ANSWERS - IES Master · TypeA TypeB TypeC , TypeD , TypeE , , 22. (b) In a PWM method of chopper switching, the frequency is kept constant and T ON is varied. Accordingly T OFF is

ESE-2018 PRELIMS TEST SERIESDate: 22nd October, 2017

ANSWERS

1. (a)

2. (b)

3. (c)

4. (a)

5. (b)

6. (c)

7. (b)

8. (b)

9. (b)

10. (b)

11. (a)

12. (b)

13. (a)

14. (d)

15. (c)

16. (d)

17. (a)

18. (b)

19. (b)

20. (c)

21. (b)

22. (b)

23. (d)

24. (b)

25. (d)

26. (b)

27. (c)

28. (c)

29. (b)

30. (d)

31. (c)

32. (a)

33. (c)

34. (a)

35. (c)

36. (c)

37. (b)

38. (a)

39. (d)

40. (d)

41. (b)

42. (c)

43. (c)

44. (b)

45. (b)

46. (d)

47. (b)

48. (a)

49. (b)

50. (b)

51. (d)

52. (b)

53. (d)

54. (d)

55. (d)

56. (a)

57. (b)

58. (b)

59. (b)

60. (a)

61. (a)

62. (d)

63. (a)

64. (c)

65. (d)

66. (d)

67. (c)

68. (a)

69. (a)

70. (b)

71. (d)

72. (a)

73. (b)

74. (d)

75. (a)

76. (c)

77. (c)

78. (a)

79. (d)

80. (b)

81. (b)

82. (c)

83. (b)

84. (b)

85. (a)

86. (c)

87. (d)

88. (c)

89. (a)

90. (c)

91. (c)

92. (c)

93. (c)

94. (a)

95. (d)

96. (b)

97. (a)

98. (d)

99. (a)

100. (b)

101. (a)

102. (d)

103. (a)

104. (d)

105. (d)

106. (c)

107. (c)

108. (b)

109. (c)

110. (c)

111. (d)

112. (c)

113. (b)

114. (c)

115. (d)

116. (b)

117. (b)

118. (b)

119. (a)

120. (c)

121. (b)

122. (b)

123. (b)

124. (b)

125. (c)

126. (b)

127. (b)

128. (b)

129. (c)

130. (b)

131. (b)

132. (b)

133. (d)

134. (c)

135. (d)

136. (d)

137. (b)

138. (c)

139. (c)

140. (d)

141. (b)

142. (a)

143. (a)

144. (d)

145. (d)

146. (a)

147. (d)

148. (a)

149. (a)

150. (b)

IES M

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1. (a)

P 30 1.5 45W

Thermal resistance = 0.6 °C/WTemperature rise = 45 × 0.6 = 27°CJunction temperature = 27° + 90° = 117°C

2. (b)

A series inductor is generally placed in orderto protect SCR for high rising current.

3. (c)Germanium has high leakage current ascompared to silicon SCR at room temperature.

4. (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

5. (b)

A GTO can be turned on and off by positiveand negative gate signal but it takes moretime to turn off. It has reverse blockingcapacity.

6. (c)

Safety factor=m

PRVV

[PRV =Peak Recovery Voltage]

= rms

PRV2V

PRV = 2.5 400 2 = 1414 V = 1.4 kV

7. (b)

2rmsVR

=2 22 V 1 V

R R2

P with SCR =1

2mV sin222

Vrms =

2 V V2 22

P =2 2V 1 V

2 R 4R

Ratio = 2 2 1V V

4R R 4

8. (b)

The output of a Full Wave (FW) rectifier is

This waveform contains an average dccomponent and an ac component.

9. (b)

Output is proportional to cosine of firing angle.Output cos

If increases cos decreases Output decreases.

10. (b)The waveform of a 3 phase half-waverectification is

50 Hz

150 Hz

input

output

11. (a)When the thyristor conducts, J1 and J3junctions are forward biased, while J2 is reversebiased.

12. (b)When the load current is ripple free, thenload current is constant. The supply waveform will be square.

13. (a)For proper turn off OFF of SCR, the anodevoltage should be reversed and the anodecurrent should be less than holding current.

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14. (d)

In first half cycle, D, will conduct from 0 to and D2 will not. So, option (a) and (b) areeliminated after , D1 is reverse biased andD2 is forward biased and starts to conduct.Current being almost constant.

15. (c)The above graph where, it shows that it canblock the voltage in bidirectional and can becontrol led bidirectionally, this is thecharacteristic of triac.

