1. (c)

2. (a)

3. (a)

4. (b)

5. (a)

6. (c)

7. (c)

8. (b)

9. (d)

10. (c)

11. (c)

12. (b)

13. (d)

14. (d)

15. (d)

16. (a)

17. (b)

18. (a)

19. (c)

20. (b)

21. (c)

22. (c)

23. (a)

24. (d)

25. (b)

ESE-2019 PRELIMS TEST SERIESDate: 25th November, 2018

26. (d)

27. (c)

28. (c)

29. (c)

30. (b)

31. (c)

32. (b)

33. (c)

34. (b)

35. (b)

36. (b)

37. (b)

38. (b)

39. (b)

40. (a)

41. (b)

42. (b)

43. (a)

44. (c)

45. (c)

46. (c)

47. (c)

48. (c)

49. (b)

50. (d)

51. (a)

52. (a)

53. (a)

54. (c)

55. (d)

56. (d)

57. (c)

58. (d)

59. (d)

60. (d)

61. (b)

62. (a)

63. (b)

64. (a)

65. (b)

66. (c)

67. (c)

68. (b)

69. (b)

70. (a)

71. (c)

72. (c)

73. (a)

74. (c)

75. (d)

ANSWERS

76. (c)

77. (b)

78. (d)

79. (b)

80. (b)

81. (d)

82. (b)

83. (d)

84. (c)

85. (c)

86. (c)

87. (b)

88. (d)

89. (a)

90. (c)

91. (b)

92. (b)

93. (d)

94. (b)

95. (d)

96. (d)

97. (b)

98. (d)

99. (c)

100. (d)

101. (b)

102. (a)

103. (b)

104. (d)

105. (a)

106. (c)

107. (b)

108. (b)

109. (b)

110. (b)

111. (a)

112. (a)

113. (c)

114. (b)

115. (d)

116. (d)

117. (c)

118. (c)

119. (b)

120. (a)

121. (a)

122. (b)

123. (d)

124. (a)

125. (a)

126. (b)

127. (d)

128. (a)

129. (a)

130. (b)

131. (a)

132. (b)

133. (d)

134. (b)

135. (d)

136. (a)

137. (d)

138. (a)

139. (c)

140. (c)

141. (c)

142. (d)

143. (b)

144. (c)

145. (c)

146. (c)

147. (d)

148. (d)

149. (a)

150. (a)

IES M

ASTER

(2)

1. (c)

3 2C(s) 10s 40U(s) s 4s 3s

Denominator : s3 + 4s2 + 3s + 0 (with negativesign)

0 1 0 0A , B0 0 1 0

0 3 4 0

Numerator : 10s + 40 (with sign intact)

C 40 10 0

y = 40x1 + 10x2

1 2

2 3

3 1 2 3

x xx xx 0x 3x 4x 4

0 1 0 0

A , B , C0 0 1 0 40 10 00 3 4 1

2. (a)

oi

i f

0 VV 0R R

or o f

i i

V R 1V R k

1Tk

Therefore sensitivity

TK

KT KS (T)K TK T

TK

K1or SK K 1/ K

= 2

21 ( K ) 1

K

TKor 1S

3. (a)Let y = x1 ... (1)

2 1dy x xdt

... (2)

2

3 22d y x xdt

... (3)

3 2

2 2d y d y dyu 6 11 6y

dtdt dt

3 3 2 1x u 6x 11x 6x ... (4)Using eqns (2), (3) and (4)

1 1

2 2

3 3

x x0 1 0 0x x u0 0 1 0x x6 11 6 1

1y x1 0 0

0 1 0 0

A , B , C0 0 1 0 1 0 06 11 6 1

4. (b)

1 2x y x u ... (1)

2x y u

As y u u 2y y

= u – 2x2 – 2u – x1

= –x1 – 2x2 – u

2 1 2x x 2x u

1 1

2 2

x x0 1 1u

x x1 2 1

5. (a)

+–

L1

i1i2

L2

R1 R2V

e

1 2 3

C

KCL at node 2

1 2dvi i C 0dt

KVL in left loop

11 1 1

diL R i e v 0dt

KVL in right loop

22 2 2

diL R i v 0dt

1dvx (t)dt

IES M

ASTER

[EE], ESE-2019 PRELIMS TEST SERIES PAPER-II (TEST-10) (3)

12

dix (t)dt

23

dix (t)

dt

1 2dv 1 1i idt c c

... (1)

1 11

1 1 1

di R1 1v i edt L L L

2 22

2 2

di R1 v idt L L

Therefore,

1 1 1

2 2 2

0 1/ C 1/ C1/ L R / L 0A

1/ L 0 R / L

6. (c)

Bode plot of lead compensator can be used toexplain the increase in bandwidth.

dB

0 1/ 1/ log

20dB/dec120 log

Thus lead compensator increases the crossoverfrequency. As crossover frequency is related tobandwidth hence, bandwidth also increases.

