ANSWERS - IES Master · R (Test - 06)-29 October 2017 (3) Mobile : E-mail: 8010009955, 9711853908 [email protected], [email protected] Regd. office : Phone : F-126,

ESE-2018 PRELIMS TEST SERIESDate: 29thOctober, 2017

ANSWERS

1. (a)

2. (a)

3. (c)

4. (b)

5. (c)

6. (b)

7. (a)

8. (a)

9. (c)

10. (a)

11. (a)

12. (b)

13. (d)

14. (c)

15. (a)

16. (b)

17. (d)

18. (d)

19. (a)

20. (a)

21. (b)

22. (d)

23. (c)

24. (b)

25. (a)

26. (a)

27. (a)

28. (c)

29. (c)

30. (c)

31. (a)

32. (a)

33. (d)

34. (a)

35. (a)

36. (d)

37. (a)

38. (c)

39. (b)

40. (d)

41. (b)

42. (b)

43. (d)

44. (b)

45. (d)

46. (c)

47. (d)

48. (a)

49. (a)

50. (b)

51. (c)

52. (c)

53. (a)

54. (d)

55. (b)

56. (c)

57. (c)

58. (a)

59. (c)

60. (b)

61. (d)

62. (c)

63. (c)

64. (b)

65. (a)

66. (c)

67. (d)

68. (c)

69. (c)

70. (d)

71. (c)

72. (b)

73. (d)

74. (a)

75. (b)

76. (d)

77. (d)

78. (a)

79. (a)

80. (b)

81. (c)

82. (c)

83. (c)

84. (b)

85. (b)

86. (d)

87. (d)

88. (a)

89. (c)

90. (c)

91. (b)

92. (d)

93. (c)

94. (b)

95. (a)

96. (d)

97. (a)

98. (a)

99. (c)

100. (b)

101. (c)

102. (d)

103. (b)

104. (c)

105. (a)

106. (d)

107. (a)

108. (a)

109. (c)

110. (c)

111. (d)

112. (a)

113. (a)

114. (c)

115. (c)

116. (b)

117. (c)

118. (c)

119. (c)

120. (a)

121. (a)

122. (c)

123. (d)

124. (d)

125. (c)

126. (d)

127. (c)

128. (a)

129. (b)

130. (a)

131. (c)

132. (b)

133. (c)

134. (c)

135. (b)

136. (b)

137. (b)

138. (b)

139. (d)

140. (b)

141. (a)

142. (d)

143. (c)

144. (c)

145. (c)

146. (c)

147. (d)

148. (b)

149. (d)

150. (a)

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1. (a)

Bandgap energy (Eg) of different materials is givenbelow :

Metals : gE 0eV as both the conduction band

and valance band overlapInsulators : Eg > 3 eV

Semiconductors: gE 1 eV

2. (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 compared topyroelectric materials.

Material Ferroelectric Para electric

SpontaneousPolarization

Strong Zero Zero

Pyroelectric

Weak

3. (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, depends ontemperature and is given by the Arrenhius

expression 0 aexp E kT .

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

4. (b)The relation between dielectric loss and frequencyis

P f 2 r

rP v 2 fC

so as frequency is doubled, the power loss isdoubled.

5. (c)

There is always dielectric losses occurring in acondenser containing lossy dielectric and thislosses may be accounted as ohmic lossoccurring in resistance R. So, the condensercan be represented as

R C

6. (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

7. (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

8. (a)

Magnetic material

Magnetic susceptibility

DiamagneticParamagneticFerromagneticAntiferromagneticFerrimagnetic

Negative and smallPositive and smallPositive and very largePositive and smallPositive and large

9. (c)

Permanent magnetic materials are difficult todemagnetize. Properties of permanent magneticmaterials :-• High retentivity• High coercivity• High permeability

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• Tall and wide Hysteresis loop

• High saturation magnetization

• High Curie temperature

• High hysteresis loss because of large area ofhysteresis loop

10. (a)

• Paramagnetic material are characterized byrandom arrangement of dipole. Due to randomdipole arrangement the value of spontaneousmagnetization is zero.

• When these materials are placed in a magneticfield, they acquire weak magnetization in samedirection of the field.

11. (a)

Hydrogen molecule is a diamagnetic materialwhere as hydrogen atom is paramagnet.

In principle any system that contains atoms,ions or molecules with unpaired spin can becalled a paramagnet. A system with unpairedspins that do not interact with each other aresystem with minimal interaction. Each hydrogenatom has one non-interacting unpaid electron.

12. (b)

Magnetic susceptibility

= Relative permeability – 1

= 5500 – 1 = 5499

13. (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 i

c xm

Np 1N3K T

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

m = 2pNp

3K

14. (c)Face centred cubic :

a

r

rr

r

where r = radius of an atomIn face-centred cubic lattice, atoms are locatedat each corner and at the centre of each face ofunit cell.

