1. (b)

2. (b)

3. (b)

4. (a)

5. (c)

6. (c)

7. (d)

8. (c)

9. (d)

10. (d)

11. (d)

12. (b)

13. (a)

14. (b)

15. (b)

16. (c)

17. (b)

18. (c)

19. (b)

20. (a)

21. (a)

22. (c)

23. (b)

24. (c)

25. (c)

26. (c)

27. (d)

28. (d)

29. (b)

30. (c)

31. (a)

32. (d)

33. (d)

34. (d)

35. (c)

36. (c)

37. (a)

38. (c)

39. (b)

40. (b)

41. (b)

42. (d)

43. (a)

44. (c)

45. (b)

46. (b)

47. (c)

48. (b)

49. (d)

50. (c)

51. (d)

52. (a)

53. (a)

54. (c)

55. (b)

56. (b)

57. (a)

58. (b)

59. (c)

60. (b)

61. (c)

62. (b)

63. (a)

64. (a)

65. (b)

66. (b)

67. (a)

68. (a)

69. (a)

70. (d)

71. (d)

72. (b)

73. (b)

74. (a)

75. (b)

76. (b)

77. (b)

78. (a)

79. (c)

80. (d)

81. (c)

82. (c)

83. (d)

84. (c)

85. (d)

86. (b)

87. (c)

88. (c)

89. (c)

90. (a)

91. (a)

92. (d)

93. (c)

94. (b)

95. (d)

96. (c)

97. (d)

98. (b)

99. (d)

100. (c)

101. (b)

102. (d)

103. (b)

104. (c)

105. (a)

106. (d)

107. (a)

108. (b)

109. (c)

110. (c)

111. (c)

112. (a)

113. (d)

114. (a)

115. (a)

116. (d)

117. (a)

118. (c)

119. (a)

120. (a)

Objectives Question Practice ProgrameDate: 2nd April, 2016

ANSWERS

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(2)

1. (b)For obtaining equivalent resistance,we haveto short circuit the independent voltagesource and open circuit indepdent currentsource. Therefore the circuit is,

2 2

1

1

RTh

RTh = 2 2 2

= 2 1

= 3

2. (b)Due to only current source of 30A,

2 630AIa

I1

12

4

2 30A

4 Ia

4

I = 2 30

2 8 = 6A

I1 = 6 6A

6 12 = 2A

Due to only voltage source of 30V,

2 6

30V

12

4

Using SourceTransformation

6 5A 612

3 5A 12I2

I2 =

3 5

12 3 = –1A

3. (b)Applying superposition therom, it is knownthat if all current source value are doubled,then node voltages also are doubled.

4. (a)Source transformation :

R RVR

VR

R2

2VR

R2V

5. (c)From linearity we known that s LV I ratioremains constant.Now power absourbed is 2.5W, so

2L LI R = 2.5W

IL = 0.5AFrom Linearity,

20V200mA = SV

500mAVs = 50V

6. (c)When port 2 is short circuit, V2 = 0. The circuitis

1F

1/31

1´

V1

From the circuit,

V1 =

11I

s 3

Z1 = 1

1

V 1I s 3

natural frequency is the pole given bys + 3 = 0

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(3)7. (d)

Conditions for two port reciprocal network,Parameter Condition

Z Z12 = Z21Y Y12 = Y21h h12 = –h21

ABCD AD–BC = 1

8. (c)

Z12 =1

1

2 I 0

VV

When I1 = 0 , V2 =

12 V

3

I2 = 1 122V V3

I2 = 14 V3

1

2

VI =

34

9. (d)V1 = AV2 – BI2I1 = CV2 – DI2

Z11 =2

1

1 I 0

VI

when I2 = 0,V1 = AV2 and I1 = CV2

Z11 =AC =

23

10. (d)

A

B

CinCin

Cin = in

in

C C CC C

CAB = in

1 5C C2

11. (d)

Z =0 R R

R 0 R

[Z] =R RR R

Z =R R

0R R

Z = 0

Hence Y does not exist.

12. (b)For a passive network, the output powercannot be greater than the input power.

