1. (a)

2. (c)

3. (c)

4. (c)

5. (c)

6. (a)

7. (b)

8. (d)

9. (b)

10. (a)

11. (d)

12. (c)

13. (d)

14. (c)

15. (c)

16. (c)

17. (a)

18. (a)

19. (b)

20. (b)

21. (b)

22. (a)

23. (b)

24. (d)

25. (b)

26. (c)

27. (c)

28. (b)

29. (a)

30. (a)

31. (d)

32. (b)

33. (a)

34. (a)

35. (d)

36. (b)

37. (c)

38. (d)

39. (b)

40. (c)

41. (c)

42. (d)

43. (d)

44. (c)

45. (c)

46. (d)

47. (a)

48. (c)

49. (a)

50. (b)

51. (d)

52. (b)

53. (d)

54. (c)

55. (b)

56. (c)

57. (d)

58. (d)

59. (c)

60. (a)

61. (a)

62. (c)

63. (b)

64. (c)

65. (c)

66. (c)

67. (a)

68. (c)

69. (a)

70. (d)

71. (c)

72. (c)

73. (c)

74. (a)

75. (c)

76. (b)

77. (b)

78. (d)

79. (d)

80. (c)

81. (a)

82. (c)

83. (a)

84. (d)

85. (a)

86. (a)

87. (c)

88. (d)

89. (d)

90. (c)

91. (b)

92. (b)

93. (a)

94. (d)

95. (b)

96. (c)

97. (a)

98. (d)

99. (b)

100. (b)

101. (d)

102. (d)

103. (b)

104. (a)

105. (c)

106. (b)

107. (b)

108. (c)

109. (a)

110. (a)

111. (b)

112. (a)

113. (a)

114. (a)

115. (c)

116. (c)

117. (a)

118. (c)

119. (a)

120. (d)

Objective Question Practice ProgrameDate: 9th April, 2016

ANSWERS

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(2)

1. (a)

G(s) can be written as 2Ks

s s1 14 10

,

As 4 and 10 are corner frequencies and at 4there is single pole and at 10, there is doublepole.at 4 , gain = 0dB

G 4 = 2

100K 44 10

= 1

K = 1/4

2. (c)Phase margin = 90 20 70

Gain margin = 1 20.5

3. (c)• Bandwidth is a frequency response

specification. In time domain, it determinesthe speed of time response.

• Phase margin is related to damping ratioas

P.M. = 1

2 4

2tan2 1 4

• Response peak (Mr) is a frequency domainspecification. In time domain, it isequivalent to overshoot of the system.

• Gain margin determines the stability ofthe system.

4. (c)

GM =1

GHAfter doubling gain,

GM´ =1 0.5GM

2 GH

5. (c)From R-H criteria and Nyquist plot thenumber of right half of s-plane poles can becalculated.

6. (a)As the system is starting from 0 at –40dB/decade, the origin has 2 poles, so the type is2 and 2 1 3 2 .

7. (b)By observing the figure,At 0 the plot is 180 .At the plot is 0 360 As the final angle is 360 .The difference of poles and zeros is 4 andinitial angle is –180° so type 2 system.

8. (d)A pole adds to –20dB slopeA zero adds to +20 dB slopeNo. of poles = 6 (2+4)No. of zeros = 2Slope at high frequency

= 20dB decade6 2 = –80 dB/decade

9. (b)

H(s) = ss s1 1

10 10000

= 11 1 s1s 10 10000

= 1010 s1 1s 10000

= 1010 j1 1j 10000

For 10 10000

H(s) = 20log10= 20 dB

10. (a)

G(s) = Ks a

G(0) = K 20dB 10a

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(3)From plot, a = 10 K = 10 × 10 = 100

11. (d)

G(s)H(s) = 2

s s 1

G Hj j = 2

j j 1

Phase cross over frequency can be calculatedas

p = –180°

= 190 tan

190 tan = –180°

1ptan = 90°

p =

12. (c)

Phase margin = g180 , where g =value ofphase at gain crossover frequency.

