answers - ies masteriesmaster.org/public/archive/2016/im-1477234702.pdf · the emf of a weston...

ESE-2017 PRELIMS TEST SERIESDate: 23rd October, 2016

1. (c)

2. (a)

3. (c)

4. (b)

5. (b)

6. (d)

7. (c)

8. (a)

9. (c)

10. (d)

11. (c)

12. (d)

13. (d)

14. (b)

15. (b)

16. (b)

17. (d)

18. (a)

19. (b)

20. (b)

21. (c)

22. (b)

23. (b)

24. (d)

25. (b)

26. (c)

27. (c)

28. (d)

29. (c)

30. (c)

31. (b)

32. (a)

33. (d)

34. (b)

35. (c)

36. (a)

37. (a)

38. (b)

39. (c)

40. (a)

41. (a)

42. (b)

43. (c)

44. (b)

45. (d)

46. (d)

47. (a)

48. (c)

49. (d)

50. (d)

51. (c)

52. (a)

53. (c)

54. (c)

55. (d)

56. (a)

57. (a)

58. (c)

59. (b)

60. (c)

61. (b)

62. (a)

63. (b)

64. (d)

65. (b)

66. (d)

67. (a)

68. (a)

69. (c)

70. (a)

71. (d)

72. (d)

73. (c)

74. (c)

75. (b)

76. (b)

77. (b)

78. (c)

79. (d)

80. (c)

81. (d)

82. (b)

83. (c)

84. (d)

85. (d)

86. (d)

87. (c)

88. (d)

89. (b)

90. (c)

91. (d)

92. (b)

93. (b)

94. (b)

95. (a)

96. (a)

97. (c)

98. (b)

99. (c)

100. (a)

101. (d)

102. (a)

103. (d)

104. (d)

105. (b)

106. (b)

107. (c)

108. (d)

109. (a)

110. (c)

111. (a)

112. (b)

113. (a)

114. (c)

115. (a)

116. (b)

117. (c)

118. (b)

119. (c)

120. (d)

ANSWERS

121. (b)

122. (b)

123. (d)

124. (c)

125. (a)

126. (b)

127. (b)

128. (b)

129. (d)

130. (a)

131. (d)

132. (b)

133. (b)

134. (d)

135. (c)

136. (a)

137. (c)

138. (a)

139. (c)

140. (b)

141. (c)

142. (d)

143. (c)

144. (a)

145. (d)

146. (c)

147. (a)

148. (b)

149. (c)

150. (d)

151. (c)

152. (b)

153. (d)

154. (b)

155. (d)

156. (a)

157. (c)

158. (c)

159. (a)

160. (b)

IES M

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Office : Phone : F-126, (Lower Basement), Katwaria Sarai, New Delhi-110016 011-26522064

8130909220, 9711853908 [email protected], [email protected]. : E-mail:

(Test-4 Solution) 23rd Oct 2016 (3)

1. (c)A rectifier type instrument measures theaverage value and it is calibrated for rmsvalue of sinusoidal ac.

Harmonic component is

= 1/22 212.1 11.1 100

11.1

= 43.4 %2. (a)

For balanced power, we can measure powerin one phase and multiply by 3 for 3 phasepower. For unbalanced load, we requireminimum 2 wattmeter.

3. (c)Frequency compensation is not used ininduction type meter. Other three types ofcompensation given in options are used.

4. (b)Energy meter will have creep error. A slighttorque developed by the light load adjustmentmay cause the disc to rotate slowly when thepotential coils are exited, but with no loadcurrent flowing. This is as creeping. This isprevented by cutting two holes or slots inthe discon opposite sides.

5. (b)Energy consumed by load in 100 sec,

=

1003603600

1000

= 1 kWh100

Meter constant= RevolutionskWh

= 51

100

= 5006. (d)

The major cause of creeping is theovercompensation for friction.

7. (c)

To make angle between P and voltage V,

90°, voltage coil winding should be sodesigned that it is highly inductive.

8. (a)

Power consumed = VIcos

= 240 20 0.5

= 3840 W

speed N = 500Revolution 3.84 kWkWh

= 1920 revolutions

hour

= 32 rpm9. (c)

Actual energy consumed

= 508kW3600

= 0.111 kWhEnergy under test,

=100

1000 = 0.1 kWh

% error =0.1 0.111

0.111

= – 9.9 %10. (d)

The meter constant =

number of revolutions

KWh

= 31056

10220 5 1 2

= 48011. (c)12. (d)13. (d)

tan =

1 2

1 2

3 W WW W

= 250 503250 50

= 2.6

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(4) (Test-4 Solution) 23rd Oct 2016

cos = 2 21 0.359

1 2.6

14. (b)Hay’s bridge is used for Q > 10

As, Q = 0LR

and, time constant, LR

So, for high time constant coil, Hay’s bridgeis used.

15. (b)Speed of the shaft = 900 rpmi.e. one minute 900 rotation 900 × 60 teeth 900 × 60 pulses

So, in one second 900 60

60

pulses

= 900 pulsesSo, frequency of output pulses = 900 persecond.

