ANSWERS - IES Masteriesmaster.org/public/archive/2016/IM-1480343039.pdf · R Office : F-126, (Lower...

1. (d)

2. (c)

3. (a)

4. (b)

5. (c)

6. (d)

7. (d)

8. (a)

9. (b)

10. (d)

11. (d)

12. (b)

13. (a)

14. (c)

15. (a)

16. (d)

17. (d)

18. (d)

19. (a)

20. (b)

21. (b)

22. (c)

23. (b)

24. (b)

25. (a)

26. (c)

27. (d)

28. (d)

29. (b)

30. (d)

31. (b)

32. (d)

33. (b)

34. (d)

35. (c)

36. (a)

37. (c)

38. (d)

39. (c)

40. (c)

41. (b)

42. (d)

43. (c)

44. (c)

45. (b)

46. (a)

47. (c)

48. (c)

49. (d)

50. (a)

51. (c)

52. (c)

53. (a)

54. (a)

55. (d)

56. (a)

57. (a)

58. (b)

59. (b)

60. (d)

61. (c)

62. (b)

63. (a)

64. (d)

65. (a)

66. (b)

67. (b)

68. (b)

69. (c)

70. (a)

71. (d)

72. (c)

73. (d)

74. (b)

75. (b)

76. (a)

77. (d)

78. (b)

79. (d)

80. (d)

81. (c)

82. (d)

83. (b)

84. (b)

85. (b)

86. (c)

87. (a)

88. (a)

89. (b)

90. (c)

91. (b)

92. (b)

93. (b)

94. (d)

95. (d)

96. (c)

97. (d)

98. (b)

99. (d)

100. (c)

101. (a)

102. (c)

103. (d)

104. (b)

105. (c)

106. (c)

107. (d)

108. (a)

109. (c)

110. (c)

111. (b)

112. (d)

113. (b)

114. (c)

115. (d)

116. (c)

117. (a)

118. (a)

119. (c)

120. (d)

ESE-2017 PRELIMS TEST SERIESDate: 27th November, 2016

ANSWERS

121. (b)

122. (c)

123. (b)

124. (c)

125. (c)

126. (c)

127. (d)

128. (b)

129. (c)

130. (c)

131. (d)

132. (a)

133. (d)

134. (b)

135. (b)

136. (b)

137. (b)

138. (b)

139. (c)

140. (b)

141. (c)

142. (b)

143. (b)

144. (c)

145. (c)

146. (d)

147. (b)

148. (c)

149. (b)

150. (b)

151. (d)

152. (a)

153. (b)

154. (b)

155. (a)

156. (b)

157. (d)

158. (a)

159. (d)

160. (d)

IES M

ASTER

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

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

(Test-12) 27th Nov. 2016 (3)

1. (d)

2. (c)

3. (a)

The sequence of request of block size:

300, 50, 200, 50

Memory regions20 250 200 350 100

1. Using first fit policy :

(a) Req. for 30020 250 200 350

300

100

(b) Req. for 5020 250 200 350

30050

100

(c) Req. for 20020 250 200 350

30020050

100

(d) Req. for 5020 250 200 350

30020050 50

100

All request of block size is allocated toexisting memory.

2. Using best fit policy :20 250 200 350 100

(a) Req. for 30020 250 200 350

300

100

(b) Req. for 50

20 250 200 350

300

100

50

(c) Req. for 20020 250 200 350

300 50

100

200

(d) Req. for 5020 250 200 350

300200 5050

All request of block size is allocated toexisting memory.

4. (b)

5. (c)

Virtual address space = 34 – bit

No. of addressable unit = 234 bytes

Size of physical memory = 256 MB

= 256 × 220 bytes

= 228 bytes

Page size = 4kB = 212 bytes

No. of pages in physical memory

= 28

1612

2 22

Page number offset16 12

16-bit is used to represent page number.

No. of pages in virtual address space.

= 34

2212

2 22

Page number offset20bit

12bit

Since page table entry contains framenumber.