16. (d)f(z) = c0 + c1 z–1

1o 1

Unitcircle

1 C c z dzz

= o2

c

(1 c )z cdz,c :|z| 1z

= c

(z)dzPoles of f(z) are z = 0,0. i.e. z = 0 pole oforder2 by residue theorem

c(z)dz = 2i { Res ((z) at z = 0)}

= 2i 2 0

2z o

(1 c )z c1 d z2 1 dz z

= 2i { 1. (1 + co)}

c(z)dz = 2i (1 + co)

17. (a)

m 0 Cpeak 0 smaxCI I I I VL

=0.110 2001m

= 12 A

18. (b)

DC chopper can be controlled by two methods,one is fixed frequency and varying ON andOFF time, other is variable frequency.

19. (b)Multiple PWM technique is used to reduceharmonic content as compared to single pulse.

20. (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 AA

21. (b)

Type Voltage CurrentType ATypeBTypeC ,TypeD ,TypeE , ,

22. (b)

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

23. (d)

Per unit ripple

0.5 Duty ratio

The graph of a typical ripple versus duty ratio.

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

V

I

working area of type-B chopper

25. (d)In a voltage source inverter, voltage is fixed

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and the output voltage wave form dependsonly on the switching sequence. Currentwaveform depends on the load impedance.

26. (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

= 166.86V27. (c)

Output voltage is given asV0 =

s n 1,3,52V

s

n2V sin n 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 volt

28. (c)

Load 880

+

–

DL

220 CH

Since the given chopper is a step up chopper

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 =

34

ONT = 200 sec

OFF in sec

200200 T

=

34

800 = OFF in sec600 3 T

2003 = OFF secT

(TOFF) = 66.66 sec

29. (b)When RLC is under damped then XC > XL andwhen diode conducts then power flow will befrom load to source side and when thyristorconducts then power flow will be from sourceto load. By considering only fundamentalcomponent on output side.

V0

i0

P0

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

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

V0 = i1V

1

1

1 =0

i

VV =

4.50150 = 3

1 =13

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=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

31. (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

32. (a)Expn.

T =1f

Duty cycle = ONTT

= onT f

= 2ms × 200= 0.4

33. (c)

Vs = 50V – 2V = 48VMaximum rms output voltage at fundamentalfrequency,

V0 =

s4V2 =

4 482 = 43.2 V

34. (a)Power = 100 KW

Vd = 500 V

P = d dV I

Id =100000

500 = 200

RMS value of thyristor current = dI 2003 3

= 115.47A

35. (c)

A 3 square wave (symmetrical) invertercontains only odd harmonics.

36. (c)

Given output power = 24W and efficiency =0.8

Input power = 24 30W0.8

Output voltage = 240V and 0.95

Input voltage = 240 (1–0.95)= 12 V

Input current =30 2.5A12

37. (b)

Output Voltage (V0) for a step down chopperis

V0 = sD V

0

s

VV = D

38. (a)

For elimination of 5th harmonic.,Pulse width = 2 d

sinnd = 0 [n = 5]

5d = 0, ,2 .....

2d =2 40, ,5 5

= 0 , 72 , 144 .....

39. (d)

A cyclo-converter is a one-stage frequencychanger. Cyclo converters are of two types(a) step-down (a) step-up.

40. (d)

Vph =4002 = 200V

V1 = phmr V sin

m

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155 =3r 200 sin

3

r = 0.937

41. (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

42. (c)

Either by increasing anode voltage or increasingin gate current, SCR can be made to turn ON.

43. (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.

44. (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

45. (b)

Only when firing angle is greater than loadphase angle, output power can be controlled,other wise, the output cannot be controlled.Conduction angle should be lower than .

46. (d)

The average power, maximum allowable gatedrive power is related by

avggatedrivepower

PP

so, Pgdp = avgP

=1000.4 = 250W

47. (b)

An opto-coupler is used to isolate any twocircuits. The control circuits will work at lowervoltages and the power circuit will work athigher voltages. Any abnormalities in powercircuit would damage control circuit, so anopto coupler is used to isolate to protect controlcircuit.

48. (a)

Applying KVL, we have

A AV 1 0I 50

AA

V 1I50

; given IA = 2 × 10–3 for OFF..

32 10 = AA

V 1 V 1.1V50

49. (b)The turnoff time is the minimum value of timeinterval between the instants when the on statecurent has decreased to zero and the instantwhen the thyristor is capable of with standingforward voltage without turning-ON.

