answers - ies master · of the ammeter under calibration and inserted in circuit. similarly to...

1. (d)

2. (d)

3. (b)

4. (c)

5. (c)

6. (c)

7. (c)

8. (d)

9. (d)

10. (c)

11. (c)

12. (b)

13. (b)

14. (c)

15. (c)

16. (d)

17. (d)

18. (b)

19. (c)

20. (d)

21. (c)

22. (d)

ESE-2019 PRELIMS TEST SERIESDate: 30th September, 2018

ANSWERS

23. (b)

24. (a)

25. (d)

26. (b)

27. (c)

28. (c)

29. (b)

30. (b)

31. (d)

32. (c)

33. (b)

34. (b)

35. (c)

36. (b)

37. (c)

38. (a)

39. (d)

40. (b)

41. (a)

42. (d)

43. (d)

44. (c)

45. (a)

46. (c)

47. (d)

48. (a)

49. (d)

50. (d)

51. (b)

52. (b)

53. (a)

54. (a)

55. (c)

56. (b)

57. (d)

58. (b)

59. (b)

60. (d)

61. (b)

62. (d)

63. (b)

64. (c)

65. (b)

66. (d)

67. (d)

68. (c)

69. (b)

70. (a)

71. (d)

72. (c)

73. (d)

74. (b)

75. (c)

76. (a)

77. (a)

78. (a)

79. (c)

80. (b)

81. (a)

82. (a)

83. (b)

84. (a)

85. (b)

86. (c)

87. (b)

88. (b)

89. (c)

90. (a)

91. (b)

92. (a)

93. (b)

94. (a)

95. (d)

96. (a)

97. (c)

98. (a)

99. (c)

100. (b)

101. (b)

102. (c)

103. (c)

104. (d)

105. (d)

106. (c)

107. (b)

108. (a)

109. (b)

110. (b)

111. (d)

112. (c)

113. (d)

114. (c)

115. (a)

116. (b)

117. (d)

118. (c)

119. (b)

120. (d)

121. (d)

122. (d)

123. (d)

124. (c)

125. (a)

126. (c)

127. (d)

128. (c)

129. (d)

130. (a)

131. (b)

132. (d)

133. (b)

134. (b)

135. (a)

136. (d)

137. (c)

138. (a)

139. (b)

140. (a)

141. (b)

142. (a)

143. (a)

144. (a)

145. (c)

146. (b)

147. (c)

148. (d)

149. (c)

150. (c)

IES M

ASTER

(2) Electrical Circuits + Measurements + Basic Electronics Engg.

1. (d)

The value of measured resistance (RM) for whichthe two different circuit measurements usingvoltmeter of resistance (RV) and ammeter ofresistance Ra, will give same error, is given by

M a vR R R

MR 200 0.02

MR 4 2

2. (d)

Drop across the CC is neglected.

VP

CC

CCRP

+

–

2A

0.05A

VP

P = 250 W

IC = 2A

PP 250V V 125VI 2

P

PP

V 125R 2500I 0.05

3. (b)

As, P PL 0.01R

orP

P

L 0.01R

P

P

Ltan 0.01R

0.573

Power factor = cos = 0.5 = 60°.

% error = tan tan 100

= (tan 60° × 0.01) × 100

= 1.732%

4. (c)

A three phase watt meter consists of two pressurecoils and two current coils.

The connection of a 3-phase watt meter can beas shown

3-phase wattmeter

CC

PC

PC

CC load

5. (c)

Bifilar windings in a standard resistor are used toeliminate residual inductance.

6. (c)

Guaranteed accuracy error (E) = 1150

100

(E) = 1.5 V

% limiting error = E 100

Reading

= 1.5 100

Reading

So, the % limiting error will decrease with anincrease in the reading. So the reading withhighest accuracy will be 130 V.

7. (c)

The percentage resolution of a DVM is given by

% Resolution = n10010

where n = number of full digits.

So, the resolution of the DVM only depends uponfull digits and hene it will be same for both 3 and

132

digit DVMs.

8. (d)

In modern standard, one ampere of current isdefiend as the constant current, which, ifmaintained in two straight parallel conductors ofinfinite length of negligible circular cross-sectionand placed 1 meter apart in vacuum, wouldproduce between these conductors, a force equalto 72 10 newton per meter of length.

IES M

ASTER

[EE], ESE Prelims Test Series |Test - 2| 30th September 2018 (3)

9. (d)

The rectifier type instruments are averagemeasuring rms responding type instruments.

They measure the average value of awaveform, and their scale is such that itsreadings are shown after multiplying with formfactor of the desired waveform. Therefore,waveform errors are present in rectifier typeinstruments when a waveform different thancalibrated is applied to them.

The semiconductor diodes used in rectifiertype instruments are very sensitive totemperature changes. Therefore temperatureerrors are preent in these instruments.

The rectifier exhibtis capacitive properties andtends to bypass the higher frequency.Therefore, meter readings may be in error byas much as 0.5% decrease for every 1 kHzrise in frequency.

10. (c)

The calibration of dc voltmeter and ammeter isdone with the help of potentiometer.

The value of current through the ammeter to becalibrated is determined by measuring thepotential difference across a standard resistorand then calculating the current. The result ofthis calculation is compared to the actual readingof the ammeter under calibration and inserted incircuit.

Similarly to calibrate the dc voltmeter, the voltageacross standard resistor R is accuratelymeasured with a potentiometer. The meter to becalibrated is connected across the same twopoints, and readign of this voltmeter is comparedwith the value obtained through potentiometermethod.

11. (c)

The indication system of a d-arsonval typegalvanometer has a small mirror upon which abeam of light is cast. The beam of light is reflectedon to a scale at some distance from theinstrument. This optical effect is that of a pointerof greater length but zero mass, so the timeresponse is quite fast.