Page number represents as 16-bit in Physical

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(4) (Test-12) 27th Nov. 2016

memory = 2 byte

No of pages : 222

Page table size : 2 byte × 222

= 2 × 4 × 220 byte

= 8 × 220 byte

= 8 MB

6. (d)

7. (d)8 1 2 3 48 8 8 3 3

1 12 2 4

1 1

1

5 1 2 2 23 3 3

4 4 25 1 1

8812

1 4 3 1 13 4 3

Number of page fault : 9

8. (a)7 22 31 32 35 45

= (32 – 31) + (35 – 31) + (45 – 35) + (45– 22) + (22 – 7)

= 1 + 4 + 10 + 23 + 15= 53

9. (b)

10. (d)

Give, disk capacity : 200 GB

Number of direct block address = 8

Number of indirect block address = 1

Number of doubly indirect block address = 1

Size of each disk block = 144 bytes

Size of each disk block address = 8 bytes

Each block contains : 144

8 = 18 pointer..

Maximum possible size :

= 8 × (size of each disk block) + (Indirectblock address) + 1x (doubly indirect blockaddress).

= 8 × 144 + 1x (18 × 144) + (18 × 18 ×144)

= 8 × 144 + 18 × 144 (1 + 18)= 8 × 144 + 18 × 19 × 144= 50400 bytes

11. (d)

12. (b)

13. (a)

14. (c)

15. (a)

16. (d)

No. of 1’s in network Id = 24

Total 1’s = 26

NID + SID = 26

Subnet Id = 26 – 24

= 2

Number of subnets = 22 = 4

17. (d)

Given IP address : 10.1.5.7 [class A]

Given subnet mark : NID Host ID

255.255.255.192

Number of 1’s in Host Id = 18

Number of subnet Id bits = 18

Number of subnet = 1218

18. (d)

19. (a)

m 0 Cpeak 0 smaxCI I I I VL

=0.110 2001m

= 12 A

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(Test-12) 27th Nov. 2016 (5)

20. (b)DC chopper can be controlled by twomethods, one is fixed frequency andvarying ON and OFF time, other isvariable frequency.

21. (b)Multiple PWM technique is used to reduceharmonic content as compared to singlepulse.

22. (c)

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

E2V R

= 50 50

4 100 10

= 58 = 0.625 A

23. (b)

Type Voltage CurrentType ATypeBTypeC ,TypeD ,TypeE , ,

24. (b)

Max steady state ripple = sV4fL

This can be decreased by :(i) Reducing voltage(ii) Increasing frequency(iii) Increasing inductance

25. (a)

Modulation index = r

c

VV

= 100150 = 2

3 = 0.66

cfN 12f if zero of triangle wave

coincides with zero of reference.

N = 1500 1500 = 2

26. (c)

half

full

PP = 1

4

Pfull = 4 × Phalf= 4 × 20= 80 W

27. (d)Line commutated inverter gives fixedvoltage and frequency. It requires at itsoutput an existing ac supply which is usedfor its commutation. Forced commutatedinverter provides an independented acoutput voltage of adjustable voltage andfrequency.

28. (d)A current source inverter is obtained byinserting a large inductance in series withsupply voltage.

29. (b)In a PWM method of chopper switching,the frequency is kept constant and TON isvaried. Accordingly TOFF is also varied.

30. (d)

Per unit ripple

0.5 Duty ratio

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(6) (Test-12) 27th Nov. 2016

The graph of a typical ripple versus dutyratio.

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

V

I

working area of type-B chopper

32. (d)In a voltage source inverter, voltage is fixedand the output voltage wave form dependsonly on the switching sequence. Currentwaveform depends on the load impedance.

33. (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

34. (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

35. (c)For first quadrant operation, a halfcontroller is most preffered. It is simpleand cheap compared to other converters.

36. (a)

If the firing anle is less than , thenstill the thyristor is in reverse biased, so itwill not turn on . At , thecontroller cannot be controlled and controlis lost.

37. (c)A motor will require only fundamental andany harmonics present would degrade theperformance of motor. Heater can takefundamental and harmonics to convertelectrical energy to heat energy.

38. (d)Integral cycle control is advised where theload has very high time constant likeheating.

