answers - ies master · transistor biasing. e b c b type of operation forward (f) reverse (r)...

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)

IES M

ASTER

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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

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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 Ê 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 Ê 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 Ê.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)

answers - ies master · transistor biasing. e b c b type of operation forward (f) reverse (r)...

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