12. (b)The sensitivity of an ammeter is given by,

s = FSD

1I

where, IFSD = full scale deflection current.

As, with the use of shunt with ammeter the fullscale deflection current of the ammeter increase,so its sensitivity will decrease.

13. (b)

A data acquisition system doesn’t contain ademultiplexer.

When the distance between transmitting andrecieving point is large, then to increase theefficiency of transmission of a data acquisiton.

14. (c)

The relation between resistances at two differenttemperatures for a thermistor is given by

1 21 2

1 1T T

T TR R e

where,1TR = Resistance at T1°C.

2TR = Resistance at T2°C.

So, we can say that the resistance of thermistorshas an exponential relation with temperature.

15. (c)

The voltage current characteristics of a thermistoris as shown below:

V

I

As can be seen from above figure, the voltagedrop across a thermistor increases with increasingcurrent, untill it reaches a peak value beyondwhich the voltage drop decreases as the currentincreases.

the reason of this is, when current flows, theheat is generated, resulting in a decrease inresistance of thermistor. At lower value of currents,the heat generated is not sufficient. With anincrease in current, when larger current flows,sufficient heat is generated, resulting in negativeresistane characteristics of thermistor.

IES M

ASTER


16. (d)

A tachogenerator is a device, which convertsmovement of shaft to an electrical output voltage/

So, it is a transducer, analog transducer andsince only one stage of conversion is there it isneither primary nor secondary type of transducer.It generates electrical energy so we can say thatit is an active transducer.

17. (d)

In a power factor, the moving system is perfectlybalanced, and the controlling forces are zero.

18. (b)

An electrodynamometer type power factor meterhas a current coil which acts as fixed coil andthe moving system consists of two pressure coils,one highly resistive and other highly inductive.

19. (c)

If the number of horizontal line intersection withcurve are nx and number of vertical line intersectionwith curve are ny, then

yx

y x

fnn f

x

y

n 25 1n 50 2

The above condition is only satisfied by curveobtained in option (c).

20. (d)

For an oscilloscope, the product of rise time andbandwidth is constant.

Bandwidth × rise time = 0.35

tr = 0.35

Bandwidth

= 60.35 35 nsec.

10 10

21. (c)PMMC type of instruments have eddy currentdamping as operating magnetic field whichproduce deflecting torque is very large.

22. (d)Let R1 and R2 be in parallel

1R =

1 2

1 1R R

1R =

1 110 5

R = 3.33

2R

R

=1 2

2 21 2

R RR R

RR

=1

1 1 2 2

RR R RR R R R

=3.33 3.335 1010 5

= 8.33 %

23. (b)0.001A

5A 5A

R = 100A

RSH4.999A

Voltage across meter = 3 21 10 10

= 0.1VCurrent through shunt = 5 – 0.001 = 4.999 A

Shunt resistance = (RSH) = 0.1

4.999

RSH = 1

49.99

24. (a)

8V

0V 20 ms

Time period of oscillator = 31

400 10 No. of pulse counted in 20 ms

= 3

3

20 101

400 10

= 8000

25. (d)

A logarithmic scale compresses the scale, sowide range can be accomodated.

26. (b)

At higher frequencies, there would be effects ofstray elements.

IES M

ASTER


27. (c)

In Ferrodynamometer type instrument,

Td 1 2 cos In D´ arsonval galvanometer instrument,

Td KiIn electrodynamometer type instrument,

Td 1 2I I cos

28. (c)

Let us take 132 digit display in 1V scale

then display range will be0.000 to 1.999 (over ranging)In 3 digit display 1V is displayed as 0.999V.If 1/2 digit is switched ON the range ofdisplay increases.

29. (b)

Let the two resistance are R1 = R2 = R= 100 1%

parallel combination of R1 and R2,

Req =R2

eqR

R

=

12

Standard deviation of the parallel combination,

eqR = 1 2

2 2eq eq2 2

R R1 2

R RR R

where 1R and 2R are standard deviation of R1and R2 respectively.

eqR =2

eq 2R

R2

R

=2

212 12

= 1 %2

30. (b)

Temperature error and Friction error are observedon DC measurement also.

31. (d)

In a moving iron instrument 2I

1

2

=2

1

2

II

=22mA

1mA

= 4

2 = 14

= 200V

4 = 50V

32. (c)

1I 5A

2I 12A 3I

2I

leads by 90° from 1I

.

23I = 2 2

1 2I I ; 3 1 2I I jI

2 23I 12 5 13A

33. (b)

(i) High torque to weight ratio indicates higher isthe sensitivity.(ii) Accuracy does not depend on torque/weightratio.

34. (b)Both Rayleigh’s current balance and lorentzmethod are absolute measurement method butRayleigh’s current balance is used for absolutemeasurement of current and Lorentz method isused for measurement of resistance.

35. (c)

Absolute error = 2300 ±6 V=

100So, When the voltmeter reads 222Vthe actual voltage will be

= 222 6V

= 216 to 228 VSo, option (c).

36. (b)

Transducer senses the quantity and producesan electrical signal then an amplifieramplifies this electrical signal. After this,the signal is modulated so that it can betransmitted to longer distance.

37. (c)

IES M

ASTER


Im

Ish

50 A, 2000

RSince, m shI I I Now, I = 2Im m2I = m shI I

Ish = Imand, m mI R = shI .R

6 650 10 2000 50 10 R

R = 2000

38. (a)

For the instrument having weak magnetic field,air friction damping is used as it does not affectthe magnetic field of the instrument. In all othersystem of damping, weak magnetic field mayget distorted and gives serious error.

39. (d)

40. (b)In dual slope integration A/D convertor, aswitch is used to first connect the input tointegrator and completely charge thecapacitor and then by help of switch,reference voltage is connected to integratorto discharge the capacitor.

