commutation is possible only when
0
p
II 1
Ip I0 ...(1) Ip = peak current through inductor andcapacitor
Ip = sCVL
Putting value of Ip in equation (i), we get
sCVL I0
650300 10L
100
63 50 10
L
1
L 63 50 10
L 9 × 50 × 10–6
L 450 H3. b
Transistor biasing.
E B C B Type of operationForward (F) Reverse (R) Active region
F F SaturationR R Cut offR F Inverted
4. aLet us assume initial voltage across capacitoris –V0 volt.where V0 = –5 volt
i(0) = s 0V VR
ic(t) = t RCs 0V V eR
Voltage across capacitor
Vc(t) = t0V V V e
Vc(t) = t RCs 0 sV V e V
= t RCs 0 s 0 0V V e V V V
Vc(t) = t RC0 s 0V V V 1 e
At infinity or steady state conditionVc(t) = –V0 + Vs + V0 = Vs = 10 volt
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5. (a)Thermal equivalent circuit of a thyristor isgiven as
Pav
Tj Tc Ts TA
jc cs sA
given Tj = 125º CTs = 70º C
Pav =j s
jc cs
T T
Putting all values, we get
Pav =125 70
0.16 0.18 = 229.17 W
6. (c)Reverse recovery charge is given as
q = 1 1300 4 s 300 2 s C2 2
= (600 × 10–6 + 300 × 10–6)C= 900 × 10–6 C
q = 900 C7. c
Load 880
+
–
DL
220 CH
Since the given chopper is a step up chopper
V0 = s1 V
1
880 = 1 220
1
4 =1
1
1 = 14
34
The output voltage pulse width is = TOFF
ON
ON OFF
TT T = 3
4
ONT = 200 sec
OFF in sec
200200 T
= 3
4
800 = OFF in sec600 3 T
2003 = OFF secT
(TOFF) = 66.66 sec
8. (b)9. (b)10. (a)11. bWhen RLC is under damped then XC > XL and
when diode conducts then power flow will befrom load to source side and when thyristorconducts then power flow will be from sourceto load. By considering only fundamentalcomponent on output side.
V0
i0
P0
T T1 2 D D1 2 D D3 4T T3 4
12. dSince the given chopper is step up chopper
V0 = i1V
1
1
1 = 0
i
VV =
4.50150 = 3
1 =13
=23
= ONTT =
23
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Conduction time of thyristor = 125 sec
TON = 125 sec
T = ON3 T2
= 63 125 102
T = 187 sec
f = 1T = 6
1187 10
= 5333.33 Hz =5.33 kHz
13. a
Vs
2t
Vd
t 2
200V
–200VT T3 4 T T1 2 T T3 4
when T1 T2 conduct Vd = Vswhen T3 T4 conduct Vd = –Vs
(Vd)avg = 0
1 200 d t 200 d t
= 1 200 200
= 1 200 400
is in radian = 45º or 4 radian
=1 200200
2
(Vd)avg = 100 V14. c
1TI Peak value of current through mainthyristor (T1) = IP + I0 ...(1)Where IP is peak value of current throughcapacitor
so Ip = sCVL
= 3020020
= 244.9489 A 245 A
and I0 is the constant load current.GivenI0 = 100APutting all values in equation (1), we get
1TI = 245+100 = 345 A• Peak value of current through auxiliarythyristor TA is ATI = I0 = 100 A
15. d
Peak value of current through T1 = s s
1 2
V 2VR R
=250 2 25010 90
= 30.55 A
Peak value of current through T2 = s s
1 2
2V VR R
=2 250 250
10 90
= 52.77 A
16. (b)The thermal equivalent of the circuit is
TJTC TS TA
Pav
Given, TC = 100º CTJ = 125º C
CA = 0.5º C/W
SA = 0.4º C/WTA = 40º C
PAV = C A
C S
T T
Pav = C A
CS
T T
Pav =100 40
0.5
(by putting value of
TC, TA and CS )
Pav =600.5
Pav = 120W...(1)
Pav = S A
CA
T T
(Test - 10)-26 November 2017 (5)
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120 = ST 40º0.4
[from equation (1)]
TS = 120 × 0.4 + 40º Ts = 48º + 40º Ts = 88º C
17. a String efficiency =
total rating of circuit
n individual rating of thyristor ...(1)
I01 = 230 115A
2For series connected SCRs
Individual voltage rating = 1800 VOverall rating = 11.34 kV
Using equation (1)
0.9 =
311.34 10n 1800
n = 7For parallel connected SCRs
Individual current rating = 1000 AOverall rating = 4.5 kA
Using equation (1)
0.9 =
34.5 10n 1000
n = 518. (c)
Given 0 avgV =0d
1 V2 ...(1)
0dV = m
0
2V 1 cos
0dV = m2V 2
...(2)
0 avgV = m2V cos 1
...(3)
From equation (1), (2) and (3)
m2V 1 cos
= m2V 22
1 cos = 1 cos = 0 = 90º
Power factor = 1s
s
Icos
I
...(4)
Is = 0180ºI
180º = 0
1I2 ...(5)
1sI = 02 2 I cos 2
= 02 2 I cos 45º
= 02I
...(6)
so using equation (4), (5) and (6)
P.F. = 0
0
2I cos 45º
1I2
= 2 2 cos 45º
=
2 = 0.636
19. (b)For firing angle = 0°, average outputvoltage of 3 semi converter is equal to
0V = m3V 1 cos
2l
= m3V 1 cos0
2l
= m3V .2
2l
0V = 3Vml
Given
0V 400V
400 =m3V l
mV l = 4003
= 418.87 V
sV =418.87 296.192V
3
per phase voltage
phV = sV3
=296.192V
3= 171.0 V at 1500 rpm
20. (a)
GivenT = 1 1msec1kHz
Ton = 400 sec
=
6on
3T 400 10T 10
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= 0.4Average output current is
0I = sV E 0.4 220 44R 1.5
0I = 29.33 AmpInput power =
sV average supply current
= 220 11.943
iP = 2.62746 kW
Power absorbed by load emf = 0E I
= 44 29.33 1.290 kWPower loss in resistor R
= 2.62746 1.29052 = 1.287 kWAlso, power loss in resistor
= 2rmsI R 1.287kW = 1287 W
rmsI = 12.87 29.29 Amp1.5
21. a• For self commutation, circuit should beunderdamped.
