answers - ies masteriesmaster.org/public/archive/2016/im-1476625666.pdf · in dry-cell battery ,...

ESE-2017 PRELIMS TEST SERIESDate: 16th October, 2016

1. (b)

2. (a)

3. (c)

4. (b)

5. (d)

6. (b)

7. (b)

8. (a)

9. (b)

10. (a)

11. (c)

12. (a)

13. (c)

14. (c)

15. (d)

16. (c)

17. (a)

18. (c)

19. (c)

20. (c)

21. (d)

22. (b)

23. (c)

24. (a)

25. (b)

26. (c)

27. (d)

28. (c)

29. (b)

30. (b)

31. (a)

32. (b)

33. (c)

34. (d)

35. (c)

36. (a)

37. (c)

38. (a)

39. (a)

40. (d)

41. (b)

42. (b)

43. (b)

44. (a)

45. (d)

46. (d)

47. (d)

48. (c)

49. (d)

50. (c)

51. (c)

52. (b)

53. (c)

54. (c)

55. (c)

56. (c)

57. (c)

58. (a)

59. (b)

60. (c)

61. (c)

62. (b)

63. (c)

64. (c)

65. (b)

66. (a)

67. (b)

68. (d)

69. (b)

70. (d)

71. (a)

72. (b)

73. (d)

74. (d)

75. (d)

76. (a)

77. (c)

78. (c)

79. (d)

80. (a)

81. (a)

82. (c)

83. (d)

84. (c)

85. (c)

86. (c)

87. (d)

88. (c)

89. (b)

90. (a)

91. (c)

92. (b)

93. (b)

94. (d)

95. (c)

96. (a)

97. (c)

98. (d)

99. (b)

100. (a)

101. (a)

102. (c)

103. (b)

104. (c)

105. (b)

106. (c)

107. (c)

108. (c)

109. (d)

110. (b)

111. (d)

112. (c)

113. (b)

114. (d)

115. (c)

116. (d)

117. (d)

118. (b)

119. (a)

120. (c)

ANSWERS

121. (a)

122. (b)

123. (b)

124. (c)

125. (d)

126. (c)

127. (a)

128. (b)

129. (b)

130. (d)

131. (d)

132. (d)

133. (b)

134. (b)

135. (c)

136. (b)

137. (a)

138. (a)

139. (a)

140. (d)

141. (d)

142. (b)

143. (c)

144. (d)

145. (d)

146. (b)

147. (c)

148. (c)

149. (d)

150. (b)

151. (d)

152. (c)

153. (c)

154. (d)

155. (d)

156. (b)

157. (d)

158. (c)

159. (a)

160. (d)

161. (b)

162. (d)

163. (d)

164. (a)

165. (a)

166. (b)

167. (c)

168. (c)

169. (c)

170. (d)

171. (c)

172. (b)

173. (a)

174. (b)

175. (a)

176. (a)

177. (b)

178. (a)

179. (a)

180. (b)

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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-3 Solution) 16th Oct 2016 (3)

1. (b)No load loss consists of core loss, frictionand windage loss and no load copper loss.But shunt branch resistance Rc in theinduction motor equivalent circuit does notrepresent the no load losses, it representsonly the core loss of the induction motor.

2. (a)According to Faraday's law, when a currentcarrying conductor is placed in magneticfield it will experience a mechanical force.To get the mechanical force, a stationaryconductor should cut the rotating magneticfield. In order to produce rotating magneticfield (RMF), it requires a 3-phasesymmetrically distributed winding and a3-phase supply is given to it. The 3-phasecurrents produce 3-phase flux in the 3-phases and they interact and produce anet flux which is a constant value butchanges its axis, which is called as rotatingmagnetic field (RMF).

3. (c)Synchronous speed is defined as, for astandard frequency and for a given polesrotating magnetic field rotates at astandard speed called as

Synchronous speed Ns = 120fP

500 = 120 ×50

P

synchronous speed (Ns) P = 12.4. (b)

The flux always moves from a leadingphase to lagging phase. The direction ofinduction motor depends on rotatingmagnetic field direction. Therefore byinterchanging any two phases, (but not allat the same time) direction of rotatingmagnetic field can be reversed, and thiswill reverse the direction of inductionmotor.

5. (d)Leakage flux in machine will depend onthe air gap between stator and rotor. Ifair gap is more leakage flux is less andvice versa. In open slot type machines airgap between stator and rotor is more i.e.reluctance is high. Therefore leakage fluxis less. In closed slot type machines airgap between stator and rotor is less i.e.

reluctance is low. Therefore leakage fluxis high. In semi closed slot type machinesair gap between stator and rotor ismoderate i.e., reluctance is moderate.Therefore leakage flux is moderate. So, allthe combinations are true.

6. (b)Operating power factor depend’s on theair gap between the stator and rotor. Ifthe air gap is more it will require moreamount of magnetizing current, this willdecrease the operating power factor andvice versa. In open slot type machines airgap between the stator and rotor is more.Therefore operating power factor is poor.In closed slot type machines air gapbetween the stator and rotor is less.Therefore operating power factor is more.In semi closed slot type machines air gapbetween the stator and rotor is moderate.Therefore operating power factor is alsomoderate. Therefore the operating powerfactor in closed slot type machines is morethan the other one.