So, 4.r = a 2

r =a 2

4 =

a2 2

15. (a)

We know that

C

D

| J || J |

9 9

101100 10 10

36

= 1010 36

= 1036 10 1

hence it is a good conductor.

16. (b)

17. (d)Ferromagnetic materials obey Curie - Weisslaw,

m = C

Twhere, C is the curie constant and is calledthe paramagnetic curie temperature.Above the curie-temperature ferromagneticmaterials, behaves l ike paramagneticmaterials.

18. (d)

Half-coefficient of semiconductor materials

RH =1

ne

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where n is the carrier-concentration,

A

B

RR

=

21B

21A

n 4 10 4 : 1n 1 10

19. (a)Superconductor shows diamagnetism. Hence,it expels magnetic flux from within its volume.A superconductor becomes a normal metalwhen a critical magnetic field is applied.

Superconductor

(T < T )C

BB = 0

H

HC

0 TC T

Normal StateSuperConductingState

20. (a)At equilibrium,

yqE = x zqV .B

Vx =y

z

EB ....(1)

and, Jx = NqVx .....(2)So, Hall coefficient, RH

= y x

z x x

E V 1B J Nq.V Nq

21. (b)

Hall effect occurs when a transverse magneticfield is applied to a conductor carrying current.Due to the current, a longitudinal electric fieldnormally present in the direction of flow ofcurrent.

22. (d)The conductivity at any given temperature(except at absolute zero) is due to bothelectrons and holes, and it is given by

i = i e hn e

=

e h2

ie h

n em m

where

i = conductivity of intrinsic material

e h, = relaxation time for electrons andholes

e hm ,m = mass of electron and holes

e h, = mobility of electron and holes

23. (c)

• The semiconductor in its purest form iscalled intrinsic semiconductor e.g. Si, Ge

• When some impurity is added to theintrinsic semiconductor, i t is calledextrinsic semiconductor.

• N-type semiconductor and P-typesemiconductor are classified on the basisof type of impurity added.

• If impurity is a donor (or Penta valent) e.g.P, Ar, Sb, etc. it is cal led N-typesemiconductor.

• If impurity is a acceptor (or tri valent) e.g.B, Al, Ga etc.

24. (b)

Let x is the displacement of electron and Eis applied electric field across conductor.

F =

2

e 2d xm eEdt

dxdt =

e

eE drift velocitym

As, mobility,

µ =drift velocity

Electric field intensity

=

e

eEm

E

=

e

em

25. (a)For a superconductor, critical field HC varieswith temperature as,

HC =

2

0C

TH 1T

where, H0 is the critical field at absolute zeroand TC is the transition temperatureSo, at T = TC; HC = 0

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26. (a)A good electric contact should offer lowresistance to flow of current. This reducesthe voltage drop and also heat generated atcontact points.

27. (a)• Superconductors are used for generating

very strong magnetic field.If a superconductor caries a current I, thendeveloped magnetic field

H =I

2 r

where r is the radius of superconductorwire.

• Magnetic bubble memories are built usingferrites.

28. (c)The temperature below which a normalmaterial behaves as a superconductor, iscalled transition temperature of that material.Experimentally, it is found that transitiontemperature of mercury is 4.12 K.

29. (c)For any intrinsic semiconductor, the intrinsicconcentration is directly proportional to T3/2.As conductivity of the instrinsic semiconductoris directly proportional to intrinsic

concentration. So, 3/2T .

30. (c)For the semiconductor material,

The Hall’s coefficient, Rx =

y

x z

E 1J B ne

As, = ne

So, RH =

1 1

ne

31. (a)

Since, Hall coefficient, RH = 1

ne

Where n is number of charge-carrier per unitvolume; m–3.and e is the charge of electron; coulomb. Sounit of RH : (meter)3 (coulomb)–1

So, option (a)

32. (a)

The current required to flowing through a superconductor of radius ‘r’ to produce a criticalmagnetic field Hc is

Ic = c2 r H

= cd2 H2

=

3

322 102 7 107 2

= 22 Ampere

33. (d)

As, fermi level for P-type semiconductor,

EF = VV e

A

nE KT log

N

where,EV = maximum energy of valence band in eV.

nV = no. of electrons in valence band,NA = concentration of acceptor atoms.

i.e.

Conduction band

Fermi level for P-type semiconductor

Valence band

34. (a)

35. (a)According to the Messener’s ef fect,superconductors expels the magnetic fieldflowing through it.So, relative permeabil i ty is zero forsuperconductor.

36. (d)Transformer oil acts as insulator and coolingmedium for transformer cores and chokes. Itsdielectric strength should be high to work as agood insulator. Flash point and viscosity arealso important parameters for transformer oil.Flash point is the lowest temperature at whichit vaporize. For better temperature stability it

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should be high. Viscosity is a measure of itsresistance to gradual deformation by stress soits should also be high for transformer oil.