Condition of underdamped RLC series
circuit is 21 R 0
LC 2L
4LRC
...(1)
Putting value of L and C
4LC =
6
64 20 10
40 10
= 2 ...(2)
From equation (1) and (2) R 2
R 1.414
Since given in option (a) R 1 is thepossible value of R for which self or loadcommutation is possible.
22. d• For current commutation the current ininductor is in reverse direction of load currenti.e. current flow through main thyristor sothat current through thyristor reduces andbecome zero. This process takes time but involtage commutation as TA is ON, Tmainbecome ‘OFF’ because of reverse voltageapplied by capacitor. Voltage commutation is faster.• In current commutation, when TA is ‘ON’,there should be some current flow throughload and the same through main thyristor.
Loading is required in currentcommutation but this is not the case withvoltage commutation.• Duty cycle for voltage commutation,
cmmax
T 2tT
max can never be equal ‘1’ because‘2tcm’ time is required to make auxiliarythyristor ‘OFF’ from ‘ON’ state.
23. (b)Since there is no other source of energy in thecircuit except capacitor. Then energydissipated in the ckt will be equal to initialenergy stored in the capacitor. Energy stored in capacitor =
20
1 CV joules2
= 261 10 1002
= 0.5 × 10–2 = 0.005 joules Energy dissipated in circuit = 0.005 joules.
24. (a)25. (c)26. (c)27. d
A resistor connected across the gate andcathode of an SCR increases its dv/dt rating,holding current, noise immunity.
28. bThyristor can be fired when input voltagebecome greater than back emf voltage E.
mV sin t > E
msin t E V
1mt sin E V
1 100t sin
200
1 1t sin
2
t 30º 29. c
Isn =
02 2 I ncos
n 2For fundamental n = 1
1sI = 02 2 I cos2
= 2 2 10 cos 60º 2
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= 2 2 10 cos 30º 7.796A
30. d
230V50Hz
+
–
D1
R
+ –
+–100
+–50
a
b2
VD
When ‘a’ is positive with respect to ‘b’ diodeD1 conducts and diode D2 is reverse biased.
Applying KVL, 2DV 150V rms
So, diode D2 is subjected to reverse voltage of150V (rms).similarly, during negative half cycle, D1 issubjected to reverse voltage of 150 V/rms.Thus, for diode D1 and D2 peak reverse voltageis 150 2V (maximum value of the voltageacross the diodes)
31. (b)
Ripple factor = 2FF 1
FF(form factor) =rms
avg
II
Irms = s I
Iavg= s I (Is - supply current)
FF = s
s
II
=
1
ripple factor = 1 1
= 1 10.3
= 3.333 1 = 2.333 = 1.527532. cFor half bridge inverter
(V1) rms output voltage across load = DCV2
Output power (P) = 21V
R
P = 2OCV4R
...(1)
For full bridge inverterrms output voltage, V2 = VDC
Output power, P1 = 22V
R = 2DCVR
2DCV4
4R
From equation (1)P1 = 4.P
33. c
id
iL
t
t
TON TOFF
id = OFFL
ON OFF
TiT T
OFF T 1 T
id = Li 1 = 8 × (1 – 0.4) = 8×0.6 = 4.8 A
34. bBy drawing the (O/P) voltage
V0VRN VYN VBN
t
Each diode conducts for 120º only. There are threepulses of output voltage during one cycle ofinput voltage.
so, frequency of ripple = 3f = 3×100 = 300 Hz35. (c)
The worst possible fault current should beconsidered for calculating the fault clearancetime.Maximum fault current occurs when sourcevoltage is at peak = 400 VWhen terminal ‘A’ is short circuited to ground,the resistance offered to the source is 1 .
Maximum fault current = 4001 = 400A
Assuming maximum fault current to remainconstant during fault clearance time tc we get
ct
2
0i dt = 100
i2.tc = 100
tc = 2100400 =
100 sec.
400 400
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=
100 1000400 400
tc = 0.625 msec.36. (b)37. (a)38. (b)39. (d)
A medium is said to be dispersive when phasevelocity of an EM wave depends on frequencyin that medium.
40. cFrom given figure, current through thyristoris
i = tV 1 eR
LR
i =
RtLV 1 e
Rwhen latchi I during t = tON then SCR willbecome ‘ON’
so
ONR t
LV 1 eR ILatch
ONtRL1 e
LatchI RV
LatchI R1
V ONR t
Le
Take loge both sides.
LatchI R
In 1V
ONR t
L
LatchI RIn 1
V
ONR tL
tON
LatchI RL In 1
R Vin Case-Ias R nR then
LatchI RIn 1V increase so
effect on tON is not much prominent.in Case-IIas L nL thentON will increase by almost ‘n’ time.