Phase lead compensator is equivalent to additionof a dominant zero, so transient response improvesi.e. rise time tr reduces, over shoot decreases etc.

The one disadvantage of phase lead compensationis that steady-state error is not improved.

7. (c)

To obtain the transfer function

2o

i1 2

1RE (s) sc1E (s) R R

sc

= 2

1 21 2

222

1s

R C1

1sR RR R

R CRR

Let 1 22

2

R RR C ; 1

R

, then

C

1s1 1 jG (s) 1 1 js

1 1tan tan

For 1 , phase output lags to the input i.e. it isa lag network.

Clearly, pole at 1 1 for 1

8. (b)

TF from bode plot :

2

1

sk 110G(s)s 1

20log k = 20

or k = 10

Also

1w20 20log 26 at10

120log 610

1or log 0.310

1or 20

Also

220 20log 40log 0 at10 20

22 2or 1 log log 0

10 20

222

400or log 1

10

2 2

40 40 1or log 110

2 400

9. (d)

IES M

ASTER

(4)

2skeG(s)s(s 1)(4s 1)

1 + G(s).H(s) = 0

4s3 + 5s2 + s(1 – 2k) + k = 0

By Routh’s array3

2x

1

0

s 4 1 2k

s 5 ke 1 x

5(1 2k) 4k x 1s5

s k

For stability

5(1 2k) 4kk 0 and 05

or 5(1 – 2k) – 4k > 0

or 5k

14

or k < 0.36

Maximum value of k is 0.36

10. (c)

Re

Imga

–1 0

1GM 20loga

120 20loga

1or log 1a

1or 10a

or a 0.1

11. (c)

Consider a typical G(j )H(j ) locus

u

jv

–90°

a

2

–1+j0

–180°

Unit circle(originas center)

1

As G(j )H(j ) locus approaches –1 + j0 point,the relative stability reduces. The relative stabilitycould thus be measured in terms of intercept a orthe angle .

12. (b)

Characteristic equation is

Aa 4 11 k 01s s 4 0.5s

or s2(s + 4) + 8kA(s + a) = 0

or s3 + 4s2 + 8kAs + 8akA = 0

By Routh’s criterion :

3A

2A

A1

A A0

A

s 1 8k

s 4 8ak1 2ak

s 8k 2ak

s 2ak

For stability

8kA – 2akA > 0 and 2akA > 0

If kA > 0, then

a < 4 and a > 0

if kA < 0 then

a > 4 and a < 0

which is not possible

Therefore, 0 < a < 4

13. (d)

j

–

–5 –3 –2 –1

IES M

ASTER

[EE], ESE-2019 PRELIMS TEST SERIES PAPER-II (TEST-10) (5)

There are three root locus branches starting eitherfrom pole or infinity

Number of asymptotes = |p – z| = |2 – 3| = 1

14. (d)

From the characteristic equation, the gain k isgiven by

k = –s(s + 1)(s + 2) = –s3 – 3s2 – 2s

Differentiating,

2dk (3s 6s 2)ds

For 2dk 0 3s 6s 2ds

1, 26 (36 24)

s 0.423, 1.5776

Drawing, pole-zero plot

–2 –1 0

Root locus lies between (–1 and 0) and –2 to So valid breakway point is –0.423

15. (d)

G(s)H(s) has four finite poles, and zero number offinite zeros. If root locus is drawn, then a rootlocus branch starts from a pole and terminates ata finite zero or infinity.So here for four poles, number of zeros at infinitywill be four as all the four root locus branchesterminates at infinity.

16. (a)

System gain and damping factor are related ininverse manner i.e.

1k

So, as k increases, (damping) decreases, sosystem stability reduces.

