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IES M

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EE (Test-3), Objective Solutions, 12th March 2016 (1)

1. (d)

2. (d)

3. (a)

4. (b)

5. (d)

6. (b)

7. (a)

8. (d)

9. (c)

10. (a)

11. (d)

12. (b)

13. (c)

14. (d)

15. (c)

16. (d)

17. (b)

18. (b)

19. (a)

20. (c)

21. (d)

22. (a)

23. (b)

24. (a)

25. (a)

26. (b)

27. (c)

28. (b)

29. (d)

30. (a)

31. (d)

32. (d)

33. (d)

34. (a)

35. (c)

36. (b)

37. (a)

38. (d)

39. (c)

40. (a)

41. (a)

42. (d)

43. (c)

44. (C)

45. (d)

46. (b)

47. (b)

48. (d)

49. (b)

50. (d)

51. (d)

52. (d)

53. (a)

54. (d)

55. (c)

56. (a)

57. (a)

58. (b)

59. (b)

60. (b)

61. (b)

62. (c)

63. (a)

64. (d)

65. (b)

66. (b)

67. (c)

68. (b)

69. (c)

70. (d)

71. (d)

72. (d)

73. (a)

74. (d)

75. (b)

76. (b)

77. (a)

78. (d)

79. (a)

80. (d)

81. (a)

82. (c)

83. (d)

84. (a)

85. (d)

86. (c)

87. (a)

88. (b)

89. (a)

90. (d)

91. (c)

92. (c)

93. (b)

94. (d)

95. (c)

96. (c)

97. (c)

98. (b)

99. (b)

100. (d)

101. (d)

102. (c)

103. (b)

104. (a)

105. (d)

106. (c)

107. (b)

108. (c)

109. (a)

110. (a)

111. (a)

112. (d)

113. (a)

114. (b)

115. (d)

116. (a)

117. (d)

118. (b)

119. (a)

120. (a)

Conventional Question Practice ProgrameDate: 12th March, 2016

ANSWERS

IES M

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(2) EE (Test-3), Objective Solutions, 12th March 2016

1. (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.

2. (d)

F = q(V B)

F

is perpendicular to both V

and B

and V

and B

can make any angle.

3. (a)

L =NI

=LIN

=3 38 10 5 10400

= 10–7 wb.

4. (b)Applying amperes circuital law, we get a magneticfield outside and no magnetic field inside

PipeCurrent flowing

No magnetic filed

5. (d)

e =ddt

= 2d (6t 7t 8)dt

= – (12t + 7)= – (12 × 2 + 7)= –31 mV.

6. (b)

2L N

2

1

LL =

22

1

NN

L2 =2

21

1

NL

N

=2500108mH

600

= 75 mH.

7. (a)The electric field induced depends on area,magnitude and rate of change of flux and number

of turns, dBe NAdt

. These are same in bothcases, both will have same induced voltage.

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

= 30 r.p.s

t =1 sec

30

e = t

=425 10

1/ 30

0.23 V.

9. (c)The path indicates that charge is negative.

10. (a)

Leakage coefficient = i

g

.

11. (d)2a r = 2 6 231.25 10 4.91 10 m

Bair =5

60.6 10 1.22T

a 4.19 10

Hair = air7

0

B 1.224 10

= 59.71 10 AT / m .

12. (b)Magnetic field density at the centre due tocircular loop.

IO

B

lId

IES M

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EE (Test-3), Objective Solutions, 12th March 2016 (3)Applying Biot-Savart law,

dB = l02I d

4 r

B = 02I .2 r

4 r

= 0I2r

B 1r

So, LoopA B

LoopB A

B r 2a 2 :1B r a

2a

O aI

A

B

I

13. (c)

x

y

z

According to Ampere’s Law.

H

=I a

2

where is the perpendicular distance of pointfrom conductor.

y

x

(–3, 4) 4

30I

a

Hence, 2 2 20 53 4

So, H

= 10 a2 5

= ˆ1.a

= x yˆ ˆsin a cos a

= x y4 3ˆ ˆa a A m5 5

.

14. (d)y

x

zr

(0, 1, 0)B

(1, 0, 0)AHA

HB

(1/2, 1/2)

Since the point 1 1, , 02 2

comes at the mid

point on the line joining (0, 1, 0) and (1, 0, 0)where conductors are located,

Magnetic field at 1 1, , 12 2

by conductor ‘A’’

and that of conductor ‘B’ are equal and oppositeto each other.