41. a
V03=
s4V sin 3 t3
=
4 230 sin 3 t
3 V03= 97.6150 sin (942.47t)
Z3 = 1R j 3 L
3 C=
36
14 j 3 2 50 35 103 2 50 115 10
= 4 + j(32.97 – 9.23) Z3 = 4 + j(23.756)
3Z = 2 24 23.756
Z3 = 16 564.34Z3 = 24.09
(I03) = 03 rms
3
VZ =
97.61502 24.09
(I03) = 2.86 A42. d
t
V0 V01
–Vs
Vs
0
i01
Im t
iT1
t
T ,T1 2 T ,T3 4
Im
iD
t
0
0
0
2
2
1
Rms value of diode current, ID1 = 043. (a)Gate pulse width is given as:
T =duty cycle
triggering freq. = 0.1400 = 250 s
As, T > 100 s , so d.c. data is applyHence, Vg Ig = 0.5 watts (average power)
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or (0.5 + 8 Ig) Ig = 0.5
2g8 I 0.5 Ig 0.5 = 0
So, Ig = 0.22 Amp.During the pulse-on period:
Es = Rs Ig + VgSo, 12 = Rs Ig + (0.5 + 8 Ig)or 12 = Rs Ig + (0.5 + 8×0.22)or 12 = Rs × 0.22 + 2.26
Rs
Es+–
Ig
Vg
A
K
Trigger circuit
or Rs =12 2.260.22
Rs = 44.2744. bUnder rated operating condition
Vt = Ea + Iara
or Vt = m m a aK I r
Given, Vt = 200V,
m2 1000
60Ia = 10 Amp and ra = 2
200 =
m2 1000K 10 1
60or Km= 1.81 V–s/radFor rated motor torque, armature current = 10A V0 = Vt = m m a aK I r
2 2 230 cos =
2 5001.81 10 160
or
650.53 cos = 104.77
or cos =104.77
650.53cos = 0.505
= 59.61º45. (b)46. (b)47. (b)
Given r 9
Refractive index, rn 9 3
48. (a)Phase velocity,
p
22
2
p
49. aFor a motor, Vt = V0 = Ea + IaraThe minimum possible speed of dc motor is zero.
This gives motor counter emf Ea = 0 sV = V0 = 0 + Iara
or 220 = 25×0.2
=5
220 = 144
Maximum possible value of duty cycle 1. sV = Ea + Iara
1 × 220 = 0.8 × N + 25 × 0.2220 – 5 = 0.8 N
or N =2150.8 = 268.75 rpm
So range of duty cycle is 1 144 and range of
speed is 0 < N < 268.75 rpm.50. aMinimum braking speed is
mmW =a a
m
I rK =
300 0.21.2
mmW = 50 rad/s or 477.46 rpmMaximum braking speed is
mxW =s a a
m
V I rK =
400 300 0.21.2
mxW = 383.33 rad/s or 3660.06 rpm
Note : 0 sV Vwhen 0 then V0 = 0 then minimumbreaking speed occur.when 1 then V0 = Vs then maximumbreaking speed occur.
51. (b)
Circulation of a vector is given by A.d l
602 2
q0 0 0
ˆ ˆ ˆA.d a A.d a A.d a
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2 2
0 0
cos d 0 cos d
2 22 2
0 0cos0 cos60
2 2
12 – 2 12
52. (d)A vector field A is said to be solenoidal.
Divergenceless if A 0
and irrotational orpotential if A 0
. For a scalar field V..
Laplace equation is 2V 0 .53. (c)
The field of curl F is purely solenoidal as . F 0
. Thus a solenoidal field A
can
be expressed in terms of vector F
as :
.A
= 0 [solenoidal field]
S
A.ds 0
and F A
54. (c)
If A
is irrotational, then A 0
L
A.d 0
l & A V for a scalar field V..
55. (c)Total charge Q will be uniformly distributedover the surface of the conducting sphere.
S2Q S1
R
ar
For Gaussian surface S1,
enclosedQE.ds 0
E 0
for 0 < r < R.
For Gaussian surface S2,
enclosedQE.ds Q
2QE
4 r
2
1E for R rr
56. (d)
Electric field exists from high potential to lowerpotential.
VP > VQ and hence VPQ is negative (VPQ =VQ – VP)Therefore VPQ is negative, there is a loss inpotential energy in moving unit charge from Pto Q and hence work is being done by theelectric field.
57. (d)Equipotential surface is a surface which is cutby the electric field perpendicularly.
Sphere
+q
The equipotential surface due to a point chargewill be concentric spheres as shown above.
58. (d)
W = F.ds
W = q E.ds
W = 9 ˆ ˆ ˆ ˆ ˆ ˆ10 4i 3j 2k . 10i 2j 7k m
W = 40 6 14 nJ
W = 20 nJ59. (b)
Energy stored in the field is given by
U =1 2
0
Q Q14 r
U =
129
2 2 2
4 1 109 101 2 3 1 1 5
U =336 10 36 mJ
749
U = 5.14 mJ60. (b)
Electric flux through the complete sphere willbe :
=0 0
encQ QE.ds
Therefore through the hemisphere will be 0
Q2 .
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61. (d)Z
E1
E2
r1 = 4
r2 = 3
Normal component of 1 inE E 3
.
E = n tE E
1tE = 1 nˆ ˆE E 5i 2j
Tangential component of E is continuous
across the boundary.
E1t = E2t
E2t = ˆ ˆ5i 2j
Also, normal component of electric flux densityis equal across the boundary.
E1t = E2t
E2t = ˆ ˆ5i 2j
Also, normal component of electric flux densityis equal across the boundary.
D2n = D1n
r2 2nE = r1 1nE
2nE =4 3 43
2 2t 2nˆ ˆ ˆE E E 5i 2jk 4k kV m
62. (c)For a dielectric -dielectric interface
(a) nD
is continuous
(b) nE
is discontinuous
(c) tD
is discontinuous
(d) tE
is continuous63. (a)
Tangential component of E is continues across
boundary.
E1t = E2t
1E sin = 2E sin ...(1)
Normal component of D is continues across
boundary.