37. (a)Power loss of a dielectric in alternating electric

field 20 0 r

1 E2

So power loss

38. (c)Temperature has no effect upon ionic polarizabilityand electronic polarizability.

39. (b)Sodium is paramagnetic.

40. (d)Ferrites from magnetic would be in form of

MO.Fe2O3 (M = divalent metal)it can be represented as

AB2O4 (A, B = metals)41. (b)

Neel temperature is the transition temperaturebetween anti-ferromagnetism and para-magnetism behaviour of a material.

42. (b)Bohr magneton is the unit of permanent dipolemoment due to spin of electrons in orbit.

1 Bohr magneton = eh

4 m= 9.27 × 10–24 ampere.m2

where, h = plank’s constant= 6.62 × 10–34 joule sec.

m = mass of electrone = charge of electron

43. (d)

As we know, mean free path, d= v and

31T

44. (b)According to wiedemann – Franz law

=TKThermal conductivity

Electrical conductivity =

22 k .T3 e

where, k = Boltzmann’s constant

e = Electronic charge

T = Temperature

The ratio of thermal conductivity and electricalconductivity is proportional to temperature andthis law holds good above debye temperature.

45. (d)If a super conducting material is placed inmagnetic field (H) it remains superconductorupto a critical field (Hc) above that it turns intonormal state. Similarly it remains superconductorbelow a transition temperature and above thattemperature it converts into normal state.

Magnetic impurities can also contribute towardscritical field.

46. (c)Rochelle’ salt is a piezoelectric crystal, whichis used as piezoelectric transducer.Piezoelectric transducers convert mechanicalmotion to electrical signal.

47. (d)

Diamagnetic material is the only magneticmaterial which has no dipoles.

48. (a)

Vab = ab cd1

d i d iL m

dt dt

= 38 10 5 100 cos t

315 10 5 100 cos t

= 4cos t 7.5cos t volt.

= 11.5cos t volt.

Vcd = cd ab2

d i diL M

dt dt

= 324 10 5 100 cos t 315 10 5 100 cos t

= 12 cos t 7.5cos t

= 22.5cos t volt.

Where 100rad / s.

49. (a)Here, the instantaneous voltage andinstantaneous current are in phase whichindicates that the circuit is in resonance so,the net impedance z = R.

R =m

m

VI

= 75 515

At resonance,

20 =

1LC

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L = 20

1C

= 2 6

1

100 400 10

=14

= 0.25 H

50. (b)

Since the current flowing in the loop is ‘i’,

Hence,

3i – 5 = i

2i = 5

i = 2.5A

So, the voltage across the 5 resistor..

VL = 5i

= 5 × 2.5

= 12.5 volt

51. (c)

From the circuit, we can observe that voltageat point ‘b’ is 20V

bV 20 V

Also, Va–10 = Vb

Va = Vb + 10 = 20 + 10 = 30V

i.e. aV 30 V

and, Vb–5 = Vc

VC = 20 – 5 = 15V

CV 15 V

52. (c)Given circuit is,

5

31

3i4A

i(i–i )L

1

Current through 5 resistor, i 4A

Applying KVL is loop (1), we get

3i + (i – iL) – 3iL = 0 i = iL iL = 4A

So, voltage across 3 resistor = 4 × 3 = 12V

53. (a)

R R

B BY

Y

rr

Let RY YB BRV V V V

If all the resistance are present is loadPower, P = 3 VPh·IPh cos

= 3·V·VR

cos 1, resistive

=23V

RIf one of the resistances of the load gets opencircuited,Power P = 2 VPh IPh cos

= 2 V VR

=22V

R% reduction in power

=

P P 3 2100 100

P 3= 33.33%

54. (d)

In a power measurement of 3 unbalancedload,

1 L L

2 L L

W V I cos (30 )W V I cos (30 )

and,

2 1

2 1

W Wtan 3W W

When, = 0 ie. Pf is unity W1 = W2 > 0When, = 60° ie. pf = 0.5 (lag) W1 = 0 but W2 > 0When, = 90° ie. pf is zero W1 = –0.5 Vl ILand W2 = 0.5 VL ILie. W2 = –W1 or, W1 + W2 = 0.

55. (b)A. For RL circuit voltage loads current by an

angle 1 Ltan

RB. For RC circuit voltage lags current by an

angle

1 1tanRC

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C. For a pure capacitive circuit current leadsvoltage by an angle 2 .

D. For a pure inductive circuit current lags

voltage by an angle 2 .

56. (c)

50V Choke10AHeater

200V, 50Hz

Given : I 10A, VH (drop across heater) = 50VRC ( Resistance of choke) 5Vs = 200V

RH (Resistance of heater) = 50 510

Hence, net resistance of the circuit is,

R = RH + RC = 5 + 5 = 10

HRV (drop across RH) = RH × I = 5× 10 = 50V

CRV = RC × I = 50V

LRV (drop across XL, the inductive reactance

of the choke) = I XL = 10 XL V.