16. (b)The pressure coil of a dynamometer typewattmeter is made highly resistive, other-wise it gives error. The error due to in-ductance of pressure coil is equal to

tan tan , where is the power factorof load and is the angle by which pres-sure coil current lags the load voltage.

17. (d)The emf of a Weston standard cell can bemeasured by a potentiometer only becauseat balance the current drawn from the cellis zero.

18. (a)

In a D’arsonval galvanometer, damping torqueis provided by eddy currents in the metalformer on which the coil is mounted.

19. (b)

(1) High torque to weight ratio leads to higher‘accuracy’.

(2) In PMMC jewels are used in bearing toreduce wear and tear.

(3) Swamping resistance is used to negatethe effect of temperature in an ammeter

as it has low thermal emf, hence itprovides thermal stabilisation.

(4)Overdamped galvanometer is sluggish innature. Hence it should be either criticallydamped or slightly under-damped.

20. (b)

V

I Im

R = 0.01 sh

I sh

R = 1 km

For FS, V = 0.5 V

ISh = Sh

V 0.5 50 AR 0.01

Im =

3

3m

V 0.5 0.5 10 AR 1 10

I = Im + Ish

= 50.005 A Ans.

Now if I = 10A (given), using I = Im + Ish

10 = m shI I

10 = 0.5 × 10–3 + Ish

Ish = 10 – 0.5 × 10–3

= 9.9995 A

Now,

Rsh = SH

sh

V 0.5I 9.9995

= 0.05 Ans.

21. (c)

In MI instrument, deflecting torque

Td = 21 dLI

2 d

and controlling torque TC = K

At equilibrium Td = Tc

21 dLI

2 d = K …(1)

For a linear scale,

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I = C (i.e. I ) …(2)

from (1) and (2)

2 21 dLC2 d = K

dLd = 2

2 KC

( = constant)

Hence plot of

dLd constant will be a

rectangular hyperbola.

22. (b)In order to obtain a waveform on a CRO, thewaveform of voltage under test is applied toY-plates and a voltage obtained from asawtooth generator is applied to the X-plates.Deflection is proportional to the voltageapplied to the deflection plates. The verticaldeflection is proportional to the voltageapplied to the Y-plate at any instant andhorizontal deflection is proportional to thevoltage applied to X-plates at any instant.

23. (b)

X-signal = V sin tY-signal = V cos t

= V sin 90 t

= V sin t 90

= V sin t 90

So the phase difference between X-signal andY-signal is 90°. So, the resulting Lissajouspattern will be a circle.

24. (d)Since, both the input are same. So there isno phase displacement between them andhence a straight line will appear on screenof CRO.

25. (b)

Since, y

x

ff =

y

x

23

y = x23

= 2 60003

= 4000 rad/sec.

Hence, y(t) = 2k sin 4000t 15º

26. (c)

R C

R , Ls

True value of Q = 0

s

LR

Measured value of Q i.e. Q´ = 0

s

LR R

So,QQ´ = 0 s s

s 0 s

L R R R RR L R

Q´Q =

s

RR R

Q´ Q

Q

= s s

s s

R R R RR R R R

27. (c)Since, the diameter of screen is 10 cm. Solength of waveform over one cycle should beequal or, less than 10 cm.

Setting of CRO = 1 sec cm

So, for 10 cm, 10 sec is required.i.e. the time period of waveform should beless than or equal to 10 sec .

T 10 sec

f = 61 1T 10 10

f 100 kHz

28. (d)For Q-meter

f = d

12 L C C

where C is tunning capacitance and Cd isdistributed capacitance.

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(6) (Test-4 Solut ion) 23rd Oct 2016

Given, f1 = 1 d

12 L C C ...(1)

and 2f1 = 2 d

12 L C C ...(2)

So,1

1

f2f =

2 d

1 d

C CC C

2 d

1 d

C CC C =

14

Cd = 1 2C 4C3

29. (c)The relation between nodes, branches andlinks is

l = b n 1

l = 7 – (4 – 1)= 7 – 3= 4

No. of links l = 4

30. (c)The given circuit is parallel connected, so

1

Z s = 1 1 111 s 1s

= 1 1 s1 s 1 s

=

22s s 1s 1 s

Z(s) = 2

s 1 s2s s 1

31. (b)Steady state voltage means voltage as t .

CV = Cs 0LimsV s

= s 0

s 5s 10Lims s 2

= 102

= 5V

32. (a)

Z(s) =

K s 3

s s1 j 1 j

= 2K s 3

s 2s 2

Given Z(0) = 5 = 3K2

K = 3.33

33. (d)For an immittance function, poles and zerosmust interlace on negative real axis. In thegiven function, poles and zeros are on thenegative real axis but they are not interlacedso none is an immittance function.

34. (b)Quality factor in terms of inductor, capacitorand resistance values is

Q0 = CRL

200 =6

3 1010L

L = 25 H

35. (c)

Band width (BW) = 2 1

= 2 K rad sec82 72

= 2 10K rad sec

= 20 K rad sec

BW = 1RC

C = 3 311

R BW 1 10 20 10

= 1 F20

36. (a)In Cauer first form L is always in series andC is always in shunt. 1Z , L is first element

2Z 0, C is first element

37. (a)

When 3 8s 0 Z 8s 3

When S = 0 network is,

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R3

1Z(s)

8 = 3 + R + 1R = 4

38. (b)In a dual circuit, the elements are replacedasR with GL with CV with IParallel with seriesMesh with node

39. (c)Statement 3 is wrong. The entity nearest toorigin is a zero and entity that is nearest toinfinity is a pole.

