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(1)

1. (b)

2. (a)

3. (c)

4. (d)

5. (c)

6. (d)

7. (d)

8. (d)

9. (c)

10. (a)

11. (b)

12. (c)

13. (b)

14. (a)

15. (b)

16. (b)

17. (b)

18. (b)

19. (c)

20. (d)

21. (b)

22. (b)

23. (d)

24. (c)

25. (a)

26. (b)

27. (d)

28. (d)

29. (b)

30. (a)

31. (a)

32. (c)

33. (b)

34. (b)

35. (d)

36. (a)

37. (d)

38. (b)

39. (a)

40. (b)

41. (a)

42. (b)

43. (c)

44. (d)

45. (d)

46. (d)

47. (c)

48. (c)

49. (d)

50. (c)

51. (b)

52. (a)

53. (c)

54. (a)

55. (b)

56. (a)

57. (b)

58. (d)

59. (c)

60. (c)

61. (a)

62. (b)

63. (c)

64. (a)

65. (d)

66. (b)

67. (d)

68. (a)

69. (b)

70. (d)

71. (d)

72. (b)

73. (d)

74. (a)

75. (b)

76. (a)

77. (c)

78. (b)

79. (d)

80. (c)

81. (b)

82. (a)

83. (a)

84. (a)

85. (a)

86. (d)

87. (c)

88. (c)

89. (d)

90. (c)

91. (c)

92. (b)

93. (a)

94. (b)

95. (c)

96. (d)

97. (c)

98. (c)

99. (d)

100. (c)

101. (c)

102. (d)

103. (c)

104. (b)

105. (a)

106. (c)

107. (d)

108. (c)

109. (a)

110. (c)

111. (a)

112. (a)

113. (b)

114. (d)

115. (a)

116. (b)

117. (a)

118. (a)

119. (c)

120. (d)

Objective Question Practice ProgrameDate: 19 May, 2016

ANSWERS

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(2)

1. (b)+

–

vt

If the diverter resistance is reduced, flux reduces, thus for a constant load torque, Ia

will increase. We know that speed mm

1

,

so speed will increase.

2. (a)

a tE 250V, V 248V

Ia =a t

a

E VR

Ia =250 248 40 A

0.05

m =3000 2 100 rad sec

60

T = a aE I 250 40

100

= 31.83 Nm

3. (c)

Current in each lamp 160I 0.5A

120 .

Total load curent IL = 0.5 × 50 = 25A

Field curent Ish = 120 2A60

Armature current = 25 + 2 = 27AA wave winding has 2 parallel path, so curent

through each conductor = 27 13.5A2

4. (d)The pole shoe performs all the functions listedin the options.

5. (c)When the speed of synchronous motor is morethan synchronous speed, then the motor

damper winding will act like inductiongenerator.

6. (d)Universal motor is used for high torque.Capacitor start is used for relatively constantspeed at a wide variety of loads. Split-phase forlow starting torque applications and Hysteresismotor for operation at synchronous speed.

7. (d)All the statements are correct regardingtertiary winding in a star-star transformer.

8. (d)

Hysteresis Loss = xmaxKhB f

Eddy curent losses = 2 22KB max f t

From the two equations we can see that theselosses depend on maximum density and supplyfrequency.

9. (c)

1 1E 4.44Bm Ac f N Volts

4m400 4.44 B 50 10 50 350

Bm = 4400

4.44 50 10 50 350 = 1.0296 T= 1.03T

10. (a)Voltage-to-frequency is maintained constant tokeep the flux constant. By this we can maximizethe use of magnetic circuit.

11. (b)

From the phasor, it is clear that ar opposesf hence demagnetizing effect.

E = Vtf90°

Ia

f

aVr

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(3)12. (c)

In three limb construction, there is no path for3rd harmonic flux, and the magnetic reluctanceis not balanced. In five limb construction, themagnetic reactance is balanced.

13. (b)

Total number of conductors = 3 14 24 1008

Total number of torque = 1008 5042

Turn per phase Tph = 500 108

3

Induced emfEph = Ph4.44K fT

= 4.44 1 0.0295 50 108 = 1100 V

14. (a)For the universal motor speed with DCexcitation is higher than with AC.