41. (a)

The main error in a dynamometer type wattmeteris due to pressure coil inductance due to whichcurrent in pressure coil is not in phase with appliedvoltage.

42. (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

43. (d)

tan =

1 2

1 2

3 W WW W

= 250 503250 50

= 2.6

cos = 2 21 0.359

1 2.6

44. (c)If the voltage across a capacitor is constant thencurrent through it is zero but energy stored is

21 CV2

current through a capacitor is given by

i = dVCdt

if dt = 0, then i = (impossible)

45. (a)

For maximum power delivery to 9 resistor, RTh

across it should be 9 .

To obtain RTh across 9 resistor, the circuit willbe as shown below

Therefore, Rs + 3 = 9 , Rs = 6

46. (c)

T-equivalent of a linear transformer

Thus, M = 5H

IES M

ASTER


L1 – M = 4H L1 = 9H

L2 – M = 6H L2 = 11H

47. (d)If only XL is variable, then for maximum powertransfer

XL = – XTh

zTh = RTh + jXTh

zL = RL + jXL = RL – jXTh

zTotal = RTh + RL

Thus, PF = 1 Phase difference between Vand I is zero

If RL is also variable, then

*Th Lz z

RL = RTh, XL = –XTh

48. (a)Using h-matrix, the equations for V1 and I2 canbe written as

V1 = 5I1 + 2V2 ...(1)

I2 = – 0.5I1 + 0.1V2 ...(2)

Using equation (2)

V2 = 1 21 0.5I I

0.1V2 = 5I1 + 10I2 ...(3)

Substuting the value of V2 in equation (1)

V1 = 5I1 + 2(5I1 + 10I2)

V1 = 15I1 + 20I2 ...(4)

From equation (3) and (4)

(z) =

15 205 10

49. (d)At t = 0+

iL(0+) = iL(0) = 0A, current in an inductor can’t

change instantaneously, At t = 0+ , all 2A passes100 resistor, thus

VL(0+) = 2 × 100 = 200V

VL(0+) × iL(0

+) = 200 × 0 = 0

At t =

inductor will reach in steady state i.e. inductorcan be treated as short circuit. Hence

iL(0) = z 1A2

VL( ) = 0

VL( ) × iL( ) =0

50. (d)

+

i

–

– +

+ –

Applying kVL in the circuit

120 – 30i – 2vA + vA = 0

or i = A120 v

30 ...(i)

Also, vA = – 15i ...(ii)

(voltage drop across 15 resistor)

Therefore, i = 120 15i

30

or 30i – 15i = 120

i = 120 8A15

Power absorbed by dependent source is

dep A2v i

= 2 × (–15i) × i = – 30i2

= –30 × 82 = – 1920W

51. (b)Network is an interconnection of two or moresimple circuit elements. It is not necessary for anetwork to contain atleast one closed path. If anetwork contains a closed path, then it is alsoan circuit.

Every circuit is a network, but not all networksare circuit

+–

IES M

ASTER


52. (b)in order to find the type of filter, two general rulesare

Inductors are assumed to behave as shortcircuit and capacitors are assumed to behave asopen circuit at very low frequencies.

Inductors are assumed to behave as a opencircuit and capacitors as short circuit very highfrequencies

Now, the equivalent circuit at low frequencies is

v0 = iv2

Equivalent circuit at high frequencies is

v0 = 0

So, we can say that the circuit block highfrequencies and passes low frequencies, henceit is low pass filter.

53. (a)For the system having AD – BC = 1, the systemis reciprocal. So, the system will have h12 = –h21.

or, 12 21h h 0

54. (a)

Self inductance of coil 2, L2 = 2

2

NI

= 3800 0.35 10

5

2L 56 mH

Now, since they are wound on same core, so

2L N

12

LL =

212

NN

L1 = 2100 · 56 0.875 mH

800

55. (c)

L1 = 20 mH, L2 = 10 mH

since 50% of flux from coil 1 links coil 2,

So, k = 0.5.

M = 1 2k L L 0.5 20 10 0.5 14.14

M = 7.07 mH

Net equivalent inductance

Leq. =

21 2

1 2

L L ML L 2M [When the inductors are in

parallel aiding]

=

21 2

1 2

L L ML L 2M [When the inductors are in

parallel opposing]

Leq. =

210 20 7.0710 20 2 7.07

= 150 3.4 mH4.14

56. (b)Number of nodes (n) = 500

Number of branches (b) = 700

Number of equations in loop current (kVL)method = b – n + 1 = 700 – 500 + 1 = 201

Number of equations in node voltage (kCL)method = n – 1 = 500 – 1 = 499

So, loop current method will have lesser numberof equations.

57. (d)

A B

IES M

ASTER


The resonant frequency for a tank circuit as shownabove is

f = 2

21 1 R

2 LC L

=

2

6 3 23

71 12 20 10 10 10

=

1000 159.15 Hz2

58. (b)

Given all the resistance of the circuit are1 .

1 1

11

22A B

C

D

11 1

11

A B

C

D1

1

Using delta to star conversion.

A B

1

1

14 1

2

A B

12

14

121

2

14 1

2

14

RAB = 1 1 = 0.559. (b)

+

–

RTh

IL =

th1/22 2

th L th L

V

R R X X

PL =

2L2 th

LL 2 2th L th L

V .RI R

R R X X

For maximum load power, LL

dP0

dX

2

L th Lth22 2

th L th L

–2V .R . X X

R R X X

= 0

Xth = – XL

or L mX –X For maximum power transfer.

60. (d)Phase angle of current

= L

–1 C

1–tan

R

= –3

–6–1

1500 80 10 –500 30 10tan

15

= –1tan –1.78 – 60.6

So, the phase angle of current is 60.6° leading.