Here, g = –115°P.M. = 180° + (–115°)

= 180° – 115°= 65°

13. (b)Nyquist plot is a parametric plot. In Cartesiancoordinates, the real part of the transferfunction is plotted on x axis and imaginarypart on y-axis. The given T.F. is 10 j as varies from 0 to , we get graph in option(b).

14. (c)Corner frequencies are 10 and 1000

Slope = 40dB 2decades = 20 dB/decade

Initial slope is zero and magnitude = 0 dB.So K = 1.

H(s) =

1

2

s s1K 110

s s1 11000

= 1 0.1s

1 0.001s

15. (c)A system with gain margin close to unity ora phase margin close to zero would beoscillatory in nature.

16. (c)According to Nyquist theorm, the minimumrate of sampling should be double themaximum frequency of signal.

17. (a)

H(z) =

5Y z 1 zX z

y[n] = x[n] – x[n–5]

18. (a)

H(s) =K 1 s a

1 s b

H j =K 1 j

a

1 jb

Phase response of compensator is

h = 1 1tan tana b

= 1

2b atan

ab

Lead compensator h 0

b – a > 0b > a

19. (b)Lead compensator = high pass filterLag compensator = low pass filter

20. (b)A phase lead network has

G(s) = 1 sT ; 11 s T

21. (b)Transfer function of a lead compensator

GC(s) = 1 aTs ; a 11 Ts

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(4)Zero should be nearer to the origin.

–1/T–1/aT

22. (a)G(s) of PID controller is

G(s) = iP D

KK sKs

= i DP

P P

K sK1K K s K

= DPi

11 T sK T s

= Di

11 T sK T s

23. (b)Maximum phase shift =

m = 1 1sin1

where =Location of zeroLocation of pole

=5 3 15 3 3

m = 1

113sin 113

= 1 1sin2

= 30°

24. (d)In phase lead compensator, zero is nearer toorigin. In a phase lag compensator, pole isnearer to origin.

25. (b)An RC combination can give a lag network

R1

R2

C

This is the circuit of a lag compensator.26. (c)

PD controller may accentuate noise at higherfrequency. It does not effect the type of sytemand it increases the damping. It also reducesthe maximum overshoot.

27. (c)Zero phase sequence component in each

phase = 13 [Current in the neutral wire]

= 1 18A3

= 6A

28. (b)Percentage resistance of line

= 2BaseKVA Resis tan ceof line 100

KV

=

3

23

10000 10 1 10010 10

= 0.1 × 100= 10%

29. (a)From the experience in the operation of powersystem it can be observed that single line toground has (70 to 80)% of fault and 3-phasefaults are only (3 to 5)%.

30. (a)Positive sequence depends on time interval,i.e. subs transient, transient and steady state.

31. (d)Only in a 3 phase short circuit fault, we havesymmetrical fault currents.

32. (b)Zero phase sequence component of current inR-phase is

ROI

= R Y B1 I I I3

= 1 12 j6 12 j12 15 j103

= 1 9 j43

= A3 j1.33

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(5)33. (a)

In a star circuit, the zero sequencecomponents of phase voltages are all in phase.But the line voltage being the vectordifference of two phase voltages, the zerosequence component cancels out when linevoltage is determined. In delta, the zerosequence component of phase currents areall line phase and consequently form a localcirculating current inside the mesh. There canbe no zero sequence in line.

34. (a)The impedance of the given single linediagram is less upto point F1 than upto pointF2. So, the fault would be severe at F1.

35. (d)

Short circuit MVA = base1

MVAZ

= 1 5 40.2

= 1 200.2

= 100 MVA

36. (b)The rating of reactors is expressed in termsof the MVA that it is designed to carry atrated current and voltage. The reactance isexpressed in per unit and is the ratio ofvoltage drop across the reactor at the ratedcurrent to the line to neutral voltage of thesystem.