So, Net magnetic field at 1 1 0, ,12 2

.

15. (c)

Since, c dH J J

So, the unit of H

will be similar to currentdensity cJ or, dJ i.e. (Ampere)/(meter)2.

16. (d)Magnetic field experienced by the loop due toinfinite conductor.

B = 0 2I2 r

where r is the distance from infinite conductorand force, l lF i B i Bsin

a

a

a

F1

F2F3 F4

I1

I2

B

here, 1F = 0 1 2I I a2 a

[away from conductor]

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and 3F = 0 1 2I I a

2 2a

[towards the conductor]

Since, 2F and 4F are equal in magnitude and

cancel each other.

So, NetF

= 0 1 21 3

I IF F4

[away from conductor].

17. (b)According to Ampere’s law, the line integral of

H

and ld

over a closed path is equal to thecurrent enclosed in that loop.

i.e. lH.d

= Inclosed

= –10AHere, negative sign appears as the dot product

of H

due to current enclosed by loop and ld

is negative.

18. (b)According to Biot-Savart Law.

dH = l

2I d

r

lId

90

So, magnetic field intensity at the centre ofsemicircle.

So, dH =

2id sin90

4 Rl

dH = 2i.d

4 Rl

So, H = 2idH d

4 Rl

= 2i R

4 R

=i

4R .

19. (a)

H =1

2 r

r =I

2 H

=1

2 1 = 1 m2 .

20. (c)For dielectric material,

D = 0E P

0 rE = 0E P

P = 0 r E1

0 eE = 0 r E1

e = r 1 .

21. (d)

• All ferroelectric materials are piezoelectric andalso, it has pyroelectric effect after a certainor critical temperature.

Ferro electric

Pyro-electric

Piezo-electric

• Ferroelectric materials are characterized byparallel alignment of dipoles. Due to paralleldipole arrangement, these materials are havinglarge amount of spontaneous polarization.

• Ferroelectric material posses centre ofsymmetry.

22. (a)

Crystal system Unit cell dimension Angle between AxisOrthorhombic a b c 90

Triclinic a b c 90Hexagonal a b c 90 , 120

Cubic a b c 90

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

P f

2 r

rP v 2 fC

so as frequency is doubled, the power loss isdoubled.

24. (a)

Bandgap energy (Eg) of different materials is givenbelow :

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Metals : gE 0eV as both the conduction band

and valance band overlapInsulators : Eg > 3 eV

Semiconductors: gE 1 eV .

25. (a)

2aSimple cubic structure

Atom is placed at the corner of each cubic shell.

APF = Volume of atom in one cubic shell

Volume of unit shell

=

3

3

1 48 r8 3

a

= 0.526

where r = radius of atoma = side of unit shellHere a = 2r.

26. (b)

Steps to find Miller indices :i) Find the intercepts of the plane on the three

crystal axes (OX, OY, OZ) as (pa, qb, rc);where a, b, c are corresponding primitivesand p, q, r be integer.

ii) Write the reciprocal of the number p, q, r as

1 1 1, ,p q r

.

iii) Find the LCM of their denominator.iv) Multiply the reciprocals by the LCM, to get

Miller Indices.Herei) Intercepts are (4, 1, 2)

ii) reciprocals : 1 1, 1,4 2

iii) LCM of 4, 1, 2 is 4

iv) Miller indices are 1 14, 1 4, 4 1,4,24 2

27. (c)

tan (loss tangent) is given by formula

tan = r

r0.004

Given that r = 5

r = rtan = 0.004 × 5= 0.02

r = 5 – j0.02.

28. (b)

Since, p = Ewhere p is dipole moment is polarizabilityE is electric field intensity

= p q.dE V d

= 2q .dV

= C.d2

= (Farad) .(meter)2.

29. (d)• All ferroelectric materials are piezoelectric

materials but all piezoelectric materials arenot ferroelectric materials.

• Quartz is only piezoelectric material, it isnot a ferroelectric material.

30. (a)Energy stored in polarizing the dielectric

W = 2 20

1 1CV C V2 2

where, C is capacitance after polarization,& C0 is capacitance before polarization

W =

20 r 0A A1 V2 d d

= 20 r

1 A V 12 d

=

2

0 r1 VAd 12 d

Energy stored per unit volume

WAd = 2

0 r1 E 12

Energy storedVolume = 1 EP

2 .