61. (b)

s = j

= –1, 5

s = –1 + j5

So, the time function is given by,

V(t) = t100e cos t 30

tV t 100e cos 5t 30 V

62. (d)The time constant = Req. Ceq.

Req. = 10 5 15

Ceq. = 0.5 1.5 2 2 2 1 F0.5 1.5 2 4

Time constant ( ) = Req. Ceq.

= 15 × 1 × 10–6

15 sec.

IES M

ASTER


63. (b)For t = 0–, the circuit can be drawn as–

3020

Current iL(0–) is given by,

iL(0–) =

205 2A20 30

Since the current in an inductor can’t changeinstantaneously, so iL(0

+) = iL(0–) = 2A.

64. (c)

Redrawing the circuit with 1R 0

+

A

F

B

E

C

D

–

1k 50i

R1

i 50i

Using KVL in ABEFA–

10 × 10–3 =103 i

i = 10 × 10–6

i = 10 µA

Now, V = –50i.R

= –50 × 10 × 10–6 × 103

V 0.5 V

65. (b)For a capacitor, the voltage v(t) is given by,

v(t) = 1 i t dtC

The area under current curve from t = 0 to t =

6.5 sec. is–

1 i t dtC = –6 3 –91 4 10 5 10 5 2.5 10

2

v(t) =–9

–61 22.5 10

0.3 10

v t 75V

66. (d)The relation for voltage and current for a capacitoris–

Vc(t) = –t /RCmV1 e dtC R

Vc(t) =

–t/RCmV e. k

–1RCRC

Vc(t) = –t/RCm–V e k

for t = 0, Vc(t) = 0 (since charge store at t = 0is zero)

0 = –Vme0+k

k = Vm

Vc(t) = Vm(1–e–t/RC)

So, maximum value of voltage = Vm

Maximum stored energy = 2m

1 CV2

67. (d)

The single voltage source may be considered tobe equivalent to two identical sources in parallel.So option (1) and option (4) are equivalent.Option (4) and option (3) are equivalent since aconnection from the positive terminals of the twosources does not affect the network becausethere would be no current in such a connection.Note : The single current source may beconsidered to be equivalent to two identicalsources in series.

68. (c)

1

1L 1 1 2

di t dV t L M i t i tdt dt

2

1L 2 1 2

di tdV t L i t i t Mdt dt

• Multually induced voltage occurs due to othercurrent .

• Self induced voltage occurs due to self current.

69. (b)

Z(s) =1sL R

sC

I s =

V sZ s

IES M

ASTER


=

22

V s .sC sC. V ssRC Is LC sRC 1 LC sLC LC

Characteristic equation is

2 R 1s sL LC

= 0

Comparing the equation with2 2

n ns 2 s = 0

n n1 ; 2Lc

=RL

=n

R 1 R 1. . LCL 2 2 L

=R C.2 L

for system to be underdamped 1

i.eR C 12 L

LR 2.C

70. (a)

System transfer function,

H s =2s

s 1

H j =2.jj 1

H j = 2

2.

1

& H j = o 190 tan

for input sint 1rad / s

H j =2 22

& H j = 90 – 1tan 1

= 90o – 45o = 45o

Steady state response

= 1. H j sin t H j

= o2 sin t 45

71. (d)

At t = 0–, switch is open,

1

10V

1H

i t

iL(0–) = 0

iL(0+) = L t 0i 0i0 t

[ Inductor does not allow sudden change ofcurrent]

At t switch is closed and inductor shortcircuit,

1

10V

i

i =10 10A1

Time constant of R-L circuit, L 1 1secR 1

i(t) = t /i ei i0

= t10 e0 10

i(t) = t10 1 eat t = 1 sec.

i(1) = 110 6.32A1 e

72. (c)

2R

1R

V

At t 0 At t 0

1 1L L

1

Vi 0 i 0R

2 2L L

1

Vi 0 i 0R

c cV 0 V 0 0V

1 21

VV iR i R 0R

1

ViR

2L

V

SC

1L

1R

2R1

ViR

i V/R1

V/R1

1LI

2LI

IES M

ASTER


So si = 0

2Ri =1

Vi 0R

1Ri =1

ViR

73. (d)

The given network is a balanced wheatstonebridge. The above circuit can be viewed as:

3

6

6

3

A

B3

ABR = 9 9

= 4.5

74. (b)

T0

A

t

f t

T0

A

t

1Af t tT

T0

1

t

2f t u ut t T

= ×

f(t) = 1 2f .ft t

= A t u ut t TT

= A Atu tut t TT T

= A Atu ut t T T t TT T

= A Atu u Aut t T t T t TT T

75. (c)

Thevenin’s equivalent of the circuit is.

10V V 1 F

1M

So V t = toV 1 e

where oV = 10V,

= 6 61x 10 x 1x 10 1s

dv t

dt=

toV .e

= toV e as 1 s

t 1s

dv tdt = 1

0V e

= 01V . 10 x 0.367e

= Vsec .3.67

76. (a)

Voltage across current source = 10V

Voltage across 1 resistor = 10V

Current in 1 =101

= 10A

77. (a)

Irms = T

2

0

1 8t dtT

=31 64 T

T 3

Here T = 1

Irms =643

Average Power loss in 9 resistor

Pav = 2rmsI 9

=64 93

= 192 W

78. (a)

i = 2 AApplying KVL, we getVab– 3i +10 = 0

Vab = 3i – 10

= 3 2 10

= –4 V

79. (c)

L1 = 5 mH, L2 = 20 mH

L N2

It the number of turns are doubled the new valueof

IES M

ASTER


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

80. (b)vc (0) = vc (0

+) = vc (0–) = 10V

= RC = 1105

R 20 20 10 = 2 sec

vcn = t t2Ae Ae

vcf = 2010 5V40

vc =

t2Ae 5

At t = 0, vc = 10 V10 = A + 5

A = 5

vc = t25e 5

= t25 1 e

81. (a)At t = , inductor behaves as short circuit

Req = 1

i(t) =601

= 60 AA

Hence IL(t) =60 2 30 A4

For finding time constant, we short circuit thevoltage source. Hence circuit becomes like