40. (a)Hurwitz polynomial has poles only in the lefthalf of the s-plane.

41. (a)These are the various forms of networkrealization.

42. (b)

43. (c)On the application of the field E, the modifiedfield due to polarization P is given by

EL =0

PE

For solids and liquid having cubic symmetry,

= 13

So, EL =0

PE3

44. (b)

As, Loss tangent, ´´r´r

tan

So, ´´r = ´

r.tan

= 2.4 × 15 × 10–4

= 3.6 × 10–3

45. (d)

46. (d)Some examples of ferroelectric materials:Rochelle salt : NaKC4H4O6 4H2OKDP (Potassium : KH2PO4dihydrogen phosphate)ADP (Ammonium : NH4H2PO4dihydrogen phosphate)Barium Titanate : BaTiO3Calcium Titanate : CaTiO3Ammonium iron alum : NH4Fe(SO4)2.12H2O

47. (a)Energy stored in polarizing the dielectric

W = 2 20

1 1CV C V2 2

where, C is capacitance after polarization,& C0 is capacitance before polarization

W =

20 r 0A A1 V2 d d

= 20 r

1 A V 12 d

=

2

0 r1 VAd 12 d

Energy stored per unit volume

WAd = 2

0 r1 E 12

Energy storedVolume = 1 EP

2

48. (c)Only orientational polarization depends upontemperature, as

P0 =2pNp

E3KT

49. (d)Since, Loss tangent

tan =

"r'r

"r = '

r tan= 2.5 × 0.004= 0.0100

As, complex relative dielectric constant is,

*r = ' "

r rj= 2.5 – j 0.01

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50. (d)Total polarization of a polyatomic gas,

P = Pe + Pi + Po

0 eE =2p

e iNp E

N E N E3KT

0 r 1 E =2p

e ip

N E3KT

0 r

y

1=

2p

e ic xm

Np 1N3K T

i.e. y = mx + cThis is an equation of straight line, having

slope m = 2pNp

3K

51. (c)Face centred cubic :

a

r

rr

r

where r = radius of an atomIn face-centred cubic lattice, atoms arelocated at each corner and at the centre ofeach face of unit cell.So, 4.r = a 2

r = a 24

= a

2 2

52. (a)

2aSimple cubic structure

Atom is placed at the corner of each cubicshell.

APF = Volume of atom in one cubic shell

Volume of unit shell

=

3

3

1 48 r8 3

a

= 0.526

where r = radius of atoma = side of unit shellHere a = 2r

53. (c)

Diamond Cubic :Diamond cubic unit cell having contribution of18 atoms.i) 8-atoms are corner atomii) 6-atoms are face-centred atomiii)4-atoms are completely inside

No. of atoms per unit cell

= 1 18 6 48 2

= 8

54. (c)

For dielectric material,

D = 0E P

0 rE = 0E P

P = 0 r E1

0 eE = 0 r E1

e = r 1

55. (d)

• All ferroelectric materials are piezoelectricand also, it has pyroelectric effect after acertain or critical temperature.

Ferro electric

Pyro-electric

Piezo-electric

• Ferroelectric materials are characterized byparallel alignment of dipoles. Due to paralleldipole arrangement, these materials arehav ing large amount of spontaneouspolarization.

• Ferroelectric material posses centre ofsymmetry.

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56. (a)

Bandgap energy (Eg) of different materials isgiven below :

Metals : gE 0eV as both the conduction band

and valance band overlapInsulators : Eg > 3 eV

Semiconductors: gE 1 eV

57. (a)

The property of material by which material getspolarized in the direction of external field andremain polarized even after removal of externalfield is called spontaneous polarization. Amongdielectric materials only ferroelectric andpyroelectric materials show spontaneouspolarization; But ferroelectric materials havestrong spontaneous polarization as comparedto pyroelectric materials.

Material Ferroelectric Para electric

SpontaneousPolarization

Strong Zero Zero

Pyroelectric

Weak

58. (c)

Elements lose (or gain) and share electrons toform ionic and covalent bonds respectively andthese bonds are stable bonds.Conductivity in ionic crystals depends oni) Carrier concentration

ii) Carrier mobility, : This in turn, dependson temperature and is given by the Arrenhius

expression 0 aexp E kT .

Where Ea = activation energyT = Temperaturek = Boltzmann’s constant

59. (b)

The relation between dielectric loss andfrequency is

P f

2 r

rP v 2 fC

so as frequency is doubled, the power loss isdoubled.