17. (b)

Closed loop transfer function of the system is

2C(s) kR(s) s(s s 1)(s 4) k

Thus characteristic equation is

s(s2 + s + 1)(s + 4) + k = 0

or s4 + 5s3 + 5s2 + 4s + k = 0

By Routh’s criterion :

4

3

2

1

0

s 1 5 k

s 5 4 0

s 21/ 5 k

(84 / 5) 5ks21/ 5

s k

For stability

84 5k5 0 and k 0(21/ 5)

84or 5k 0 and k 05

84or k and k 025

Thus 840 k25

18. (a)

For unit ramp input

ssv s 0

1 1ek lim sG(s)

Here, e10(1 sk )G(s)

s(s 2)

Thus, es 0 s 0

10(1 sk ) 10lim sG(s) lim 5s 2 2

Therefore, ss1e 0.25

19. (c)

kGT1 GH

Sensitivity, TH

T HSH T

= HkG1 GH

H kG1 GH

= 2

2kG H GH

1 gHkG 1 GH(1 GH)

If GH >> 1, TH 1S

System is less sensitive to variation in H i.e.better performance.

IES M

ASTER

(6)

TK

T k kkGS . 1 GHk T k kG1 GH

= G k 11 GH1 GH kG

20. (b)

P + D control impacts transient response, i.e.,damping ratio (increases), peak overshoot (reduces)etc.

P + I control action improves steady state responsei.e. reduces the steady state error.

Derivative feedback control action :

+

–R(s) 2

n

ns(s 2 )

C(s)

sK

E(s)+

–

2n

2 22nn n

C(s)R(s) s s2 k

(No addition of

zero)

n tk2

(damping ratio increases)

21. (c)

ss s 0

sR(s)e lim1 G(s)H(s)

Type of system increases i.e. number of integration(1/s term) increases in G(s). Thus as

sss 0, G(s) , hence e 0

22. (c)

Steady state errors for various inputs and systemtypes.

p

v

a

Type of input Steady state errortype 0 type 1 type 2

1Unit step 0 0(1 k )

1Unit ramp 0k

1Unit parabolick

p s 0k lim G(s)

b s 0k lim sG(s)

2a s 0

k lim s G(s)

23. (a)

Using the freebody diagram of mechanical system

t t

11 1 1 1 1 2 1 2

2 1 2

dvf v k v dt M k dtv vdt

f (v v ) 0

and t

22 2 2 2 12 1

dvM k dt f (v v ) 0v v

dt

= F(t)

From electrical network

t t1 1

1 1 1 21 1 2

1 2

2

e de1 1e dt C (e e ) dtR L dt L

(e e )R

and

t2 2 1

2 2 12 2

de (e e )1C (e e ) dt i(t)dt L R

Comparing the mechanical and electrical equations

1 1 1 1 1 11 1

1 1R , L , C M , e vf k

2 2 2 22 2

1 1L , R , e vk f

C2 = M2, i(t) = F(t)

24. (d)

Heat flow rate = Heat energy

time

Current = Charge flow

timeAs current flow is due to potential difference orvoltage gradient, in the same manner, heat flow isdue to temperature difference or temperaturegradient.

25. (b)

26. (d)

Transfer function =

C s T GR s

Sensitivity of open transfer T with respect to G isgiven by

IES M

ASTER

[EE], ESE-2019 PRELIMS TEST SERIES PAPER-II (TEST-10) (7)

TGS =

dT T dT G G. 1. 1dG G dG T G

...(i)

Transfer function =

C Gs TR 1 GHs

Sensitivity of closed loop transfer T with respectto G is given by

TGS =

dT T dT G.dG G dG T

= 21 G 1

G 1 GH1 GH1 GH

...(ii)

Hence equation (i) and (ii), clear that with anegative feedback in the closed loop control system,the system sensitivity to parameter variationdecreases.

27. (c)For stability of negative feedback closed loop controlsystem output linearly varies with input, whenany non-linearity should not be introduced by thecomponent of control system.

28. (c)Using feedback of a system,• Bandwidth increases• Distortion decreases• Gain decreases• Accuracy of the system increases• It used to stabilized the unstable system but

some cases it unstablize the stable system.

29. (c)• If a system is sensitive to parameters (i.e

temperature, pressure etc.), it will give differentoutput for same input at different time,temperature or pressure.

• If a system is insensitive to input command, itwill give same output at different inputs.Hence for a good control system, it insensitiveto parameters and sensitive to input command.

30. (b)The transfer function is applicable to linear time-invariant (LTI) system with initial conditions arezero.

31. (c)

2G R s 1G

C(s)

2

2

G1 G R s 1G

C(s)

Transfer function =

1 2

2

G GC sR 1 Gs

32. (b)Note: Transient response depends more on thepoles near to the origin and steady state dependson all poles.