D1n = D2n
r1 inE = r2 2nE
r1 1 r2 2E cos E cos ...(2)
Dividing (1) by (2), we have
r1
tan =
r2
tan
tantan
=
r1
r2
64. (c)Electric field inside a conductor is always zeroand electric field lines can cut the conductorsurface at right angle only.Electric field exists inside a dielectric material.
65. (b)
Net potential at 0 = 0 0
2QQ4 a 4 b
V =0
Q 1 24 a b
66. (c)
Applying Gauss’ law to a arbitrary Gaussiancylindrical surface of radius a b , wehave
Q = E.ds E 2 L
E =Q a
2 L
V =a a
b b
Q ˆ ˆE.d a .d a2 L
l
V =a
b
Q d Q bn2 L 2 L a l
Capacitance, C = Q 2 L
bV lna
67. (d)Given xE 8cos t
yE 24cos( t 90 )
x| E | 8
y| E | 24
90
If x y| E | | E |
and 90
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Then it is elliptically polarised.
68. (a)
4
P(–4,6,4)
y=2 4 yz
r
x
Electric field due to infinite uniform line,
E = o2 r
Here, x yˆ ˆr ( 4 0)a (6 2)a
= x yˆ ˆ4a 4a
Hence, 9
x y9
ˆ ˆ( 4a 4a )10 10E4 2102 4 2
36
= x yˆ ˆ22.5( a a ) V/m
69. (b)
Since, vD
yx zv
DD Dx y z
= (1) (xy) (x)x y z
= 0 + x + 0
v at (1, –2, 4) = 1 c/m3
70. (d)
Work done, W = B
A
Q E dl
= B
A
5 (y dx x dy)
= B
A
5 [(x 1) dx x dx]
= x 2
x 1
5 (2x 1) dx
=
22
1
2x5 x2
= –[(4 – 1) + (2 + 1] × 5
= –30 Joule
71. (c)
x dx
dI
Taking an elemental circular area of width dr ata radius r,
Current through element,
dI = J 2 r dr
Total current, I = r
0
J 2 r dr
=
22 10
0
10 2 r drr
= 220 2 10
= 240 10 A
= 400 mA
72. (d)
y
z
x
(0,2,–1)x=0, z=0
x=0, z=–2
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As the point (0, 2, –1) lies exactly between thegiven two lines. Hence, the magnetic fieldintensity by each line are opposite to each other.
Hence, net magnetic field intensity at point (0,2, –1) will be zero.
73. (d)
Given, r r9, 1
Since, o r
o r
EH
E 112015 9
E 12015 3
E 600 V/m
74. (b)
Average pointing vector,
22oo
E1 1P H2 2
= 21 120 (0.15)2
= 460 225 10 = 213.5 W/m
75. (b)
Conduction current density, Jc = E
cJ 2sin100tE100
= 0.02sin100t
Hence, displacement current density
d o rdEJdt
= o 4.5 0.02 100cos100t
= o9 cos100t
76. (d)
For a loss-less line,
Electrical path length = l LC l
=
6 6 9 22 10 10 40 10 25 10 25 10
= 7 122 10 10 0.25
= 7 62 10 10 0.25
= 5 radian
77. (b)
E = dVdx
= 2d (5x 10x 9)dx
= –[10x + 10]= –[10(1) + 10]= – 20 V
78. (d)
x
y
z
1A1
21A
Right Hand Thumb Rule :Thumb is in the direction of current, fingerswill represent the direction of magnetic field.Due to wire 1 , H field will be in the xadirection and due to wire 2 , H field will bein the
za direction. Thus, x and z
components are non zero.79. (b)
Zk
X
Y
y = 6
s
02
2s
1 C m6000s
02
i
j
Given s =
21 C m6000
Thus, associated electric field strength
E =s
n0
a2
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=
6
9
1 106000 j V m12 10
36
E = ˆ3j V m
80. (d)The net capacitance can be considered astwo capacitances in series.
1
1AC
d 2
and 2
2AC
d 2
C =
21 2 1 2
21 21 2
C C 4 A2AC C dd
C =1 2
1 2
2Ad
81. (a)
I
RdH
P
dl
Magnetic field
dH at point P due to currentelement Idl is given by:
3
I dl RdH
4 R82. (d)
(0, 2, 0)
(1, 0, 0)
Y
IX
The direction of current is given by righthand curl rule.Right hand thumb is placed along directionof current and fingers are curled anddirection of magnetic field is along thetengents of this circle.Direction of field is as shown in figure
At (1, 0, 0) it will be j
At (0, 2, 0) it will be i83. (a)
dsP
According to Stoke’s theorem, which relatesline integral to surface integral,
H.dl =
H.ds
Thus,
dsP =
P.dl
84. (c)
X
Y
–
+
++++++++++++
Z
According to Gauss’s Law
s
E.ds
= 0
charge enclosed
E.2 R. l =l
0
.
E =02 R
This is the electric field intensity by aninfinitely long line charge at a distance R,and its direction will be radially outward.