13. (a)

V2 = 22 1

I4 I I2

I2 = 21

V2I2

...(1)

I1 = 2 21 2I VV V2 2

= 2 11 2

V VI V4 2

V1 = 1 234I V2

...(2)

14. (b)

V1 = 2 2AV BI

A =2

1

2 I 0

VV

By KVL, at I2 = 0V1 = V2

A =2

1

2 I 0

V 1V

15. (b)

10k10k

10k

Vb

Let Vb be the voltage at the middle node

Vb =

aV2010K 10K

1 1 110K 10K 10K

= aV203 3

a a a bV V 20 V V 4mA5K 2.5K 10K

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(4)

aa

V207V 80 403 3

Va = 19 V

16. (c)The given measurements are for the terminal.Open circuit voltage = 10VShort circuit current = 1.5A

thR = OC

SC

V 10V 6.66I 1.5A

17. (b)

j810 0º

1 4Va

–j2

a a aV V V 10 0 04 j8 j2 1

Va =10 0 V

1.05 j0.4

i = aV 1 84.23A4 j8

i(t) = cos A2t 84.23

18. (c)

Power absorbed= L LV i

P = 2LL .i24 4i

Power absorbed is maximum when

L

dPdi = 0

3L L

L

d 024i 4idi

2L24 12i 0

2L24 12i = 0

2Li = 2

iL = 2 A

19. (b)A dynamometer type wattmeter responds toaverage value of active power = V Icos .

20. (a)

When the load power factor exceeds 60°, thenone wattmeter show negative reading.

W1 = V12 I1 cos (30° + )When cos crosses 90°, then it is negative

cos 30 = negativeIf 30 > 90°

> 60°

21. (a)

Power factor = 1 1 2

1 2

3 W Wcos tan

W W

= 1cos tan

= cos(90°)= 0

22. (c)Frequency compensation is not used ininduction type meter. Other 3 type given inanswer are used.

23. (b)Energy meter will have creep error. A slighttorque developed by the light load adjustmentmay cause the disc to rotate slowly when thepotential coils are exited, but with no loadcurrent flowing. This is as creeping. This isprevented by cutting two holes or slots in thediscon opposite sides.

24. (c)

At 0.5 power factor, = 1cos 0.5 60

Total power = 20 KW

= L L3 .V I cos

= L L3.V I 0.5

so, L L3 V I

2= 20 KW

W1 = VL IL cos(30° – 60°)

= L L3 V I

2= 20 KW

W2 = VLIL cos (30° + 60°)= 0

25. (c)

To make angle between P and voltage V,90°, voltage coil winding should be so designedthat it is highly inductive.

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(5)26. (c)

Actual energy consumed

= 508kW3600

= 0.111 kWhEnergy under test,

=100

1000 = 0.1 kWh

% error =0.1 0.111

0.111

= – 9.9 %

27. (d)A combination of voltmeter and ammeter isneeded to measure power. As the circuit isbalanced one pair of voltmeter and ammeteris sufficient to measure 3 phase power.

28. (d)When power factor is 0.866, then W1 = 2W2

29. (b)

LOAD

CC

PC

Figure (a)

LOAD

CC

PC

Figure (b)In fig. (a), if load current is high, the errorintroduced due to voltage drop across currentcoil will be large. So, connection of fig. (a) isnot employed for large load current.In fig. (b), if load current is high the currentflowing through pressure coil will be verysmall as compared to the load current andhence power loss in pressure coil will be verysmall as compared with the load power.Hence, connection of fig. (b) is preferable.

30. (c)The pressure coil circuit in low power factorwattmeter is designed to have a low value ofresistance, so that current is increased to givean increased operating torque. But due to this,the wattmeter will give a serious error as

power loss in the pressure coil may accountsa large percentage of the power beingmeasured.So, to compensate the power loss in pressurecoil, compensating coil is employed.

31. (a)The main error in a dynamometer typewattmeter is due to pressure coil inductancedue to which correct in pressure coil is not inphase with applied voltage.

32. (d)Braking torque,

TB Flux of the permanent magnet,

Magnitude of the eddy current, ie

Effective radius of the disc, R

i.e. TB ei R

eE Rr

n Rr

2nR

where Ee = Emf generated in the discn = speed of the discr = resistance of the eddy current path

33. (d)

tan =

1 2

1 2

3 W WW W

= 250 503250 50

= 2.6

cos = 2 21 0.359

1 2.6

34. (d)

The meter constant =

number of revolutions

KWh

= 31056

10220 5 1 2

= 480

35. (c)The power being measured in a low powerfactor circuit is small and current is high on

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(6)account of low power factor. So, both theconnection of wattmeter gives error owing tolarge load current or due to large power lossin voltage coil. To compensate this,compensating coil is used.