39. (c)Output voltage is given asV0 =

s n 1,3,52V

s

n2V sinn sin n tn

where n = 1 cosntansinn

rms value of nth harmonic voltage is

Vn = s2V sin n2n

for, first harmonic voltage is

V1 = s2V sin

V1 =

2 200 2sin 18052

= ON

OFF ON

tt t = ONt

total time period

=25

V1 = 85.625 volt40. (c)

Load 880

+

–

DL

220 CH

Since the given chopper is a step upchopper

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(Test-12) 27th Nov. 2016 (7)

V0 = s1 V

1

880 = 1 220

1

4 =1

1

1 = 14

34

The output voltage pulse width is = TOFF

ON

ON OFF

TT T = 3

4

ONT = 200 sec

OFF in sec

200200 T

= 3

4

800 = OFF in sec600 3 T

200

3 = OFF secT

(TOFF) = 66.66 sec

41. (b)When RLC is under damped then XC > XLand when diode conducts then power flowwill be from load to source side and whenthyristor conducts then power flow will befrom source to load. By considering onlyfundamental component on output side.

V0

i0

P0

T T1 2 D D1 2 D D3 4T T3 4

42. (d)Since the given chopper is step up chopper

V0 = i1V

1

1

1 = 0

i

VV =

4.50150 = 3

1 =13

=23

= ONTT =

23

Conduction time of thyristor =125 sec

TON = 125 sec

T = ON3 T2

= 63 125 102

T = 187 sec

f = 1T = 6

1187 10

= 5333.33Hz = 5.33 kHz

43. (c)

Pulse width =25

= 72º = 2d

d = 36º

V7 = s4V sin 7d sin 7 tn

V7 rms = s2 2V sin 7 36º7

V7 rms = 0.122 Vs

44. (c)

Given fI = 2A

For firing angle 30°

tV = mV 2 230(1 cos ) (1 cos30 )

tV = 193.20V

tV = b a aE I r

or tV = f s a a(K.I ) I r

or 193.20 = a2 1500(0.5 2) I 0.8

60

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(8) (Test-12) 27th Nov. 2016

or aI 45.15 Amp

Torque e f aT (K.I ) I = (0.5 2) (45.15)

eT 45.15N – m

45. (b)

Ripple factor = 2FF 1

FF(form factor) = rms

avg

II

Irms = s I

Iavg = s I (Is - supply current)

FF = s

s

II

=

1

ripple factor = 1 1

= 1 10.3

= 3.333 1 = 2.333 = 1.527546. (a)

V0 = s2V

1

for the given case

when 0.3 then

V0 = s0.3V

1 0.3

= s

3V7

V0 < Vs

Thus working as step down chopper.When 0.7 then

V0 = s0.7V

1 0.7

V0 = s7V3

V0 > Vs thus working as step upchopper.

47. (c)For half bridge inverter

(V1) rms output voltage across load = DCV2

Output power (P) = 21V

R

P = 2OCV4R

...(1)

For full bridge inverter

rms output voltage, V2 = VDC

Output power, P1 = 22V

R = 2DCVR

2DCV4

4R

From equation (1)P1 = 4.P

48. (c)

id

iL

t

t

TON TOFF

id = OFFL

ON OFF

TiT T

OFF T 1 T

id = Li 1 = 8 × (1 – 0.4) = 8×0.6 = 4.8 A

49. (d)

V03 =

s4V sin 3 t3

=

4 230 sin 3 t

3 V03 = 97.6150 sin (942.47t)

Z3 = 1R j 3 L

3 C

=

36

14 j 3 2 50 35 103 2 50 115 10

= 4 + j(32.97 – 9.23)

Z3 = 4 + j(23.756)

3Z = 2 24 23.756

Z3 = 16 564.34Z3 = 24.09

(I03) = 03 rms

3

VZ =

97.61502 24.09

(I03) = 2.86 A

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(Test-12) 27th Nov. 2016 (9)

50. (a)Minimum braking speed is

mmW =a a

m

I rK =

300 0.21.2

mmW = 50 rad/s or 477.46 rpm

Maximum braking speed is

mxW =s a a

m

V I rK =

400 300 0.21.2

mxW = 383.33 rad/s or 3660.06rpm

Note : 0 sV V

when 0 then V0 = 0 then minimumbreaking speed occur.when 1 then V0 = Vs then maximumbreaking speed occur.

51. (c)For any intrinsic semiconductor, theintrinsic concentration is directlyproportional to T3/2. As conductivity of theinstrinsic semiconductor is directlyproportional to intrinsic concentration. So,

3/2T .