37. (c)Zero-sequence reactance diagram.

Xg0 = 0.2 Xg0 = 0.2

1 2 3XT0 XTL0

0.15 0.05 A

0.2

B

3

XAB = 0.15 0.05 0.2

= 0.2 0.2

= 0.10

38. (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.

39. (b)Fault MVA or, Fault level = (prefault voltage)

× (Post fault current)

or =pu

Base MVAZ

40. (c)

41. (c)Zero sequence component of fault current willflow only in those faults which include ground.

42. (d)Here, Ipositive + Inegatve + Izero = 0So, the fualt will be double line to groundfault.

+–

Z1

1 0º

Z2 Z0

1aI

2aI 0aI

1 2 0a a aI I I = 0

43. (d)Y22 = sum of admittance connected to the bus

2, they are 1 1 1, ,

j0.1 j0.1 j0.1 .

Y22 =1 1 1

j0.1 j0.1 j0.1

= –j30

44. (c)Since a great majority of transient faults are

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(6)line to ground in nature, by selecting singlepole operating and reclosing method thetransient stability is effectively improved. Ifa transient line to ground fault had to occuron the generator bus it is immediately seenthat during the fault there will now be adefinite amount of power transfer. If the CBpole corresponds to the faulty line is opened,the remaining power is intact. Thus thetransient stability is improved.

45. (c)The steady state limit is EV/XTransient stability can be increased byincreasing the steady state limit, so when wereduce X, we can increase transient stabilitylimit. X can be reduced by adding seriescapacitance.

46. (d)The admittance diagram is

1 2

510

10.1 = 10 and 1 5

0.2

Bus admittance matrix = 10 5 5

5 5

=15 5

5 5

47. (a)Inertia constant

H = Stored energy in Joule at synchronus speedVA rating of the machine

Energy stored = H × VA rating= 64 100 10

= 6400 10 J

= 400 MJ

48. (c)Equal area criteria can be used for only atwo-machine system or one machineconnected to infinite bus.

49. (a)Let a system have steady state power limit of500MW, but when carrying 300 MW, a suddenaddition to the load of 100 MW might causeloss of synchronism. This explains transient

stability limit will be always less than thesteady state stability limit.

50. (b)From the equal area criteria

d ddt dt

= aP d2M dt

-

For the stability, speed must become zero atsame time.

ddt = 0

51. (d)Critical clearance time : The maximum valueof time allowed for protective gear to operatewithout loss of stability is called criticalclearing time.

52. (b)When the system was operating at ‘1’ motorwill be running at synchronous speed s . Butas the mechanical load increased, first themotor 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 .

53. (d)YBUS Matrix :• Diagonal elements of matrix is formed by

adding the line admittance connected tothat node.

• OFF diagonal elements of matrix is formedby taking negative of line admittanceconnected between them.

• YBUS matrix is easier to construct andmodified.

• Some off diagonal elements of matrix arezero i.e. spanse. The spansity increases inlarger system.

ZBUS Matrix :• ZBUS = –1

BUSY i.e. inverse of YBUS matrix• Z-Bus is a full-matrix i.e. zero elements in

Y-Bus become non-zero elements in thecorresponding ZBUS.

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(7)• The formation of ZBUS requires either

matrix inversion or, use of involvedalgorithms.

54. (c)The process of convergence due to G-S methodis slow i.e. it takes larger number of iterationsbefore a solution is obtained. The process ofconvergence can be speeded up if the voltagecorrection during consecutive iteration ismodified to

accK 1PV = K K 1 K

P P PV V V

where is known as acceleration factor andit is a real number. 1.6 is a generalrecommended value for most of the systems.