7. (b)Depending on the type of rotor used,induction motors are classified as squirrelcage induction motor and slip ringinduction motor. The first motors aresquirrel cage induction motors which havelow starting torque. For high startingtorque requirement slip ring inductionmotors also called as wound rotors aredeveloped. In slip ring induction motor,the rotor core consists of slots whichcontain 3- phase winding similar to stator.The rotor winding is essentially connectedin star and the three terminals end at 3slip rings which are mounted on the shaft.A suitable value of external resistance isadded equally into the rotor 3-phasewinding through stationary slidingcontacts known as brushes which areplaced on the slip rings. When the motoris started with suitable external resistanceit produces the high starting torque.

Starting torque stT E×I×cos

stRT E×I×Z

Therefore the starting torque is directlyproportional to resistance value, which willincrease the starting torque of slip ringinduction motors.

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(4) (Test-3 Solution) 16th Oct 2016

8. (a)In slip ring induction motor brushes arepresent. Due to this, frequent maintenanceis required. Their design is complicateddue to winding, slip rings and brushes andalso expensive. Slip ring induction motoris comparatively having more weight thansquirrel cage induction motor. Theconstruction of squirrel cage inductionmotor is simple and robust and it is cheapas compared to slip ring induction motor.Therefore squirrel cage induction motor iswidely used.

9. (b)

Slip S = s r

s

N – NN

Ns =120 – f

p

= 120 ×50 =1500rpm4

Percentage Slip = 1500 ×1470 =0.02%1500

10. (a)

Slip S = s r

s

N – NN

Slip at standstill = s

s

N – 0N

1 speed of rotoratstandstill is = 0

11. (c)

Reactive power Q V × Ecos – VX From

the above equation, the reactive powergenerated or delivered is significantlydepends on excitation. When the excitationis increased (over excited), then E cos >V, which means generator delivers orsupplying reactive power to the load andoperates at lagging power factor.

12. (a)During the charging and discharging time,the value of specific gravity of electrolytevaries. For example, during the chargingtime, the value of specific gravity increasesfrom the current level and during thedischarging time, the value of specific

gravity decreases from the current level.But, the rest of all factors which shown inabove are not changed. Hence, the state ofcharge of a battery is given by the specificgravity of electrolytic.

13. (c)

P E × V × sin

X

maxP maxE × X sin =1

V

Maximum power is developed when loadangle = 90°. Therefore maximum powerdeveloped will depend only on voltage andreactance values.

14. (c)Direct axis synchronous reactance Xd=Vmax/Imin,Quadrature axis synchronous reactance Xq= Vmin/Imax.Therefore, direct axis synchronousreactance Xd is greater than quadratureaxis synchronous reactance Xd.

15. (b)When the rotor and stator rotatingmagnetic fields align along with thequadrature axis, air gap reluctance ismaximum. Therefore more magnetizingcurrent is drawn by the stator to maintainthe flux and the voltage is minimum. Quadrature axis synchronous reactance Xq= Vmin/ Imax.

16. (c)17. (a)

In dry-cell battery , the electrolyte is inthe form of a paste. And there is nopresence of nickel and iron material. Dueto that, the internal resistance of the drycell is very less. The value is approximatelyin between 0.2 to 0.4?

18. (c)In Potier Triangle or ZPF method, voltagequantities are calculated on emf basis andmmf quantities are calculated on mmfbasis. Therefore, ZPF method is mostaccurate among all these four methods.

19. (c)As the armature flux is in phase with themain field flux it is additive in nature andincreases the net flux. Therefore the emfwill be increasing with the load. Voltage

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regulation is zero or negative value on theleading power factor load or capacitive loadas the voltage rises due to increase in load.

20. (c)For capacitive load, the armature flux isin phase with the main field flux, it willadd and increases the net flux in the airgap. This effect is known as magnetization.The induced emf will be increasing withthe load. Therefore under leading loadcondition the flux is greater than the actualvalue.

21. (d)The sulphation in a lead acid battery occursdue to sudden discharge in a full chargedbattery. So, it produces more internalresistance and thereby the battery can'tbe reached to the fully charged condition.Since, sulphation occurs continuously dueto the incomplete charge.

22. (b)The capacity of the cell is measured bythe discharging at a constant current untilit fully drains out for that particular time.Hence, the capacity of the cell/battery ismeasured by the Ampere-hour rating.

23. (c)Deep-cycle lead acid (DCLA) batteryprovides a steady current over a long timeperiod and these are all connected togetherto form a battery bank. The technology oflead acid battery is uncomplicated andmanufacturing costs are low compared toother type of battery . So, it is cost effectiveand it has good discharging characteristics.

24. (a)When connecting the cells in series,according to the Kirchhoffs law the voltageacross the cells can be increased and inthe same time, the current capacity of thecells can be the same capacity of the singlecell.

25. (b)The iron content in the electrolyte leadsto make a conductive path from positiveplate to negative plate. Short circuit occursin between the positive and negative plate.Hence, we consider iron as the commonimpurity in battery electrolyte.

26. (c)Frequency of Rotor emf = Slip × SupplyFrequency

Frequency of Rotor emf = 0.02 × 50 1Hz.

27. (d)The rotating magnetic field will rotate atsynchronous speed.

Ns = 120 × f 120 ×50 =750 rpm

p 8

Therefore the speed of rotating magneticfield is 750 rpm for 8 pole 50 Hz machine.

28. (c)

Starting torque Tst E×I× cos E ×

E RR + jX R + jX

Maximum starting torque will obtainedwhen R = X i.e. the rotor resistance shouldexactly equal to the reactance at standstill.Under such condition rotor power factor is0.707 lagging (presence of inductance).