2 H2

2

=L 2 2secR 1

82. (a)

(100 ) (0.5A 0.25A) 25V

83. (b)

Nodal analysis

a aV V 2

1 2= 2

aa

VV 12

= 2

a3V2

= 3

Va = 2 V

84. (a)

As

is Laplace transform of step function

2As

is Laplace transform of ramp function

2 2s

is Laplace transform of sint

2 2s

s is Laplace transform of cost

85. (b)

N1 N2

Applying cutset here

i1

i2

4

2

– +x

– +8V

i1 = i28V4 =

xV2

x = – 4V

86. (c)

Rearranging the resistors

A B1

3

1

3

3

RAB = 3

IES M

ASTER


87. (b)

Inductor will not change the current suddenly. Li 0 = LI 00

88. (b)

2V = 2 12

di diL Mdt dt

V2 = 1did 04 3dt dt

= 1di3dt

= 2t3 e V2

= 2t6e V

89. (c)

Applying KVL around the loop

+–

+–

–+3V

10V

1V

V1+ –

13V 1V V = 10V

V1 = 6V

90. (a)

Now given that l l0n and the volume remainssame

l l0 0 0a an or 0a a n

0

RR =

ll

0 2

0

a n n na

R = n2R0

91. (b)JFETs operate in depletion mode only i.e.VGS < 0V. This condition is necessary so as toreverse bias the gate junctions and thus createdepletion region in the channel which controls thedrain current flowing through the channel.

92. (a)

Ebers Moll model is used to represent the largesignal model of BJT, that involves two ideal diodesplaced back-to-back with reverse saturation currents

–IEO and –ICO and two dependent current controlledcurrent source shunting the ideal diodes.

EIE IC

–(IE0) –(IC0)

IIC IIC

B

VE VC

IB

C

93. (b)

Fermi-Dirac probability function, f(E), is theprobability that an energy level E being occupiedby an electron and it is given by

f(E) = FE E /KT1

1 e

Probability that an energy level E is occupied by ahole is given by

1 f E = FE E /KT11

1 e

=

F

F

E E /KT

E E /KTe

1 e

= FE E /KT1

1 e

94. (a)

Graph IC versus VCE for constant IB represent theoutput characteristics of a BJT in common-emitterconfiguration.

From this graph, can be directly determined byfinding ratio of the change in collector current tothe change in base current.

95. (d)

For the Bipolar Junction Transistor (npn type orpnp type) to operate in cut-off, both the Emitterjunction (JE) and collector junction (JC) must bereverse biased.Circuit in option (d) uses a p-n-p BJT and biasingused reverse biases both JE and JC. Thus, this circuitrepresents the cut-off of p-n-p transistor.

96. (a)

Given,Collector current, IC = IE + ICO

This relationship holds good for the BJT operatingin active region of operation.

IES M

ASTER


97. (c)

In a MOSFET, source region supplies the chargecarriers to the drain region through the channelformed between the two regions.Thus, the inversion layer formed between the sourceand drain has the polarity same as that of themajority carriers in the source.

=CE

C

B V constant

II

98. (a)

Carrier concentration in a semiconductor is afunction of both time and distance and continuityequation is derived to establish the relationshipbetween carrier concentration and time & distance.It is given by

p p p0 n

n

dn n n dJ1dt q dx

For a p typesemiconductor

2

pp p p0 pn n2

n

d n Edn n n d nD

dt dxdx

...(i)

pn p n n

dnJ n q E q D

dx

When l ight is made to fal l on a p-typesemiconductor continuously with electric field (E)= 0, Equation (i) becomes

2p p po p

n 2n

dn n n d n0 D 0

dt dx

2

p2

d n

dx=

p po

n n

n nD

2

p2

d n

dx= 0

n

Since recombinationlifetime is infinite,

np = A·x + BThus, minority carriers varies linearly with distance.

99. (c)

Diffusion length (L) is defined as the averagedistance which an injected charge carrier travelsbefore recombination and it is given by

Lp = p pD (for holes)

and Ln = n nD (for electrons)

The relationship between diffusion constant (D)

and mobility is given by the Einstein equationas

p nT

p n

D D kT Vq

Where VT is the Volt-equivalent of temperature.

At equilibrium, the product of free negative andpositive charge concentrations is a constantindependent of the donor and acceptor impuritydoping. This relationship is called the Mass-actionlaw and is given by

2inp n , where

ni is the intrinsic carrier concentration.

Charge neutrality equation: A semiconductor isneutral, that is, the magnitude of positive chargedensity is equal to that of the negative chargeconcentration, or

p + ND = n + NA,

where NA (ND) is the acceptor (donor) impurityconcentration.

100. (b)

The effective mass of a charge carrier is given by

2*

2 2

h 2m

d E dk

Thus, the curvature of the band determines theeffective mass.

From the graph shown,

2 2

2 2conduction valenceband band

d E d Edk dk ...(1)

Free electrons in the conduction band and holes inthe valence band serves as carriers of current.

2 2* *n p2 2 2 2

Conduction ValenceBand Band

h 2 h 2m m

d E dk d E dk

101. (b)

Here ND >> ni, thus free electron density inequilibrium is given by

n = ND = 1014/cm3

According to mass action law,

np = 2in

p = 2 2i i

D

n nn N

= 210

141010

= 106/cm3

IES M

ASTER


102. (c)

Indirect transition via a discrete level :E

Et

k2

1

In an indirect bandgap semiconductor, electrontransition from the conduction band to the valence bandoccurs in two steps : One, change in momentum &energy and second, change in energy.