However, the supply voltage being the vectorsum of the drops VR and VL.

V = C HR R LV V jV

(200)2 = 2 2L50 50 V

2LV = 2002 – 1002

VL = 173.21 V

XL =173.21 173.21 17.23

I 10

This gives the impedance of the choke as

Z = 2 2 2 2C LR X 5 (17.32) 18

57. (c)Given voltage source, v = 1+sint

v 2

1

t

we see that, 0 0,i 10v

Here, v = the voltage across the non linearVoltage across the non linear resistor

= v – 1 × i = v – i

i = 10 (v – i)

i = 10 v – 10 i

11i = 10 v

i =10 v11

= 0.91 (1+sint)

58. (a)The circuit shown in question, can be redrawnas,

b

d

2

2

2 2

I

a n c

3

10V

Here, Rbn = Rcn = Rdn =

6 6

6 6 6

= 36 218

Req = [(4) || 5] + 2

= 20 3829 9

Current, =eq

VR

= 10

(38 / 9)

= 90 2.34 A38

59. (c)Given circuit is,

4A +– 2V

5Va

+ –

b1

2

(I–1)

(4–I)

1AI

Current in 2 resistor

= 2 1A2

Applying kVL in loop (i), we get

– 5 + I × 1 = 0 I = 5A

So, current in section a–b,

abΙ = I – 1

= 5 – 1 = 4A

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60. (b)Given circuit is,

6V +– 6A 2

1

(I+6)+

I4

–Let current delivered by voltage source = IApplying kVL in loop-1, we get6 – 4I – 2(I + 6) = 0

6 – 4I – 2I – 12 = 0

6I = –6 I = –1A

So, power delivered by voltage source = 6I

= 6 × (–1)

= –6W

Now, voltage across current source

= 2 ( 6)

= 2 × (–1 + 6)

= 10 volt

So, power delivered by current source

= 6 × 10

= 60 W

61. (d)Since,

cos2t = cos²t – sin²t

= 1 – 2sin²t

sin²t = 1 (1 cos2t)2

L[sin²t] = 2 21 1 s2 s s 2

=2 2

21 s 4 s2 s(s 4)

= 21 42 s(s 4)

= 22

s(s 4)

62. (c)Since, capacitive reactance

Xc =1c

= 1

2 fc

ie. capacitive reactance Xc increases if both Cand f decreases.

63. (c)Given circuit is under resonance, so currentwill be maximum,

Irms = rmsVR

= (10 / 2)

100Average power will be dissipated only inresistor.

So, Pavg = 2rms( ) R

=

210 / 2 100

100

=100

2 100= 0.5 W

= 500 mW

64. (b)Since, current will leave the dot in first inductorand enter the dot in second inductor, hence.equivalent inductance will be,

Leq = L1 + L2 – 2M= 6 + 4 – 2 × 2

= 6 H

So, resonance frequency, 0 = eq

1L C

0 =

1163

= 12

= 0.707 rad/sec.

65. (a)

For RTh

2

2 2a

b Rab = (4) 2

=86

= 1.33

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i.e. RTh = 1.33

For VTh:By transformation theorem, the circuit can beredrawn as,

–

+

20A

2 2a

b

4V

+–

2

I

I =

20 4

4A6

Vab = 2I – 4 = (2 × 4) – 4 = 4V

i.e. VTh = 4V66. (c)

For lattice network, the z-parameters are,

z11 = z22 = 1 2z z2

and z1 = z11 = 2 1z z2

So, z11= z22 =

6 4 5

2

and, z12 = z21 =

6 4 1

2 |z| = z11 z22 – z12 z21

= (5 × 5) – (1 × 1) = 24Then,

h11 =

22

z 24z 5

h21 = 21

22

z 1z 5

67. (d)For T-network,

Z1 Z2

V1 V2

+ +

––

Z2

11 12

21 22

Z ZZ Z

=1 2 3

3 2 3

Z Z ZZ Z Z

=40 R R

R 30 R

Fiven, Z12 = Z21 = 20ie. R = 20So, Z11 = 40+R = 40 20 60

and Z22 = 30 + R = 30 20 50

68. (c)For the given circuit

V1 = (I1 + I2) Za

And, V2 – 2 b 1 2 aI z (I I )z 0

V2 – 2 b 1I z V 0 V1 = V2 – I2 zb

V1 = V2 – I2 × 2 …(1)and V2 = I2 (za + zb) + I1 za

I1 = 2 2 a ba

1 V I z zz

I1 = V2 – 3I2 …(2)For ABCD parameters

1

1

VI

=

2

2

VA BIC D

A BC D

=

1 21 3

69. (c)The frequency at which the voltage across thecapacitor is maximum,

maxcf =

2

21 1 R

2 LC 2Lwhich is below the resonance frequency, f0 =

1

2 LC

The frequency at which the voltage across theinductor is maximum

maxLf =

2 2

1 12 C RLC

2which is above the resonance frequency, f0 =

1

2 LC

70. (d) Quality factor,

Q =resonant frequency

Bandwidth =

0ff

=

6

320 10

100 10 = 200

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71. (c)