Rotor power factor cos =

R RR + jX R + jX

= 0.707 lagging.29. (b)

Therefore the condition for maximumrunning torque is S = R / X or R = S.X Thecondition for maximum running torque isrotor resistance should be equal to sliptimes the rotor reactance at standstill.

30. (b)

Tr = S × 2R

R + jX

The condition for maximum running torqueis R = S.X By substituting the above valuewe can get the maximum torque. Tmax 1 /(2X). Therefore the maximum torque willdepend on the rotor reactance X only. Therotor resistance cannot vary the maximumtorque developed in the motor but the slipor speed changes due to the rotorresistance.

31. (a)The torque is proportional to slip only inthe stable region and it is inverselyproportional at higher operating slips. Theentire operating region is divided into two

regions. Running torque Tr = S × 2R

R + jX

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1. Low slip region: S.X >> R Tr S/R Tr S(for a given R value) Therefore under lowslip region motor operates in stable region.2. High slip region: S.X << R Tr R/S.X2

Tr 1/S (for given R and X values)Therefore under high slip regions motoroperates in unstable region.

32. (b)Basically in the circle diagram for aninduction motor, the diameter of the circlerepresents the rotor current. By using thiscircle, the current per phase with blockedrotor and full voltage on stator and theline current in amperes for any given loadcan be found out.

33. (c)

At t = 0¯, capacitor will be open circuited.

So, VC(0–) = 5 × 0.3 = 1.5V

CV (0 ) = CV (0 ) 1.5V

At t , capacitor will be open-circuited,

Hence, CV ( ) =0.65 0.3

0.3 0.6

= 0.65 0.30.9

= 1V

Time constant; = eqR .C =

0.3 0.6 10.3 0.6

= 0.18 10.90

= 0.2 sec. = 1 sec5

Voltage across capacitor,,

{Vc(t) = V( ) [V( ) t /V(0)]e }

= 1 – [1 – 1.5]e–5t

= 1 + 0.5e–5t Volt

34. (d)

At t = 0–, capacitor will be open-circuited.

VC(0–) = (12 × 6) = 72V = cV (0 )

At t = 0+, circuit will reduce to,

6

3

72V12A

I

+–

I = 723

= –24A

35. (c)

At t , capacitor will be open-circuited.

2

2

I V (C

cV ( ) = 2 = 1 2 2V

Given, initial voltage,

VC(0) = 4V

So, voltage across capacitor,

VC(t)= { V( ) [V( ) t /V(0)]e }

= eq.t /(R C)2 [2 4].e

=

1t/ 422 2e

= t /22 2e Volt.

36. (a)

When switch K was open for long time,inductor gets short-circuited and capacitorgets open-circuited.

So, at t = 0–

V(0–) = R

= 1 × 1

= 1 volt

and, VC(0–) = VC(0+) = V = 1 volt

Now, current through capacitor at t = 0+ will be

zero so, ci (0 ) = 0; [As capacitor is open-

circuited]

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CdVdt

= 0

t 0

dVdt

= 0

37. (c)

When switch ‘S’ is closed at t = 0, inductorwill be open-circuited and capacitor will beclosed-circuited.

At t = 0+; the circuit will reduced to,

10V

2

i(0 ) =102

= 5A

At t , inductor gets short-circuited andcapacitor gets open-circuited. The circuit willbe,

10V

2

i( ) = 10 5A2

38. (a) V(t) = A t u(t)T

A (t T)u(t T)T

A u(t 2T)

Taking Lapalce transformation, we get,

V(s) = sT2 2

A 1 A 1 eT Ts s

2sTA es

= sT 2sT2

A A1 e esTs

39. (a) I(s) = 2s

(s 2) = 2s 2 2(s 2)

= 2 2(s 2) 2(s 2) (s 2)

= 21 2

(s 2) (s 2)

Taking Inverse Laplace transform, we get,

i(t) = e–2t – 2t.e–2t

= 2te (1 2t)

40. (d) In a series R–L circuit,

V

R L

+– i(t)

i(t) = Rt/LL

V (1 e ) i (t)R

Voltage across inductor L,

VL(t) = di(t)Ldt

= Rt/LV RL eR L

= Rt/LV e

Instantaneous power supplied to theinductor,

p(t) = L LV (t) i (t)

= Rt/L Rt/LV(V e ) (1 e )R

= 2

Rt/L 2Rt/LV [e e ]R

41. (b)

For the above circuit

i = Rt/Lm mI sin t I sin e ;

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where, 1 LtanR

When, 0 or, ie. when voltageapplied is passes through its maximum, therewill be no transient.

42. (b)

Characteristic equation in the s-domain forthe given circuit is,

2 R 1s sL LC = 0

then, 2n =

1LC

1LC

and, n2 =RL

= 1 R LC2 L

= R C2 L

R = L2C

= 1 222 8

=12

= 0.5

43. (c)

Given, Z(s) =(S 1)S

z( j ) =

j 1j =

j 1j [ 1]

z(j ) =

2 21 1 21

z( j ) = 1–90 tan (1/ 1)

=

–90

4

= –45°

so, steady state response

f(t) = K z(j )

cos(t z( j ))

= 2 2 cos(t 45 )

= 2cos(t 45 )

= –2sin(90 t 45 )

= –2sin(t 45 )

44. (a)

When switch ‘S’ is at position 1,

i (0–) =VR

Now, when switch ‘S’ is at position 2,

i (0+) = i (0–) = VR

and, i ( ) = 0

So, current i = t Li ( ) i( ) i(0) e

= Rt/L 3Rt LV V0 0 e .eR R

where, time constant l 3r

45. (d)

v(t) =1 r(t)