50. (b) Let f(x,y) = 2x – x2 + my2

for f(x, y) to be harmonic2 2

2 2f f 0

x y

– 2 + 2m = 0 m

51. (d)2

c

(z 8)dz ;(0.5z 1.5i)

(C is the circle x2 + y2 = 16 or

|z| = 4)

= 2 2

c

(z 8) dz(z 3i)

=

cf(c)dz

Re

Im

4i3i

4

C

f(z) has only one pole z = zi and lies within c.Hence, by residue theorem

cf (z)dz 2 i{Res.of f(z)atz 3i}

= 2i {2 (3i)2 + 8)) = – 4i

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52. (b)x3 = iIts one root is given by

x3 = i /2e

x = i /6e

= cos i sin6 6

= 3 i

2 2

53. (d) = ax2 y – y3

2 = 2 2

2 2( ) ( )

x y = (2ay) + (–6y)

2 = 0 2 ay – 6y = 0

2y (a – 3) = 0

y 0, a 3 54. (d)

F(s) = ln 2s

s 9

2 2dF(s) 1 2s–

ds s s 9

We know that

L(t f(t) = – d F(s)ds

L–1 d F(s)ds = – tf(t)

L–1 2

1 2ss s 9

= – f(t)

1 – 2 cos 3t = – f(t)

f(t) = 2cos3t 1

t

u(t) ; ( t > 0)

55. (d)The given family of curves

y – 2x = c

dy 2dx

= 0

its orthogonal trajectories are given by

1 2dydx

= 0

dx + 2dy = 0

x + 2y = a

56. (a)

21 sinx

D D 1

= 21 sinx

1 D 1

= 1 sin xD

= sinx dx= – cos x

57. (b)

3 5y'' (y sin x) y' y = 3cosxSince this equation contains a termcontaining y15 y´, hence it is non linear.

58. (b)32

42

d y dy ydtdt

= e–t

Order of the differential equation= order of highest order derivative= 2

59. (b)The given differential equation

2

2d x dx6 8x

dtdt = 0 ... (i)

Its auxiliary equation ism2 + 6m + 8 = 0(m + 2) (m + 4) = 0m = – 2, – 4

General solution of (i) is

x(t) = 2t 4t1 2c e c e ...(ii)

Given initial conditions are(a) x(0) = 1

1 = c1 + c2 ... (iii)

(b)t 0

dxdt

0 = – 2c1 – 4c2

c1+ 2c2 = 0 ... (iv)Solving (iii) and (iv), we getc2 = – 1c1 = 2

The solution isx(t) = 2 e–2t – e–4t

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60. (a)Given the differential equation

dydx = 1 + y2 ... (i)

2dy

1 y = dx

Integrating, we gettan–1 y = x + c

y = tan (x + c)

y = tan (x+3); (c = 3) is a paraticularsolution

61. (a)The differential equation is

dyx ydx

= x4

dy 1 ydx x

= x3 ----(i)

I.F. = l1dx nxxe e

I.F. = xHence, solution of (i) is

y.(x) = 3x .x dx c

yx = 5x c

5

4x cy5 x

Given y(1) = 6/5

65 =

1 c c 15 1

y = 4x 1

5 x

62. (d)Given the differential equation

2 dyx 2xydx =

2 log xx

dy 2 ydx x

= 32log x

x

Integrating factor (I.F) = 2 dxxe

.l2 n xe = x2

Solution of (i) is

y.x2 = 2

32log x x dx c

x

= 2log x dx c

x

yx2 = 22(log x) c

2

y = 2

2log x c

x x

y(1) = 0 - 2log1 c

1

c = 0

y(x) = 2log x

x

y(e) = 21e

63. (a)

2

2u

t

=

22

2uc

x

here {u(x, t): displacement of a point of thestretched string at time ‘t’ and at position ‘x’. ‘c’: velocity.}

64. (c)

dz z log zdx x

= 2z (logz)x

21 dz 1 1

dx x log zz(log z)

=

1x

let 1

logz = u

21 1 dz

z dx(logz) =

dudx

du 1 udx x

= 1x

I.F. = 1dxxe

= lnxe

= 1x

65. (d)

Given, A =2 2

a bc d

We know, A (adj A) = | A | I

Option (b) is true.

|adj A|=|A|n–1 where n is order of matrix A.

|adj A| = |A| ( A is2 2)

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|Adj A–1| = |A–1| |A|

|AA–1| = 1

|A| = |A–1| = I

if |A| 0

then |A–1| = |A|–1

i.e., option (c) is true.