72. (b)

11 22 12 21

11 22 12 21

11 22 21 12

21 12

Symmetry Reciprocity1. Z parameter Z Z Z Z2. Y parameter Y Y Y Y3. h parameter h h h h

h h 14. ABCD parameter A D AD BC 1

73. (d)

Transmission Parameters Network ‘b’

=1 a 1 a3 2 3 2

=

1 3a 3a A B9 3a 4 C D

For reciprocity, [AD – BC] = 1

9 13a 4 1 3a 3a

29a 12a 4 = 1

a = 1 or 1/3

74. (a)

For z-parameter, we get

V1 = 1 225I 10I ...(1)

and V2 = 1 2200I 200I ...(2)

For the given network,

V2 = 250I ...(3)

From (1) and (3),

V2 = 21

V200I 5050

V2 = 1100I

100I1 = –50I2 [For eqn. (2)]

–2I1 = I2

V1 = 1 225I 10 2I

[For eqn. (1)]

= 45I1

Zin = 1

1

V45

I

75. (b)

For a passive network, the output power cannotbe greater than the input power.

76. (d)

For Z11 and Z21, put I2 = 0, then

Z11 = 11

1

V ZI

Z21 = 21

1

V ZI

For Z12 and Z2, put I1 = 0, then

Z12 = 11

2

V ZI

Z22 = 21 2

2

V Z ZI

Z-parameters are

1 1

1 1 2

Z ZZ Z Z

77. (d)

A

B

CinCin

Cin = in

in

C CC

C C

CAB = in1 5C C

278. (a)

When in series, we have to add the Z-parametersto combine the two, two network.

79. (a)

L-C network has only energy storage element.Power dissipation elements ‘R’ is not there, sothe poles and zeros lie on the imaginary axis.Network function should be determined by zeroinitial condition.

80. (b)

2Ai

6A 1A

4A3A

a

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Applying KCL at node a, we get

6A 1A 4A 3A 0

i = 2A

81. (c)For obtaining equivalent resistance,we haveto short circuit the independent voltage sourceand open circuit indepdent current source.Therefore the circuit is,

2 2

1

1

RTh

RTh= 2 2 2

= 2 1= 3

82. (c)

Using superposition theorm, current in 10resistor due to 10A current source is

I1 = 1010A20

= 5A

5

5 10 10A

Current in 10 resistor due to 20A currentsource is

5

5 1020A

I2 =

520A20 = 5A

Total current = I1+I2= 5+5 = 10A

83. (c)

From linearity we known that s LV I ratioremains constant.

Now power absourbed is 2.5W, so2L LI R = 2.5W

IL = 0.5AFrom Linearity,

20V200mA = SV

500mAVs = 50V

84. (b)Applying superposition therom, it is knownthat if all current source value are doubled,then node voltages also are doubled.

85. (b)

Power obserbed by RL is,

PL =

2s L

2s L

V R

R R

from the expression, power is maximum whenRs = 0.

86. (d)

Thevenin equivalent of unknown circuit is

RV

RTh

VTh

by voltage division V =

ThTh

RVR R

substituting values from table,

6 =

ThTh

3VR 3

...(1)

8 =

ThTh

8VR 8

...(2)

Solving the equation (1) and (2), we get

RTh = 2 and VTh = 10V

87. (d)

[Y] = 1Z

11 12

21 22

Y YY Y =

22 12

21 1111 22 12 21

Z Z1Z ZZ Z Z Z

=

22 12

21 11

Z Z1Z ZZ

[Where Z = Z111Z22 – Z12Z21]

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88. (a)When switch was open, total charge q= 1 1C V

Let after closing the switch; voltage acrossparallel combination is V.

t 0

VC1 C2

As charge will be conserved.

Now, q = eqC .V

ie. 1 1C V = 1 2C C V

V =1 1

1 2

C VC C

89. (c)

Power P = 2rmsi R

Here 2rmsi =

T2

0

1 i t dtT

= 3

2

0

1 3t dt 273

2rmsi R = 270W

R =270 1027

90. (c)

1

1L 1 1 2

di t dV t L M i t i tdt dt

2

1L 2 1 2

di tdV t L i t i t Mdt dt

• Multually induced voltage occurs due toother current .

• Self induced voltage occurs due to selfcurrent.

91. (b)

T0

A

t

f t

T0

A

t

1Af t tT

T0

1

t

2f t u ut t T

= ×

f(t) = 1 2f .ft t

= A t u ut t TT

= A Atu tut t TT T

= A Atu ut t T T t TT T

= A Atu u Aut t T t T t TT T

92. (d)The single voltage source may be consideredto be equivalent to two identical sources inparallel. So option (1) and option (4) areequivalent.Option (4) and option (3) are equivalent sincea connection from the positive terminals of thetwo sources does not affect the networkbecause there would be no current in such aconnection.Note: The single current source may beconsidered to be equivalent to two identicalsources in series.