So, E at (R, 0, 0) = x
0a
2 R
85. (d)The electric field at a point due to line chargeis,
E = L
20
R.2 R
where, x y zˆ ˆ ˆR 1 1 a 2 2 a 3 5 a
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i.e. R = y zˆ ˆ0 4a 2a
R = 2 24 2 20
Hence, E
= y z9 9ˆ ˆ4a 2a
10 10 9 1020
E
= y zˆ ˆ18a 9a
86. (b)Let the required point is 3P 0, y, 0
Then, 13R
= x y zˆ ˆ ˆ0 4 a y 2 a 0 7 a
= x y zˆ ˆ ˆ4a y 2 a 7a
and 23R
= x y zˆ ˆ ˆ0 3 a y 4 a 0 2 a
= x y zˆ ˆ ˆ3a y 4 a 12a
Now, electric field at P3 X-direction,
Ex =
9
323 20
10 25 44 4 y 2 7
322 2
60 3
3 y 4 12
As Ex = 0, we get20.48y 13.9y 73.12 0
y = –6.89 or –22.11i.e. P3 = (0, –22.11, 0)
87. (b)z
x
y
y = 3m
E+E–
y�a
The electric field intensity at a point due toplane
E = sn
0a
2
3, 3,3E = 8 9
y10 36 10 a
6 2
= yˆ30a V m
88. (a)
4 cm
12
34
4 cm
Total potential energy stored,
W =4
n nn 1
1 q V2
= 1 1 1 2 3 3 4 41 1 1 1q V q V q V q V2 2 2 2
Here, potential at charge ‘1’ is,V1 = 21 31 41V V V
= 0
q 1 1 14 0.04 0.04 0.04 2
= 9 9 1 11.2 10 9 10 20.04 2
= 730.89Here, V1 = 2 3 4V V V V
So, total potential energy stored.
= 14 qV2
= 92 1.2 10 730.89
= 1.75 J
89. (d)Magnetic field due to a solenoid with ‘n’ termsper unit length and current I will be:
B = µ0 n I
=
72
1004 10 510 10
B = 3 b2
W2 10m
90. (d)Magnetic field lines always form a closedpath. The net field lines having a volume toalways equal to the net field lines enteringthe volume. Hence, net magnetic flux
B ds 0.For magnet 1 and 2, net flux leaving thevolume is equal to net flux entering thevolume. Hence, net flux is zero.
91. (a)
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Given,
23 2x y z31 2F a a a3x K zK xy K z 3xz y
F =
x y z
23 231 2
a a a
x y z3x K zK xy K z 3xz y
= 2 2x y z3 12a a a1 K 6x K x3z K 3z
= 2x y z3 2 1a 3z a x aK 1 K 1 6 K
Given vector field F is irrotational i.e.
F = 0
2x y z3 2 1a 3z a x a 0K 1 K 1 6 K
if 3 2 1K 1 0, K 1 0, 6 K 0
3 2 1K 1, K 1, K 6
23 2x y z x y z31 2a a a a a a.F . 3x K zK xy K z 3xz yx y z
23 231 2 3x K zK xy K z 3xz yx y z
= 1K 0 6xz
At point (1, 1, –2)
.F = 1K 6 1 2
= 112 K
= 12 6 6 1K 6
92. (d)Radius of uniformly charged sphere = R
Charge density = 2C m
Now, as per Gauss’s law,
D.ds
= charge enclosedNow, at r = 2R,
21r 2R
D dS 4 R
21D 4 2R = 24 R
1D4
, 11E
4
...(1)
at r = R2 ,
2D ds = 0
D2 = 0 E2 = 0
Now, 2
1
RE r E2 0E Er 2R
93. (d)Given,
2rD r a 2sin a
in spherical coordinatesystemwhere D = electric flux densityand Dr = r2, D 2sin , D 0According to Gauss’s law,
D.ds =
.D = v [by divergence theorem]
.D in spherical coordinate is given by
.D =
2r2
1 1 sin Dr Dr r rsin
1 Dr sin
=
2 22
1 1sin 2sinr .rr r rsin
10rsin
= 3
24r 1 02 2sin cosr rsin
=4cos cos4r 4 rr r
Point P.D =
cos60 14 4 61 11 2
94. (c)
AB
C
+Q
–2Q
+Q
a 30,2
a ,02
a ,02
0
Let the point P(0, y)Potential at point P(0, y) due to all threecharges
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V1 =0
1 Q KQ4 AP AP
V2 =0
1 Q KQ4 BP BP
V3 =0
1 2Q 2KQ4 CP CP
V =
1 2 3KQ KQ 2KQV V V 0AP BP CP
AP BP
or, 2KQAP = 2KQ
CPor, AP = CP
22a y
4 = a 3y
2
or,2
2a y4
=2
2 a 3y ay 34
y a 3 =2a
2
y =a
2 3Therefore, the coordinates of the point P is
a0,P2 3
.
95. (c)z(0,0,a)
(0,a,a)(0,0,0)
(a,0,0)(0,a,0)
y
xHere, the enclosed volume is a cube, havingsix faces. Then, the charge enclosed is,
Q =s
D ds
= 2 32xy.dy.dz x dxdz 6z .dxdy
= a a a a
0 0 0 02ay.dy.dz 2.0.y.dydz
a a a a
2 2
0 0 0 0x dxdz x dxdz
a a a a
3 3
0 0 0 06 .dxdy 6x .dxdy0
= 4 4
4 5a a a0 0 6a3 3
= 4 5a 6a
96. (a)Since, m = r 1 = 6.5–1 = 5.5then, magnetisation
M
= mH
M
= x y z5.5 ˆ ˆ ˆ10a 25a 40a
=
2x y zˆ ˆ ˆ55a 137.5a 220a A m
97. (a)
d = 1mm
A=2m2
C1
C2
r1=1.5
r2=3.5
C1 = 1 10 r 0 r.A 2 Ad d2
and,C2 = 2 20 r 0 r.A 2 Ad d2
Now, C1 and C2 are in series, so theequivalent capacitance,
Ceq =
1 2
1 2
20
r r1 2
01 2r r
2 A.C .C d2 AC C
d
= 1 2
1 2
r r0
r r
.2 A .d
= 12
3
2 8.85 10 2 1.5 3.51 10 1.5 3.5
= 937.17 10 F
= 337.17 10 F
= 0.03717 F
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98. (b)Let, Q be the charge that can be stored onthe capacitor.Then the electric field inside capacitor
E =0
QK.A. ;
where K is dielectric constantThis, electric field should not exceed thedielectric strength 7 11.9 10 Vm .