Approximated transfer function of the system

= 1000

s 20 25s 1

= 2

s s 1

33. (c)

V0(s) = iR sLI (s)R sL

0

i

V sI s =

R sLsLR 1R

=

sL11 sLR

So, the block diagram will be,

sLI (s)i V (s)0

1R

34. (b)

s 1

R(s) E(s) C(s)

0

s 1 E s

s 1

E ss 1

R(s) C(s)1s 1

IES M

ASTER

(8)

C sR s =

1s 1

35. (b)

At G j 180º gain is –3dB, so gain margin is3dB.At 0dB, phase = –150º,So phase margin = 180 – 150

= 30º

36. (b)A phase lead network has

G(s) = 1 sT ; 11 s T

37. (b)Proportional derivative (PD) controller increasesthe damping ratio of the system. Thus, it improvesthe transient response and deteriorates the steady-state response.PD controller has high sensitivity because theoutput of the system has an additional term whichvaries in proportion to its derivative i.e.

Output of the system = p D

dc tK c t Kdt

38. (b)Integral controller improves the steady stateaccuracy and derivative controller increase thedynamic response. As we need to have best steadystate then integral controller must be used.

39. (b)Resonance magnitude

Mr = 21

2 1

Given 0.5 than

Mr = 21

2 0.5 1 0.5

= 10.866

= 1.15

40. (a)When Nyquist plot of G(s)H(s) passes through

1, j0 then it implies that

gain cross over frequencyG Hs s 180

Phase margin = 180° – 180° = 0°

41. (b)

Open loop transfer function = 10

s s 2Putting s j ,

G j = 10

j j 2

G j =

1

1 1

tan 0

tan tan2

= 1 1tan tan2

= 1tan2 2

0 G j 2

0 G j

PM = to2

= 0 to2

42. (b)

–6 –4 –2

Break Point

Root Locus

43. (a)When a pole is added to system loop transferfunction then the type of system increases astype of system increases steady state errordecreases. And speed of system is proportionalto steady state error hence system becomesslower.

44. (c)

IES M

ASTER

[EE], ESE-2019 PRELIMS TEST SERIES PAPER-II (TEST-10) (9)

t s 0Limy .LimsGt s

and the input is given at instant t = 2

= 2ss 0LimsG es

= 2s

2s 0

1Lims es s 2s 2

=12

= 0.5

45. (c)Characteristic equation of system

1 G s = 0

4 3 22s 5s s 3s 1 0

Routh-Hurwitz criterion

4

3

2

1

0

s 2 1 1s 5 3s 1 5 1s 28s 1

Negative coefficient in first column of R-H criterionso system is unstable.

46. (c)

0s 6 j8 s 6 j8

2 28 0s 6

2 2s 2 6 s 10 0

n = 10

=2 62 10

= 0.6

47. (c)

48. (c)Settling time of a system

at 2% tolerance = n

4

at 5% tolerance = n

3

49. (b)No. of roots of characteristic equation in right

side of s-plane is equal “No. of sign changes inconsecutive rows in the first column of Routh’sarray”. In given Routh’s array coefficient of firstcolumn.

4 6 –2 –3 4 –7 –8

3 times sign change occurs in routh’s array firstcolumn therefore 3 roots in right side of s-plane.

50. (d)

IL =1 VdtL

=6

gV T 50 6 10L 0.2

= 1.5 mA

51. (a)Rating of thyristor = 400 V, 50 ADRF = 20 %Required overall rating = 8 kV, 800 A

DRF = 1 – String effeciency For series connection,

DRF = 1

8 10000.2 1400 n

20n 250.8

For parallel connection,

DRF = 1

8000.2 150 n

16n 200.8

52. (a)

Ambient temperature = 25 °CThermal resistance = 2 °C/WPower dissipation = 60 W

25602

120 25 145 C53. (a)

2 2j j

dvI Cdt

IES M

ASTER

(10)

2

2

3j

12j

Idv 16 10dt C 20 10

= 800 V s

54. (c)

Diode can block reverse voltage & can conductonly forward currentSCR can block bidirectional voltage & canconduct only forward currentTriac can block bidirectional voltage & canconduct bidirectional currentMOSFET can block forward voltage & canconduct bidirectional current

55. (d)

• BJT is a current driven device.• MOSFET is a voltage driven device.

56. (d)

57. (c)

58. (d)

Thyristor opens if holding current IH goes below5mA. 0.7V is the drop out point.