55. (b)

Type of Bus Unknown (not specified)

Known

1. Generator Bus or PV Bus

2. Voltage controlled Bus or slack bus

3. Load Bus or PQ Bus

P,|V| Q,

|V|,

P, Q

P, Q

|V|,

56. (c)Energy stored by the equivalent machine= Sum of the energies stored by individualmachine connected in parallel.So, GeHe = GAHA + GBHBEquivalent inertia constant,

He = 350 1.6 400 1.1500

= 1000500

= 2.00 pu.

57. (d)Steady-steady stability refers to the ability ofthe power system to regain synchronism aftersmall and low disturbance, such as gradualpower changes.The transient-state stability is the ability ofthe system to regain synchronism after a largedisturbance.Dynamic stability is concerned with smalldisturbances lasting for times of the order of10 to 30 seconds with the inclusion ofautomatic control devices.

58. (d)• Ferromagnetic materials are characterized by parallel

arrangement of dipole.

• Antiferromagnetic materials arecharacterized by antiparallel arrangement of dipole withequal dipole moment value in oppositedirection.

• Ferrimagnetic materials are characterizedby antiparallel arrangement of dipole withunequal dipole moment value in oppositedirection.

59. (c)

60. (a)• Cobalt (Co), Nickel (Ni), and Iron (Fe) are

the ferromagnetic materials.• Brass is an alloy made from copper and

zinc. They are commonly used inarchitectural applications.

61. (a)• Paramagnetic material are characterized

by random arrangement of dipole. Due torandom dipole arrangement the value ofspontaneous magnetization is zero.

• When these materials are placed in amagnetic field, they acquire weakmagnetization in same direction of the field.

62. (c)Diamagnetic materials have negativesusceptibility and almost zero relativepermeability.

m = negative and r 0

Paramagnetic materials have magneticdipoles but they are randomly arranged andtheir is no interaction among them.Ferromagnetic materials have low electricalresistivity.In Ferrimagnetic materials, hysteresis loophave low area due to which eddy currentlosses or dielectric losses are small.

63. (b)Ferrites have following features:• Magnetic moments of adjacent atoms are

aligned in opposite direction but thereexists some net magnetic moments.

• It has very high electrical resistivity (>105 m).

• It has extremely low dielectric loss.

IES M

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(8)• General Formula : MO. Fe2O3, where M

represents divalent metals. eg. Mn, Mg,Ni, Co, Cu, Zn, Cd etc.

• High permeability.• Mechanically hard brittle.

64. (c)

Power P = 2rmsi R

Here 2rmsi =

T2

0

1 i t dtT

= 3

2

0

1 3t dt 273

2rmsi R = 270W

R =270 1027

65. (c)

Series-aiding 1 2L L 2M 10

1 2L L = 10 2x1 8mH

M = 1 2K L L

1 2L L = 2 2

14

M 1 16K

So 21 2L L = 2

1 2 1 2L L 4 L L

= 28 4x16= 0

so 1 2 1 2L L 0 L L

1 2 1 2L L 8 L L 4mH

66. (c)

r

ER

EIr R

E6 ...... 1r R

E4 ...... 2r R 10

from equation 1 and 2

64

=

r R 10r R

r R = 20

E = 120V [from-(i)]

So120I'

20 20

= 3A r R 20

67. (a)

V(s) = Z s .I s

=

2 s 1.s s 2 s 3

i t 2u t

Initial Value :

t 0LimV t =

sLims.V s

=

s

2 s 1Lims. 0s s 2 s 3

Steady State Value :

tLimV t =

s 0Lims.V s

=

s 0

2 s 1 2 1Lims.s s 2 s 3 6 3

68. (c)

o30

o60

reference

i

V

Here, Vector representation of V and Ishows that circuit element will be acapacitor.

CV 100XL 10

= 10

1C

= 10

1C

10

= 5

110 x 10

6C 10 F = 1 F

69. (a)1. L(Unit impulse) = 1

1I , Pole at 1s s 1

2. L(Unit step) = 1s

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(9)

1I Pole at 0, 1s s s 1

3.

t 1L e u t s 1

21I Pole at 1, 1s

s 170.