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(6) EE (Test-3), Objective Solutions, 12th March 201631. (d)

As, fermi level for P-type semiconductor,

EF = VV e

A

nE KT logN

where,EV = maximum energy of valence band in eV.nV = no. of electrons in valence band,NA = concentration of acceptor atoms. i.e.

Conduction band

Fermi level for P-type semiconductor

Valence band

32. (d)Since, Loss tangent

tan =

"r'r

"r = '

r tan= 2.5 × 0.004= 0.0100

As, complex relative dielectric constant is,

*r = ' "

r rj= 2.5 – j 0.01.

33. (d)Some examples of ferroelectric materials:Rochelle salt : NaKC4H4O6 4H2OKDP (Potassium : KH2PO4dihydrogen phosphate)ADP (Ammonium : NH4H2PO4dihydrogen phosphate)Barium Titanate : BaTiO3Calcium Titanate : CaTiO3Ammonium iron alum : NH4Fe(SO4)2.12H2O

34. (a)

35. (c)On the application of the field E, the modifiedfield due to polarization P is given by

EL =0

PE

For solids and liquid having cubic symmetry,

=13

So, EL =0

PE3

.

36. (b)Dielectric loss in dielectric,

t = ´´ 2 30 r 0

1 E Watt m2

So, dielectric loss depends on the imaginarypart of complex dielectric constant.

37. (a)Since, polarization, P = Np

0 eE = N E

e =0

N

r 1 =0

N

r =0

N1

.

38. (d)For body centered cubic lattice;

a

3a

The atomic radius; 4r 3a

r = 3a4

Number of atoms per unit volume

=1 188

= 2Atomic packing factor :

=3

3

42 r3a

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=38 3a

3 4

= 0.68 i.e. 68%.

39. (c)The electronic polarizability

e = 304 R .

40. (a)

41. (a)

P 30 1.5 45W

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

42. (d)

IL =1 VdtL

=6

gV T 50 6 10L 0.2

= 1.5 mA.

43. (c)

Germanium has high leakage current ascompared to silicon SCR at room temperature.

44. (C)

A RC in series is placed across SCR in order toprotect if from high dv/dt. It is also called snubbercircuit.

45. (d)The maximum voltage that can be applied willbe in accordance with thyristor having lower PIV.

650V 229.8 230V

2 2

46. (b)

Secondary breakdown occurs in BJT not in IGBT.

47. (b)Anode

Gate

CathodeMCT- MOS controlled Thyristor

48. (d)

A UJT can be used for all the three purposes.

49. (b)

SCR is a current driven device.

50. (d)

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

51. (d)

Thyristor opens if holding current IHgoes below 5mA. 0.7V is the drop outpoint.

V = 3 30.7 5 10 1.5 10

= 8.2V.

52. (d)

Pmaxavg

ON

Pf T

Duty cycle, = ONON

T f TT

PmaxavgP 100 250 W

0.4

.

53. (a)

Pulse repetative rate = 31

2.5 10= 400 s

Space ratio = 1/10

Pulse width =400 36.4 s11

As the pulse width is more than 20 s , so theSCR will turn on.

54. (d)

Im = m BV V1K

=12V 4V

1K

= 8 mA.

55. (c)

56. (a)Applying KVL, we have

A AV 1 0I 50

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AA

V 1I50

; given IA = 2 × 10–3 for OFF..

32 10 = AA

V 1 V 1.1V50

57. (a)For proper turn off OFF of SCR, the anode voltageshould be reversed and the anode current shouldbe less than holding current.

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

59. (b)Gate terminal in SCR losses control over anodecircuit voltage and current once the SCR startsconducting forward current.

60. (b)A surge curent rating is one of the many ratingspecifications of an SCR. It is maximum non-repetitive current with sine wave.

61. (b)The turn-off time of line commutation is between10 & 100 sec and that of forced is between 7& 20 sec .

62. (c)For building up of voltage at the terminal of dcshunt motor the field circuit resistance shouldbe less than the critical resistance.