(T / 2)

1 Tr tT / 2 2

1 Tr tT 22

=

2 2 Tr(t) r tT T 2

2 Tr tT 2

=

2 4 Tr(t) r tT T 2

46. (d)

Final value theorem,

tLimf(t) = t 0

Lim sF(s)

=3 2

S 0lim s[3s 8s 4s

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45 1 ]s s

=4 3 2

S 0lim s[3s 8s 4s

315 ]s

= 5

=

47. (d)

RMS value of current waveform,

irms =

1/212

0

1 [i(t)] dtT

=

1/21t 2

0

1 [(1 e ) dt1

=1

2t t

0

[1 e 2e ] dt

=

12t t

0

e et 2( 2) ( 1)

= 21(1 0) (e 1)2

12(e 1)

= 21 0.5e 0.5

12e 2

= 1 22e 0.5e 0.5

= 1 21 (4e e 1)2

48. (c)

The step response of the function

f(t) = 2t t8e 3e dt

=

2t te e8 3 C( 2) ( 1)

= –4e–2t + 3e–t + C

At time t = 0,

f(t) = –4 + 3 + C = 0

C = 1

i.e. A = –4, B = 3, C = 1

49. (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)

50. (c)

Given circuit is under resonance, so currentwill be maximum,

Irms = rmsVR

= (10 / 2)

100

Average power will be dissipated only inresistor.

So, Pavg = 2rms( ) R

=

210 / 2 100

100

=100

2 100

= 0.5 W

= 500 mW

51. (c)

Since, current will leave the dot in firstinductor and enter the dot in second inductor,hence. equivalent inductance will be,

Leq = L1 + L2 – 2M= 6 + 4 – 2 × 2

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(10) (Test-3 Solut ion) 16th Oct 2016

= 6 H

so, resonance frequency, 0 = eq

1L C

0 =

1163

= 12

= 0.707 rad/sec.

52. (b)

0V shortcircuit

Z

V

HV LV

Irated

Here, Vapplied = rated Z

Given,rated

rated

ZV

= 0.06

rated Z = 0.06 × Vrated

i.e., Vapplied = 0.06 × 2200

= 132 Volts

53. (c)

In order to keep the same value of flux in thetransformer core, magnetic flux density Bm isto be kept constant.

since, V Bmf

mVBf

so,1

1

Vf =

1

1

Vf

44050 = 1V

40

1V =440 40

50

= 352V

54. (c)

55. (d)

56. (c)

57. (c)If only one network is taken

P =

2th

th

V RR R

& if two networks are taken

taken P' =

2th

th

VRR R

2

=2

th

th

V4 R2R R

comparing P and P', P' would be in betweenP and 4P.

58. (a)

VC(0)– =

10 6k6k 4k

= 6 V = CV 0

RI 0 = CV 0R

= 6V

6k = 1 mA

59. (b)Current here has 3 components

r.m.s of the total = 2 2 21 2 3rms rms rms

1rms = 4

2rms =2 2

2 = 2

3rms =4 2

2 = 4

rms = 2 2 24 2 4 = 36 = 6A

60. (c)By applying KVL in various loops, we canget voltage V1

A B

C D50V

+10V

–+

40V–

+ 70 –

+ 30 –

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Applying KVL in ABCD Loop,V1 + 100 – 10 V – 70 V – 10 + 50 = 0

V1 = –60 V61. (c)

Off the total 70V, 10V is across R2, so 60V isacross R1. 30V is across 60 resistor so,

current = 3060 = 0.5A

R = VI = 60

0.5 = 120

62. (b)

Voltage across inductor is V = diLdt

3 100mA100 102ms = –5 V (2 to 4)ms

200ms0.14ms = 5 V (4 to 8) ms

100ms0.12ms = –5V (8 to 10) ms

63. (c)By applying cut principle

4A

1

5A

1A

2

5 cut

cut

principle states that the current passing through this should be zero.

V = 5 1A

= –5V64. (c)

A capacitor does not resist any abrupt changein the current flow. It resists abrupt changein voltage.

65. (b)

The ampere’s circuital law in Integral form is

enclosedH.dl I

This is also called as maxwell’s equationfor static field.

Another equation for amperde circuital lawin integral form for time varying field is

DH.dl I dst

This is also called as maxwell’s equationfor time varying field.

66. (a)

When ampere’s law is applied todifferential surface we get

H J

This is also called as maxwell’s equationin static field.

67. (b)

We know

H J

2

i j k

Hx y z

x 2yz ( x)

=

2i ( x) (2yz)

x

22j ( x) (x )

x

22k (2yz) (x )

x

= i [ 2y] j[ 1 0] k[0]

j 2y i j

x rcos y r sin in cylindrical form

6cos45 y 6sin45 = 4.24

= 62

= 6 0.707

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x = 4.24, y = 4.24

j 2(4.24)i j

= 8.48i j

68. (d)

C

D

| J | E| J | E E

(where = frequency , are constants)s)

If 1 then 1

(Good conductors)

If 1 then 1

(Dielectrics)

If C D(J ) (J )

E E

; 2 f

f2

when C D| J | | J | then it is a quasi conductor..