Req = 6 3

=6 3 26 3

I (t) =t 5t4e 6e

2

= t 5t2e 3eTaking Laplace transform

I(s) =

2 3

s 1 s 5

=

2s 10 3s 3s 1 s 5

=

5s 13

s 1 s 5Hence poles are at

s = – 1, – 5And zeros at

s =13 ,5

71. (c)

i = t/RCi ei i 0

Req = 13

i(0) = 3 2 3A13

Ceq = 1 F2

i = 0A

i = t/ 6t1/2 1/33e 3e

72. (c)

3A

3A20

10

Li 0 = Li 3A0

3A

3A10

20

Vx = 3A 10

= – 30V

73. (c)Charge area under the graph

Charge =1 C2 5 1 12

= 5 – 1

= 4 C

74. (a)According to BLondel’s theom, in a N-wiresystem, for measuring unbalanced load powerconsumption, minimum no. of wattmetersrequired is (N–1). In the present case N = 3,and minimum wattmeters required is 2. So,2 & 3 wattmeters can be used.

75. (c)In the case of power measurement by twowattmeter method, in a balanced 3-phasesystem with a pure inductive load the powerfactor is zero and Both the wattmeters willindicate the same power value but of oppositesign.

76. (b)The pressure coil of a dynamometer typewattmeter is made highly resistive, other-wise it gives error. The error due to induc-tance of pressure coil is equal to

tan tan , where is the power factor

of load and is the angle by which pres-sure coil current lags the load voltage.

77. (b)Energy consumed by load in 100 sec,

=

1003603600

1000

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(10)

= 1 kWh100

meter constant = RevolutionskWh

= 51

100= 500

78. (d)The major cause of creeping is theovercompensation for friction.

79. (d)

As, W2 = L LV I cos 30

The wattmeter W1 reads zero, when

cos 30 = 0

i.e. 30 = 90°

= 60°

i.e. power factor, 1cos 0.52

As the 60 i.e. power factor < 0.5, thewattmeter W2 will read negative and that ofwattmeter W1 will read positive.

80. (c)Driving torque in an induction type of energymeter, d 1 2T .

If both current coil and voltage coil is wronglyconnected, deflecting torque will remain sameas

d 1 2T´

1 2

But if either of current coil or voltage coil iswrongly connected, deflecting torque will beopposite, as

dT´ 1 2.

1 2.

81. (a)

ID =20 105K

=10

5K

= 2 mALimit is 4m A, so safe.

82. (c)Centre tapped full wave rectifier require acentre tap transformer for its operation.Others can be connected directly.

83. (a)The average value of dc voltage i.e. loadvoltage can be varied by controlling the phaseangle ( ) of firing pulses.

84. (d)

dc rmsV =1/2

msin2V

2 4

dc rmsV =1/24 sin 2100

2 4

= 67.42 VThe average power delivered to the load is

Pac = 2dc rmsV

R

= 267.42 454.5W10

85. (a)

mph sdc d

3 3V 30 LV cos I

where is the overlap angle. So when theoverlap angle is increased, the cosine termdecreases and the output voltage alsodecreases.

86. (a)

90°

E2

I2

I

I1

If core-losses and leakages are neglected, then

1I = 2I I

1I =

12

2

NI IN

so, 2I = 1I I

2| I | = 2 21| I | | I |

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(11)

= 2 2(3) (1)

= 8

i.e.,1

22

N.I

N = 8A

I2 =110 8220

=84

= 2= 1.414 A

87. (c)

Transformeroil

Oil9°–

Buchholzrelay

Conservator

Breather

Tank

88. (d)

The emf equation of transformer is,

E = m2 f N

E = m c2 fB A N

As voltage remains same i.e., 2000 V

f.Bm = constant

so, Hysteresis loss, Ph 1.6mB f

0.6m m(B f ).B

0.6mB

As frequency of operation decreases from 50Hz to 40 Hz. Bm will be increased such that(Bm.f) remains constant

so, 0.6h mP B increases

andEddy-current losses, Pe 2 2mB f

(Bmf)2

constant

89. (b)