60. (c)

There is always dielectric losses occurring in acondenser containing lossy dielectric and thislosses may be accounted as ohmic loss

occurring in resistance R. So, the condensercan be represented as

R C

61. (b)

Steps to find Miller indices :i) Find the intercepts of the plane on the three

crystal axes (OX, OY, OZ) as (pa, qb, rc);where a, b, c are corresponding primitivesand p, q, r be integer.

ii) Write the reciprocal of the number p, q, r as

1 1 1, ,p q r .

iii)Find the LCM of their denominator.iv)Multiply the reciprocals by the LCM, to get

Miller Indices.Herei) Intercepts are (4, 1, 2)

ii) reciprocals : 1 1, 1,4 2

iii)LCM of 4, 1, 2 is 4

iv)Miller indices are 1 14, 1 4, 4 1,4,24 2

62. (a)

Crystal system Unit cell dimension Angle between AxisOrthorhombic a b c 90

Triclinic a b c 90Hexagonal a b c 90 , 120

Cubic a b c 90

63. (b)Kinetic energy of electron = eV

= 19 31.6 10 2 10 J= 163.2 10 J

64. (d)For body centered cubic lattice;

a

3a

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The atomic radius; 4r 3a

r = 3a4

Number of atoms per unit volume

=1 188

= 2Atomic packing factor :

=3

3

42 r3a

=38 3a

3 4

= 0.68 i.e. 68%

65. (b)Dielectric loss in dielectric,

t = ´´ 2 30 r 0

1 E Watt m2

So, dielectric loss depends on the imaginarypart of complex dielectric constant.

66. (d)The dipole moment per unit volume ispolarization, P

D = 0E P

0 rE = 0E P

P = 0 r E1

67. (a)

68. (a)Since, polarization, P = Np 0 eE = N E

e =0

N

r 1 =0

N

r =0

N1

69. (c)The electronic polarizability

e = 304 R

70. (a)

G(s) = C (s)R (s)

C (s) = R(s) G(s)

Consider impulse input (t)

r(t) = (t)

L [r(t)] = 1 = R(s)For impulse input R(s) = 1C(s) = G(s)

71. (d)g(t) = e–3t, r(t) = e–4t

G(s) = L(e–3t) = 1

s 3

R(s) = L (r(t)) = L (e–4t) = 1

s 4C(s) = G(s) R(s)

=1 1 1

(s 3) (s 4) s 3 s 4

C(t) = L–1 [C(s)] = e–3t – e–4t

C(t) = e–3t – e–4t

72. (d)• Transfer function can be defined for linear

system only. for non-linear system toolslike state space representation may beused.

• Transfer function is a property of thesystem which doesn’t depends upon theinput

73. (c)Taking laplace transform of the equations3Y(s) + 3 s2 Y(s) + 2 s Y(s) + Y(s)= s2 × (s) + 2s × (s) + 3 × (s)

Y (s)X(s) =

2

3 2s 2s 3

s 3s 2s 1

Transfer function

G(s) = 2

3 2Y(s) s 2s 3X (s) (s 3s 2s 1)

74. (c)Transfer function = Laplace transform ofimpulse responseg(t) = e–t (1 – cos 3t)= e–t – e–t cos 3tG(s) = L [g(t)] = L [e–t – e–t cos 3t ]

= 2 21 (s 1)

(s 1) (s 1) (3)

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= 21 (s 1)

(s 1) s 2s 1 9

= 2 2

2s 2s 10 s 2s 1

(s 1) (s 2s 10)

= 29

(s 1) (s 2s 10)

75. (b)

a C (s) A (s)R (s) 1 B (s) A(s)

b R(s) A(s) + R(s) B(s) = C(s)

C (s)R (s) = A(s) + B(s)

c R(s) A(s) – C(s) B(s) = C(s)

C (s)R (s) =

A (s)1 B (s)

d C (s)R (s) =

A (s)1 B (s)

76. (b)R(s) G1(s) – R(s) G2(s) – R(s) G3(s) = C(s)

C (s)R (s) = G1(s) – G2(s) – G3(s)

G (s) – 1 G (s) – G (s)2 3

R(s) C(s)

77. (b)It is a case of shifting of take off pointbefore the summing pointHere Z(s) = R(s) Y(s)Also in original case X(s) = R(s) Y(s)Z(s) = X(s)

(b) is an equivalent form* In case of shifting of take-off point after

summing point

+–

R(s) C(s)

Y(s)X(s)

+

+–

R(s)

C(s)

Y(s)Z(s)

+

+–

78. (c)G1 and G2 are in cascade and theirequivalent is in parallel with G3(s)= G1 G2 + G3

G G + G1 2 3+R(s)

C(s)

H(s)

C (s)R (s) =

1 2 3

1 2 3

G G G1 (G G G ) H

79. (d) R (s) = 0

3

3 1

G1 G H

++

H1

G1 G2

X(s)

C(s)

80. (c)

s + 2+

C(s)1

s (s 10) +

N(s)

s 2[C (s) N (s)]s (s 10)

= C(s)

N(s). s 2

s(s 10) =

s 21C(s)s (s 10)

= 2s 9s 2C(s)s (s 10)

C (s)N (s) = 2

s 2(s 9s 2)

81. (d)For block diagram (ii)

HR (s) R(s)s 3

= C(s)

HR (s) 1s 3

= C(s)

C (s)R (s) =

H1s 3

For (i)C (s)R (s) =

ss 1

For (i) and (ii) to be equivalent

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ss 1

= H1

s 3

H = (s 3)s 1

82. (b)