69. (b)

From the options we can see that option (a)

rFrt

is maxwell’s equation in time

varying field

70. (d)

when ever electromagnetic wave travelsin a free space then

0,

0

0

71. (a)

As the wave is travelling in a free space

0 , 0

0 0

? 92 10 rad / s

70 4 10 H / m

90

1 10 F / m36

9 7 912 10 4 10 1036

9 1612 10 109

=9 82 10 10

3

= 2 103

2 2 3 0.942m2 10 10

72. (b)

For a lossless dielectric

0, 0 r 0 r

2 9375MHz

0 r 0 r

7 914 10 1 10 936

7 9110 10 99

1610 1

6 82 9375 10 10

22 9375 10

2 93.75

187.5

73. (d)

r 1 (for non magnetic material)

r 2

410 / m

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f 2.5MHz

tan

4

6 9

1012 2.5 10 10 2

36

4

310 36

5 2 10

436 1010

236 10

tan 0.36 = Loss tangent

74. (d)

In free space 0 00, , , 250

0 0

7 914 10 1036

161 109

8103

8250 103

( = 250 given from equ.)

8250 3 10

8750 10

975 10

75. (d)

endH.dl I

2

0

(H a ).(rd )a 60

2

0

H rd 60

2 H r 60

60H2 r

60H2 5mm (r = 5 mm given)

B .H

= r 0H r( 1)

73

604 102 5 10

4120 105

= 2.4 mT

76. (a)

zVoltage (v)E a

dis tance between planes (d)

z2 a

0.1

(given V = 2V d = 0.1m)

z20a

J E

= 100rz(1000e )(20a )

= 100r 2z2000e a A / m

77. (c)

H.dl ( H).ds (stoker theorem)

enclosedH.dl I (Smpere’s law)

v

D.ds ( .D)dv

(Gauss divergenece theorem

Q D.ds (Gauss Law)

78. (c)

We know velocity

PV

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0 0

(as it is free space substituting in aboveequation)

P0 0

V

0 0

1

0 0

1

79. (d)

E / H , the characteristic Impedance ofthe medium.

0 r

0 r

= r

r

2120 1208

= 11202

120A

2

120 2120

A 2

PV

here 610 P1 CV

r

6

810 0.044 0.042

33 10

80. (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 E1rt x

81. (a)

The electromagnetic wave equation is givenby

22

2E EE 0t t

82. (c)

For free space and P are each zero

hence 1 and 2 are false

hence statements 3 and 4 are correct

H J, D e

83. (d)

This is a standard information

D layer - 70 KM

E layer - 110 KM

1F layer - 180 KM

2F layer - 25-400 KM

84. (c)

We know max

min

VS RV

= | V | | V || V | | V |

=

| V |1| V || V |1| V |

= 1 | |1 | |

r| | e k

= r

r

1 k1 k

85. (c)

The reflection co-efficient r(k ) , load

impedance b(z ) and characteristic

impedance 0(z ) are related as

L 0r

L 0

Z ZkZ Z

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86. (c)

Frequencies in UHF range propagated bymeans of space waves.

87. (d)

The wavelength along the wall is greaterthan in the actual direction of propagation.

88. (c)

We know

PV

0 0

(in free space r r1, 1

0 0

1

Hence velocity is independent of requency.

89. (b)

We know EH

andEH

120 for free space.

The phase of H in reflected componenthas to reverse so that of travel of thereflected wave is correctly given by thedirection E×4.

E H are perpendicular to each other inthe incident component/forward travellingcomponent and backward travell ingcomponent of wave.

ayH [120cos( t z)

120

30cos( t z)]

ay 1[2cos( t z) cos( t z)]2 2

90. (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

91. (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 andreflection co-efficient is zero.

Voltage is uniform along the line

max min| V | | V |

1 0SWR1 0

= 1 : 1

92. (b)

The given wave is travelling in the+Zdirection because of the presence ofthe term 2 z and E is entirely in x-direction.

Now E and H direction gives the directiongives the direction of travel of tower.

It is concluded that H must be in +ydirection.

x y za a a

Further

0

0

| E || H | ,

(infrespace),

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0| B | | H |

0| E || B |

00

0

| E |

0 0 | E |

7 914 10 10 | E |36

91 10 | E |9

8 8y

1 10 60cos(24 10 t 2 z)a3

8 8y20 10 60cos(24 10 t 2 z)a

8 8y20 10 cos(24 10 t 2 z)a

93. (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 thewave length therefore in lZ Z .

94. (d)

We know

2P

1

1 1tan3 3

P 30

and P 2

t 1

sinsin

t

sin30 1sin 3

t1sin 32

3sin t2

t 60

P 30 t 60

95. (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

96. (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.

97. (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 60e | |60 60

= 1

VSWR = 1 e1 e

= 1 11 1

10

98. (d)

Given xE 8cos t

yE 24cos( t 90 )

x| E | 8

y| E | 24

90

If x y| E | | E |

and 90

Then it is elliptically polarised.

99. (b)

for x2

line

in LZ Z 200

for line4

20

inC

ZZ

Z

2(100)50

100 10050

200

/ 2 line || / 4 line are parallel

in200 200Z 100200 200

for / 8 line in 0| Z | Z 100

100. (a)

We know the input impedance for

a / 4 line is

20

inL

ZZ

Z

but LZ , 0Z 230

in230Z

inZ 0

101. (a)

For a quarter wave

20

inL

ZZ

Z

20 in LZ Z Z

0 in LZ Z Z

given LZ 200

inZ 50

0Z 200 50

10000

0Z 100

102. (c)

given refractive index ( ) 1.5

frequency = 1410 HZ

Velocity of light in vacuum 8(Co) 3 10

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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

103. (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

104. (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

105. (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.