i.e.0
QKA 71.9 10
Qmax = 27 2 121.9 10 5 100 10 8.85 10
= 784.075 10
= 8.4 C99. (d)
Since, energy stored, W = 1 B.H2
W = 271 4 10 6002
= 0.226 J/m3
So, energy stored in sphere= W × Volume
= 3240.226 1 103
= 60.94 10 J
= 0.94 J100. (d)
The energy stored in the magnetic field attime t is,
U = 21 Li2
= 22 t0
1 Li 1 e2
The rate at which the energy is stored is,
P = dUdt
= 2 t/ t/0
1Li 1 e e
= 20 t/ 2t/Li e e
The rate will be maximum, if dP 0dt
or, 20 t/ 2t/Li 1 2e e 0
or, t/e = 12
So, the maximum rate,
Pmax =20Li 11
2 4
=
2
2
LV VL 4R4RR
= 210 1004 2 8
= 12.5 J/sec
101. (c)
+
R
CV0V i
+
Inverter
Differential Controller
+
R2
R1V0Vi
+
Inverter
Proportional Controller
+
C
V0Vi
+
Integral Controller
R
+
C2
V0Vi
+
PID Controller
R2InverterR1
C1
102. (b) Integral control alone introduces thehunting in the system
It improves the steady state performance
103. (c)
Given 2kG(s)
s s 20 s 30
Type –2 system
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We know steady state error
ess = t
s 0
ess =t
s 0
3
2
20 2ss
k1s s 20 s 30
=t
s 0
2
2
40
ks 1s s 20 s 30
=
40k0
20 30
ess =40 20 30
k
= 0
Now with integral controller KIs
ess = t
s 0
3
3
20 2sskk1
s s 20 s 30
I
= 40 40
110
= 0
104. (b)
Given 2 2
c1 2 2
1 R C sG (s)1 R R C s
It is in the time constant form
T1 = R2C, T2 = 1 2 2R R C ; T1<T2
xPZ
2 2
1R C 1 2 2
1R R C
Location of zero (z) = 2 2
1R C
Location of pole (P) = 1 2 2
1R R C
Z > PPlotting the location of zero and poleHence it is a lag network.
105. (b)
Given c
100 1 0.25sG (s)1 25s
Comparing with the standard transfer function
Gc(s) =
1 sT
1 sT
Here = 100 T = 0.25Then mid corner frequency
m = 1
T
=1
0.25 100
=1
10 0.25
=4
10
m = 0.4 rad/s
106.(a)Lag compensatorLag compensator adds a pole near to originIt has a dominant pole nearer to origin
Lag Compensator
Magnitude
without Compensator
As the slope added to the system and dueto increase in slopeBandwidth will get reduced.
107.(b)Lead compensator high pass filterLag compensator low pass filterLead lag compensator bandpass filter
108. (b)
109. (a)Type of the transfer function denotes theno. of poles at origin.
110. (c)
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x1
e f g
a x2 b x3 c x4 d x5
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 = dgNon-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
111. (b)
We know X(s) = –1SI A B U(s) {for intial
conditions hil}
Substituting X(s) in Y(s) we get
X(s) = C –1sI A B U(s)
sI A =s 0 2 00 s 0 4
= s 2 0
0 s 4
–1sI A = 1
s 2 s 4 s 4 0
0 s 2
X(s) = 1
s 2 s 4 s 4 0
0 s 2
11
= 1
s 2 s 4 s 4s 2
Y(s) = 4 0 1
s 2 s 4 s 4s 2
= 1
s 2 s 4 4 0s 4s 2
= 1
s 2 s 4 4 s 4 0
=
4 s 4
s 2 s 4
= 4
s 2 =4
s 2
= 2t4e
112. (b)
Transfer function = C –1sI A B D
C = 1 1 , sI A =s 1 0
0 s 3
, B =
01
–1SI A = 1
s 3 s 1 s 3 0
0 s 1
Transfer function
= 1 1 1
s 3 s 1 s 3 0
0 s 1
01
= 1 1
1 0s 1
10s 3
01
= 1 1
01
s 3
=1
s 3
113. (a)Given differental equation is
2
2d y dy2 3y(t) u(t)
dtdt
Let x1 = y
1 2x x
2 1x x y
y 2y 3y u
1 1 1x 2x 3x u
2 2 1x 2x 3x u
2 2 1x u 2x 3x
2 1 2x u 3x 2x
1 2x x
2 1 2x u 3x 2x
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1
2
xx
=
0 13 2
1
2
xx
+01
A =0 13 2
C = 01
114. (b)
A = 1 42 5
sI =s 00 s
sI A = s 00 s
– 1 42 5
= s 1 4
2 s 5
sI A = s 1 s 5 8
= 2s 4s 5 8
= 2s 4s 3
Characteristic equation is sI A 0
i.e 2s 4s 3 0
115. (b)Derivative feedback controlFor derivative feedback control the actuatingsignal is obtained as the difference betweenthe proportional error signal and derivative(rate) of the output signal. Therefore, theactuating signal for derivative feedback controlaction is
ea(t) = e(t) –
tdc t
kdt
where kt is constantHence derivative feedback control is knownas rate feedback or tachometer feedback.
116. (d)
skt
2n
ns s 2
C(s)E(s)
R(s)+
ess = lim
s 0 sE(s)
= 2lim 1
s 0 s
2 2n n t
2 2 2n n t n
s 2 k s
s 2 w k s
ss tn
2e k
ess when hence
117. (b)A potentiometer is used as a error detectiondevice. Its input is the position error andpotentiometer convert it to correrspondingvoltage.
118. (d)Impulse input yields natural responsedepending only on the parameters of thesystem and not on input.
119. (a)Steady State Error
Type of inputUnit Step
Unit Ramp
Type 01/1+K
Type 10
1/K
Type 200
Type 300
120. (c)
Pole is located in right side of s planetherefore unstable system.