52. (c)For the semiconductor material,

The Hall’s coefficient, Rx = y

x z

E 1J B ne

As, = ne

So, RH = 1 1ne

53. (a)

Since, Hall coefficient, RH = 1ne

Where n is number of charge-carrier perunit volume; m–3.and e is the charge of electron; coulomb.So unit of RH : (meter)3 (coulomb)–1

So, option (a)

54. (a)The current required to flowing through asuper conductor of radius ‘r’ to produce a

critical magnetic field Hc isIc = c2 r H

= cd2 H2

=3

322 102 7 107 2

= 22 Ampere

55. (d)As, fermi level for P-type semiconductor,

EF = VV e

A

nE KT logN

where,EV = maximum energy of valence band ineV.

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

i.e.

Conduction band

Fermi level for P-type semiconductor

Valence band

56. (a)

57. (a)According to the Messener’s effect,superconductors expels the magnetic fieldflowing through it.So, relative permeability is zero forsuperconductor.

58. (b)

bn an

cn

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(10) (Test-12) 27th Nov. 2016

The sequence is acbacb = acb

59. (b)

ZL = 40 j30

LZ = 2 240 30 50

IL = P

P

VZ

VP = L PI Z= 10 × 50= 500V

VL = P3V

= 3 500 V

60. (d)

ZA = 6 3

= 2

IP = 120 60 A rms2

IL = P3 I

= 60 3

= rms60 3 A61. (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

62. (b)

Applying KVL in 1st loop we get

36 = 3120 10 I

I1 = 336 1.8mA

20 10

+

–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

63. (a)

i = 2 AApplying KVL, we getVab– 3i +10 = 0

Vab = 3i – 10

= 3 2 10

= –4 V64. (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

65. (a)

Irms = T

2

0

1 8t dtT

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(Test-12) 27th Nov. 2016 (11)

=31 64 T

T 3

Here T = 1

Irms =643

Average Power loss in 9 resistorPav = 2

rmsI 9

=64 93

= 192 W

66. (b)

PowerPhase = 690W 230W

3

PhI = 230W230V

= 1AIL = PhI 1A

67. (b)Let R be the resistance between each phase

RT = R 2R 2 RR 2R 3

Given RT = 12

R = 18

68. (b)When one of the resistor is disconnected,the remaining two resistors behave as ifthey were combined in series across 400V

PhI = L400I 4A

50 50

69. (c)The -connected impedance can bereplaced by Y connected impedance using

ZY =Z3

ZY = 153

= 5

70. (a)

(inward)

AnticlockwiseB

(outward)

AClockwise

The straight current carrying conductorproduces the magnetic field having directionoutward perpendicular to the paper in loop Aand inward perpendicular to the paper in loopB. According to Lenz’s law, the polarity ofinduced emf will be in such a way that itopposes the cause of induced emf. Hencethe current in loop A will be clockwise, sothat its magnetic field is inward and currentin loop B will be anticlockwise so that itsmagnetic field is outward.

71. (d)

m = r 1 = 6.5 – 1= 5.5

M = mH = x y z10a 25a 40a5.5 A m

= 55ax + 137.5ay – 220az A/m

72. (c)

E =ddt

=d(BA)

dt

=2R B

T

73. (d)

F = q(V B)

F is perpendicular to both V

and B and V

and B can make any angle.

74. (b)

The induced current would be in such a wayso as to oppose the change of flux. Currentwould be anticlockwise direction, to opposethe flux due to large loop, where the currentis in clockwise direction.

75. (b)1 gauss = 10–4 wb/m2

e =ddt

=dBAdt

= –0.02 × 100 × 10–4

= –2 × 10–4 V

76. (a)

L =NI

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(12) (Test-12) 27th Nov. 2016

=LIN

=3 38 10 5 10400

= 10–7 wb

77. (d)

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

78. (b)Applying amperes circuital law, we get amagnetic field outside and no magnetic fieldinside

PipeCurrent flowing

No magnetic filed

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

= 30 r.p.s

t =1 sec

30

e = t

=425 10

1/ 30

0.23 V

80. (d)

e =ddt

= 2d (6t 7t 8)dt

= – (12t + 7)= – (12 × 2 + 7)

= –31 mV

81. (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 bethere. Hence electric lines of force will bealways perpendicular to the surface ofconductor.