Speed

ndcnac

current

15. (b)If the supply terminal gets interchanged thesequence change and the flux rotate in differentdirection causing the motor to rotate in oppositedirection.

16. (b)Slip of the induction motor is given as ratio ofrotor copper loss to rotor input power.

17. (b)Pull out torque decreases with the decrease ofexcitation. Excitation and pull out torque havea direct relationship.

18. (b)

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

19. (c)Shunt fault is characterized byi) Increase in current

ii) Fall in voltageiii) Fall in frequency

20. (d)In a string of insulator, maximum voltageacross the disc occurs in that disc which isnearest to the conductorAs, string efficiency

Phase voltage or voltage across stringVoltage across thenumber of disc disc nearest to condutor

i.e. 0.80 =m

4003

20 V

Vm =

400253 KV

0.80 20 3

21. (b)

22. (b)

23. (d)Since VS = AVR + BIR ...(i)and IS = CVR + DIR ...(ii)In equation (ii) the component of chargingcurrent is CVR.So, charging current per phase

IC = C(VPh)R

=1320.005

3

=66 A

3

24. (c)The stored energy in the rotor = GH

= 100 × 8= 800 MJ

25. (a)Fissionable materials :

238 235 238 23392 92 92 92U, U, Pu, U,

Fertile materials:

238 23292 90U, Th.

– fissionable materials can be producedartificially from the fertile materials

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(4)26. (b)

In order to reduce the sheath losses, bondingof cable is done-by bonding of cable at both theends, the equivalent resistance of the systemis increased whereas the inductance is reduced.

27. (d)

As, P = 1VI cos

Now, P = 22V.I cos

1VI cos = 22VI cos

I2 =1I2

Now, the power loss, 2

2L 2P I R

=2

1I R2

=21I R4

= 1LP4

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

29. (b)• Mho relay is used for long transmission line• Negative sequence relay is used in generator

in protection against unbalancing.• Thermal relay is basically used in motor

protection against overload.

30. (a)For economic operation

LG

G

1dF P1dP P

=

G1 150.02P 2

1 0.2

G0.02P 2 = 12 0.02PG = 10

PG =10 500 MW

0.02

31. (a)For EHV lines where total reactance is high,the shunt compensation is used mainly for

improvement of system stability.

Since, Pe =. VE sinX

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

32. (c)Load Factor

= Units generated in agiven periodMaximum demand × given period

= 1 80 60100 40 20 602

100 100

= 4000 2400 120010000

= 7600 0.7610000

So, the load factor = 76%

33. (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.

34. (b)Since, for two generating system,

PL = 2 21 11 1 2 12 2 22P B 2P P B P Bfrom here, we can observe, the unit of B-coefficient is (MW)–1

35. (d)

A deliberate connection of a resistance inparallel with the contacts of the circuit breakeris called resistance switching. When arc isshunted by the resistance R, a part of arccurrent is diverted through this resistance.This result in the decrease of arc current andan increase in the rate of deionization of thearc path. Thus, the arc. resistance is increasedleading to a further increase in currentthrough the shunt resistance. Hence resistanceswitching damp out the transient occured afterbreaking the fault current.

36. (a)

The voltage across the arc is in phase withthe arc current as the arc can be consideredas resistive in nature.

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(5)37. (d)

Simple differential protection system isinadequate due to :i) Magnetising inrush current : Since

magnetising inrush current flows onlythrough primary winding, it causesdifference in CTs output and makes the relayto operate.

ii) Mismatch in CTs characteristics : It causesdifference in respective secondary currentson occurrence of through faults.

iii) Tap changing : During tap changing voltageratio differs from the original one and affectsthe operation of the differential relay.

38. (b)Ripple frequency = 3f, where f is the supplyfrequency.Given f = 100 HzRipple frequency = 300 Hz.