36. (c)

A

V

IV

I IL

Load

Since, reading of instrument = 180 Wi.e. VI = 180

L V200 180I I

L200 180I 0.1 V200I 0.1A2000

IL = 180 0.1 0.8A200

So, the power taken by load = V × IL= 200 × 0.8= 160 W

37. (a)A monostable multivibrator can be used aspulse stretcher.

38. (c)

C = A B and D = A BThis is the expression of half subtractor.

39. (b)In asynchronous counter, as CLK is notapplied at all the flip-flops simultaneously,so there is always a propagation delay atevery stage of flip-flop which results into alarge delay for output i.e. the effect of aninput clock pulse “ripples” through thecounter, taking some time, due to propagationdelays, to reach the last flip-flop.This cummulative delay of an asynchronouscounter is a major disadvantage for high-frequency application because it limits therate at which the counter can be clocked andcreates decoding problem.

40. (b)As the ring counter utilizes one flip-flop foreach state in its sequence. It has theadvantage that decoding gates are notrequired. For this, initially, a 1 is present

into the first flip-flop, and the rest of flip-flops are cleared.

0 1 2 3CLK Q Q Q Q0 1 0 0 01 0 1 0 02 0 0 1 03 0 0 0 1

41. (b)

D-flip flop with a feedback at Q acts as atoggle mode.

In toggle mode fout = inf2

= 10KHz2

= 5KHz

42. (d)The above flip-flop indicade a trailing edgetriggered JK flip-flip.

43. (a)Truth table of JK flip-flop is

n n n 1

n

n

J K Q0 0 Q1 0 10 1 01 1 Q

equation Qn=1 = n n n nJ Q K Q satisfies. UsingK-map and characteristic of JK flip-flop. Wecan arrive at the expression.

44. (c)For 4-flip flops the count is 24 = 16, So theno. of unused or missed states are

16 – 10 = 6 counts

45. (b)

Maximum counting speed = Pd

1N t FF

= 1

4 50ns

= 5MHz

46. (b)

Output frequency= nInput frequency

2

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(7)where, n = no. of flip-flops

Output frequency = 6

1220.48 10 Hz

2

=

4

112048 10 Hz

2 2

= 41 10 Hz2

= 5 KHz47. (c)

Initially0 1 1 1 0

1 0 1 1 1

0 1 0 1 1

0 0 1 0 1

1100 1

110 0

11 0

11

CLK-1

CLK-2

CLK-3i.e. register contains (00101)

48. (b)D flip-flop :

J

K

DEN

Q

Q–

EN

1

1

D

0

1

J

0

1

K

1

0

Q

0

1

n+1

i.e. Qn+1 = D

49. (d)

50. (c)4 22

2 12

2 + 1 = 3

2×1

2×1

2×1cs

cs

cs

4×1 Multiplexer

51. (d)LDA address = 3 bytes

ORI bytes = 2 bytesMOV A, B = 1 bytes

52. (a)DMA transfers data form memory to memorywithout consuming microprocessor time. Forharddisk where lot of memory has to betransferred, DMA is most suited.

53. (a)The address start from 0000 to 1FFFH. Thechip select pins are A13, A14, A15. They haveto be zero to address memory. So we have tofix A13, A14, A15 as zero. Other address linescan be zero or one accordingly we can access0000 to 1FFF memory.

54. (c)Push H instruction has

Fetch Write Write

6 T states 3 T states 3 T statesn = 3T = 6 + 3 + 3

= 1255. (b)

The total number of memory locations for an-bit memory address register is 2n

216= 64 K bytes

56. (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).

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(8)4. Indirect Addressing Mode: - The data is

transferred 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

57. (a)XRA A instruction execution resets contentsof accumulator and hence sets zero flag.

58. (b)TRAP : Edge as well as level triggered. Itmeans signal should go high and stay highuntil it is acknowledged.RST 7.5 : Positive edge triggered. It means itcan be triggered with a short duration pulse.RST 6.5 or RST 5.5 : Level triggered. It meanssignal has to be kept high untill themicroprocessor recognises it.