82. (d)

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

83. (b)Both the new capacitances will be in parallel.

Cnew =C KC2 2 =

(K 1)C2

84. (b)

E = dVdx

= 2d (5x 10x 9)dx

= –[10x + 10]= –[10(1) + 10]

= – 20 V

85. (b)At balance,

11

4 4 2 3

11

1R j C R j L R R1R j C

1 4 4 2 3 1 1R R j L R R 1 j C R

1 4 4 1 2 3 1 1 2 3R R j L R R R j C R R REquating real and imaginary part, we get

R4 =

2 3

1

R R 0.4K 0.6K 0.24K 240R 1K

and L4 = C1R2R3

= 0.5 × 10–6 × 0.4 × 103 × 0.6 × 103

= 0.120 H

86. (c)In order to get high resolution of a wire-wound potentiometer, thin wires (i.ehaving smaller diameter) is used, whichincreases the steps of measurement.

87. (a)The current in resistor R is

I = 2VR

Vx = V I R

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(Test-12) 27th Nov. 2016 (13)

= 2VV RR

= V 2 V

= V 1 2

88. (a)Let impedance of A is, Z = R+jXunder balance condition

R4(R+jX) = 3 2 2R r j L

R+jX = 3 32 2

4 4

R Rr j LR R

X should be inductive impedance.Hence A should be unknown self-inductance having resistance r1

where, r1= 32

4

R rR

and self-inductance should be

L1 =

32

4

RLR

89. (b)

If aV 0 , bI 0 and cI 0 , it means thereis a fault occurred in line-a which reducesits voltage to zero. Hence there will be nocurent in line-b and line-C.

90. (c)Shunt fault is characterized by(i) Increase in current(ii) Fall in voltage(iii) Fall in frequency

91. (b)

92. (b)The materials used for fuse element mustbe of(i) low meeting point, so that it easily

melts at proper fault curent and openthe circuit in case of fault.

(ii) high conductivity, so that the ohmiclosses due to fuse element is low

(iii) low cost(iv) free from deterioration

93. (b)• Mho relay is used for long transmission

line

• Negative sequence relay is used ingenerator in protection againstunbalancing.

• Thermal relay is basically used inmotor protection against overload.

94. (d)The universal relay torque equation is,

T = 2 21 2 3k I k V k VIcos k

Rectance relay is obtained by puttingk1 is positive, k3 is negative and k2 = 0i.e. current produces operating torque andvoltage current directional elementproduces restraining torque.

95. (d)

Since, Load factor = Average load

Maximum loadSo, Average load = 0.5555 × 15 KWand average daily energy consumption ofthe consumer

= 0.5555 15 24 200 KWh

96. (c)For economic allocation of load,incremental fuel cost of each unit shouldbe equal.

So, 1IC = 2CI

140 0.2P = 225 0.3P

1 20.2P 03P = –15 ...(1)

and total load, P1 + P2 = 200 ...(2)Solving equation (1) and (2), we get

P1 = 90 MWP2 = 200 – 90 = 110MW

97. (d)

Since, load factor = Average load

Maximum load

Average load = 0.75 × 50= 37.5 MW

Plant capacity factor

=Actual energy

Installed capacity×time

=Average load time

Installed capacity time

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(14) (Test-12) 27th Nov. 2016

=Average load

Installed capacity Installed capacity =

Average loadPlant capacity factor

= 37.5 75MW0.5

So, reserve capacity= Installed capacity

– Maximum load

= 75 – 50= 25 MW

98. (b)The main objective of load frequencycontroller is to apply control of frequencyand at the same time of real powerexchange via the outgoing lines i.e.regulate the generations to maintain thefrequency constant.

99. (d)For EN = 0 (Chip of MUX is inactive)S0S1 = 11, D = 0, but there is no output i.e.the o/p is a high input impedance state i.e.(0000) but as these bubbled outputs then itshould be (1111).