106. (c)

given j /2 jstE (0.5x ye )e

x| E | 0.5

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

So, x y| E | | E |

and phase = 90°

hence it is elliptically polarised wave

107. (c)

For a uniform plane wave

E and H must be in a plane

But both are perpendicular to each other.

108. (c)

109. (d)

110. (b)

111. (d)

Main memory size = 16 mB

= 10 1016 2 2

= 242 B

Physical address

tag set Block offset

10 x y

24 = 10 + x + y

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x + y = 14

Cache size = Number of set * line per set *block size

= x y2 4 2

= x y4 2

= 144 2

= 162

112. (c)

A BZ

C D E

Instruction Comment1. ADD Z A B z A B

2. SUB T D E T D E

3. MPV T T C T T C

4. DIV Z Z T Z Z T

113. (b)

114. (b)

115. (c)

TK,n = K n 1

K cycles are required to complete theexecution of the first instruction and theremaining (n–1) instructions require (n–1)cycles.

= 6 + (9 – 1) = 14

116. (d)

117. (d)

CPU generates 32-bit address

No. of addressable unit = 232 bytes

Block size = 8 bytes = 23 bytes

No. of blocks in main memory = 32

293

2 22

Block size = line size

Cache size = 1024 bytes = 210

210 = 2x+3

= 2(line field of r bits)+(block number of bits)

x+3 = 10

x = 7

size of tag = 29 – 7 = 22 bits

118. (b)

1298210

5298211

2299212

0012298

0009299

210 PC

AC

1298 IR

Main memory CPU register

Fetch Cycle

1298210

5298211

2299212

0012298

0009299

210 PC

AC

1298 IR

0012

1298210

5298211

2299212

0012298

0009299

210 PC

AC

1298 IR

0024

1298210

5298211

2299212

0012298

0009299

210 PC

AC

1298 IR

Main memory CPU register

Execution Cycle

1298210

5298211

2299212

0012298

0009299

210 PC

AC

5298 IR

0024

1298210

5298211

2299212

0012298

0009299

210 PC

AC

2299 IR

0024

0012

Load ac from memory

Add to ac from memory

Store ac to memory

119. (a)

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15 clock cycles need to serv ice 5instruction.

120. (c)

K n 1

4 2000 1 Max

[4+1999] × 250

= 500.75×10–6 micro second

121. (a)

122. (b)

Not attempt to read to and write from thesame register in one time unit becausethe result would be unpredictable.

123. (b)

124. (c)

125. (d)

126. (c)

Variation of mobility with temperature:

T3/2

impurityscattering

T–3/2

T

Lattice scattering

Variation of Drift velocity (vd) with electric field(E):-

Drift velocity, dv E, where is the mobility..

For n-type Silicon,

(i) For 3nE 10 V / cm, constant i.e. vd

varies linearly with E.

(ii) For 3 4 1/2n10 V cm E 10 V cm, E

i.e. 1/2dv E

(iii) For 4 1nE 10 V / cm, E i.e.vd= constant

vd v Ed1/2

v Ed

vd = constant

E

Variation of carrier concentration withtemperature:

n or p

Intrinsic Region

Extrinsicregion

Impurity ionization

T

Fermi-Dirac distribution function: It is given by

F

1f(E)E E1 exp

kT

f(E)T=0K

T>0K

EE F

1

0.5

(Fermi Energy)

127. (a)

Two types of scattering mechanisms exist in asemiconductor: Impurity scattering 1 and Lattice

scattering 2 .

The mobil ity, when both the scatteringmechanisms exists at the same time, can be foundfrom the relation

1

= 1 2

1 1

=

1 2

1 2

=

250 500250 500

= 166.67 cm2/V-s

128. (b)

In a degenerate n-type semiconductor, theconcentration of electrons in the conduction bandexceeds the effective density of states in theconduction band i.e. ND > NC.

Also, Fermi level of an n-type semiconductoris given as

CF C

D

NE E kT nN

l

Since NC < ND, EF lies in the conduction band.

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129. (b)

For a p-type semiconductor,

p = pF VA V

E EN N exp

kT

pF VE E =V

A

NkT nN

l

As NA increases, pF vE E decreases i.e.

Fermi level moves towards the valence band.

As temperature (T) increases, pF vE E

increases i.e. Fermi level move away from thevalence band and towards the middle of the energyband gap.

130. (d)

Let the diode be assumed to be forward biased.Then replacing the diode with its equivalent model,circuit becomes

9V

600

100

ID

300

0.7V 20V1

1

Applying KCL at node (1), we get

19 V600 =

1 1V V 0.7

300 120

V1 = 2.08V

Diode current, ID =

1V 0.711.5mA

120

( ID = +ve, diode is forward biased)

131. (d)

Law of the junction gives the relationshipbetween the minority charge carrier density at theedge of the space-charge region in terms of theapplied bias voltage. It is given by

n n0T

Vp (0) p expV

132. (d)

Avalanche diode is a lightly doped pn junction.Normal diode has higher doping concentration incomparison to Avalanche diode.

Zener diode is a highly doped p-n junction withdoping level of 1:105.

Tunnel diode is a heavily doped p-n junctionwith doping level of 1:103.