In the indirect transition involving a change inmomentum, part of the energy is given as heat to thelattice rather than as an emitted photon. Thus, indirectbandgap semiconductors are not used in optoelectronicdevices.

103. (c)

The resistivity of a semiconductor is given by

= n p

1 1nq pq

D n

1N q

n D n D

for an n type semiconductorn N and p N

= 16 191

2 10 1.6 10 1300 = 0.24 cm

104. (d)

Gold is extensively used as a recombination agentin semiconductor devices. This is because, by introducinggold into silicon under controlled conditions, designerscan obtain desired carrier lifetimes.

105. (d)

The carrier concentration in the semiconductor is afunction of both time and distance. The differentialequation which governs this functional relationship iscalled the continuity equation. This equation is basedon the fact that charge can neither be created nordestroyed. Thus, continuity equation follows the law ofconservation of charge.

106. (c)

Let the reverse saturation current of a pn junctiondiode at a temperature (T1) of 25°C be I0,25. Then, Reversesaturation current at a temperature (T2) of 80°C is given

by

I0, 80 =2 1T T10

0,25I 2

=80 25

100,25I 2

= 5.50,25I 2

= 45 × I0,25

107. (b)

8V

100

R

IS

IL

IZ

V =4VZ

+

–

From the circuit,

IS =

8 4 40mA100

Load current, IL = IS – IZThus, IL,max = IS – IZK = 40m–10m

= 30mA

Minimum value of R so that the voltage across itdoes not fall below 4V is given by

Rmin =Z

L,max

V 4 133.33I 30 mA

108. (a)

Cut-in voltage of various types of diodes is givenbelow:

1. Germanium diode:- 0.2V

2. Silicon diode :- 0.6V

3. Schottky diode :- 0.3 V

4. Tunnel diode :- 0V

109. (b)

Transition region in a p-n junction diode consists ofimmobile acceptor and donor ions in p-side and n-siderespectively.

+ ++ ++ ++ ++ ++ +

p-type n-type

x = –Wp x = Wnx = 0

IES M

ASTER


Since the net charge must be zero,

e NA Wp = e ND Wn

p

n

WW =

D

A

N 1N

Wp < Wn

110. (b)

The expression for the depletion layer width is givenby

jA D

2 1 1W V ,q N N

Where Vj = Vo+VR, for reverse bias voltage of VR

= Vo–VF, for forward bias voltage of VF

111. (d)

Forward current, ID = 5mA

VT 26mV (at room temperature)

Dynamic or AC resistance of a forward biased Silicondiode is given by

rac =T

D

VI

=2 26m

5m 2 for Silicon and

1 for Germanium

= 10.4

112. (c)

Depletion capacitance:- It is due to the presence ofuncovered immobile charges present in the depletionregion. It comes into effect when the diode is reversebiased. In forward biased diode, it is negligible. It is alsocalled as Space-charge / Transition capacitance.

Diffusion capacitance:- It occurs due to the storageof minority carriers outside the depletion region. It comesinto effect when the diode is forward biased. It is alsocalled as storage capacitance.

113. (d)

Symbols for various types of diodes is shown below:

Varactor diode :-

It represents the variable capacitance of the diode.

Zener diode:-

Tunnel diode:-

Schottky diode:-

114. (c)

Small-signal equivalent of Tunnel diode:-LSRS

–RnCS

where, Rs ohmic resistance

Ls Series inductance and depends uponthe lead length and geometry of thediode package

CS Junction capacitance, depends upon theapplied bias and is measured usuallyat the valley point.

–Rn Negative resistance of the Tunnel diode.

Equivalent circuit of a Varactor diode:-

CT

Rr

RS

(Reverse resistance)Ohmic/Contactresistance,

(Transition capacitance)Small-signal equivalent of normal p-n junction diode:-

r

CD Tor C

(Dynamic or ACresistance)

D

T

C Diffusion capacitanceC Transition capacitance

115. (a)

Pinch-off voltage (VP) is the diode reverse voltagethat removes all the free charge carriers from the channel.It is given by

VP = 2Dq Na ,

2Ewhere, ND is the dopant atom concentration in the

n-channel, a is half of the distance between the twogate terminals.

116. (b)

RC

10V

5V +–

200k CB

E

+

–VCE,sat+

–VBE,sat

Applying KVL in Base-emitter loop,

5 – 200K × IB – VBE,sat = 0

IB =BE,sat5 V 5 0.8

200 k 200 k

IES M

ASTER


IB = 21 ACollector saturation current,

IC,sat = B.I 100 21 A

= 2.1 mA

For transistor to remain in saturation,IC IC,sat

CE,sat

C

10 VR

IC,sat

RC CE,sat

C,sat

10 V 10 0.2I 2.1m

RC 4.67 k

RC, min = 4.67 k

117. (d)

Doping concentration of emitter region is very highbecause it supplies the charge carriers.

Doping concentration of Base region is very low soas to prevent recombination of charge carrierssupplied by the emitter in the Base region.

Doping concentration of collector region is moderate.

Thus,18 35 10 cm Emitter region

17 310 cm Collector region7 32 10 cm Base region

118. (c)

Biasing of a transistor is done so as toestablish operating point (Quiescent point orQ-point) in the active region of operation ofBJT. This is done so as to improve thestability so that Q-point is stable and BJTcan work as an amplifier.

119. (b)

In semiconductors, mobility of electrons isapproximately 2.5 times that of holes.

In semiconductors, increase in temperatureincreases more number of electron-hole pairs(EHPs). Thus, its conductivity increases andresistiv ity decreases with increase intemperature.