100. (c)Given Ckt is a Adderby adding the bits

3 2 1 0A A A A1 0 0 1

3 2 1 0B B B B1 1 1 0

We get

3 2 1 0

3 2 1 0

3 2 1 0

3 2 1 0

1001 A A A A1110 B B B B

C 1 0111Y Y Y Y

C Y Y Y Y 10111

101. (a)From given Ckt

Output at 160KW 16kHz

10

Output at 16Kx 1kHz16

Output 1Ky 40Hz25

Output at 40Z 5Hz8

102. (c)From figure we can see it is a 3-bit Johnsonring counter and it requires (2n = 2 × 3 =6 states) to reach its initial.Method (2) :Solving by each clock pulse.

0 1 2Q Q Q1 1 0 initial value

Clock1 1 1 1Clock2 0 1 1Clock3 0 0 1Clock 4 0 0 0Clock5 1 0 0Clock 6 1 1 0

103. (d)The circuit is a –ve edge triggered clockripple counter and it is a Down counter theclockpulses are shown below.

1 2 3 4 5 6 7 8

0

0

0

1

1

1

0

1

1

1

0

1

0

0

1

1

1

0

0

1

0

1

0

0

0

0

0

For x = 0 0Q is connected to clock of TT1

Flip-flop 1Q is connected to clock of TT2 Flip-flop. Hence Down counter.For x = 1 Q0 is connected to clock of T1 Flip-flop. Q1 is connected to clock of T2 flip-flophence Up counter.

104. (b)

I0I1

I16

o/p16×1MUX

Mode 16Counter

4 data select lines

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(Test-12) 27th Nov. 2016 (15)

From the figure we can see that Mod 16 willhave 16 outputs for each output of the Mod-16 counter one data input is selected.

105. (c)We know for R-2R DAC output voltage.

V0 =n 1

ref fin

i 1

V R2 bi2 R

For Rf = R = 1We get

V0 =n 1

ref iin

i 1

V 2 b2

=3 2 1 0

4 40 2 1 2 0 2 1 25

24 24 2 2

=0 1 0 15

16 4 8 16

=1 154 16

=55

16

=2516

= 1.5625

106. (c)

We know resolution = reffn

V2

= 652

= 0.0156

% resolution= refn

V 1002

0.0156×100 = 1.56%

107. (d)It requires large no. of comparators.

108. (a)Both are correctAs the input is analog signal i t isapproximated by a step in a sample andhold Ckt at the input side.

109. (c)

110. (c)Serial in serial out shift register - (2n – 1)serial in parallel out shift register - nParallel in serial out shift register - (n – 1)Parallel in parallel out shift register - no clockpulse region.

111. (b)A D-Flip-flop changes its state at the positiveedge triggered part of the clock pulse whereas the D-latch will change its state the leveltriggered part of the clockpulse and alsoremains off once the clock pulse is removed.

112. (d)Without solving we can say that it is aJohnson counter as inverted output isconnected to input of flip-flop.

Initial Value

By Solving

Initial Value

Clock 012345678

Q 011110000

R 001111000

S 000111100

T 000011110

Y=QT 100000001

Serial in = T 111100001

As it under goes 2n states = 2 × 4 = 8 states.It is johnson ring counter.

113. (b)Given i/p is 0000 0001 = 0.05i.e. 1 step = 0.05 V

11001010= 7 6 3 12 2 2 2

= 12.8 + 64 + 8 + 2= 138 + 64= 202 steps

o/p voltage = 202 × 0.05 = 10.10 V

114. (c)By solving the states from CLK diagram

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(16) (Test-12) 27th Nov. 2016

Initial Valueunused states

Initial Value

Initial Value

Clk 01234567

Q10001000

0 Q11000100

1 Q11100010

2 Q01110001

3

3714

Unused states

1

2

4

8

Hence, Mod 4 counter

115. (d)From the given circuit diagramwhenever Z = 0 Mux gets enabledZ = 1 Mux gets disabled.

1 0

2

0

2

0

y zx y z S S F x,y,z0 0 0 1 0 I 00 1 0 0 0 I 01 0 0 1 0 I 01 1 0 0 0 I 1 x,y,z

116. (c)Drawing the truth table for 4 × 1 MUX.

x y F0 0 10 1 01 0 01 1 1

F(x, y) = m 0,3

117. (a)

We can find out from the timing diagram ofopcode fetch machine cycle.

(PC)

Address Bus

PC (PC+1)

RD = 0 Memory

Database

Database

IR decode

during 1T -state both AA15-A8 (higher order)address bus and A7-A0 (lower order) addressbus carry address.