1 2

1 2 1

G G1 G G H

+G G1 2 –

H1

G3

G4

G5

–+

+G3 – G + G4 5

1 2

1 2 1

G G1 G G H

G3 – G + G4 5 G6

C(s)R(s)

H2

1 23 4 5

1 2 1

1 23 4 5

1 2 1

G GG G G

1 G G HG G

1 G G G1 G G H

+–

1 2 3 4 5

1 2 1 1 2 3 4 5

(G G ) (G G G )1 G G H G G (G G G )

G6

C(s)

C(s)R(s) = 1 2 6 3 4 5

1 2 1 1 2 3 4 5

(G G G ) (G G G )1 G G H G G (G G G )

83. (c)For block diagram

C(s)R(s) =

1 ss(s 1) (s 1)

1 s61s(s 1) s 1s

=

2

2

1(s 1)

61s(s 1)

= 2s

s 2s 5

84. (d)R(s). G1G2 + R(s) G3 = C(s)

C(s)R(s) = G1G2 + G3

85. (d)In first case we are shifting a summingpoint after the block

X1G–

+

(X1– X )2 (X1 – X )G2

X2

X1

G

+–

X2 G

1G

1

1

XG

12

X X G.G

X2

So not equivalent. Actual equivalent is

X1 G+–

X2

(X – X ) G1 2

G

For second diagram it is moving take offpoint after block

X1 G X G1

X1

X1 X G1/

X1

1G

G

Not equivalent.Actual equivalent is

X1 X G1

X1

1G

G

86. (d)Regenerative feedback implies positivefeedback

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R(s) ++

CG(s)

H(s)

[C(s) H(s) + R(s)] G(s) = C(s)

C(s)R(s) =

G(s)1 G(s) H(s)

• It may lead to unbounded response orunstability in system

• eg in OPAMP circuit below feedback ispositive

+–Vin Rin

RF

Vout

87. (c)For an ideal OPAMP

V+ = V–

–+

R

1Cs

V (t)1 V (t)2

+

–

Nodal equation for

1 2O V s O V sR 1 Cs

= 0

1V sR

= +CsV2(s)

2

1

V sV s =

1RCs

88. (d)Closed loop gain for positive feedback

=

G s1 G s H s

G(s) = 163

= 2

H(s) =16

CR

=2 3

11 26

89. (b)

1

1

C50,

R

2

2

C 10 10R 1 10 11

10% reduction in system 1 = 0.1×50 = 510% reduction in forward path of system2 = 0.1×10 = 1Forward path gain in first system now= 50 – 5 = 45;Forward path gain for system 2 = 10 – 1= 9Transfer function of system 1 = 45Transfer function of system

2 = 9 9

1 9 1 10

% reduction in transfer function of system

1 = 50 45 100 1%

50

% reduction in transfer of system

2 =

10 911 10 100 10%

1011

K1 = 10, K2 = 1K1 – K2 = 9

90. (c)Through variable passes through theelements and has the same value at inputand output of the element e.g., current ina resistor.

elementOutput

iiInput

91. (d)Variables that appear across the twoterminals of an element e.g. voltage acrossa resistor.

VOutputInput

92. (b)With given values block diagram

+–

Y(s)10 2s 50

s 6s 500

1R s

s

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Y sR s =

2

2

10 s 50s 6s 500

10 s 501s 6s 500

Y sR s =

210 s 50

s 16s 1000

Y(s) =

2

10 s 50s s 16s 1000

Final value theorem

yss = t s 0lim y t lim sY s

=

2s 0

s 10 s 50lims s 16s 1000

=12

93. (b)

10s 5

1s

5

Y(s)R(s)+

–

+

+

Loop1Loop2

Forward path P1

No non-touching loop.Forward path gain

P1 = 10

s s 5

L1 = 10 5s 5

, L2 =

10s s 5

= 1 – (L1+L2), 1 = 1

By mason gain formula t(s) = 1 1P

= 50 101

s 5 s s 5

=

2s 5s 50s 10s s 5

T(s) =

2

10s s 5

s 55s 10s s 5

= 210

s 55s 10

94. (b)Tachometer converts mechanical energyinto electrical energy. It works as voltagegenerator with output voltage proportionalto angular velocity

et ddt

et(t) = tdKdt

Et(s) = tK s s

Et ss = Kt(s)

95. (a)• Transfer function of DC motor

s

E s

= m

m

Ks sC 1

1s implies DC motor is integrating device

• DC motor has a built in feedback loopbecause of by back emf. It tends toimprove the stability of the motor.

96. (a)for given network,

Z(s) = 1R SL

SC1R SL

SC

=

2

S R L1R 1C S sL LC

...(1)

from the pole-zero diagram

Z(s) =

K S 2

S 1 2j S 1 2j

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=

2

K S 2s 2s 5

...(2)

Z(jo) = K 2 1 K 5 2

5

Comparing equation (1) & (2), we get

1 5 2C 0.4FC 2 5

1 1 15 L 0.5H2LC 5C 55

R 2 R 2L 2 0.5 1L

97. (c)from the pole-zero diagram

zeros : s = 0, –2

poles : s = –1, –3

So, Z(s) =

K S S 2S 1 S 3

=

2

2

2K S 1s

1 3S 1 15 s

Z ( ) =K 1 2 (Given)1 1

K = 2

Hence, Z(s) =

2S S 2S 1 S 3

98. (b)

G(s) = 1

s 2

G j = 1

2 j

For 3-dB frequency 1G j2

of

maximum megnitude.