V = 3 30.7 5 10 1.5 10

= 8.2V

59. (d)

Ripple factor = ripple voltage rms

dc voltage

= 2 1

502= 0.028

60. (d)In half wave rectifier

Idc =

m

L

E 1 cos2 R

Idc =200 31

2 50 2

= 1.1879

Power = 2dc LI R

= 70.6 W

61. (b)

tan =L

R

= 3100 12.73 10

4 1

= 45º

should be greater than , i.e. greater than45º

> 45º

62. (a)

tna =L 10 1

R 10

= 45°So, firing angle ‘ ’ must be higher than 45°,thus for for 0 45 , V0 is uncontrollable.

63. (b)The SCR has a typical turn off t ime inmilliseconds. So for higher frequencies thenegative half cycle will be of less time than therequired turn-off time. So upto 300Hz ac supply,turn-off is possible using this method.

64. (a)IL = Ih × 3

18 mA = Ih × 3

Ih =18mA

3 = 6 mA

65. (b)

f =1T

T = lT T1R C n

1

= 470×103×0.01×10–6 ln1

1 0.7

= 47×10–4×1.2= 5.64 ms

f = 31

5.64 10 = 177 Hz

66. (c)The switch given can pass the current in eitherdirection and block it in forward direction.

67. (c)

IES M

ASTER

[EE], ESE-2019 PRELIMS TEST SERIES PAPER-II (TEST-10) (11)

Irating =

ON0

TIT

=

0 ONV E TR T

=

sV ER

rdId =

s2 V ER = 0

=s

E2V

r maxI =

s

s

s

EV E2V E

R 2V

=2

s

E4V R

=

50 50

4 100 10

=58 = 0.625 AA

68. (b)

Type Voltage CurrentType ATypeBTypeC ,TypeD ,TypeE , ,

69. (b)

Max steady state ripple = sV4fL

This can be decreased by :i) Reducing voltageii) Increasing frequencyiii) Increasing inductance

70. (a)

Modulation index =r

c

VV

=100150 =

23 = 0.66

cfN 12f if zero of triangle wave coincides

with zero of reference.

N = 1500 1500 = 2

71. (c)Input voltage of chopper is continous, where asoutput voltage is discontinouse

V0 Vin

72. (c)Output Voltage = Input voltage × duty ratio

= 250 × 0.14= 35V

73. (a)

T =1f

Duty cycle = ONTT

= onT f

= 2ms × 200= 0.4

74. (c)

half

full

PP =

14

Pfull = 4 × Phalf

= 4 × 20= 80 W

75. (d)Line commutated inverter gives fixed voltage andfrequency. It requires at its output an existing acsupply which is used for its commutation. Forcedcommutated inverter provides an independentedac output voltage of adjustable voltage andfrequency.

76. (c)Vs = 50V – 2V = 48V

Maximum rms output voltage at fundamentalfrequency,

IES M

ASTER

(12)

V0 =

s4V2

=

4 482

= 43.2 V

77. (b)In a PWM method of chopper switching, thefrequency is kept constant and TON is varied.Accordingly TOFF is also varied.

78. (d)

Per unit ripple

0.5 Duty ratio

The graph of a typical ripple versus duty ratio.

79. (b)A type-B chopper is a step up regenerativechopper.

V

I

working area of type-B chopper

80. (b)The maximum ripple occurs at D = 0.5, and isgiven by

mxI = sV4fL

= 3100

4 1 10 0.1 = 0.25 A

81. (d)In a voltage source inverter, voltage is fixed andthe output voltage wave form depends only onthe switching sequence. Current waveformdepends on the load impedance.

82. (b)

45 2 t

2 400 2 200

Output voltage is

V0 =

45

m2 m10 45

1 V sin t V sin t2

= 45 1800 45

2 200 cos t 2cow t2

= 166.86V

83. (d)Current i(t) through the thyristor

i(t) = 00

400 sin tL

0

1LC

VL(t) = Ldi t

dt= 0400cos t

VC(t) = VC – VL

= 0300 400cos t

VL(0) = 400VVC(0) = –100V

84. (c)The following is the power electronic controllerand application.

85. (c)For first quadrant operation, a half controller ismost preffered. It is simple and cheap comparedto other converters.

86. (c)1) Bridge type - 8 thyristors2) mid point type - 4 thyristors3) Three-phase to single phase - 6 thyristors4) 3 phase to 3 phase, 3 pulse type uses 18

tyristors

87. (b)Static VAr does not have any effect on the loadreactive power requirement. It helps in improvingvoltage profile and power factor.

88. (d)A cyclo-converter is a one-stage frequencychanger. Cyclo converters are of two types (a)

IES M

ASTER

[EE], ESE-2019 PRELIMS TEST SERIES PAPER-II (TEST-10) (13)

step-down (a) step-up

89. (a)

If the firing angle is less than , then stillthe thyristor is in reverse biased, so it will not

turn on . At , the controller cannot becontrolled and control is lost.