Hence most appropriate option is (d).

66. (d)

Given A = 2

1 1 1 2 2 21 1 1 2 2 21 1 1 2 2 2

By definition on of scalar multiplication

A2 =

2 2 22 2 22 2 2

2 2 22 2 22 2 2

= 4 4 4 4 4 4 4 4 44 4 4 4 4 4 4 4 44 4 4 4 4 4 4 4 4

= 6 2 2 22 2 2 6A2 2 2

2A 6A

67. (c)

Given A A,= defn of skew-Hermitianmatrix

Now, iA iA=

i A=

= iAi.e. iA is hermitian matrix

68. (a)8(0 – 12) – x (0 – 24) = 0

x 4

69. (a)

Since i = cos 2 + i sin 2

i = i 2e

(i)i = ii 2e = 2i /2e = 2e

70. (b)

Given f(t) =t ;0 t kk1;t k

L {f(t)} = k st st0 k

te dt e . dtk

=

kst st

20

1 e etk s s

+

st

k

es

= sk sk

2 21 e e 1kk s s s

+ 0 + skes

= sk

21 e

ks

71. (d)f(t) = cos (Jat)

= J( jat) J( jat)e e

2

f(t) = at ate e

2

L{f(t)} = 1 1 12 s a s a

= 2 2s

s a

72. (a)Formula

t

0L f( )d =

1 F(s)s

73. (b)(Elimination of options)Trace (A) = 2 – 1 + 0 = 1Sum of eigen values = 1, is satisfied by onlyoption (b).Note : Be careful. It is only necessarycondition but not sufficient.

74. (d)

Let F(s) = 2 21

s 4 s 9

= 2 21 1 15 s 4 s 9

f(t) = L–1 {F(s)}

= 15

1 1sin2t sin3t2 3

= 1 1sin2t sin3t10 15

75. (a)

I = 2

0( )

( ) (2 ) a f x dx

f x f a x …(i)

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I = 2

0(2 )

(2 ) (2 (2 ))

a f a x dx

f a x f a a x

I = 2

0(2 )

(2 ) ( )

a f a x dx

f a x f x …(ii)

(i) + (ii) 2I = 2

0( ) (2 )(2 ) ( )

a f x f a x dx

f a x f x

2I = (2a – 0)

I a

76. (c)

2

21

x x dx

Let f(x) = 1 x x

=

( 1); 0( 1); 0 1( 1); 1

x x xx x xx x x

=

1 2 ; 01 ; 0 12 1; 1

x xx

x x

12345

–3–2

–3/2–1 –1/2 1 3/2 3

f(x) = 1 – 2x

f(x) = 2x – 1

f(x) = 1

f(x)

2

2( )

f x dx =

3/2 1 1/2 0 1 3/2 22 3/2 1 1/2 0 1 3/2

4 3 2 1 1 1 2

dx dx dx dx dx dx dx

= 3 3 1 3 1 34 2 3 1 2 1 1 2 22 2 2 2 2 2

= 32 1 2 1 7.52

77. (c)

I = 3/2/2

3/2 3/20sin

sin cos

x dx

x x…(i)

I =

3/2/2

0 3/2 3/2

sin2

sin cos2 2

x

dxx x

I = 3/2/2

3/2 3/20cos

cos sin

x dx

x x…(ii)

(i) + (ii) 2I = /2

01

dx = 2

4

I

78. (a)

I = t dt1

02sin

2 4

= t1

0

2 cos/2 2 4

= 4 cos cos

4 2 4

= 4 1 1 0

2 2

79. (d)

( )b

af x dx = ( / ) , 0bc

acf x c dx c

Let xc = t

dx = cdt

( / ) ( )bc b

ac af x c dx f t cdt

= ( )b

ac f x dx (change of variable)

1( ) ( / )b bc

a acf x dx f x c dx

c

1c

80. (b)

I = 2 /

0 0

x x yx e dy dx

y x =

y

y x =

xy = 0

Region of integration

After changing the order of integration

I = 2 /

0x y

x yxe dx dy

x2/y = t

2x dx dty

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xdx = 2ydt

y

y x =

x

x =

I = 0 2t

t yydte dy

= 012

ty

y e dy

= 012

yye dy =

01 ( ) ( )2

y yy e e

I = 12 (0 + 1) =

12

81. (b)