Efficiency =output

output losses

In the case of two winding and auto-transformer, the losses remain same. But as

the output MVA of an auto-transformer ismore than two winding transformer, theefficiency increases

90. (c)

91. (b)

% resistance = 2 2

Full Load copper loss 100Full Load V I

1.5 = Full load copper loss 100300 1000

Full Load copper loss = 4500W

Given, 173.2 = Iron loss3004500

Iron Loss = 1500WTotal Loss = 4500 + 1500 = 6000W

92. (b)

For parallel operation of 3 transformers, therelative phase displacement of secondary ofall transformers must be zero.

Here Y/Y (0º) - 0º

Y 30º - 0º

Y 30º - Y 30º(depends on group)

0º - 0º

93. (a)The value of resistance 2R 1.2 referredto primary side,

'2R =

21

22

N RN

=

21

2

N 1.2N

for maximum power transfer to 1.2 , '2R

should be equal to source resistance ie. R1 =10.8

i.e.

21

2

N 1.2N = 10.8

21

2

NN = 10.8

1.2 = 9

1

2

NN = 3

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(12)

N2 = 1N3

= 483 = 16

94. (d)

I0

I1I2

I3

y F

P Q

S1 S0

4×1 MUX0 1 1

1

0

0

Output, F can be written as

F = PQ PQ = P Q= XNOR (P,Q)

95. (b)D can be written as

D = AB ABoutput, f can be written as

f = D.0 D.C DC

So, f = AB AB C

= ABC ABC

96. (c)Counter, flip-flop & shift registers have amemory storage element. Therefore, they aresequential circuits. Only adder is acombinational circuit.

97. (a)To reset all the flip-flopsQ1 = 1, Q2 = 0, Q3 = 1

(101)2 = (5)10The given circuit is a Mod-5 counter becauseafter the counter state is 101, the counterresets and starts counting again from 000.

98. (d)Cases should be replaced by ceases PIPOmode is not done by a shift register.

99. (b)For 8×1 multiplexer :

S2

00001111

S1

00110011

S0

01010101

Y

I0

I1

I2

I3

I4

I5

I6

I7

For the output Y to be a copy of input I5,S0 = 1, S1 = 0 and S2 = 1.

100. (b)8 bit serial-in/serial-out shift register,So, four clock pulses are required for outputto be Q3. time delay, td = 4T

= 3

14 sec150 10

= 30.02667 10 sec= 26.67 s

101. (d)In a 8085 micro processor, On receiving aninterrupt from an I/O device, the CPUBranches off to the interrupt service routineafter completion of current instruction.

102. (d)Direct memory access (DMA) is a feature ofcomputerized systems that allows certainhardware subsystems to access main systemmemory independently of the centralprocessing unit (CPU).Without DMA, whenthe CPU is using programmed input/output,it is typically fully occupied for the entireduration of the read or write operation, andis thus unavailable to perform other work.With DMA, the CPU first initiates thetransfer, then it does other operations whilethe transfer is in progress, and it finallyreceives an interrupt from the DMA controllerwhen the operation is done. This feature isuseful at any time that the CPU cannot keepup with the rate of data transfer, or whenthe CPU needs to perform useful work whilewaiting for a relatively slow I/O data transfer.DMA is the fastest mode of data transfer.

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(13)

103. (b)1. Immediate Addressing Mode: - An

immediate is transferred directly to theregister.Eg: - MVI A, 30H (30H is copied into theregister A)MVI B,40H(40H is copied into the registerB).

2. Register Addressing Mode: - Data is copiedfrom one register to another register.Eg: - MOV B, A (the content of A is copiedinto the register B)

MOV A, C (the content of C is copied intothe register A).

3. Direct Addressing Mode: - Data is directlycopied from the given address to theregister.Eg: - LDA 3000H (The content at thelocation 3000H is copied to the registerA).