90. (c)A motor will require only fundamental and anyharmonics present would degrade theperformance of motor. Heater can takefundamental and harmonics to convert electricalenergy to heat energy.

91. (b)

Given firing angle 30 , Load phase angle

1 10tan 4510

. As the load phase angle

is greater than firing angle, the voltage controlleris not controlled.

Current =V 230 23Z 10 2 2 A

92. (b)Only when firing angle is greater than load phaseangle, output power can be controlled, other wise,the output cannot be controlled. Conduction angleshould be lower than .

93. (d)Integral cycle control is advised where the loadhas very high time constant like heating.

94. (b)

=a a a

v f

V R Ik I

= 220 0.5 44

0.7 2

= 141.4 rad/sec

95. (d)The average power, maximum allowable gate drivepower is related by

avggatedrivepower

PP

so, Pgdp = avgP

=1000.4 = 250W

96. (d)

Vph =4002 = 200V

V1 = phmr V sin

m

155 =3r 200 sin

3

r = 0.937

97. (b)

Vor =nV

m n

given n = 5 & m = 5, V = 230V

Vor =5230

5 5

= 162.63V

Pload =2

orVR

= 2162.63

10= 2.64 kW

98. (d)Vt = Ea + Iara

= m m a aK I r

600 = m2 1500K 80 1

60

Km = 3.33 V-s/rad (or) Nm/A

a3 2 400 3.33 2 12001 cos45 I 1

2 60

Ia = 45A

99. (c)given that Km= 2 Nm/A

Torque Te = KmIaTe is given as 70 Nm

Ia =e

m

TK

= 70Nm2Nm

A

= 35A

100. (d)

The relation between electric flux density and

IES M

ASTER

(14)

polarization is given by,

D .P

1

D = 2.5 2 a1.5

D = 2

3.33 nCam

at 5 cm,

D = 3.33 a0.05

D = 66.6 a 2nCm.

101. (b)

Let the potential of equipotential surface is V0.

So, V0 = 100 ln tan2

(or) 0ln tan 0.01V2

01 0.01V2tan e ...(1)

now, as V0 is constant, so the equation (1) will

be of the form, constant , which represents

a conical surface.

102. (a)

v .D

; assuming spherical coordinates

421 . 5r

rr

v 20r

at r = 0.06m

v 20 0.06

3v 1.2mC / m

103. (b)

The vector A is given by

x y zA 1 0 a 2 0 a 3 0 a

= x y za 2a 3a

Vector B is given by,

x y zB 2 0 a 3 0 a 2 0 a

x y zB 2a 3a 2a

x y zA B a a 5a

Unit vector in direction of (A –B)

= x y za a 5a

27

= x y z0.19a a a0.19 0.95

104. (d)

The relation between cylindrical and cartesianco-ordinate is given as

e = 2 2x y ,

e cos x,

e sin y

G = x y21 ecos a esin a

e

G = x y1 cos .a sin ae

again Ge = eG.a

= x e y e1 cos a .a sin .a .ae

ex aa . cos

ey aa . sin

2 2

eGcos sin 1

e e

G G.a

= x y1 cos .a .a sin .a .ae

IES M

ASTER

[EE], ESE-2019 PRELIMS TEST SERIES PAPER-II (TEST-10) (15)

= 1 cos . sin sin .cose

= 0

z z y zz x1G G.a cos a .a sin .a .a 0e

e ze zG G a G a G a

eG1 a 0 0e

105. (a)

The volume charge density is given by,

19

v 15ep.e 0.27 1.6 10V 10

v 3.Ce 43.3

m.

106. (c)

When battery is removed, the charge won’tchange with change in plate seperation

So, the surface charge density (es) will remainsame (as the plate area is constant)

The electric flux density D = es, will remainunchanged.

107. (b)

The charge density is given by.

x y zx

e .D D D Dy z

x3

xe C1

m.

The net charge enclosed within the surface is,

Q = e.V = 1 × 34 0.0033

= 1.13 × 10–7C

Q = 113 nC

The flux passing through surface of shpere,Q 113nC

108. (b)

v 0e D .E

= 3 3

202 2

0

2a1 . e3

= 202

0

1 1. 3 e3

e v = – e0

109. (b)

The relation between polarization and electricfield intensity is given by

P = 0P 1 .E

9

120

EP 2 10

1 8.85 10 3.5 1

VE 90.4 m

110. (b)

Force on electron is given by,

Fe = qe E = m. a

31 12

e19

eE

m a 9.1 10 9.67 10q 1.6 10

E = 55Vm.