63. (a)

Eb = V – IaRa

At starting Eb = 00 = V – IaRa

V = IaRa

Ra =240 12020

Ra = Rse + (Ra+ Rf)120 = Rse + 2Rse = 120 – 2

= 11864. (d)

In a dc motor, due to the presence of commutatorand brush system, the current flowing in the

armature winding is alternating and hence eddycurrent losses takes place in both armature andpole face.

65. (b)For dc series motor

T = aK I

Under linear region of BH curve2aT I

But under saturation, does not change with IaHence aT I

Therefore, 10 % increase in Ia will result is 10 %increase in torque.

66. (b)For lap wound machine = No. of poles

= No.of parallel pathsNo. of armature turns = 200 2 400

Nol. of armature turns per path 400 1004

Resistance of each path 100 0.04 4

Resistance of armature circuit4 14

.

67. (c)When DC machine runs as motor,

1bE = V – IaRa

= 240 – 20 2= 200 V

When DC machine runs as generator,

2bE = a aV I R

= 240 20 2 = 280 V

2 1b bE E = 280 200 80V .

68. (b)

For reversal of direction of a dc shunt motoreither field flux or armature flux has to bereversed. Hence either field or armature terminalneed to be interchanged.

69. (c)

2a aT I I (series motor aI )

2

1

TT =

2

1

2a2a

I

I

T2 =2

1 220T10

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EE (Test-3), Objective Solutions, 12th March 2016 (9)= 20 Nm × 4= 80 Nm.

70. (d)

71. (d)Ia = IL + If

=200100100

= 102 AEg = V + IaRa

= 200 + 102 × 0.01= 201.02 V.

72. (d)

Commutator in a d.c. machine is used to converta.c. to d.c. The number of segments decide thed.c. output wave smoothness.

73. (a)Brushes in the dc machines are used to collectthe current. Current produced in the armaturewinding is passed on to the commutator andthen to the external circuit by means of brushes.Brushes are made of carbon.

74. (d)

220 V DCM Rsh

IshIRa

Ia

Ish = sh

220 220 1AR 220

Ia =a

220 210R

=10 20 A0.5

I = Ish + Ia= 20 + 1 = 21 A.

75. (b)Armature reaction per pole

=

current in a no. of turnsarmature conductor

pole

=

aTotal No. of conductorI A

2Pole

=

50 7206 2

6[ lap winding, A= P]

= 500ATAnd the shape of armature reaction mmfis triangular as shown in figure below,

armature mmfwave

pattern

N S

76. (b)Shunt motor is a constant speed drive. Seriesmotor can be considered as constant power drive

N 1T

TN = Constant• Cummulative compound,

= sn se

and as load increases, Ia increases, so seincreases as N 1 , when increases speeddecreases. Cummulative differential :

= sn sed

As load increases, se increases and decreases, so speed increase.

77. (a)Compensating winding is used to reduce theeffect of armature reaction by balancing thearmature mmf. So, these winding should beconnected is series with the armature winding.The direction of the current in the compensatingwinding must be opposite to that in the armaturewinding just below the pole faces.The interpole winding is also connected in serieswith the armature winding, so that the armatureflux in the commutating zone which tends to

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(10) EE (Test-3), Objective Solutions, 12th March 2016shift the magnetic neutral axis, is neutralised byan appropriate component of interpole flux.

78. (d)

Due to the problem of interruptiong highlyinductive field current and time needed for thefield current to build up in opposite direction, itis common practice to reverse armatureconnections.

79. (a)

2

1

NN =

a2 1

a1 2

EE

1 a1 2 a2I I Torque constant

Ia2 = 40x 1

2x

Ea1 = 220 40 0.5 200

Ea2 = t a2V I Ra 220 40x 0.5

= V200 20x

Speed is increased by 50%

2

1

NN =

32

32 =

220 20x x200

2x 11x 15 = 0

x = 1.6

1 2

1

1.6 1100 100 37.5%1.6

.

80. (d)Compensating winding in a dc machine is usedfor neutralizing crosmagnetising effect.

81. (a)Pout = VIa

Armature copper loss = 2a aI R

Total inoput power = 2i a a a CP VI I R P

Efficiency =out

in

PP

=a

2a a a C

VIVI I R P

For to be maximum C

u a

Pd 0 IdI R

.

82. (c)Wave winding would have some slots vacantand this would creat mechanical imbalances.So dummy coils are placed to balance the rotormechanically.

83. (d)Complete rotation of mechanical angle = 360°

Electrical angle = P3002

So, elec = mechP2

.