Here maxG j = 12

at 0

cG j = c

12 j =

1 122

1 22c

1

4 =

12 2

Solving c = 2 rad/sec.

Close loop system T(s) =

11s 2

1 s 31s 2

maxT j = 13 at 0

1cT j = 1c

1 1 1j 3 32

1 22c1

1

9 =

13 2

2c1 9 = 18

1c = 3 rad/sec.

• For a negative feedback system,bandwidth increases vis-a-vis open loopsystem.

99. (c)Overall transfer function of inner loop =

1

G s1 G s G s

+–

C(s)

1

G s1 G s G sR(s)

C sR s =

1

1

G s1 G s G s

G s11 G s G s

=

1

G s1 G s G s G s

G1(s) = 1

C sR s =

G s

1 G s G s = G(s)

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100. (a)

To find

C sD s , Put R(s)=0, Block diagram

reduces to

G CK+

–D

Here forward path gain = GFeedback = K

C s GD s 1 KG

101. (d)There are two forward pathP1 = 2×3×4 = 24P2 = 5There are four loops with a pair of non-touching loop.L1 = –2L2 = –3L3 = –4L4 = (5)(–1)(–1)(–1) = –5L1, L3 = non touching loop

= 1 – (–2 –3 –4 –5) + (–2 × (–4))= 1 + 2 + 3 + 4 + 5 + 8 = 23

1 1, 2 1, 3 4

T(s) = 1 1 2 2P P

= 24 1 5 4 44 1.9

23 23

102. (a)Type of a system Nos. of poles atoriginOrder of system Degree ofdenominator polynomial or Nos. of totalpoles.A Type 1, Order secondB Type 0, Order secondC Type 2, Order threeD Type 0, Order three

103. (d)• Feedback introduces the posibility of

instability as it controls the dynamics ofsystem by adjusting the poles.

• Feedback reduces sensitivity to parameter

variation, improves transient response,minimises the effect of disturbance signal.

• It increases band width of signal.• Add more components and make it more

complex.• Reduces the gain of the system.

104. (d)

•XR

ratio is kept low so that maximum

torque is obtained at start.Torque

RotorSpeedSyn.

Speed

XR

• Drag cup design is to keep the inertia lowerand obtain good accelerating characterstic.

105. (b)• It is a two phase induction motor having

two winding at stator oriented 90º to eachother.

106. (b)In servo application, velocity feedback isused to improve stability or dampling ofclosed loop system.

107. (c)Voltage is proportional to angular velocity e t t

e(t) = tK t = tdKdt

Laplace transformE(s) = tK s s

Transfer function

E ss

= Kts

108. (d)

H(s) = 20.8

s s 2

= 20.8

s 2s s 2 =

0.8s 1 s 2

One pole lie in RHS of s-plane.So output would be unbounded of form

y(t) = K1et + K2e

–2t

109. (a)

The capacitor is employed in main winding a

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auxiliary or both to produce required anglebetween store field s and rotor field V asbefore capacitor insertion both no fields arein phase followed by zero starting torque.

110. (c)

Ohmic losses are s Pairgap

Output power = Pm = (1 – s)Paigap

Efficiency =

airgap

airgap

P1 s1 sP

The value near to (1 – s) is 1 s1 s

111. (a)Speed of rotor flux = ns (at stand still)and speed of rotor flux at nr is sns. Thenw.r.t. stator speed of rotor field is

s s shr sh n sh n1 s

112. (b)

agP = LossP 300 7500WS 0.04

s = 120f 120 50P 4

= 1500Vpm

Td = agP 75002s 150060

= 47.75 Nm

113. (a)Slip at which maximum torque occurs isgiven as

Sb = 2

2 2

R 0.03 0.03 0.060x x 0.5

hs = 120 50 250r.p.m.24

Rotor speed at maximum torque = ns(1 – sb)

= 250 235 r.p.m.1 0.06

114. (c)When the speed of synchronous motor is morethan synchronous speed, then the motordamper winding will act like inductiongenerator.

115. (a)

qX = min

max

V 96 8I 12

dX = max

min

V 108 10.8I 10

116. (b)Ns is same for both

1

1

120 fP

= 2

2

120 fP

120 2510 =

2

120 60P

P2 = 24

117. (c)

KVA

KW

KVAR

KVAR = KVA sin

= 12400 sin cos 0.69

= 1825 KVA

118. (b)

Open circuit emf f

Neglecting saturation If

1

2

VV = 1

2

f 1

f 2

I fI f

V2 = 2

1

f 21

f 1

I fV

I f

= 2 25 3203.2 50

= 100V

119. (c)Electric field at a distance ‘r’ from an infiniteline charge of density is

E =02 r

=0 0

4 12 2

120. (d)Magnetic field experienced by the loop due

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to infinite conductor.