4. Indirect Addressing Mode: - The data istransferred from the address pointed bythe data in a register to other register.Eg: - MOV A, M (data is transferred fromthe memory location pointed by the regiserto the accumulator).

5. Implied Addressing Mode: - This modedoesn’t require any operand. The data isspecified by opcode itself.Eg: - RAL, CMP

104. (a)Interrupt

type Trigger Priority Maskable Vector address

TRAP

RST 7.5

RST 6.5

RST 5.5

INTR

Edge

Level

Level

Level

1st

2nd

3rd

4th

5th

No

Yes

Yes

Yes

Yes

0024H

003CH

0034H

002CH

105. (c)Programmable counter timer 8253 operatesin six modes :Mode 0 : Interrupt on terminal countMode 1 : Programmable one-shotMode 2 : Rate generatorMode 3 : Square wave generatorMode 4 : Software wave generatorMode 5 : Hardware triggered mode

106. (b)

107. (b)Execution of HLT instruction takes 5Tstates instead of 4T states. One extra T-state is used to provide high impedancestate i.e. tristate.

Since, time period, T = 1f =

61

2 10

= 0.5 secSo, for 5T state, t = 5 × T

= 5 × 0.5 sec= 2.5 sec

108. (c)• Phase angle of a system is independent of

gain of the system. Thus, assertion is true.• Variation in gain varies the gain crossover

frequency. Thus, it affects the phase marginof system, which is calculated at the gaincrossover frequency. Reason is false.

109. (a)

110. (a)

SSC =g v

1 1S X X

If the number of sections (n) is large. Actuallyin a tie line.

SCMVA = g n

g g v

X XS1X X Xn

as n , 1 0n and

SCMVA =g v

1 1S X X

111. (b)Modern generators are designed to havesufficiently large reactance to protect themeven in event of dead-short. For old generatorsa generator reactor is used.Reason is true but it is not the exactexplanation.

112. (a)Zero-sequence curent can never flow in thelines connected to a delta-connected winding

IES M

ASTER

(14)as no return path is available for zero-sequence curents. Zero-sequence curents can,however, flow through the delta connectedwinding themselves if any zero sequencevoltages are induced in delta.

a

b

c

I = 0a0

Iab0Ica0

I =0b0

I =0c0

Ibco

113. (a)Since, in a power network each bus isconnected only to a few other buss (two orthree), the YBUS of large network is verysparse, i.e. it has a large number of zeroelements.

114. (a)The magnetic properties of grain-orientedsteel are best along the direction of rollingand hence it requires smaller cross-sectionalarea of core to produce same amount ofmagnetic flux density. Hence motor made ofgrain-oriented steel are smaller in size as thatmade of non grain-oriented steel.

115. (c)Addition of small quantity of silicon ironimproves several of qualities of iron eg. theresistivity increases, the coecive forcedecreases etc. So, the iron with addition ofsilicon is used as soft magnetic material.

116. (c)As the power being meaured in a low powerfactor circuit is small and current is high onaccount of low power factor. So, it is absolutenecessary to compensate for the pressure coilcurrent by using compensating coil.

117. (a)Tappings in a transformer are provided on hvside due to following reasons:1. No. of turns on hv side is large. Hence a

fine voltage regulation can be obtained2. The hv winding is placed on the outer side3. The current in the hv winding is low.

118. (c)

For 3 transformers operated in parallel, therelative phase displacement between thesecondary line voltage of all the transformersmust be Zero. So the transformers to beconnected in parallel must belong to samevector group.

Diferent vector group transformers havedifferent phase displacement of secondary linevoltage but have same phase sequence.

119. (a)DMA is faster form of data transfer. The DMAtechnique does not make use of the interruptmechanism.

120. (d)Port B can work in mode 0 and 1 with eitheras a input port or output port.Port A can work in mode 0, mode 1 and mode2 with either a input port or output port.