93. (c)L1 = 5 mH, L2 = 20 mH

L N2

It the number of turns are doubled the newvalue of

1L = 212 L

= 4 5 20 mH

2L = 222 L

= 4 20 80 mH

Total InductanceLeq = L1 + L2 – 2 M

(For negative polarity)

M = 1 2K L L

= 0.5 20 80= 0.5 40= 20

Leq = 20 + 80 – 40= 60 mH

94. (b)

Applying KVL in 1st loop we get

36 = 3120 10 I

I1 = 336 1.8mA

20 10

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+

–5 k

10 I2

vI2

20 kI1

36V

1

1

Applying KCL at node 1I2 = I1 + 10 I2

I2 = 1I 0.2mA9

Voltage across 5K resister

v = 325 10 10I

= 3 35 10 10 0.2 10= 10 V

95. (a)i = 2 A

Applying KVL, we getVab– 3i +10 = 0

Vab = 3i – 10

= 3 2 10

= –4 V96. (d)

20V

2

i+

–v

Given4i = v2 – 8 ...(1)

Applying KVL we get20 – 2i = v ...(2)

From equation (1) and (2), we get4i = (20 – 2i)2 – 8

4i = 400 + 4i2 – 80i – 8 i = 100 + i2 – 20i – 2 i2 –21i + 98 = 0 (i – 7) (1– 14) = 0 i = 7, 14 A

97. (a)

Irms = T

2

0

1 8t dtT

=31 64 T

T 3

Here T = 1

Irms =643

Average Power loss in 9 resistorPav = 2

rmsI 9

=64 93

= 192 W98. (a)

Electric field is given by :

1 21 22

0 1 0 2

Q Qˆ Ê r a a4 r r 4 r r

where 1a and 2a are unit vectors along 1r r

and 2r r

.

Here, 1ˆ ˆ ˆr r 0 1 i 0 2 j 0 2 k

ˆ ˆ î 2j 2k

2 2 21r r 1 2 2 3

And, 2ˆ ˆ ˆr r 0 3 i 0 6 j 0 6 k

ˆ ˆ ˆ3i 6 j 6k

2 2 22r r 3 6 6 9

9 6 ˆ ˆ î 2j 2k3 9 10 10E 0,0,0

9 3

6 9 ˆ ˆ ˆ3i 6 j 6k27 10 9 1081 9

3 3ˆ ˆ ˆ ˆ ˆ Ê 10 i 2j 2k 10 i 2j 2k V m

3ˆ ˆ ˆ ˆ Ê i i 2 j 2 j 2k 2k 10 V m

ˆ Ê 4 j k kV m

99. (c)

C1

C2

C3

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Here, C2 = 0 03 A 3 Ad d22

C3 = 0 04 A 4 Ad d22

C1 = 0 02 A 4 Ad d2

C2 and C3 are in parallel : 2 3C C C

C = 07 Ad

Now, C and C1 are in series.

Ceq =

2

001

10

A28ACC 28d

AC C 11 d11d

100. (b)

Magnetic flux is given by B A weber

2WeberB

mA.

Magnetic field intensity unit is Ampere/m.

101. (c)

Ampere law states that line integral oftangential component of H around a closedpath is the same as net current I enclosed bythe path.

encH dl I

Applying Stokes Theorem:

Ienc =

s

H dl H dS

Ienc =

S

J dS

H =

J (in point or differential form)

102. (d)

Gauss Law is given by :

eneD ds Q

VB p

Magnetic field lines always exist in closed pathsand magnetic monopole does not exists.

E 0Electric field lines never exist in closed path

Hence,

E 0

103. (b)

Magnetic monopoles does not exist and hencemagnetic field lines are always closed. So, ithas no sinks or sources.

S

B ds = or

B 0

Hence,

B is solenoidal but it is non-

conservative as

H 0.

104. (c)

×

M N

H2H1

Here, = 30°

and H1 = 2I 5H

2 a 2

H1 is due to 10 A at M and H2 is due to 10Aat N.

H1 cos and H2 cos will cancel each other..