59. (c)Control word register for PPI 8255 :

7 6 5 4 3 2 1 0BIT No.

1, if A,B,C are I/P or O/P0, if C are to be set or reset

Mode of Port A

Port Clower

Port BMode of Port B

Port Cupper

Port ASo, for the given configuration of ports,

1 0 0 0 1 0 0 0

8 8 88 H

60. (b)Accumulator :FFH =+01 H =

1111 1111

0000 00010000 00001

Carry (CY)So, S(sign) = 0; as MSB is 0

Z (Zero) = 1; as result is zeroCY (Carry) = 1; as carry is produced.

61. (c)[C] FFHi.e. 115 16 15 16

= 1515 16

= 255So, the register ‘C’ will decrement 255 timesuntil content of register C = 0.So, number of times the loop will execute

= 255

62. (b)Software Interrupts :RSTn;where n = 0, 1, 2, 3, 4, 5, 6,7.i.e. RST 0 : RST 1, RST 2, RST 3, RST 4, RST5, RST 6 and RST 7.Hardware Interrupt :TRAP, RST7.5, RST 6.5, RST 5.5 and INTR

63. (a)“RLC” means rotate accumulator left withoutcarry.

1cy

A7 A6 A5 A4 A3 A2 A1 A0

0

7 4i.e. CY = 0

[A] = 74 H

64. (a)Some instructions of 8085 take 6T states foropcode fetch cycle. These areINX rP

DCX rP

PCHLSPHLPUSHRXCALL

65. (b)8254 consists three counters,each of 16-bit.8254 can be programmed in 6-modes.

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(9)66. (b)

95H 1001 0101AA9H 1010 1001

is compolement of A9H 010101101

2'scomplement of A9H 0101 0111

So, 1001 0101A9HA0101 01111110 1100

Here, cy = 0 i.e. cy 1For borrow, replace CY with CYSo, CY flag = 1After execution of CPI A9H, accumulator willremain same.i.e. A = 95H

CY = 1

67. (a)TRAP :i) highest priorityii) only non-maskable interruptiii) vectored interruptiv) both positive edge an level triggeredv) a hardware interrupt

68. (a)

Power =2V

RLimiting error = 2 5% 0.2%

= 10% 0.2% = 10.2%

69. (a)The time response depends on dampingsystem (i.e. over damped, under damped andcritically damped)

70. (d)

Absolute error, A = m tA A= Measured value – True value

71. (d)

The lower current full scale deflection willdecide the total current

25I125

= 1 mA

I = 5 mA

72. (b)A linear output and fast response is a desiredfeatures of any measuring instruments. Otherparameters like dead zone, dead time wouldcreate nonlinearity, which is undesirable.

73. (b)

Absolute error in Voltmeter = 20 1100

= 0.2 VNow, error as percentage of true value.For 2V,

Error as % of true value = 0.2 1002

= 10%

For 5V,

Error as % of true value = 0.2 1005

= 4%

For 10V,

Error as % of true value = 0.2 10010

= 2%

For 20V,

Error as % of true value = 0.2 10020

= 1%

74. (a)For balanced power, we can measure powerin one phase and multiply by 3 for 3 phasepower. For unbalanced load, we requireminimum 2 wattmeter.

75. (b)Dead Zone : The largest change of inputquantity for which there is no output of theinstrument.Dead Time : The time required by aninstrument to start to respond.

76. (b)% accuracy in V = 2%and, % accuracy in I = 2%

Since, resistance, R = VI

% accuracy in R = (% accuracy in V + % accuracy in I)= 2 2 = 4%

Since, 3

V 25R 500I 50 10

Limit in R = 4500 20100

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(10)

77. (b)For a normal spring controlled moving ironinstruments

Deflection, 2 dL1 I2 d

dLd 2

2I

For scale to the linear, i.e. I

Then, dLd 2

2

dLd

2

dLd

= constant

78. (a)

• Propagation constant, = j j

For highly conducting medium, >>

= j

= j j j2 2

= 2

• Radiation intensity 2U r P,

=2 2r E2

• Wave propagation, P = H

79. (c)Angle of reflection = 90°, for total reflection.

sin isinr = r2

r1

sin i = r2

r1sin90

=12

i = 1 1sin 452

80. (d)

Propagation Vector, P

= E H

E

and H

. Both are perpendicular to the

direction of propagation and E

and H

alsoindividually perpendicular to the directionof propagation.