84. (a)

When switch was open, total charge q= 1 1C V

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

t 0

VC1 C2

As charge will be conserved.

Now, q = eqC .V

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

V = 1 1

1 2

C VC C .

85. (d)

Current in the above circuit depends on currentsource only.

LR LV I x R = 2 x 5 10V .

86. (c)

Let diode is ON; Applying KVL in the loop.

V 4i 3 = 0

V = 4i 3

Slope = 4

for V 3V i = 0

for V 3V i = 0.

87. (a)

For the given Circuit,

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thR =2 x 1 22 1 3

So time constant,

=2CR C sec3

Then, V t = t

V 1 e

=

t1 x 3 1 e1 2

= ti e

at t = t0 oV t = ot1 e

=2

toe

= 1

Now, dv t

dt=

ot t /1 1e et

odv tdt

=ot1 1.e 10

t

=1 2 C

10 3

110

C =3 0.15F

20 .

88. (b)

Let individual time constant are 1 2&

1 21 2

1 2

L LandR R

After series combineation,

eq = 1 2L L

and eqR = 1 2R R

equivalent time constant

eq =

1 2

1 2

L LR R

Given,

1 2

1 2

L LR R

=

1 2

1 2

L LR R

2

=1 2 1 2

1 2

L R R L2R R

1 2 1 1 2 22R R L 2R R L

= 2 21 1 2 1 2 1 2 2 1 2L R R L R R R L R L

1 2 1 2 2 1 2 1R L R R R L R R 0

1 2 1 2 2 1R R R L R L = 0

1 2R R 0 i.e. 1 2R R

1 2 2 1R L R L 0 1 1

2 2

R Li.e.R L

.

89. (a)

30

60V

i t

20mH

M 10mH

2V t

1i t

2V t = 1di tM.

dt

=

33x10 t3 2d 6010x10 x 1 edt 30

= 3 3 1500t310 10 2 10 .e volts2

= 1500t30e volts .

90. (d)

At t 0 , Capacitor acts as short circuitand inductor acts as open-circuit

R

V

I

I =VR

At t capacitor acts as open-circuit andinductor acts as short circuit.

R

V

I

I =VR

.

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(12) EE (Test-3), Objective Solutions, 12th March 201691. (c)

1K2KV

1 2C CV V

At t 0 , combined voltage across two

capacitors 1 2c cV V =1 1V V

1 2 3

Charge must be equal when capacitor areconnnected in series.

q = 1 21 C 2 cC V C V

for equivalent capacitance

q = 2 2

1 2C C

1 2

C CV V

C C

=1 2

1 2

C C V.C C 3

1C1

1VC

1C. 2

1 2

.C V 2 V 2V. . voltsC C 3 3 3 9

&2

1 2C

2 1 2

C .C1 V 1 V VV . . . voltsC C C 3 3 3 9

.

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

L N2

It the number of turns are doubled the new valueof

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.

93. (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.

94. (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.

95. (c)f(t) = 1[u(t–1)–u(t–2)]+2[u(t–2)– u(t–3)] + 1[u(t–3)–u(t–4)]

= u(t – 1) + u(t – 2) – u(t – 3) – u(t – 4).

96. (c)V = IR R

= 200 0.02 30

= 4 30

IL =4 30

200 0.5 j

=4 30100 j

= 0.04 60

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Ic =

6

4 3010

200 50j

= 0.04 120 I = IR + IC + IL

= 0.02 30 0.04 60 0.04 120

= 0.02 30°.

97. (c)

a

Since charge density = 0According to Gauss’s Law, D.ds Q

For r < a

2D.4 r = 30

4 r3

D = 0r3

For r a

2D.4 r = 30

4 a3

D =3

02

a3r

.

98. (b)

Q

Conducting plates

No. of image charges = 360 1

=

360 130

= 12 1 11 .

99. (b)

+Q10cm

20 cm

1 2

• As the sphere of radius 10 cm encloses thetotal charge Q. So the concentric sphere ofradius 20 cm will definitely enclose the totalcharge Q. According to Gauss’s Law

(flux)2 = total charges enclosed= (flux)1= 20 V.m.