I = 2 2 21 1 1

0 0 0

x x y

xyz dz dy dx

=

2 22 121 1

0 002

x yx zxy dy dx

= 21 1 2 2

0 01 12

xxy x y dy dx

=

212 2 2 410

0

12 2 2 4

x

y

y x y yx dx

= x xx x dx

2 2 21 20

1 (1 ) (1 )(1 )2 2 4

= 1 2 40

1 [1 2 ]8

x x x dx

I = 1 1 2 1 18 2 4 6 48

82. (c)5/22

2

2

dy1dx

d ydx

= c

52dy1dx

=

222

2d ycdx

order = 2, degree 2

83. (b)2

2d y dy5 6y

dxdx = 0

Its auxiliary equationm2 – 5m + 6 = 0

m = 2, 3 General solution,

y = c1e2x + c2e

3x

y(0) = 2

c1 + c2 = 2

y (x) = 2x 3x1 23c e2c e

y (0) = 5

2c1 + 3c2 = 5

c2 = 1, c1 = 1

y = e2x + e3x

84. (b)A family of straight lines through origin (non-vertical lines)

y = m x

dydx = m

y =dy xdx

xdy – ydx = 0

85. (a)Since complex numbers do not follow orderrelation i.e. Z1,< = > Z2 don’t hold.

86. (c)Given the relationxn +1 = xn (N – ln xn) ...(i)Comparing(i) it with

xn +1 = xn – n

n

f(x )f '(x ) ..(ii)

We get

– n

n

f(x )f '(x ) = xn (N – ln xn) – xn

n

n

f(x )f '(x ) = xn {(1 –N) + ln xn}

n

n

f(x )f '(x ) =

n

n1

x

nx (N –1)l

f(xn) = ln xn –(N – 1) = 0 xn = e(N – 1)

87. (d)Given the differential equation

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dydx = xy2 = f(x, y); (say)

y(1) = 1, h = 0.1Euler’s method,y(1.1) = y(1) + h f(1,1)

= 1 + (0.1) (1 × 12) = 1.1y(1.2) = y (1.1) + h (1.1, 1.1)

= (1.1) + (0.1) (1.1) (1.1)2

= 1.2331

88. (c)Order of convergence in newton raphson isquadratic (i.e.2).

89. (a)Let f(x) = x log10 x – 1.2 = 0Newton-raphson method

xn + 1 = xn – n

n

f(x )f '(x )

xn + 1 = xn – n 10 n

10 n 10

x log x 1.2log x log e

..(i)Given x0 = 2.5

x1 = 2.5 – 10

10 10

2.5log 2.5 1.2log 2.5 log e

x1 = 2.7465x2 = 2.7406 ; putting n = 1 in (i)x3 = 2.7406 ; putting n = 2 in (i)

90. (c)

91. (c)Givenx3 – 6x2 + 11x – 6 = 0 has 1 & 3 as its tworoots. (x – 1) (x – 3) is a factor

(x3 – 6x2 – 11x – 6)x3 – 6x2 + 11x – 6 = 0

(x2 –4x + 3) ( x – 2) = 0; x = 1,2, 3.

92. (c)By simpson’s 1/3 rule

31

1 dxx = 0 2 1

h[(y y ) 4y ]3

y(x) = 1x

y0 = y(1) = 1

y2 = y(2) = 12

y2 = y(3) = 13

3

11dxx =

1 1 11 43 3 2

= 109 = 1.111

93. (c)Let f(x) = x3 + 3x – 7 = 0

f’(x) = 3x2 + 3Newton raphson iterative scheme is

xn +1 = xn – 3n n

2n

x 3x 73x 3

xn +1 = 3n2n

2x 73x 3

Given x0 = 1

x1 = 3

22 1 7 9 1.5

63 1 3

94. (a)

i 3–i

z i dz i

4

i

1 (z i)4

= 4 41 (i i) ( i i)4 =

14 . 24 i4 = 4

95. (d)

I = (2,4)

2

(0,3)2 d (3 )d ,y x x x y y

along the parabolax = 2t, y = t2 + 3dx = 2dt, dy = 2t dtI =

1

2 2 2

t 02t 6 4t (2dt) (6t t 3)(2tdt)

= 1

2 2 3

012t 12 12t 2t 6t) dt

= 1

3 2

0( 2t 24t 6t 12)dt

= 2 24 6 12

4 3 2

= 33

296. (b)

Let f(x,y) = 2x – x2 + my2

for f(x, y) to be harmonic2 2

2 2f f 0

x y

– 2 + 2m = 0

m 1

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97. (a)

2D1 ds

(s 1)Poles are –1 & 1 and only s = 1 lies within D.Hence by residue theorem

1 1–1 1 2–i2

iw

i2D

0

2D

1 ds(s 1) = 2i { res of 2

1s 1 at s = 1}

= 2i { 1

1 1 }

= i98. (d)

ii = ii /2e {one of the values} = e–/2

99. (a)

f(t) = t1(1 e )t

tL(1 e ) =1 1s s 1

t1 eL

t

= s1 1 dss s 1

= s( ns n(s 1))

=

s

sn1 s = 0 –

sns 1

s 1n

s

100. (b)

Capacitance, C = 0

e

F mdlogr

As the length of the line increases, C increases.