Metals have positive temperature coefficientof resistance and are of much smallermagnitude i.e. + 0.4%/°C

120. (d)

Collector leakage current in common-emitter

configuration, ICEO = 500 A

ICEO = CBO1 I ,where ICBO is the collector leakage current incommon-Base configuration

500 = (1 + 99) × ICBO

ICBO = 5 A

121. (d)Given, = 0.98

=0.98 49

1 1 0.98

Collector current, IC is given by

IC = B CBOI 1 I

= 49 100 (1 49) 5

= 4900 250

= 5150 A 5.15mA

122. (d)

There are two types of breakdown possiblein a transistor : Avalanche breakdown andPunch through. In the Avalanche breakdownmechanism, the Avalanche multiplicationfactor depends on the voltage VCB betweencollector and base and is given by

nCB CBO

1M ,1 V V

where VCB is the collector-to-base voltage,

VCBO is the maximum reverse biasing voltagethat can be applied before breakdown betweenthe collector and base terminals

and n is the parameter that controls thesharpness of the onset of breakdown andranges between 2 to 10.

123. (d)

When the emitter-base junction (JE) is forwardbiased and the collector-base junction (JC)is reverse biased, transistor is said to beworking in the active region of operation. Inthe active region, transistor is used as anamplifier.

Transistor is said to be working in cut-offregion of operation, when both JE and JC arereverse-biased.

Transistor is said to be working in saturationregion of operation, when both JE and JC areforward biased.

IES M

ASTER


NOTE:- Transistor is used as a switch,when it is made to operate in thesaturation and cut-off region.Transistor is operating in inverse-active region,when JE is reverse biased and JC is forwardbiased. In this region of operation, it worksas a low gain amplifier.

124. (c)

Input static characteristic:- Plot of inputvoltage versus input current for different valuesof output voltage.

Output static characteristic:- Plot of outputcurrent versus output voltage for differentvalues of input current.

Transistor Input Output Input static Output staticConfiguration Side Side Characteristics Characteristics

Plot of V Vs. Plot of I Vs.cEBCommon Base Emitter Collector I for different V for differentE CBconfiguration values of V . vCB

alues of I .EPlot of I Vs. VPlot of V Vs.Common C CEBE

emitter Base Collector I for different for differentBconfiguration values of Ivalues of V BCE

Plot of V Vs. PCommon CBCollector Base Emitter I for differentBconfiguration values of VCE

CElot of I Vs. V forEdifferent valuesof I .B

125. (a)

If IC,sat and IB can be determinedindependently f rom the circuit underconsideration, the transistor is said to be in

saturation if C,sat BI · I

126. (c)

The increase in magnitude of collector reversevoltage increases the space-charge width atthe output junction diode. Such an actioncauses the effective Base width to decrease.This phenomenon is called as Early effect orBase width modulation Consequences ofEarly effect :-

(i) Less chance of recombination within the Base

region. Hence, the transport factor * and

increases with an increase in magnitudeof junction voltage.

(ii) The charge gradient is increased within thebase and consequently, the current ofminority carriers injected across the emitterjunction increase.

(iii) At a certain collector voltage, the transitionlayer covers whole of the base region and

thus collector and emitter are effectivelyshorted. This phenomenon is called punch-through or Reach-through and due to this,the transistor action ceases.

127. (d)1. Mobilities due to two or more scatteringmechanism add up inversely

1 =

1 2 3

1 1 1 ........

2. True

3. = HH

RR

and RH = 1

q carrier concentration

128. (c)

(P) The basic principle of the experiment is thatexcess minority carrier pulse drifts in theelectric fieldand spreads out by diffusion

(Q) By monitoring concentration at different pointsand at different times we can independentlymeasureminority carrier mobility anddiffusion co-efficient D

129. (d)

1. Traps acts as inter mediater of recominationagent

2. For low level injection excess carrier populationis given be product of carrier lifetime andgeneration rate

3. In equilibrium there is no net flow of carriers

130. (a)For Ga EF > Ga(–) so singly negativeFor Zn Zn–2 > EF > Zn(–) so again single (–)veFor Au Au(–) > EF > Au(+) so neutral

131. (b)

At equilibrium I = 0 J = 0

ndnE qDdx

= 0

n ndnnq E qDdx

= 0

E = –

axon n

axn n o

N a eD Ddn / dxn N e

IES M

ASTER


KTE aq

132. (d)

At equilibrium J = 0 n ndnnq E qD 0dx

E = –

axo

axn o

N a eDn dn / dx KT KT an q qN e

1. Since E.F will be directed in x-direction e–(s)

will be less kinetic energy and more potentialenergy as we move in x-direction. So option A, B,C are wrong2. At equilibrium we have constant formfermilevel so option B and C are wrongFrom (1) and (2)

133. (b)

n(x)

Electric field

J diffusionn

J driftn

(As area is constant I J)

Jn diffusion = ndnqD doublesdx

Jn drift = nnq E doubles

134. (b)There is no channel present between thesource and drain in an Enhancement typeMOSFET (E-MOSFET). Thus, no drain currentflows at VGS = 0V Transfer characteristic ofE-MOSFET :-

–ve +ve

ID

VTp VTn

p-channelE-MOSFET

n-channelE-MOSFET

VGS

135. (a)A photon with energy less than Eg is unable toexcite an electron from voilence band to theconduction band, that’s why there is neglisible

absorption of photons with hv < Eg

136. (d)

Sensitivity of an ammeter is defined as s = FSD

1I

Since, with the use of ammeter, the full scaledeflection current increases, so the sensitivity ofammeter will decrease.

Again, when the sensitivity is low, that impliesfull load current is high. So, for a given deflection,more power will be consumed by the instrument.Hence loading effect is more for low sensitivityinstruments.

137. (c)

In galvanometer, damping may be provided byconnecting a resistor across the coil. When thecoil rotates in the magnetic field, a voltage isgenerated in the coil which circulates a currentthrough the coil and external resistor. Thisproduces an opposing or retarding torque thatdamps the motion.