81. (c)Waves Frequency (f) Wavelength (C/f)Visible light 1210 Hz300 3000 710 m1 10

TV, FM 610 Hz30 300 m1 10Microwave 910 Hz3 300 310 m1 100

Gamma Rays 1510 Hz300 3000 1010 m1 10

82. (c)When an EM wave travels in a conductingmedium, its amplitude is attenuated by thefactor ze . The distance ‘ ’ through whichthe wave amplitude decreases to a factor 1e(or 37% of the original value) is called skindepth of the medium.

AmplitudeE0

0.368E0

z(depth)

83. (d)

84. (c)Poynting vector, P E H

So, the direction of P

will be perpendicularto both E

and H

.

85. (d)

= j

= 45º

E

leads B

by 4

86. (b)According to Maxwell’s equation

.D = P

and H = E E j Ej

= E Et

= DEt

where E = j t0E e

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(11)87. (c)

As P E H

and E E HH

P =

x y z

x y z

x y

ˆ ˆ ˆa a aE E EH H 0

= y z x y y xx y zz x0 H E E H E Hˆ ˆ ˆa a a0 E H

= x y y xy z x z x y z E H E Hˆ ˆ ˆH E a E H a a

But wave is propagating in Z-direction,So, EZ = 0 ...(1)

and x y y xE H E H 0 ...(2)

In all four options given, only option (c) holdsthe both equation (1) and (2) true.

88. (c)Maxwell’s curl equation for static magneticfield,

H = J

0

= J

B = 0J

Note :Maxwell’s curl equation for time-varying field.

H

= DJt

89. (c)

Elements lose (or gain) and share electrons toform ionic and covalent bonds respectively andthese bonds are stable bonds.Conductivity in ionic crystals depends on(i) Carrier concentration

(ii)Carrier mobility, : This in turn, depends ontemperature and is given by the Arrenhius

expression 0 aexp E kT .

Where Ea = activation energyT = Temperaturek = Boltzmann’s constant

90. (a)

Conductor : Valence band andconduction band overlaps toeach other.

Semiconductor : Small gap between valanceband and conduction band.

Insulator : Wider gap between valenceband and conduction band.

91. (a)According to Clausius-Mossotti equation,

r

r

12

=

0

N3

= 0 r

r

3 1N 2

= 0 e

e

3N 3

since, e = r 1

92. (d)When the side of square plate = l

C1 = 0 Ad

= l2

0d

Now, as the side of square plate = l/2

C2 ='

0 Ad

= l 20 2

d

C2 = l204d

C2 = 1C4

So, the capacitance will become one fourth.

93. (c)Unlike the ferroelectrics, antiferroelectricshave no permanent moment, because ofgeometry of the spontaneous polarization. Inferroelectric material, polarisation occur in theabsence of an applied electric field.

94. (b)

50012V

VZ

KVL for input, we get

VCC = 500 B C z BEI I V V

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(12)

12 = 500 EI 5 0.7

EI = 12.6 mA IC = 12.47 mAKVL for output, we get

VCC –VCE = 500 × IE

CEV = C B12 500 I I

= 12 – 500 × 12.6 × 10–3

= 5.7 V

95. (d)

LR

222

20V 10VzR 0

Lii

zi

iz, (max) = z max

z

PV

= 400mW 40mA10

From the circuit,

i = 20 10 45mA222

and i = iL + iz

L(min)i = i – iz, (max)

= 45mA 40mA= 5mA

RL =

z

L, min

Vi

= 10 2K5mA

96. (c)In p-type holes are mojority carriers. Electrondensity changes with temperature.

97. (d)

IDS =

2GSQ

DSSP

VI 1 V

=

2430 18

=

2130 1 2IDS = 7.5mA

VGS = –IDRS

Rs = 34

7.5 10= 533

98. (b)In RC coupled amplifier, the gain decreasesdue to coupling capacitance in the lowfrequency response and junction capacitanceat high frequency response.

99. (d)We can see that both BE and BC junctionsare forward biased. So the BJT is operatingin saturation.