B = 0 2I2 r

where r is the distance from infiniteconductor and force, l lF i B i Bsin

a

a

a

F1

F2F3 F4

I1

I2

B

here, 1F = 0 1 2I I a2 a

[away from conductor]

and 3F = 0 1 2I I a

2 2a

[towards the conductor]

Since, 2F and 4F are equal in magnitudeand cancel each other.

So, NetF

= 0 1 21 3

I IF F4

[away from conductor]

121. (b)According to Ampere’s law, the line integralof H

and ld over a closed path is equal to

the current enclosed in that loop.

i.e. lH.d

= Inclosed= –10A

Here, negative sign appears as the dotproduct of H

due to current enclosed by loop

and ld is negative.

122. (b)According to Biot-Savart Law.

dH = l

2I d

r

lId

90

So, magnetic field intensity at the centre ofsemicircle.

So, dH = l2

id sin90R

dH = l2

i.dR

So, H = l2idH d

R

= 2i R

R

= iR

123. (d)The electric field intensity due to a chargedplane,

E = n

0a

2

where is the uniform charge density.So, the electric field intensity E

at origindue to a change sheet placed at Z = 10 m.

0,0,0,E

= z0

a2

= 9

z8

20 10 a12 10360

= zˆ360 a

= zˆ360 a V m

124. (c)

Let ia be the current through 5 resistor.Va = 5ia

a aa a0.8V 2 5i 2.5i 16 i 10

a4i = a a4.5i 5i 57

a5.5i = 57

Va = 57 5 51.8V5.5

125. (a)By using thevenin theorem at left and rightside, and by replacing with thereninequivalent.

+–+–

1 2 2 2

4 12VV1

+

–5V

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1 11 V 5 V 12V 04 3 4

V1 = 5.6 V

126. (b)

Vin = L C

22 22X XR V VV

= 223 14 10

= 2 23 4= 5

127. (b)The circuit can be redrawn as

Req

5 20

105

5

25

15

15

eqR = 10 30 30 30 = 5

128. (b)When potential difference between B and Dis zero, wheatstone bridge is balanced.

1 1 2 26 6 4 x1 1 2 2

1122

= (4 + x)

x = 2

129. (d)Resistance depends on resistivity of thematerial and also on the physical dimensionsof the resistor. In the above problem physicaldimensions are not given, so it is datainsufficient.

130. (a)

For the given Circuit,

thR =2 x 1 22 1 3

So time constant,

=2CR C sec3

Then, V t = t

V 1 e

=

t1 x 3 1 e1 2

= ti e

at t = t0 oV t = ot1 e

=2

toe

= 1

Now, dv t

dt=

ot t /1 1e et

odv tdt

=ot1 1.e 10

t

=1 2 C

10 3

110

C =3 0.15F

20

131. (d)

At t = 0–, switch is open,

1

10V

1H

i t

iL(0–) = 0

iL(0+) = L t 0i 0i0 t

[ Inductor does not allow sudden change ofcurrent]

At t switch is closed and inductor shortcircuit,

1

10V

i

i =10 10A1

Time constant of R-L circuit, L 1 1secR 1

i(t) = t /i ei i0

= t10 e0 10

i(t) = t10 1 eat t = 1 sec.

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i(1) = 110 6.32A1 e

132. (b)

Let individual time constant are 1 2&

1 21 2

1 2

L LandR R

After series combineation,

eq = 1 2L L

and eqR = 1 2R R

equivalent time constant

eq =

1 2

1 2

L LR R

Given,

1 2

1 2

L LR R

=

1 2

1 2

L LR R

2

=1 2 1 2

1 2

L R R L2R R

1 2 1 1 2 22R R L 2R R L

= 2 21 1 2 1 2 1 2 2 1 2L R R L R R R L R L

1 2 1 2 2 1 2 1R L R R R L R R 0

1 2 1 2 2 1R R R L R L = 0

1 2R R 0 i.e. 1 2R R

1 2 2 1R L R L 0 1 1

2 2

R Li.e.R L

133. (b)

Z(s) =1sL R

sC

I s =

V sZ s

=

22

V s .sC sC. V ssRC Is LC sRC 1 LC sLC LC

Characteristic equation is

2 R 1s sL LC

= 0

Comparing the equation with2 2

n ns 2 s = 0

n n1 ; 2Lc

=RL

=n

R 1 R 1. . LCL 2 2 L

=R C.2 L

for system to be underdamped 1

i.eR C 12 L

LR 2.C

134. (d)

At t 0 , Capacitor acts as short circuitand inductor acts as open-circuit

R

V

I

I =VR

At t capacitor acts as open-circuit andinductor acts as short circuit.

R

V

I

I =VR

135. (c)

Thevenin’s equivalent of the circuit is.

10V V 1 F

1M

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So V t = toV 1 e

where oV = 10V,

= 6 61x 10 x 1x 10 1s

dv t

dt=

toV .e

= toV e as 1 s

t 1s

dv tdt = 1

0V e

= 01V . 10 x 0.367e

= Vsec.3.67

136. (a)

System transfer function,

H s =2s

s 1

H j =2.jj 1

H j = 2

2.