118. (a) MVI A, 04 H

A 04H

RLC (Rotate left with out carry)

0 0 0 0 1 0 0 0cy0 0

A 08H

MOV B, A

B 08H

RLC

0 0 0 1 0 0 0 0cy0 0

A (10)H

119. (c) LXI SP, 2100 H

SP 2100 (top of the stock)

LXI B, 0000H

B 00H

C 00H

PUSH B

20FF B

SP

SP

20FE20FF2100

0000xx

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(Test-12) 27th Nov. 2016 (17)

20FF (00)H

20FF E

20FE (00)H

POP PSW

Flag register (20FE)

Flag register (00)H

A (20FF)

A (00)H

120. (d)

8254 programmable interval timer/counter is functionally similar to thesoftware designed counters and timers.

it generates accurate time delay and itcan be used for applications such

(1) real-time clock

(2) event counter

(3) digital one shot

(4) square wave generator

(5) complex waveform generator

121. (b)

It is short abbreviation for the operation tobe performed.

122. (c)

We know that

Address bus : It is a uni-directional bus

Data bus : Bi-directional Bus

Control Bus : Bi-directional bus.

123. (b)

MVI, 80 H

A80 H

A1000 0000

RAR (Rotate accumulator right with carry)

0 1 0 0 1 0 0 0cy0

A 40 H

SIM (Set interrupt mask)

SOD

0 1 0 0 0 0 0 0

SDE R7.5

M7.5

M6.5

M5.5MSE

Serialoutput bitSerial data enable=1

Reset MSE = 0 (Mask set disabled)

124. (c)

MVI A, 18 H

A 18 H A = 1×16+8 = (24)10

A 0001 1000

RLC (Rotate left without carry)

0 0 1 1 0 0 0 0cy0

A 30 H

A = 3×16’+0×16°

A = (48)10

125. (c)

(1) MVI A, C5H

A C5H

(2) ORA A

A 1100 0101

1100 0101

A 1100 0101

A C5H

(3) RAL (Rotate accumulator Left with carry)cy1

1 0 0 0 1 0 1 0

AA (8A)H

8

(4) RRC (Rotate right with out carry)cy1

0 1 0 0 1 1 0 1

AA (45)H

8

126. (c) (1) MVI A, 07 H

A 0000 01111 A = (07)H

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(18) (Test-12) 27th Nov. 2016

Rotate Left = (07)10

(2) RLC (without carry)

cy0 0 0 0 0 1 1 1 0

E0

A (OE)H

(3) MOV B, A

B (OE)H

(4) RLC

cy0 0 0 0 1 1 1 0 0

CA (1C)H

1

(5) RLC

cy0 0 0 1 1 1 0 0 0

8A (46)HA (38)H

3

]

(6) ADD B

A (46)H = (70)10

127. (d)

When ever a CALL instruction isapproached the next instruction addressafter the CALL instruction present in theprogram counter is loaded on to thestack.

Now the program counter is stored withthe address mentioned in the callinstructions.

128. (b) RST 7 7 × 8 = (56)10

= (0038)H

56

8

16

3

129. (c)

Whenever the RET instruction isexecuted at the end of the subroutine.

The data stored on the top of stockduring the call instruction is transferedback to the program counter.

130. (c)

Stack :

It is a R/w memory reserved for storinginformation temporarily.

131. (d)

EI (Enable interrupts)

MVI A, 08 H

A -08 H

SOD

0 0 0 0 1 0 0 0

SDE R7.5

M7.5

M6.5

M5.5MSE

Serialoutput Data

Serial data enable

7.5 isON asit is ‘0’

“o” available

“1” Non available

R

Hence all the interrupts RST 7.5 RST 6.5and RST 5.5 are enabled.

132. (a)Convert the feedback sytem to unityfeedback system, the new Ge is

Ge =

G s

1 G H Gs s s

eG s =

2

2

s 2s s 1

s 2K 11s s 1

= 3 2s 2

s s s 2K 2K 1

Now Ge has no pole at origin so type is 0.