The smaller the enternal resistor, larger will bethe damping torque.

138. (a)

In order to minimize errors in measurement, theoperating current of thermistors must be kept aslow as possible.

At higher value of currents, self heating effectoccurs, and since the resistance of thermistorchanges considerably with temperature, the selfheating effect can cause a significant error.

139. (b)

Due to the shunt resistance, there is some errorintroduced in measurement of Q-facor of the coil.

Qtrue = shuntmeas.

RQ 1

R

Here, R = actual resistance of coil.

So, we can say that the measured value of Q-factor is more than the true value.

140. (a)

At high value of load current, a self braking torqueis produced.

The disc revolves continously in the field of theseries magnet under load conditions and therefore,

IES M

ASTER


there is a dynamically induced emf in the discbecause of this rotation. This emf causes eddycurrents which interacts with the field of the seriesmagnet to produce a braking torque. So theenergy meter registers lower than the actual valueof energy consumed.

In order to minimize self braking action, the fullload speed of the disc is kept as low as possible.

141. (b)

At steady state position, Td = TC

Due to balance of torques needle stop at steadystate position and underdamping is used toreduce the oscillation of pointer around steadystate position.

142. (a)

To overcome the frictional torque, we need largecoil with strong magnetic field.

Td = NBAITo increase Td, large coil and strong magneticfield are used.

143. (a)

Voltage across an inductor, if i is the currentflowing through it,

v = diLdt

if dt = 0

v = Thus correct option is (a)

144. (a)

According to power convention, an elementabsorbs power, when the current enterspositive terminal of the element and deliverspower when current enters negative terminalof the element.

This can be shown as below–

Since the resistor can only absorb thepower, so always the voltage drop indirection of current flow.

Inductor and capacitor can absorb or deliverthe power, so voltage may either rise ordrop in the direction of current flow.

145. (c)

146. (b)Explanation to assertion:• The mobility of electron and holes for Si

and Ge is given as follows.Si Ge

Electron mobility;n :1300 cm2/v-s 3800 cm2/v-s.

Hole mobility;p :500 cm2/v-s 1800 cm2/v-s.

Electron mobility is higher than holemobility because effective mass of electronis less than effctive mass of hole.Explanation to reason:• The general graph of recombination in

an extrinsic semiconductor is given as.

P

t = t

Pn

t = 0P0 P0 x

P = P + P (0) e–t/

0p

P = P = P (0)0

The excess concentration decreasesexponentially to zero (P= 0 or P = P0) witha time constant equal to mean life time.

147. (c)Explanation to Assertion: In intrinsicsemiconductor, a covalent bond is formedbetween two adjacent atoms constituting thesemicionductor. when (due to rise intemperature or high potential) these covalentbond breaks, electrons released from valenceband to conduction band leaving hole withits parent atom. Hence with each bondbreaking a pair of electron and hole isproduced.Explanation to Reason: For an intrisnsicsilicon.n = 1300 cm2/v-s, p = 500 cm2/v-s

n

p = 1300

50 = 2.6

IES M

ASTER


Hence, n = 2.6 times p .

148. (d)

Explanation to Assertion: As thetemperature of a transistor increases, thebase spreading resistance rbb' in the hybrid-model at high frequencies increases becausethe mobility of the carrier reduces. Henceassertion is wrong.Explanation to Reason: With the increasein temperature mobility of the carriersdecreases due to increase in scattering.

149. (c)

In a BJT, the doping level of emitter isextremely high as compared to the basebecause it is emitter which will providecharge carriers for the flow of current.More over the area of base is minimal sothat recombination is as low as possible.So, the region of emitter is immaterial solong as doping is kept lower.

The emitter injection efficiency, isdefined as [p-n-p transistor is taken]

= EP

E

II

= ECurrent injected carriers at JTotal emitter current

If NE >> NB, then EPI IE

1

So clearly, emitter injection efficiency, does not depend on base region width.Hence reason is false.

150. (c)

Hybrid Model : h-Parameter modelI1 I2

V1 V2

+

–

+

–

Transistor in anyConfiguration

V1 = h11I1 + h12 V2

I2 = h21 I1 + h22 V2

V1 and I2 is dependent variableI1 and V2 is independent variable

1

2

VI

=

11 12

21 22

h hh h

1

2

IV

Important Point of Hybrid Model:1. We use h-parameter in the equivalent

circuit of a transistor since theseparameters are relatively constant (realnumber) with respect to audio frequencyand temperature.

2. We find the h-parameter under differentconditions, i.e., by open circuit the inputor short circuit the output. Since theseparameter have mixed units, these arecalled hybrid parameter.

3. For a given configuration of transistorthe form h-parameter is always specifiedby the manufactures.

4. They are easy to measure.5. They can be obtained from the transistor

static characteristic curves.6. Hybrid model is particularly convenient

to use in the circuit analysis and design.7. Hybrid model is parameter holds good

for any configuration specified. Type oftransistor that is npn or pnp, Si or Geat low frequency.

Output admittance when inpt is opencircuit(s)

1V = h11I1 + h12V2(KVL)

I2 = h21I1 + h22V2(KCL)For the reason part, the value of h-parameteris always specified by the manufacturerhence it changes as per configuration.For example, if hie, hre, hfe, hoe is specified bythe manufacturer we need to use approximateconversion formulae ie(i) for common collector

hic = hie

hrc = re1 h 1hfc = –(1 + hfe)hoc = hoe

(ii) for common base

hib = ie

fe

h1 h , hrb =

ie oe

fe

h h hre1 h

hfb =

fe

fe

h1 h , hob =

oe

fe

h1 h

answers - ies master · of the ammeter under calibration and inserted in circuit. similarly to...

Documents