39. (a)

V+ = iV V

Let V1 be the voltage at node of T network

1V V VR R = 0

V1 = i2V 2V

1 0i 1 V VV V V 0R R R

3V1 = 0V V

6Vi =0

i 0i

VV V 5V

40. (b)

0 Z BEV V V 12 0.7 11.3V

CE i 0V V V 20 11.3 8.7V

Current through 220 resistor is

I1 =20 12 8 36.4mA

220 220

Current in 1K resistor is

I2 =0V 11.3 11.3mA

1K 1K

IB =CV 11.3mA 0.226mA

50

IZ = 1 BI I 36.4 0.226

= 36.17 mA

41. (a)Apply KVL, we get

VCC – C C CEI R V 0

IC =CC CE

C

V VR

=5 0.21K

= 4.8 mA

IB =CI 4.8mA 48 A

100

42. (b)

z maxI =400mW 40mA

10

L Zi i =20 10 45mA

222

L mini = 45mA 40mA 5mA

RL =10 2K

5mA

43. (c)At a given value of VBE, increasing the reverse-bias voltage on the collector-base junction andthus increase the width of the depletion regionof this junction. This in turn results indecrease in the effective base width W. SinceIS is inversely proportional W, IS increases andthat iC increases proportionally. This is calledearly effect.

44. (d)

i21k

iL

9ki1

3.6k

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(6)

Li = 11 2 s s2

Ri i i iR

= 1s2

R1i R

=90.5mA 11

= 5 mA

45. (d)A Q-point in the middle region will give adistortion less output. It has more space forinput variation if in the middle.

VCE

IC

Q

46. (d)

Condition for saturation is C BI I .

47. (c)In n-state amplifier gain of overall amplifierincreases whereas bandwidth decreases.

Bn = 1/nB 2 1

For nn 1, B B

48. (c)From the Fermi-dirac statistics, the probabilityof electron occupation of an anergy level equaltot he fermi level is given by

f(E) =f

1E E1 exp

KT

For E = Ef,

f(E) = 1

1 exp 0 = 0.5

49. (d)

AV =0

i

VV

AV = m L Cg R R = mg 10K 10K

= 0.5mA 5K25mA

= –100

50. (c)In saturation region

ID = 2´

n GS THK V V

gm =D

GS

IV

= n GS THK´ 2 V V

gm = n OX dW2 C IL

mg WL

if Id = constant

gm2 = m12 g

51. (b)

f = 1T

T = lT T1R C n

1

= 470×103×0.01×10–6 ln1

1 0.7

= 47×10–4×1.2= 5.64 ms

f = 31

5.64 10 = 177 Hz

52. (a)

Vdc = r P P

mV

V2

= 628 2 = 25V

Drop in the regulator = 25V – 12V= 13V

Power dissipated = voltage drop × current= 13V × 2A= 26W

53. (c)Total propagation delay, Td = 4 × 20 ns

= 80 ns

So, counting time T 80ns

So frequency of counting, 1f T

i.e. counting speed,

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(7)

f 91 1T 80 10

61000 10

80

12 MHzSo, maximum counting speed in the givenoption = 10 MHz.

54. (a)Number of pulses counted

=

Time taken by ramp voltage to fall to zero

Time period of oscillator

=3

3

25 101

600 10

= 600 × 25= 15000

55. (b)

FX

Let X = 0, then F = 0and if X = 1, then F = 0So, F = 0

56. (a)For dual slope integrating type digital voltmeter

a r1 1V .dt V .dt 0

RC RC a 1 r 2V .t V .t 0

t2 =a

1r

V tV

where t1 and t2 are time duration of firstintegration and second. So total conversion time,

t = 1 2t t

= a1 1r

Vt tV

= a1r

V1t V

= ar

V1 1n. Vf

= ar

Vn 1Vf

.

57. (b)For counter type ramp ADC, maximumconversion time for n-bit

= n CLK2 t

Here n = 8 Maximum conversion time = 28 × tCLK

= 8 312

40 10

= 256 msec.40= 6.4 ms

58. (d)Sequence of conversion time :Flash type < SAR Type < Dual-slope type

59. (c)RAM is a volatile memory because it cannotretain stored data when power is turned off.Thats why it is typically used for short-termdata storage. Static RAM, Dynamic RAM andcache memory are types of RAM.