The electron moves in a direction opposite tothe field,so the field is from B to A.

111. (a)

The relation between applied electric field andpolarization is given by

P = 0 1 .E

9

123.54 101

8.85 10 200

3

0

0

Permittivity of dielectric 3Permittivity of air

112. (a)

IES M

ASTER

(16)

The component of 1E

, normal to surface x = 0,

is given by. x x1E .a a

1NE

= x y z x x80a 60a 30a .a a

= x80a

1 1T 1 NE E E

1 y zTE 60a 30a

As, the tangential component of electric field iscontinous, so

2 1 y zT TE E 60a 30a

113. (c)

The cpacitance of a co-axial cables type capacitoris

12C

bn a

ll

for unit length, 12C

bn a

l

The capacitance of a parallel plate capacitor is,

2AC

d d

l

for unit length, 2Cd

As the two capcitors store the same energy whenapplied with same voltage, so C1 = C2

2

b dn a

l

(or) b 2dna

l

2db ea

114. (b)

The charge in a capacitor is

Q = C.V = 0 A.V

d

Q = 0E . A

The charge stored in the capacitor is directly

proportional to the product of and E

For air : E 30kV / cm

For barium titanate : E 36000kV / cm.

For silicon dioxide : E 604.8 kV / cm.

For polyetheline: E 604.8 kV / cm.

So, we can say that barium titanate will storemost amount of charge for given plate area.

115. (d)

As A and B are at the same potential, the workdone will be zero.

116. (d)

34n r3

= 34 R3

R = 1/3rn

1qVr

, 2nqVR

2

1

VV =

nrR

= 1/3nr

n r

=2/3n1

117. (c)

According to gauss law

= Qenclosed

= Q1 + Q2 + Q3

118. (c)

Point B and C are at equipotential.

So, ABW = ACW

119. (b)

IES M

ASTER

[EE], ESE-2019 PRELIMS TEST SERIES PAPER-II (TEST-10) (17)

Equipotential surface

Lines of force

So, they are perpendicular.

120. (a)

v =r r

c

=83 10

4.5 2

= 81 10 m s

121. (a)

122. (b)According to Maxwell’s equation

H = J

So, J =

x y zˆ ˆ ˆa a a

x y z10x z 12

=

x y zz z 10x12 10x12ˆ ˆ ˆa a a

y z x yx z

= xa

123. (d)

Resistance, R = VI =

lE d

J ds

=

lE d

E ds

Capacitance, C =QV =

l

E ds

E d

So, RC =

l

l

E d E ds

E ds E d

i.e. RC =

124. (a)

++++++++++++++++++

E

r

According to Gauss’s Law,

E ds =

0

Q

lE 2 r = l

0

E = r

0a

2 r

y

(0,1)

1x

Line-1

Line-2 (0,–1)

0 (0,0)

–

So, electric field at the origin by line-1,

1E = y

0a

2 r

= y0

a2

[ r = 1]

and electric field at the origin by line-2,

2E = y

0a

2 r

= y0

a2

[ r = 1]

so, resultant electric field intensity at theorigin

E =

1 2E E

= y0

2 a2

IES M

ASTER

(18)

= y0

a

125. (a)Potential at a point due to an electric dipole,

V =

20

ˆp a4 r

Here, ra = x y zˆ ˆ ˆ2 0 a 2 0 a 1 0 a

= x y zˆ ˆ ˆ2a 2a a

r =

ra

= 2 2 22 2 1= 3

Now, potential

V =

x y z x y z 9

20

ˆ ˆ ˆ ˆ ˆ ˆ3a 2a a 2a 2a a10

4 4 14 4 4 1

V = 9 99 10 6 4 1 10

9 3

=

9 39 3

= 1 volt

126. (b)Energy stored in the field,

E = 1 1 2 21 1q V q V2 2

E =1 2

0 2 1

q q4 r r

= 9 6 6

2 2 29 10 1 10 4 10(1 2) (3 1) ( 1 5)

= 5.14 mJ

127. (d)Total flux passing through the whole cube

= Total charge enclosed= 60 mC[According to Gauss’s Law]

The flux passing through one face of the cube

=Total flux

6

=60 mC6 = 10mC

128. (a)

H = I

2 r

r =I

2 H =

12 1 =

1 m2

129. (a)1. Kirchoff’s current law : Rate of flow of charge

entering at a node is equal to rate of flow ofcharge leaving the node.