Conjugated pole on imaginary axis thereforeresponse of system would be sustainedoscillations.
Single pole at origin therefore response of
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system would be step response.
Conjugated pole on left side of s plane thereforeresponse of system would be decayingoscillation.
121. (c)
Impulse = d step responsedt
= 2t td3e 4e 2dt
= t 2t4e 6e
122. (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
123. (c)A signal flow is used to obtain the transferfunction of the given system using masongain formula.
124. (b)
Total system transfer function is 1 3
2
G GG
errors are 1 3 2
= 1 2 3 125. (a)
Feedback control system is sensitive to bothfeedback path parameters and forward pathparameters, but it is more sensitive tofeedback path parameter compared to forwardpath parameters.
126. (a)Open loop gain= G2Feedback gain = HG1
TF = 2
1 2
G1 HG G
127. (d)
C sR s = G G s 11
s 1 s 1
G s 1
s 1
= s 2s 1
G + s + 1= s + 2G= 1
128. (a)Nyquist stability test can give inputs on thestability and how to stabilize.
129. (c)A system with gain margin close to unity ora phase margin close to zero would beoscillatory in nature.
130. (d)Forward paths: Only one and the gain isG1G2G3
Loops: L1 = – G1 G2 H1,
L2 = – G1 G2 G3 H3,
L3 = – G2 G3 H2
Non-touching loops : NIL
Using mason gain formula C(s)R(s)
= 1 1P
= 1 – (L1 + L2 + L3) = 1 + (G1G2H1 +G1G2G3H3 + G2 G3 H2)
1 = 1
1 2 3
1 2 1 1 2 3 3 2 3 2
G G GC(s)R(s) 1 G G H G G G H G G H
131.(d)Given
C(s)G2G1
H1
++
–
+
–R(s)
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C(s)G2G1
H1
++
–
+
–R(s)
H1
By shifting the summing point to left of G1
C(s)G2G1
H1
R(s)
H /G1 1
Feedbackpaths
Inner loop
++
– –
By merging feedback paths and solving forinner loop
C(s)G2+
–R(s) 1
1
G(1 – G )
11
1
HHG
C(s)R(s)
=
1 2 1 2
1 1 2 1 2 111 1 2 1
1
G G G G(1 – G ) G G H G HH(1 – G ) G G H
G
1 2
2 1 1 1
G GC(s)R(s) G H (1 G ) (1 – G )
132.(c)Given r(t) = 2cos (2t – 45)Comparing withr(t) = A cos ( t – )
A = 2, 2,
By puttings = j j2
closed loop transfer functionC(s)R(s)
= G(s)
1 G(s) H(s)
T(s) =
C(s) 10 10 10R(s) s 11 j 11 2j 11
Since input is sinusoidal, so steady stateoutput is
C(t)ss = 2| T ( j ) | T( j ) · R(t)
,
2T(j ) 0.894 10.3
C(t)ss
= (0.894 –10.3) · 2 cos(2t – 45 )
ssC(t) 1.788 cos(2t – 53.3 )
133.(c)
Given
s3 + (K + 0.5)s2 + 4Ks + 50 = 0
The Routh’s array is
3
2
1
0
s 1 4K
s K 0.5 5050s 4K – X
K 0.5
s 50 X
For stability
K + 0.5 > 0 K > –0.5 ...(1)
and504K –
K 0.5> 0
or 4K (K + 0.5) > 50or K(K + 0.5) > 12.5or K2 + K/2 – 12.5 > 0
K > – 3.8 AND K > 3.3...(2)from equation (1) and (2)
Therefore, the condition for stability isK > – 3.8, K > –0.5 and K > 3.3Thus, if K > 3.3, then all conditions aresatisfied.If we put K = 3.3 then s1 row is zero row.The subsidiary equation for s2 row for K =3.3 is3.8s2 + 50 = 0
S = ± j 3.63 Frequency of oscillation is 3.63 rad/sec.
134.(d)
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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)
By putting different values of , we get
= 0 1 180 = 0.01 1 –178.85 = 0.1 1 –168.57 = 1 1 –90 = 10 1 –11.521 = 100 1 –1.145 = 1000 1 –0.1145 = 1 0
G (j ) G(j )
= 0 = u
jVGH plane
135.(b)
G(s) = e10(1 sK )s(s 2)
c
R
(s)(s)
= e2
e
10 10 sKG(s)1 G(s) H(s) s s(2 10K ) 10
H(s) = 1on comparing with standard second ordertransfer function
2n = 10 n 3.16 rad/sec
n2 = 2 + 10 Ke
Ke = n2 – 210
=
2 0.6 3.16 – 210
Ke = 0.18
136.(a)Coorresponding nyquist plot by mapping inwhole Right half of s-plane is
j I G(j )mg
=
–1 + j0
Real G(j )
from this plot it is observed that (–1 + j0)point is encircled twice in the clockwisedirection.
Therefore N = –2.