60. (c)Tristate buffer in a memory allows the datalines to act as either input or, output lines andconnect the memory to the data bus in acomputer. These buffers have three outputstates : HIGH (1), LOW(0), and HIGH-Z (open).

61. (a)(–19)10 =

+19100111

For negative SignNow, 1’s complement form of –19 is producedby taking 1’s complement of +191’s complement : 011001

1's complement of +19

2’s complement : 011001+1

1011012’s complement of –19 is 101101.

62. (b)The first cycle of any instruction is fetch cyclefollowed by execution cycle. Fetch cycle includesthe transfer of content of program counter inmemory then memory places the opcode on thedata bus so as to transfer it to the CPU.

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(8)63. (c)

CA H = 1100 1010RIM :

SID I7.5 I6.5 I5.5 IE M7.5 M6.5 M5.5

Interrupt pending status of RST7.5

So, RST 7.5 interrupt is pending.

64. (a)

65. (d)XRA A means XOR operation of accumulatorwith content of accumulator.The output of XOR operation is low if boththe inputs are same. So, the output will bezero. Hence, the zero flag will be set.

66. (b)Capacity of RAM = 8 k

= 23 × 210= 213

So, number of address lines = 13starting address,

1000H

Finaladdress

2FFFHF F F2

0001 0000 0000 00000001 1111 1111 11110010 1111 1111 1111

67. (d)After execution of call instruction, stackpointer will be decremented by two to loadthe content of program counter. So, stackpointer contains 08D4H.Then program counter is loaded with theaddress of subroutine address i.e. 02AF H.

68. (a)The trap has the highest priority amonginterrupts. It is a non-maskable interrupts.It is unaffected by any mask or, interruptenable. Hence, it does not need active lowsignal INTA.

69. (b)

70. (d)

8 E

37H 0011 0111A56H 0101 0110

CY 1 0000 00011000 1110

8EH

So, Accumulator, A = 8EH

71. (d)The programmable interrupt controller 8259is used when several I/O devices transfer datausing interrupt.

72. (b)STA instruction is used to store the data intothe accumulator from the memory whoseaddress is specified in the instruction. So, it isdirect addressing instruction.

73. (d)• PUSH instruction is used to load the 16-bit

data into stack.• POP instruction is used to read the 16-bit

data from stack.• In CALL instruction, the 16-bit address of

next instruction to be executed is first storedin program counter and then stored in stackfrom program counter.

74. (a)JNC is conditional jump instruction in whichprogram jumps to the instruction specified bythe address if the specified condition is fulfilled.The program proceeds further in the normalsequence if the specified condition is not fulfilled.It is an immediate addressing mode instruction.

75. (b)Antiparallel thyristor can conduct current inboth the direction and the diode can blockvoltage in only one direction.

76. (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

77. (c)

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

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(9)78. (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.

79. (d)All the other three except MOSFET are currenttriggered devices. A MOSFET is a voltagetriggered device.

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

81. (b)Output Voltage (V0) for a step down chopper is

V0 = sD V

0

s

VV = D

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

83. (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 .....

84. (a)

R 0.12

I = 10APower loss when MOSFET is ON I2R =

2 0.1210

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

= 12W2

= 6 W

85. (a)

m 0 Cpeak 0 smaxCI I I I VL

=0.110 2001m

= 12 A

86. (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

87. (c)For first quadrant operation, a half controlleris most preffered. It is simple and cheapcompared to other converters.

88. (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.