2. Ampere’s Law : H.d I

l

3. Faraday’s Law : dd BAEdt dt

4. Gauss’ Law : D.ds q

130. (b)

The ampere’s circuital law in Integral form is

enclosedH.dl I

This is also called as maxwell’s equation forstatic field.

Another equation for amperde circuital law inintegral form for time varying field is

DH.dl I dst

This is also called as maxwell’s equation fortime varying field.

131. (a)

When ampere’s law is applied to differentialsurface we get

H J

This is also called as maxwell’s equation instatic field.

132. (b)

We know

H J

2

i j k

Hx y z

x 2yz ( x)

IES M

ASTER

[EE], ESE-2019 PRELIMS TEST SERIES PAPER-II (TEST-10) (19)

=

i ( x) (2yz)x z

2j ( x) (x )

x z

2k (2yz) (x )

x z

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

j 2y i j

x rcos y r sin in cylindrical form

6cos45 y 6sin45

= 62

= 6 0.707

x = 4.24, y = 4.24

j 2(4.24)i j

= 8.48i j

133. (d)

C

D

| J | E| J | E

(where = frequency , are constants)

1

(Good conductors)

1

(Dielectrics)

If C D(J ) (J )

E E

; 2 f

f2

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

134. (b)

From the options we can see that option (a)

BEt

is maxwell’s equation in time

varying field

135. (d)

when ever electromagnetic wave travels in afree space then

0,

0

0

136. (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 3 0.942m

2 10 10

137. (d)

endH.dl I

2

0

(H a ).(rd )a 60

2

0

H rd 60

IES M

ASTER

(20)

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

138. (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 2z20000e a A / m

139. (c)

H.dl ( H).ds (Stokes theorem)

enclosedH.dl I (Ampere’s law)

v

D.ds ( .D)dv

(Gauss divergenece theorem

Q D.ds (Gauss Law)

140. (c)

We know velocity

PV

0 0

(as it is free space substituting in aboveequation)

P0 0

V

0 0

1

0 0

1

141. (c)

For free space and are each zero

hence 1 and 2 are false

hence statements 3 and 4 are correct

H J, D 0

142. (d)

Ampere Law states that

H dl Ienclosed.

H dl = – 30 + 10 =– 20A

–30A is taken because the path taken is oppositeto the direction of magnetic field due to 30A current.

143. (b)

Magnetic monopoles does not exist and hencemagnetic field lines are always closed. So, it hasno sinks or sources.

S

B ds = or

B 0

Hence,

B is solenoidal but it is non-conservative

as

H 0.

144. (c)

The tangent to an electric field line at any pointgives the direction of electric field at that point.

Electric field lines always cut a conductor in thedirection normal to the surface.

145. (c)

IES M

ASTER

[EE], ESE-2019 PRELIMS TEST SERIES PAPER-II (TEST-10) (21)

C1

C2

C3

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

146. (c)

Potential across the capacitor V = Q/C

Now, when dielectric is filled, capacitance of thecapacitor increases to rC C .

Potential r r

Q Q VVC C

Potential decreases.

Electric Field, E = Vd

Electric field also resduces to r

VEd

147. (d)

V

AQ –Q

1 2

Electric field near plate 2 due to plate 1 will be :

E =0 0

Q2 2A

Therefore, force on plate Z due to electric field willbe :

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

148. (d)

The divergencen of gradient of a scalar is knownas its laplacian

2 2 22

2 2 2V V VV

x y z

= 0 + 4xz3 + 12xy2z 4xz (z2+3y2) at

P(1, –2, 1) = 2V

(1, –2, 1) = 4 (1) (1) [12 + 3(–2)2]

= 52

As 2V 0 at P (1, –2, 1), so potential fielddoes’t satisfy laplace equation.

149. (a)

IES M

ASTER

(22)

The angle between two vectors A and B

, isgiven by

A BsinA . B

A.BcosA . B

now, if sin = 0.56

= 34° and 145°, both satisfy the above equation,so there is an ambiguity.

But cos = 0.56

55.94 or 55.94

So, by use of dot product, the angle betweentwo vectors can be determined without anyambiguity.

150. (a)

Considering a simple, open loop transfer function

C(s) kG(s)R(s) s 1

With unity feedback, closed-loop transfer function

c

C(s) k k / (1 k)R(s) s (1 k) s 1

where c / (1 k)

To determine the bandwidth :

22b b

22b c b

c

b

b c

11 or (OL)211 or (CL)2

(CL)(1 k)

(OL)

c clopsed loop system response is faster..

Note : BW × rise time = constant