Collector current IC = 12 0.2 5.3 mA

2.2K

100. (c)In CB and CC amplifier, output is in phasewith input. In CE amplifier, output is out ofphase with input.

101. (b)A GTO can be turned on and off by positive andnegative gate signal but it takes more time toturn off. It has reverse blocking capacity.

102. (d)

An optocoupler consists of infra red light emittingdiode and a photo diode or photo transistor.They are used for providing isolation in gatetriggering circuits.

103. (b)A series inductor is generally placed in orderto protect SCR for high rising current.

104. (c)B2

RB2

RB1

B1

EP

B2

B1

n

B2

B1

Equivalent diagram of UJT

105. (a)For inductive load,

V = Ldi LIor Vdt T

T = LIV =

30.1 100 10100

= 0.1 ms

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(13)106. (d)

The sharing of voltage depends on the leakagecurrents of thyristor.

107. (a)The tun-off time, tq of a thyristor is definedas the time between the instant anode currentbecomes zero and the instant thyristor regionsforward blocking capability.The circuit turn-off time, tc is defined as thetime between the instant anode currentbecomes zero and the instant revere voltagedue to practical circuit reaches zero.Time, tc > tq, for reliable turn-off the thyristor,otherwise the device may turn-on at undesiredinstant.

108. (b)Plug setting multiplier,

PSM = / CTratioPr imary fault current

Secondary relay current

= 200400 5 15

= 2000 5400

109. (c)Transmission lines are classified aselectrically short, medium and long on thebasis of analysis of its capacitance. As thelength increases, capacitance becomesconsiderable. But as the frequency or,wavelength of the line is high, capacitancecomes into picture even for short length ofline. Hence, the concept of electrically short,medium and long lines is primary based onthe frequency or wavelength.

110. (c)At the time of interruption of current, themagnetic energy will be converted intoelectrostatic energy.

i.e. 21 Li2 = 21 CV

2

V = Li.C

V = 36

0.253 3 5 100.01 10

V = 15 KV

111. (c)

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

= 500 0.6 350 1.2 = 300 + 420= 720

Then, Heq = 720 7.2100

112. (a)In a parallel ac circuit, we take voltage asreference as all the elements connected acrosswill have same voltage. Reference quantityshould be common for all the elements, sovoltage is taken as reference.

113. (d)

114. (a)Eddy currents are induced in the conductorof current coils and other solid metal partsby alternating magnetic field of the currentcoil. These eddy currents become considerablespecially when current coil carries heavycurrent.To reduce the eddy current loss occuring inconductor of current coil, the conductor isstranded.

115. (a)A carry-look ahead adder (CLA) is a type ofadder used in digital logic. A carry-lookaheadadder improves speed by reducing the amountof time required to determine carry bits. Itcan be contrasted with the simpler, butusually slower, ripple carry adder for whichthe carry bit is calculated alongside the sumbit, and each bit must wait until the previouscarry has been calculated to begin calculatingits own result and carry bits (see adder fordetail on ripple carry adders). The carry-lookahead adder calculates one or more carrybits before the sum, which reduces the waittime to calculate the result of the larger valuebits. A parallel carry adder generates sumdigits directly from the input.

116. (d)

Sequential circuits :

• It gives outputs depending on the presentinput and past input also.

• It contains atleast one feedback path.

IES M

ASTER

(14)

• It contains some memory.

• It exhibit cyclic nature i.e. its outputsequence repeats after some clock cycle.

117. (a)In a measuring instrument, the dampingtorque is necessary to bring the movingsystem to rest to indicate steady reflection ina reasonable short time.It exists only as longas the pointer[disambiguation needed] is inmotion.Under the absence of damping torquethe pointer oscillates for a short period oftime and comes to steady position and thissituation is called under damping. If thedamping force is too large, then the pointerwill come to rest slowly and this is called asover damping.

118. (c)Swamping resistance is made of manganinhaving neglegible temperature coefficient is

connected in series with the coil. It is used toeliminate the error due to change intemperature.

119. (a)There are four contributors to totalpolarization of a material. These contributorsare electronic, ionic, orientational and spacecharge polarization. At optical frequencies(upto 1016 Hz), only electrons can respond toalternating electric field as ions and moleculardipoles cannot respond to frequencies morethan 1013 Hz and 108 Hz, respectively.

120. (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.