So, CX =1

2 fC decreases.

101. (a)

Voltage across string = phase voltage

=33 kV

3

102. (d)

Corona effect produces ozone gas, reducestransmission efficiency and causes line to drawnon-sinusoidal current.

103. (a)The corona effect depends upon the shape andconditions of the conductor surface. An irregularsurface gives rise to more corona effect.

104. (d)W = mass × g

= 3 210 10 Kg 9.81m s

Power developed = overallW H = (104 × 9.81) × (25) × 0.8= 1962 kW

105. (d)

Higher load factor means greater average load,resulting in greater number of units generated fora given maximum demand. Thus the standing(fixed) charges, which are proportional tomaximum demand and independent of numberof unit generated, can be distributed over a largernumber of units supplied and therefore, overallcost per unit of electrical energy generated willbe reduced.With the given number of consumers, the higherthe diversity factor of their loads, the smaller willbe the capacity of the plant required andconsequently the fixed charges due to capitalinvestment will be much reduced.

106. (c)

Transmission lines are classified as electricallyshort, medium and long on the basis of analysisof its capacitance. As the length increases,capacitance becomes considerable. But as thefrequency or, wavelength of the line is high,capacitance comes into picture even for shortlength of line. Hence, the concept of electricallyshort, medium and long lines is primary basedon the frequency or wavelength.

107. (c)

Sag, S =2WL

8T

108. (b)

aI =1 2 0a a aI I I

aI = 12 0

ba c2

II I

[Since, 1 1 0 0

2b a a cI I and I I ]

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109. (c)

Since, GeqHeq = 1 1 2 2G H G H

= 500 0.6 350 1.2 = 300 + 420= 720

Then, Heq =720 7.2100

110. (c)Since, for static device e.g. transmission line,transformers etc., the positive sequenceimpedance and negative sequence impedanceare equal. Hence, for transmission line, thepositive sequence current is equal to thenegative sequence current.

111. (d)

Since, the fault occured is line-line type, itmeans, it does not involve ground. So, therewill be no zero-sequence current.

112. (c)Fault current =

Voltage at the fault point

Total impedance upto fault

113. (b)Fault MVA or, Fault level = (prefault voltage)× (Post fault current)

or = pu

Base MVAZ

114. (c)

A

B

C

V = 0A

V = 0B

I = 0C

i.e. IC = 0, VA = VB = 0So, this is a double line to ground faultoccurred between phase A and B.

115. (d)

p.u. fault current for 3 fault = pu

1

EZ

p.u. fault current for line to ground fault

=pu

1 2 go

3EZ Z Z

Here,pu

1

EZ =

pu

1 2 go

3E2

Z Z Z

1

j0.15 =go

6j0.15 j0.15 Z

Zgo = j0.90 j0.30 j0.6

0 nZ 3Z = j0.6

Zn = 0j0.6 Z j0.6 j0.053 3

=0.55j j0.18 pu

3

116. (b)

For zero-sequence circuit in 3 transformer..

Primary Secondary

S1 S3

S4S2

Switch S1 or S3 will closed when it is stargrounded.Switch S2 or S4 will closed when it is delta.Here primary side is star-grounded. So S1will close, and secondary side is star withoutground, so no switch on secondary will close.

117. (b)

When the system was operating at ‘1’ motorwill be running at synchronous speed s .But as the mechanical load increased, firstthe motor will deaccelerate and hence willincrease. Motor should have stop at ‘2’ but itwill not stop because of moment of inertiaand continued deaccelerate upto 3 (due toequal area criteria). Now, as electrical poweris more than load at ‘3’, again it startsaccelerating and moves to 1.So, at point ‘2’, while oscillating from 1towards point 3, the speed of motor will beless than s .

118. (b)

Expression for the restriking voltage

V = maxt1 cosVLC

So, rate of rise of restriking voltage

RRRV = maxVdV tsindt LC LC

RRRV will be maximum when tsin 1LC

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or t LC2

.