100. (d)

X

Y

Z

2

3

1

1

0 1

1

Circulation F

around closed path

= F.d

l

=1 2 3

F.d F.d F.d l l l

For curve (1), Z = 0

So, F = x yˆ ˆxa ya

Then, 1

F.dl = x y x y zˆ ˆ ˆ ˆ ˆxa ya dxa dya dza.

=0 1

1 0

xdx ydy

= 0 12 2

1 0

x y2 2

=1 1 02 2

The value of l

F.d will be same for curve (2) and

curve (3). So,

F.dl = 0

101. (d)

Applying image of charge method+q

V=0d

IES M

ASTER

(14) EE (Test-3), Objective Solutions, 12th March 2016+q

d

–q

d

(image charge)

So, force,

F =

2

2 200

q.q q16 d4 . 2d

.

102. (c)

Exp nullpoint

Q+ Q+

103. (b)

Apply gauss law at the surface of first sphere.

s

E.ds

=0

Q

E = 2

0

Q 14 R

= 20

Q4 R .

104. (a)

105. (d)The electric field intensity due to a charged plane,

E = n

0a

2

where is the uniform charge density..

So, the electric field intensity E at origin due to

a change sheet placed at Z = 10 m.

0,0,0,E

= z0

a2

= 9

z8

20 10 a12 10360

= zˆ360 a

= zˆ360 a V m .

106. (c)Tst = 12 NmTL = 8 Nm

Tacc = st LT T

= 12Nm 8Nm 4Nm

= 1 rad/sec2

I = accT 4Nm1rad sec

= 4 Kgm2.

107. (b)An SCR is not a self commutating device. SCRrequires a separate circuitary for its turn offpurpose. An IGBT can be turned ON and OFFby its gate pulse. IGBT is self commutatingdevice.

108. (c)

I = ripple current

= dc battery ONV V T

L

= 3

30.3 1 1050 10

10 10

=40 0.3 1.2A

10

.

109. (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 .....

110. (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.

IES M

ASTER

EE (Test-3), Objective Solutions, 12th March 2016 (15)111. (a)

R 0.12

I = 10APower loss when MOSFET is ON I2R = 2 0.1210

= 12WThe device is ON for half time and OFF for otherhalf, so average power loss is

=12W

2= 6 W.

112. (d)String efficiency

=Voltage across the string

n voltage across the unit nearthe power conductor

=1 100

4 0.30

= 83.33 %.

113. (a)

newpuX =

2Old newold B Bpu New Old

B B

V SXV S

=

213.2 500000.213.8 30000

= 0.305 pu

114. (b)

The HVDC circuit breaker are quite expensiveas there is no zero crossing as in case of ac sothat the contacts could be seprated at thatinstant.

115. (d)

For an open ended line,Reflection coefficient of voltage

=0

0

00

ZR 1R Z R 1ZR Z R 1R

as R

i.e. reflected voltage, r iV 1 VV

Reflection coefficient of current

=

0

0

R Z1

R Z as R

i.e. reflected current, Ir = –1.(Ii) = –I.

116. (a)A

L

A

G

KInductor has the basic property that it opposesthe change in the current. So, when a seriesinductor is placed with SCR, it reduces the valueof (di/dt) by increasing the ‘dt’.

117. (d)During reverse recovery time, the device continuesto conduct in the reverse direction (not in forwarddirection) because of the presence of storedcharges in the depletion region and thesemiconductor layers.• While designing any circuit, a dead time or,

commutation interval must be given betweenremoval of outgoing device and addition ofincoming to device; otherwise source wouldexperience a direct short-circuit throughoutgoing and incoming devices.

118. (b)Armature reaction Shifts the Magnetic NeutralAxis (MNA) axis of dc axis. Hence shifting thebrushes would result in smooth commutationbut it is the effect of shifting of brushes that fluxper pole gets reduced.

119. (a)

For dc machines, bEN

...(1)

When the field winding gets disconnected, therewill be only residual magnetic field flux which isvery small. So from the equation (1) it is clearthat the machine would run upto dangerouslyhigh speed.

120. (a)

For a series motor, aIAt no-load, Ia is very low

As, speed, b b

a

E ENI

As Ia is very low at no-load, a series motorattains dangerously high speed. To avoid this,we never run dc series motor at no-load.As swineburne’s test is a no-load test, it cannotbe done on dc series motor.

answers r - ies masteriesmaster.org/public/archive/2016/test-paper... · 1 1 cv c v 2 2 where, c is...

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