Net field intensity = H1 2sin H sin

= 2H1 sin 30°

=

5 122 2

H =

5 Am2 .

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105. (a)

P

Qa

Fig.2

Fig.1

'EP

Qa

E

E = 2

KQa

...(1)

Applying Gauss’s law for fig-2

E ds =

0

Q

' 2E 4 a = 0

Q

E' = 2 2

0

Q KQ E4 a a

E' = E = 100 mV/m

106. (d)

V =

zV 1 V Vˆ ˆ â a a

z

= 2z 2z 2z

zˆ ˆ ˆ5e sin a 5e cos a 10 e sin a

2, 3,0V =

z3 1 3ˆ ˆ ˆ5 1 a 5 1 a 20 1 a

2 2 2

= zˆ ˆ ˆ4.33a 2.5a 17.32a

107. (a)

W = B

A

Q E dl

= Q VAB = Q (VB – VA)

=

2 210 10sin90ºcos60º sin30ºcos120º104 1

= 100 1 1 11 1616 2 2 2

= 28.125 Joule

108. (a)

zVoltage (v)E a

distance between planes (d)

z2 a

0.1

(given V = 2V d = 0.1m)

z20a

J E

= 100rz(1000e )(20a )

= 100r 2z20000e a A / m

109. (c)

H.dl ( H).ds (stoke’s theorem)

enclosedH.dl I (Ampere’s law)

v

D.ds ( .D)dv (Gauss divergence theorem

Q D.ds (Gauss Law)

110. (c)We know velocity

PV

0 0

(as it is free space substituting in aboveequation)

P0 0

V

0 0

1

0 0

1

111. (d)

E / H , the characteristic Impedance ofthe medium.

0 r

0 r

= r

r

2120 1208

= 11202

120A

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120 2120 1/ 2

A 2

PV

here 610 Pr r

1 CV

=8 83 10 3 10

42 8

6

810 0.044 0.042

33 10

112. (a)

One may arrive at the correct choice bycomparing the two sides dimensionally.

Since 1 has the dimension of velocity

squared the pertinent choice is:

2 2y y

2 2

E E1

t x

113. (a)The electromagnetic wave equation is given by

22

2E EE 0t t

114. (c)

For free space and are each zero

hence 1 and 2 are false

hence statements 3 and 4 are correct

H J, D

115. (c)

given refractive index ( ) 1.5

frequency = 1410 HZ

Velocity of light in vacuum 8o(C ) 3 10

Velocity of light in glass = C

Refractive index oC( )C

83 101.5C

83 10C1.5

82 10

We know Cf

8

142 1010

62 10

2 m

116. (b)

f(x) V t)0- forward

represents a wave motion in forward direction

f(x) V t)0+ reverse

represents a wave motion in reverse direction

117. (c)dvEdr

given 5

0

6rV

, r 1m

5

0

d 6rEdr

4

0

30r

We also know 20

1 qE4 r

(by Coloumb’s law)

equating both

4

20 0

30r 1 q4 r

6q 30r 4

6q 120 r ( r 1)

q 120 coloumbs

118. (c)

given j /2 jstE (0.5x ye )e

x| E | 0.5

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2 2y| E | cos / 2 sin / 2 =1

So, x y| E | | E |

and phase = 90°hence it is elliptically polarised wave

119. (c)

For a uniform plane wave

E and H must be in a plane

But both are perpendicular to each other.

120. (a)

• Intrinsic impedance, =

j

j

• For free space, 0 ,

=

0 r 0

0 r 0

jj

r r1, 1

• For conducting medium, ; so,

=

j 45

• For perfect dielectric; 0 ,

=

jj

121. (a)

1.Kirchoff’s current law : Rate of flow of chargeentering at a node is equal to rate of flow ofcharge leaving the node.

2.Ampere’s Law :

H.d Il

3.Faraday’s Law :

dd BAEdt dt

4.Gauss’ Law : D.ds q

122. (c)

As the wave is propagating in Y-direction. Sothe component of electric field will be either inx-direction or in z-direction but never in Y-direction.

123. (d)

When a charged particle is moving itexperiences force due to both magnetic andelectric field.

124. (d)Number of revolutions = 1800/60

= 30 r.p.s

t =1 sec

30

e = t

=425 10

1/ 30

0.23 V

125. (c)

Since, there is no electric field inside theconductor, hence there will be no tangentialcomponent at the surface of conductor. Onlynormal component of electric field will be there.Hence electric lines of force will be alwaysperpendicular to the surface of conductor.

126. (d)As A and B are at the same potential, thework done will be zero.

127. (c)

a

Since charge density = 0

According to Gauss’s Law, D.ds Q

For r < a

2D.4 r = 30

4 r3

D = 0r3

For r a

128. (a)

For a quarter wave

20

inL

ZZ

Z

20 in LZ Z Z

0 in LZ Z Z

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given LZ 200

inZ 50

0Z 200 50

10000

0Z 100

129. (b)

given LZ (75 j40)

0Z 75

f 6MHz

as LZ is capacitive in order to havecancelling effect an inductance should beconnected at load so that load matchingtakes place.

130. (a)

We know reflection co-efficient

r 2 1

i 2 1

EE

and 2r

1120

[ r 1 for non-magnetic material]

112016

1204

30

1 120 for free space

30 120150

90150

3

5

131. (c)

Transmission of power to a load over atransmission line achieves optimum valuewhen SWR becomes 1 : 1

This is possible when L 0Z Z

where no reflected wave exists and reflectionco-efficient is zero.