133. (b)

Let the transistor be assumed to be in activeregion of operation. Applying KVL in Base-emitterloop, we get

200k3k

5V

10V

IC

VC

5 – 200k × IB–VBE = 0

IB =BE5 V 5 0.7

200k 200 k

IB = 21.5 A

Collector current, IC = B coI (1 ) I

=100 (21.5 ) (1 100) 20n

= 2.152 mA

Collector voltage, VC= 10 – IC × 3k

= 10 – 2.152 m × 3 k

= 3.544 V

For the given npn transistor,

VCB = VC – VB

= 3.544 – 0.7 = 2.844

Since VCB > 0, npn transistor is working in Activeregion of operation. (Initial assumption is correct)

134. (b)

Let us assume that the MOSFET is working insaturation region. Then

Drain current, ID = 2n oxGS T

C W. V V2 L

= 22m.(1.8 1)2

= 0.64 mA

Drain voltage, VD = 3.3 – ID × 1k

= 3.3 – 0.64m × 1k

= 2.66 V

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Since VGS = 1.8 V > VT

and VDS = 2.66 > (VGS – VT) i.e.

(1.8 –1) = 0.8V,

MOSFET is working in saturation region.

135. (c)

Symbol of n-channel JFET :-

G

D

SIn JFET, the direction of arrow is pointed in

such a direction as if the Gate-source junction isforward biased.

Symbol of p-channel enhancement typeMOSFET:-

G

D

S

B(Gate)

(Drain)

(Body/Substrate)

(Source)Broken line in the symbol shows that no channel

is present during the construction of MOSFET.

Symbol of n-channel enhancement typeMOSFET:-

G

D

S

B

Symbol of p-channel depletion type MOSFET:-

G

D

B

S136. (b)

Output characteristic of JFET :-

ID

VDSCut-offregion

Ohmic/Trioderegion

Active/saturation

region

AvalancheBreakdownRegion

137. (c)In the saturation region of operation of the JFET,

drain current (IDS) is given by

IDS =2

GSDSS

P

VI 1

V

...(i)

where IDSS is the drain-to-source saturationcurrent at VGS = 0V,

and VP is the pinch-off voltageTransfer characteristic of the JFET is the plot

of output drain current, ID Vs. the input gate-to-source voltage, VGS.

From Eq. (i), in the saturation region, JFETtransfer characteristics are parabolic.138. (a)

An analysis of Avalanche breakdown for thecommon-emitter configuration indicates that thecollector-to-emitter breakdown voltage with open-circuited base, designated as BVCEO, is given by

BVCEO = nCBOfe

1BVh

where BVCBO is the collector-to-base breakdownvoltage with open-circuited emitter.139. (a)

I

RdH

P

dl

Magnetic field

dH at point P due to currentelement Idl is given by:

3

I dl RdH

4 R

140. (d)

(0, 2, 0)

(1, 0, 0)

Y

IX

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The direction of current is given by right handcurl rule.

Right hand thumb is placed along directionof current and fingers are curled and directionof magnetic field is along the tangent of thiscircle.

Direction of field is as shown in figure

At (1, 0, 0) it will be j

At (0, 2, 0) it will be i

141. (d)

Magnetic field

B due to infinitely long wire isgiven by:

0

1 2µ I

B sin sin4 a

Here, 1 290 and 0

0µ I

B sin90 sin04 a

= 0µ I k4 a

142. (b)

Magnetic flux is given by B A weber

2WeberB

mA.

Magnetic field intensity unit is Ampere/m.

143. (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)

144. (d)

Gauss Law is given by :

eneD ds Q

VB p

Magnetic field lines always exist in closedpaths and magnetic monopole does not exists.

E 0Electric field lines never exist in closed path

Hence,

E 0

145. (d)

Ampere Law states that

H dl Ienclosed.

H dl = – 30 + 10 =– 20A

–30A is taken because the path taken isopposite to the direction of magnetic fielddue to 30A current.

146. (b)

Magnetic monopoles does not exist andhence magnetic field lines are always closed.So, it has no sinks or sources.

S

B ds = or

B 0

Hence,

B is solenoidal but it is non-

conservative as

H 0.

147. (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.

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H1 cos and H2 cos will cancel each other..

Net field intensity = H1 2sin H sin

= 2H1 sin 30°

=

5 122 2

H = 5 A

m2 .

148. (c)

The tangent to an electric field line at anypoint gives the direction of electric field atthat point.

Electric field lines always cut a conductor inthe direction normal to the surface.

149. (d)

P S1

x

Ry

S2

Consider Gaussian surface S1 of radius xwith 0 < x < R

Qenclosed = 34 x3

E.ds

= encQ

E

=3

24 x3 4 x

E

= x3

E x

For the Gaussian surface S2 with radius ysuch that y R .

Qenc = 34 R3

E.ds

=enc

0

Q

E

=3 3

2 20 0

4 R R 13 4 y 3 y

150. (b)

Applying Gauss law to an arbitrary Gaussianspherical surface of radius r (a < r < b)

Q = 2rE.ds E 4 r

E = r2

Q a4 Er

Potential difference between the conductorsis :

V =a a

r r2b n

Q ˆ ˆE.d a .dr a4 r

l

V =Q 1 1

4 a b

Capacitance C = Q 4

1 1Va b

151. (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

152. (c)

C1

C2

C3

Here, C2 = 0 03 A 3 Ad d22

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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

153. (c)

Potential across the capacitor V = Q/C

Now, when dielectric is filled, capacitance ofthe capacitor increases to rC C .

Potential r r

Q Q VVC C

Potential decreases.