137. (d)Given
G(s) =
2
2 2s 8s 15
s 4s 10 s 3s 12
=
2
4 3 2s 8s 15
s 7s 34s 78s 120
=
2 3 4
2 3 4
1 8 15s s s
7 34 78 1201s s s s
signal flow graph of G(s) in phase variableform
–120
Y(s)1
U(s)
1s
x1
x2x3x4
151s
1s
1s
–78
–34–7
1
8
4x
assuming node, x1, x2, x3, x4, 4x as givenin signal flow graph above, we get
1x = x2
2x = x3
3x = x4
4x = –120 x1 – 78x2 – 34x3 –7x4 + u(t)so
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1
2
3
4
x
x
x
x
=
0 1 0 00 0 1 00 0 0 1120 78 34 7
1
2
3
4
x 0x 0 u(t)x 0x 1
138. (c)Put R(s) = 0
SoR(s) 0
C(s)N(s) =
1 1P
[using Mason gain formula]Signal flow graph corresponding to givenblock diagram is
G(s)
1 1C(s)
N(s)
–H(s)
–1
N(s)
P1 = 1
1 = 1 H(s)G(s)
= 1 + G(s) H(s) + G(s)
So R(s) 0
C(s)N(s) =
1 G(s)H(s)1 G(s)H(s) G(s)
if 1 + G(s) H(s) = 0 then output C(s)will not be affected by the disturbanceN(s) so
H(s) = 1
G(s) [By putting the value of G(s)]
H(s) =
s s 1 s 2
K s 3139. (b)
GivenG(s) = e–sT· G1(s)
G(s) = e–sT·
11s s
…(1)
Then characteristic equation is
1 + G(s) H(s) = 01 + e–sT G1(s) = 0G1(s) = –esT
1( ) s jG s = –ejT …(2)
From eqn (2), it is clear that critical pointhas been shifted to –ejT from (–1 + j0)From eqn (1) G(j) = – T – 90° – tan–1 () = (let At = gc)PM = 180° + PM = 180° – T – 90° – tan–1 ()as T increase PM decrease so systemstability reduces.
140. (d)From log-magnitude plotCorner frequency at log = –1or = 0.1Hence pole at = 0.1
and gain ; log|G| = 1, G = 10
Hence transfer function
=10 1
(1 s / 0.1) s 0.1
141. (a)
The equation of performance for the systemare
1 01 1 1 0B (X X ) K (X X ) = 2 0K X
or 1 1 1 1 1 0(sB K )X (s) (sB K )X (s)
= 2K X(s)
0
1
X (s)X (s) = 1 1
1 1 2
sB KsB K K
T =
11
1
11 2
1 2
B sK 1K
sB(K K ) 1K K
Let 1 2
1
K KK
= a
where a 1
and 1
1 2
B TK K
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Then, 0
1
X (s)X (s) =
1 1 aTsa 1 Ts
Therefore zero is nearer to origin than polei.e. lead network.
142. (d)
OLTF is given by G(s) =d
21
K(1 sT )s (1 sT )
Putting s j
G(j ) = d
21
K(1 j T )(1 j T )
Magnitude, |G(j )| =2
d2 2
1
K 1 ( T )
1 ( T )
Phase angle G(j )
= 1 1d 1180 tan ( T ) tan ( T )
0G(j ) = 180
G(j ) = 180
Since at
G(j ) = – 90°, which is possible only if
1 1d 1tan T tan T > 0
or, 1dtan T > 1
1tan T
Td > T2
143. (d)
Given
G(s) = 210
s 3s 10 ; r(t) = 8 cos (2t + 30°)
forced response will be
y(t) = A G(j ) cos ( t ) ...(1)
where = 2 and 30
= 2G(j )
G(j ) = 2
103j 10
G(j ) = 2 2
1010 9
Put = 2
2G( j ) =
1036 36
= 10 5
6 2 3 2 ...(2)
and G(j ) =
12
3tan10
2G(j ) = 1 6tan6
= –tan–1 (1)
= –45° ...(3)so by putting all values from equation (2),(3) to equation (1), we get
y(t) =58
3 2 cos (2t + 30° – 45°)
= 9.428 cos (2t – 15°)144. (a)
The breakaway points of root locus are the
solution of dK 0ds
G(s) H(s) = 2K , H(s) 0
s(s 8s 32)
Then characteristic equation is1 + G(s) H(s) = 0
2K1
s (s 8s 32)
= 0
K = –s (s2 + 8s + 32)diffrentiating w.r.t ‘s’ both sidesdKds = – (3s2 + 16s + 32)
by putting dKds = 0
3s2 + 165 + 32 = 0
s1, s2 = 8 j 4 23
since these points are not a part of rootlocus because
1 2s sG(s) or G(s) 180°so there is no breakaway points.
8 j 4 2s13
G(s)
= – 164.206°
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8– j 4 2s23
G(s) = 164.20°
145. (c)
G(s) =10
s(s 1) , H s 10
T s C.L.T.F. =G(s)
1 G(s)H(s)
THS =
2
2G(s)T H HC(s)
H T G(s)1 G(s)H(s)1 G(s) HC(s)
=
10 10G(s)H(s) s(s 1)101 G(s) H(S) 1 10
s(s 1)
=
100s(s 1)
s(s 1) 100s(s 1)
THS =
100 100s(s 1) (j10)(1 10j) 100
= 100
10j 100 100 =
100 10j10j
THS = 10
146. (d)
Converting block diagram into signal flowgraph as given below :
2 12
OutputInput
–3
–8 –1/4
Now forward path
1M 3
1 1 ( 3) 4
1 1( 1 L )
2M 2 12
2 1 (0) 1
3M 8 12
3 1 (0) 1
Loops; 11L 12 34
Using Mason’s gain formula,
OutputInput = 3(4) (2 12) ( 8 12) 21
1 3
147. (c)Givens5 + 2s4 + 3s3 + 6s2 + 10s + 15 = 0The Routh array for the above polynominalis as follows
s5
s4
s3
s3
s1
s0
120
6 15
6 15 2.5 15
6 15
15
36
2.5
2.5
15
0
0
10150
0
0
0
s2
In third row, first element become zero,then it is replaced by a small positivenumber .
As lim 0 then, the Routh arraybecomes
ssssss
5
4
3
2
1
0
1 3 102 6 150 2.5
– 2.52.5 1515 0
so there are two sign changes in the firstcolumn of Routh array There are two roots is lying in the righthalf of s-plane.Ans due to this system is unstable.
148. (b)149. (a)150. (c)
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