Given, 0 = 41 10LC

t = 41LC

2 2 10

= 41 sec.

2 10Average RRRV

= Peak Restriking Voltage

Time to reach peak restriking voltage

=6

4

100kV1 10 sec

2 10

= 2200 10 kV sec.

= 2KV sec

119. (a)

120. (c)

Plug setting Multiplier,

PSM =Fault current in primary winding

CTratio (Relay current)

=

4000400 505

5 100

= 20

121. (b)

E

L

C

CB

Applying Laplace’s for the circuit,

I(s) = E .Ys s

= 1 1E sCs R sL

=

2s LCR R sL E ssRL

= 2RLC 1 1 Es s ssRL RC LC

E(S) =

2

I ss1 1 Cs s

RC LC

The characterisitc equation is

2 1 1s sRC LC

= 0

n =1LC

and12LC

=1 1 1 LC

RC RC 2

=1 L

2R CFor critical damping, 1

So, R =1 L2 C

= 61 12 0.01 10

= 81 10 5K2

122. (b)

Due to doubling effect, the rms value ofbreaking current=1.8 × (symmetrical breaking capacity)But as making current is taken in maximumvalue or, peak value. So, the making capacity

= 2 1.8 symmetrical breaking capacity

= 2.55 symmetrical breaking capacity

= 2.55 2000

= 5100 MVA

123. (b)For protecting star/delta transformer, the CTsshould be connected as delta in HT side andstar in LT side.Let 300A current is flowing in line of LT side.So, phase current in LT side

=300 A

3and phase current in HT side

=30011 100A

33 33

Since, the HT side of transformer is connectedin star. So, line current = phase current =100AHence, current in primary of CT on HT side= 100A.

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So, for protection to work only on internalfault, the current in secondary of CT on HTside

=5 A3

So, the ratio of CT connected on HT side

=100

53

124. (b)

The adjustment of travelling distance of anelectromechanical relay is commonly knownas time setting Higher the time settingmultiplier, longer the relay will take to operate.So, for TSM 0.8, operating time is 4 sec.

For TSM 1, the operating time

=4 1

0.8

= 5 sec.

125. (c)

50Hz50Hz

T /RXX R /TX X500 kHz

Line trap unit is used for carriercommunication between two stations. Carriercommunication signals have very highfrequency from 50 kHz to 500 kHz and haveto propagate through the same line over whichpower signal of 50 Hz frequency is travelling.So, for proper transmitting and receiving ofcarrier signal, the line trap unit offers highimpedance to power frequency (i.e. 50 Hz)and low impedance to carrier frequencysignals.

126. (b)Restriking voltage is expressed as

V = m11 cos tVLC

So, rate of rise of restriking voltage

RRRV, dVdt = mV 1sin t

LC LCi.e. RRRV depends on only L and C only.

127. (b)

In biased differential relay, the bias or fixedpercentage is the ratio of the differential

operating coil current to the averagerestraining coil current.

128. (b)

If aV 0 , bI 0 and cI 0 , it means thereis a fault occurred in line-a which reduces itsvoltage to zero. Hence there will be no curentin line-b and line-C.

129. (c)

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

130. (b)

131. (b)

The materials used for fuse element must beof(i) low meeting point, so that it easily melts

at proper fault curent and open the circuitin case of fault.

(ii) high conductivity, so that the ohmic lossesdue to fuse element is low

(iii) low cost(iv) free from deterioration

132. (b)

• Mho relay is used for long transmissionline

• Negative sequence relay is used ingenerator in protection againstunbalancing.

• Thermal relay is basically used in motorprotection against overload.

133. (d)

The universal relay torque equation is,

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

Rectance relay is obtained by putting k1is positive, k3 is negative and k2 = 0i.e. current produces operating torque andvoltage current directional element producesrestraining torque.

134. (c)

Since, Load factor = Average load

Maximum loadSo, Average load = 0.5555 × 15 KW

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and average daily energy consumption of theconsumer= 0.5555 15 24 200 KWh

135. (d)

For economic allocation of load, incrementalfuel cost of each unit should be 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

136. (d)

Since, load factor = Average load

Maximum load Average load = 0.75 × 50 = 37.5 MWPlant capacity factor

=Actual energy

Installed capacity×time

=Average load time

Installed capacity time

=Average load

Installed capacity

Installed capacity = Average load

Plant capacity factor

=37.5 75MW0.5

So, reserve capacity= Installed capacity – Maximum load= 75 – 50 = 25 MW

137. (b)

The main objective of load frequency controlleris to apply control of frequency and at thesame time of real power exchange via theoutgoing lines i.e. regulate the generations tomaintain the frequency constant.