Electric Field, E = Vd

Electric field also resduces to r

VEd

154. (d)

V

AQ –Q

1 2

Electric field near plate 2 due to plate 1 willbe :

E =0 0

Q2 2A

Therefore, force on plate Z due to electricfield will be :

F

=2 2 2

0 0

Q C VQE2A 2A

C = 0Ad

F

=

2 20

20

A .V2 A .d

F

= 0 22A V

2d

155. (d)

Equivalent Admittance of the circuit,

Y = Y1 + Y2

= 1 j C

R j L

=

22

R j L j CR L

At resonance, imaginary part of admittance= 0

Im (Y) = 0

22L

R L= C

22R L = L/C

so, Equivalent impedance at resonance

Zeq =

22R L1Y R

= LC

R [From eqn (1)]

=L

RC

156. (b) For series R – L – C circuit, thecharacteristic equation is,

s2 + R 1s 0L LC

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n =1LC ;

n2 =RL

= R R CLC2L 2 L

for undamped, = 0

i.e. R = 0

for damped, < 1,

R C L1 R 22 L C

For critically damped, = 1,

R = L2C

for overdamped,

R >L2C

157. (d)

Series impedance, z = R + j (XL – XC)

At resonance frequency f = f0,

XL = XC

Circuit will be purely resistive

At lower half frequency i.e. f < f0,

XL < XC

Circuit will be capacitive in nature i.e.current leads the applied voltage.

At upper half frequency i.e. f > f0, XL > XC

Circuit will be inductive in nature i.e.current lags the applied voltage.

158. (c) For circuit to cause parallel resonance,

Y = Y1 + Y2

0 = 2

1j cz

z2 = 1

j c

= j c

= 61j500 102

= j 0.250 m

159. (a)

The frequency at which the voltage acrossthe capacitor is maximum,

maxcf =

2

21 1 R

2 LC 2L

which is below the resonance frequency, f0 =

1

2 LC

The frequency at which the voltage acrossthe inductor is maximum

maxLf =

2 2

1 12 C RLC

2

which is above the resonance frequency, f0 =

1

2 LC

160. (d) Quality factor,

Q =resonant frequency

Bandwidth

=0ff

=

6

320 10

100 10

= 200

161. (b)For a passive network, the output powercannot be greater than the input power.

162. (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

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Z12 = 11

2

V ZI

Z22 = 21 2

2

V Z ZI

Z-parameters are 1 1

1 1 2

Z ZZ Z Z

163. (d)

A

B

CinCin

Cin = in

in

C C CC C

CAB = in

1 5C C2

164. (a)When in series, we have to add the Z-parameters to combine the two, two network.

165. (a)L-C network has only energy storage element.Power dissipation elements ‘R’ is not there,so the poles and zeros lie on the imaginaryaxis. Network function should be determinedby zero initial condition.

166. (b)

2Ai

6A 1A

4A3A

a

Applying KCL at node a, we get

6A 1A 4A 3A 0

i = 2A

167. (c)Charge area under the graph

Charge =1 C2 5 1 12

= 5 – 1

= 4 C

168. (c)

Li 0 = 6 3A2

Req = 400200 200

= 400 400

= 200

Leq = 20 mH

Li = eq

eq

Rt

Lfinal initial finali ei i

= 3200 t

20 100 e3 0

= 10000t3e

169. (c)

0V 0 = C1 C2V V0 0= 10 + 15 = 25V

CV = 3 10 6V5

Req = 65

= eq eq6 0.6R C 0.15 5

Ceq = 0.1F

0V t = t/6 e25 6

= 8.33t6 19e V

170. (d)

171. (c)

A is true

But R is not true

Because curl is studied for only opensurfaces and not closed surfaces.

172. (b)

A is true because

P0 r 0 r

V

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0 r 0 r

1

Pr r

CV

0 0 0 r

2 22 f

r r

C

R is also true

173. (a)

174. (b)

175. (a)

For a good Tunnel diode, the essentialrequirements are large (IP/IV) ratio and large voltageswing.

P V

Ge GaAs SiI I 8 15 3.5

Since (IP/IV) ratio of Silicon is less, Tunnel diodeis fabricated with Germanium or Gallium Arsenidematerial and not with Silicon.

176. (a)

Below pinch-off region, when VDS is small, JFETworks as a voltage-controlled resistor i.e., the drain-to-source resistance is controlled by the biasvoltage, VGS. JFET is referred to as Voltage VariableResistor (VVR) or Voltage Dependent Resistor(VDR).

VGS3

VGS2

VGS1

VDS

ID

JFET as a VVR

JFET, as a VVR, can be used to vary thevoltage gain of a multistage amplifier, as the signallevel is increased. This action is called AGC orAutomatic Gain Control.

177. (b)

Increase in magnitude of collector reversevoltage increases the space-charge width at theoutput junction. Since base region is lightly doped,more penetration of transition layer takes place inthe base leading to the reduction of the effectivebase width. This phenomenon is referred to as Earlyeffect or Base width modulation.

As the effective base width is reduced, there isless recombination in the base region and thus,

transport factor * and increases.

178. (a)

Collector region of a BJT is made physicallylarger than the emitter region. This is because muchheat is generated at the collector junction and thusit has to withstand the temperature generated atthe collector.

179. (a)

In saturation region of operation in a MOSFET,drain current has a slight variation with respect tochange in drain-to-source voltage.

Drain current, ID = 2n oxGS T

C W (V V )2 L

DS1 V ,

where is called the channel length modulationcoefficient.

ID

IDSS

VDS

As the drain-to-source voltage (VDS) is increasedbeyond the pinch-off point, effective length of thechannel between the drain and source decreases.This phenomenon is called as channel lengthmodulation. Due to this, ID depends on the appliedvoltage VDS slightly and is represented by a slopeon the output drain characteristics.

180. (b)

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