133. (d)

Angle of asymptote is 2K 1Poles zeros

= 2K 1 2K 14 1 3

K = 0 Angle = 3

K = 1 Angle =

K = 2 Angle = 53

134. (b)Given characteristic equation

3 2s 4s s 6 0

by applying routh’s method,

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(Test-12) 27th Nov. 2016 (19)

3

2

1

0

s 1 1s 4 6

4 6s 2.5 04

s 6

There is two times sign change, so thereare two roots in right of s-plane. The totalnumber of roots being 3, the number ofroots to left are (3 – 2) = 1.

135. (b)E(s) = R(s) –C(s)H(s)

=

R G Hs s sR s 1 G Hs s

=

R s1 G Hs s

ess = s 0

LimsE s

=

2

1ss11

s 12s 6s 3

=1 3

1 413

136. (b)

1 1 2 1 2P G G 1

2 3 2P G G

1 3 2 1L G G H

2 1 2 1L G G H

3 4L G

T(s) = 1 1 2 2

1 2 3

P P1 L L L

= 1 2 2 3

1 2 1 2 3 1 4

G G G G1 G G H G G H G

137. (b)

T(s) =

G s1 G s

=

2

2K s 1s 3s1 K 2 K

All the coefficients of characteristicequation should be non zero and positive.

1 K 0 K 1

2 K 0 K 2

Common to both is K > –1.

138. (b)

For u ut t 2

2 2s

st

0

1 ex e dts s

For 2te u t 1

2t st

0

1x e e dts s 2

139. (c)All the coefficient need not be always non-zero.

140. (b)In an open loop system, the control actionis independent on the desired output. Thereis no feedback available.

141. (c)A transfer function is Laplace output byLaplace of input. It is independent of input.It is done with zero initial condition.

142. (b)Delay time is defined as the time takenfor response to reach 50% of the final valuefor the very first time.

1

0.5

tdelay tpeaktrise

143. (b)A velocity error constant can be measuredusing a ramp input.

144. (c)System is type 2 system, thus to both stepand ramp input the error is zero.

R(s) = 2s 2s

E(s) =

2

s 210 s 2s

s 2

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(20) (Test-12) 27th Nov. 2016

sse t = s 0LimsE 0s

145. (c)

H(s) = 2100

s 20s 100 2n = n100 10 rad sec

n2 = 20

=n

202

= 2020

= 1So, it is critically damped.

146. (d)Applying routh’s criteria

3 2s 5s 7s 3 0 3

2

1

0

s 1 7s 5 3

7 5 3 3232 5 5s 05

s 3

There is no sign change, all the roots lie inthe left half of s-plane.

147. (b)Absolute stability is when the disturbances,die out with time. If the magnitudeincreases, the system is unstable. If theoscillations remain same, then the systemis marginally stable.

148. (c)The characteristic equation is given by

22

K s 41 0s 2 s 5 s 6

It has zeros at –2j, +2j and poles at –2, –2 ,–5, –6

Asymptotes meet at = 2 2 5 64 2

= 15 7.52

149. (b)The characteristic equation is

2s 10s K = 0

n = x

=8K as K 0

150. (b)

The peak overshoot depends only on thedamping ratio and not on the magnitudeof input.

151. (d)

If the poles are negative distinct real thanthe system is over damped. If the rootsare on the same point on real axis, thesystem is critically damped and if it has aimaginary part it is under damped.

152. (a)

Roots on imaginary axis representmarginal stable system.

153. (b)

154. (b)

A freewheeling diode is necessary to avoidany high voltages due to sudden currentswitching in load (RL). Inductor will notallow sudden change of curent. So the freewheeling diode provides current path.

155. (a)

A inter group reactor is used in a cycloconverter to limit the circulating currentsin the cyclo converter circuit.

156. (b)

At absolute temperature, intrinsicsemiconductor behaves like on insulatoras there is no free electrons available.

157. (d)

158. (a)

For low resistance measurement, Kelvindouble bridge method is used, which elimi-nate the error due to contact resistance.

For the use of kelvin double bridge fourterminals resistors is used for proper con-nection.

159. (d)

If there are repeated roots on j axis, thesytem will be unstable and the magnitudeof the output increases.

160. (d)

Bandwidth has no effect on the noisefiltering characteristic of a system. It isthe frequencies for which input can bereproduced at output.

ANSWERS - IES Masteriesmaster.org/public/archive/2016/IM-1480343039.pdf · R Office : F-126, (Lower...

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