138. (c)

d d

Fig. Horizontal Configuration

GMD= 3 d d 2d = 3 2 d

d

d

d

Fig. Equilateral triangle configuration

GMD= 3 d d d= d

Inductance,

L = 4 GMD2 10 n n GMDGMR

and capacitance,

C = 02 1

GMD n GMDnr

As, GMD of equilateral triangle configurationis reduced. Hence, in this configurationcapacitance will increase and inductancewill decrease.

139. (c) Pin type insulator - As the name suggests,

the pin type insulator is mounted on a pinon the cross-arm on the pole. There is agroove on the upper end of the insulator.The conductor passes through this grooveand is tied to the insulator with annealedwire of the same material as the conductor.Pin type insulators are used for transmissionand distribution of communications, andelectric power at voltages up to 33 kV.Beyond operating voltage of 33 kV, the pintype insulators become too bulky andhence uneconomical.

Suspension insulator - For voltages greaterthan 33 kV, it is a usual practice to usesuspension type insulators shown in Figure,consisting of a number of glass or porcelaindiscs connected in series by metal links inthe form of a string. The conductor issuspended at the bottom end of this stringwhile the top end is secured to the cross-arm of the tower. The number of disc unitsused depends on the voltage.

Strain insulator - A dead end or anchor poleor tower is used where a straight section ofline ends, or angles off in another direction.

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These poles must withstand the lateral(horizontal) tension of the long straightsection of wire. In order to support thislateral load, strain insulators are used. Forlow voltage lines (less than 11 kV), shackleinsulators are used as strain insulators.However, for high voltage transmission lines,strings of cap-and-pin (suspension)insulators are used, attached to thecrossarm in a horizontal direction. Whenthe tension load in lines is exceedingly high,such as at long river spans, two or morestrings are used in parallel.

Shackle insulator - In early days, theshackle insulators were used as straininsulators. But now a day, they arefrequently used for low voltage distributionlines. Such insulators can be used eitherin a horizontal position or in a verticalposition. They can be directly fixed to thepole with a bolt or to the cross arm.

140. (d)As in HVDC transmission system, thefrequency of voltage or current is zero. Hence,there is no flow of charging current due tocapacitance from conductor to earth and thereis no skin effect.

141. (b)

In order to reduce the sheath losses, bondingof cable is done-by bonding of cable at boththe ends, the equivalent resistance of thesystem is increased whereas the inductanceis reduced.

142. (a)

For EHV lines where total reactance is high,the shunt compensation is used mainly forimprovement of system stability.

Since, Pe = . VE sinX

where X = XL – XCIncrease in reactance will increase the systemstability.

143. (a)Maximum demand never means the greatestinstantaneous maximum demand but thegreatest short time averaged demand occuringduring a period of time under consideration.

144. (d)Reactance relay operates only when thereactnce sensed by the relay is less than thespecified value. It never depends on faultresistance. So, reactance relay is used in thiscondition.

145. (d)In load frequency control, frequency is to bemaintained constant if there is change in loador change in generation. The other possibility isthat the loading remains constant and frequencyis varying. So, load frequency control can bedone by regulating turbine speed (i.e. frequency)or generator output (i.e. load).

146. (a)The energy in the transients are dissipated inthe form of corona loss, so it reduces thetransients produced by lightening.

147. (d)

Surge tank is located near power station asmuch as possible so that a sudden closing oropening of the gate value does not pressuriesthe penstock much.

148. (a)

Since, in a power network each bus isconnected only to a few other buses (two orthree), the YBUS of large network is very sparse,i.e. it has a large number of zero elements.

149. (a) For measuring positive and negative sequencevoltage, neutral is taken as reference busbecause neither the positive sequence currentnor negative sequence current flows fromground to neutral. So neutral is at groundpotential. But for zero-sequence voltage, ground istaken as reference because zero sequencecurrent flows from ground to neutral.

150. (b)

A freewheeling diode is necessary to avoidany high voltages due to sudden currentswitching in load (RL). Inductor will not allowsudden change of curent. So the free wheelingdiode provides current path.

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ANSWERS - IES Master · TypeA TypeB TypeC , TypeD , TypeE , , 22. (b) In a PWM method of chopper switching, the frequency is kept constant and T ON is varied. Accordingly T OFF is