Voltage is uniform along the line

max min| V | | V |

1 0SWR1 0

= 1 : 1

132. (b)

Given 8PV 4 10 m / s

8Zf 10 H

We know PVf

8

84 10

10

4m

length of line l 2m

Since the length of the line is half the wavelength therefore in lZ Z .

133. (c)

We know that

0R j LZG j C

with 0Z as pure resistance we have

L 0 , C 0 , 0RZG

Loss less line

R G 0

0LZC

Distortion less line :

R L ,G C

0LZC

hence it is not lossless line and notdistortionless line

134. (c)

We know reflection co-efficient

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L 0

L 0

Z ZZ Z

j60 60j60 60

| | =2 2

2 2

60 60

60 60

= 1

VSWR = 1 | |1 | |

= 1 11 1

10

135. (b)

for 2

line

in LZ Z 200

for line4

20

inC

ZZ

Z

2(100)50

100 10050

200

/ 2 line and / 4 line are parallel

in200 200Z 100200 200

for / 8 line, in 0| Z | Z 100

136. (b)

If a change of polarisation direction is required,a twist section may be used extending overtwo or more wavelengths.

Tapers are used to couple waveguides ofdifferent dimensions.

Bends & corners are used to change direction.

137. (b)

Average power is given by d radave 2

G PP

4 r

Here, d 10 dG 10dB 10log G

dG 10

3radP 20 10 w and 3r 10 m

3

ave 610 20 10P

4 10

=

1 W20

138. (b)

Horn antenna, an example of an apertureantenna is a tapered section of a waveguideproviding a transection between a waveguideand the surroundings. It finds its applicationin aircraft.

Parabolic dish Reflector reflects EM waves byconducting sheet. Wired antennas are led bytransmission lines.

139. (d)

The wave is propagating in +z direction. Theexpression of electric field for TM mode isgiven by :

i zZ 0

m x n yE E sin sin ea b

comparing with

i zZE 30sin 50 x sin 60 y e

i zsin 60 y e

m 50a

2m 50a 50 8 10 4

n 60b

2n 60 5 10 3

Mode is 43TM

140. (b)

A waveguide behaves as a high pass filter.There is heavy attenuation for frequenciesbelow cut-off although the waveguide itself isvirtually lossless. Such attenuation is due toreflection at the mouth of the guide and sometake place in evanescent modes.

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141. (a)

The wavelength of the wave propagating in awaveguide is given by

2m12a

where Free space wavelength.

Hence is always greater than A.

142. (d)

Phase velocity is the velocity with which thewavechanges phase at a plan foundary andnot the velocity with which it travels along theboundary this apparent velocity which is greaterthan velocity of light in space is called phasevelocity.

143. (c)

A parasitic element shorter than the drivenone or radiator tends to increase radiation inits own direction and behaves as convergentconvex lense which is called a Director.

Radiator

Director

Reflector

A parasitic element longer than the driven oneand which reduces signal strength in its owndirection and increases in opposite directionis called a reflector.

144. (c)

TEM (Transverse Electromagnetic Mode) doesnot have any cut-off frequency and hence itwill be supported at any frequency.

Cut-off frequency is given lay: 0 = m c

nd (formode m)

145. (c)

A waveguide may consist of one or moreconductors, or no conductors at all and willsupport waveguide modes.

146. (c)

For a Transverse magnetic wave propagatingin a waveguide, the electric field component

along the wave propagation is given by:

Ez =

z0

m x n yE sin sin ea b

For m = 0, n 0 , Ez = 0

For n = 0, m 0 , Ez = 0

TM01 and TM10 will not exist and TEM modecannot propagate in a rectangular waveguide.

Only TM11 can propagate.

147. (d)

Loop Antenna consists of a wire with one ormore turns. They are also called wire Antennasand used in Buildings, ships, Automobiles,etc.

Horn Antenna is an example of ApertureAntenna is a tapered section of a wave guide.It is used in Aircraft as it can be convenientlyflush mounted.

Parabolic dish reflector reflects EM wave by aconducting sheet. It is used in communication,Radar & Astronomy. Yagi Uda Array is usedfor short wave transmissions by RadioAmateurs.

148. (b)At absolute temperature, intrinsic semi-conductor behaves like on insulator as thereis no free electrons available.

149. (d)In intrinsic semiconductor, the total electricalconductivity.

e h e hne ne

Although the number of electrons inconduction band and holes in valence bandare equal in intrinsic semiconductor, theconductivity due to electrons is more thanthe conductivity due to holes. Because themobility of electrons is higher than that ofholes.

150. (a)

Alloying elements always increases theresistivity of the metal to which they are added.One reason behind this is, the atomic size ofparent atom and alloying element, are differentwhich leads to the variation in the latticeparameter and thus, the electron scatteringoccurs.

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