89. (d)

Vph =4002 = 200V

V1 = phmr V sin

m

155 =3r 200 sin

3

r = 0.937

90. (c)

91. (c)

Safety factor =m

PRVV

[PRV = Peak Recovery Voltage]

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(10)

=rms

PRV2V

PRV = 2.5 400 2= 1414 V= 1.4 kV

92. (b)

tan = LR

=4100 12.73 10

4

1 = 45º

should be greater than , i.e. greater than45º

> 45º

93. (a)

tna = L 10 1R 10

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

94. (b)

2nV = 4KTBR

T = (273 + 27) = 300 K4 23R 1000 , B 2 10 Hz, K 1.38 10 J K

2 23 3 4nV 4 1.38 10 300 10 2 10

2nV = 1433.12 10

Vn = 0.57 V

95. (c)

Pt =110KW 15KW12

Pi = 15 –10 = 5 KWAs efficiency = 0.75

DC input power = 5 6.66KW0.75

96. (d)

Frequency deviation f

= 105.007 105 0.007MHz 7KHz

Carrier swing = 2 f= 2 × 7 kHz= 14 KHz

97. (c)Sampling rate should be that of twice ofmaximum frequency.

= 2 × 500 Hz= 1000 Hz

98. (c)The difference between incoming signalfrequency (fc) and its image frequency is 2If.The RF filter may provide poor selectivityagainst adjacent channel separated by a smallfrequency difference but it can providereasonable selectivity against a stationseparated by 2If .

99. (d)

Pm = 2m t

x(t) = 0 0m cos cost tt

= 0m t cos cos2 t

2

After filter action

y(t) = m cost2

So, power = 2

m2 P cosy t 4

100. (c)

The envelope of input signal is a1 K m t that will be the output of evnvelope detector.

101. (c)

quantization noise 1

LevelsIf the number of quantization level increases,then noise will reduce.

102. (d)Total power radiated Pt is

Pt =2 21 2

CP 1 2 2

=2 20.3 0.420 1

2 2

= 22.5 W

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(11)103. (c)

The filter characteristic is shown below

2 2fCf –WC f +WCfC

C cf W 2W or f 3W

C Cf W 2W or f W

so, fC > 3W

104. (b)

Frequency modulation index = mf

= 310 10 Hzfm = 200 Hz

Modulation index = 310 10 50

200

105. (a)Power in a frequency modulation doesnotchange with modulation. The power remainssame as before modulation.

106. (c)PC = 10 KW = 0.7

Pt =2

CP 1 2

=20.710 1

2

= 12.45 W

107. (d)Nyquist rate = 2 × 30 K = 60 KHz6 level PCM, so we need at least n = 33 2 6 .

Bit rate = 60 3 180KHz = 180 Kbps

108. (c)

The band width of x(t) = 410

2 Hz

Nyquist rate = 2 × BW

=4102

2

= 104

= 10 KHz

109. (a)

It =1/22

cI 1 2

20 =1/22

18 12

21

2

=220

18

21

2

= 1.234

2 = 0.4690

= 0.68

110. (c)The direct method of determination is notpreferred as it is a wasteful, need cumbersomeloading arrangement and adequate powersupply is to be made available.

111. (a)A DC series motor develops high startingtorque, so it is used in electric fraction.

seT aI aI

Tse 2aI

Ia is high in series motor at starting, so, Tse ishigh and traction requires high torque.

112. (a)In a distribution transformer, powerfluctuations are more. It will not operate onfull load. Iron losses are kept low, so that theymay be approximately equal to copper lossesand increase efficiency.

113. (b)In this operation a synchronous machineoperates like a lagging power factor load.

114. (d)Synchronous impedance method is calledPessimistic method as it gives higher valvethan normal.

115. (a)The transient-stability can be improved in bestway by using high speed breaker because thecircuit breaker clears the fault quicker.

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(12)116. (b)

It is true that, skin effect is much smallerwith stranded conductor than with solidconductor but main reason for using strandedconductor in cable is for easy in handling andlaying.

117. (a)BJT higher transconductance gives the BJT ahigher gain than FET which has comparativelylower value of transconductance.

118. (a)In I2L logic family, transistor require lessnumber of masks and diffusion. Due to thisreason, the transistor occupy less space andhence I2L has higher density of integration.

119. (c)SIM instruction is used to selectively maskthe hardware interrupt and to provide serialoutput data.RIM is used for serial input data.

120. (d)Assertion is false and reason is true. Thebandwidth differs for FM and PM but theydon’t produce different sets of frequencies.