1

& H j = o 190 tan

for input sint 1rad / s

H j =2 12

& H j = 90 – 1tan 1

= 90o – 45o = 45o

Steady state response

= 1. H j sin t H j

= o2 sin t 45

137. (c)

1K2KV

1 2C CV V

At t 0 , combined voltage across two

capacitors 1 2c cV V =1 1V V

1 2 3

Charge must be equal when capacitor are

connnected in series.

q = 1 21 C 2 cC V C V

for equivalent capacitance

q = 2 2

1 2C C

1 2

C CV V

C C

=1 2

1 2

C C V.C C 3

1C1

1VC

1C. 2

1 2

.C V 2 V 2V. . voltsC C 3 3 3 9

& 2

1 2C

2 1 2

C .C1 V 1 V VV . . . voltsC C C 3 3 3 9

138. (a)

f =eq

12 L C

Leq = 1 2L L 2M = 4 mH

f = 6 31

2 0.1 10 4 10

= 50 KHz2

139. (c)

fr =1

2 LCI =

L L

V VX 2 f L

L rHere, f f

I =1/2CV

L

140. (b)A resonance XL = XC

Current = VR

= 1045

= 0.22

141. (c)

given i y0E E cos( t z)a

93 10 rad/s

10

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As nothing is mentioned about medium,dielectric etc it is free space.

i t rE E E

t r( E E )

i t tE E E

i tE 2E

t i1E E2

yt 01E E cos( t z)a2

142. (d)

We know skin depth

1 ,f

1 25 cm, 2 ?

1 2 1 2f ,f , , 1 2,

1f 1MHz 2f 4MHz

1 2

2 1

ff

1

2

41

1

24

1

22

12

25cm 12.50cm2 2

143. (c)

given 3b(t 2t)m

3 3(t 2t) 10

N 100

NdEdt

d| E | Ndt

3 3d100 (t 2t) 10dt

3 3t 4100 (3t 2) | 10

3100 (3 16 2) 10

3100 (48 2) 10

3100 46 10

146 10

= 4.6 V

144. (a)

We know that

C OC SCZ (Z )(Z )

given CZ 50

OCZ (30 j40)

C scZ (30 j40)(Z )

sc(50) (30 j40)Z

2sc(50) (30 j40)Z

sc2500 (30 j40)Z

30 j40 (30 j40)

2500 (30 j40)2500

(30 j40)

145. (d)

We know that

conduction current density cJ E

displacement current density D(J ) (D)t

Et

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E

given 210 m r 4

C DJ J

E E

E 2 f E

E2 E

f2

2

0 r

102

2

9

1012 10 4

36

2

9

102 109

79 102

74.5 10

645 10 Hz

146. (c)We know Maxwell’s equation for static field

H J

We also know 0B H

0

BH

Substituting H in the equation we get

0

B J

0B J

147. (a)

We have CZ (30 j40)

LZ (30 j40)

transmission co-efficient L

tL 0

2Z(T )Z Z

2 (30 j40)60 j80

2 (30 j40) (60 j80)100

=3 4j (60 j80)5 5

3 468 805 5

3 4j80 j605 5

48 64 j48 j48

= 100

148. (b)

From figure maxv 4 minv 1

max

min

| V | 4VSWR 4| V | 1

Reflection co-efficient = 1 VSWR1 VSWR

1 41 4

53

0.6 180

(0.6)

L 0 L

L 0 L

Z Z Z 50 0.6Z Z Z 50

L L0.6Z 30 Z 50

L0.4Z 80

LZ 800 / 4 200

149. (c)Statement-I is false

q

dl

(equipotential)E.dl 0E.dl V

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because when electromagnetic wave isincident on a dielectric boundary it ispartially reflected and partially refracted.Statement-II is true.

150. (d)We know that

.( A) 0 (divergence of curl)

( A) 0 (Curl of gradient)

151. (c)

S

E.dg (net f lux managing f rom

surface)

152. (b)

It is polarisation.

153. (d)

We know 1VSWR1

= 1 0.8 1.8 18 91 0.8 0.2 2

154. (b)

SWR = 1 | |1 | |

where reflection co-efficient

| | = magnitude of reflection co-efficient

L 0

L 0

Z ZZ Z

where LZ and 0Z are complex quantities

jQe

where is vector quanrtity

155. (d)Synchronous impedance method is calledPessimistic method as it gives higher valuethan normal.

156. (a)

Angle of asymptote 1

P ZIf P = no. of poles increases the angle reduces.Hence root locus tend towards right of s-plane.

157. (c)For impedance matching, short-circuited stubis preferred over open-circuited stub becausein short-circuited stub adjustment of lengthis practically convenient. Open circuit stubis liable to radiate electromagnetic power.

158. (c)For a loss-less line, attenuation constant iszero.and, propagation constant,

= j

= j 0

So, =

159. (a)There are four contributors to totalpolarization of a material. These contributorsare electronic, ionic, orientational and spacecharge polarization. At optical frequencies(upto 1016 Hz), only electrons can respond toalternating electric field as ions andmolecular dipoles cannot respond tofrequencies more than 1013 Hz and 108 Hz,respectively.

160. (b)

A is true

because PV

0 0

7 9

114 10 10

36

83 10 m / s R is true

( A) 0

but R is not explanation of A.

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