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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
In the figure shown the readings of the ammeters `A_(1) and A_(2)` are respectively. A. `(1)/(3) A and 1/3 A`B. Zero and 1 AC. zero and 0.5 AD. 0.5 A and Zero |
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Answer» Correct Answer - C The diode is reverse bias condition therefore current through it zero. Hence , reading of ammeter `A_(1)` is zero. However, reading of ammeter `A_(2)` is `I_(2)=V/R=(10)/(20)=0.5A` |
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| 52. |
For a common base configuration of `PNP` transistor `(l_(C))/(l_(E))=0.98`, then maximum current gain in common emitter configuration will beA. 49B. 98C. 4.9D. `24.5` |
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Answer» Correct Answer - A `beta=(alpha)/(1-alpha)=(0.98)/(1-0.98)=49` |
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| 53. |
In the given figure, which of the diodes are forward biased? A. `A,B,C`B. `B,D,E`C. `A,C,D`D. `B,C,D` |
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Answer» Correct Answer - B In forward bias `V_(p)gtV_(N)` |
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| 54. |
In the given figure, which of the diodes are forward biased?A. 1,2,3B. 2,4,5C. 1,3,4D. 2,3,4 |
| Answer» Correct Answer - B | |
| 55. |
In the following, which one of the diodes is reverse biased ?A. B. C. D. |
| Answer» Correct Answer - D | |
| 56. |
A `N`-type silicon sample of width `4xx10^(-3)m`, thickness and length `6xx10^(-2)m` carriers a current of `4.8mA`, when the voltage is applied across the length of the sample. The free electron density is `10^(22)m^(-3)` |
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Answer» The current density `j=(i)/(A)=(4.8xx10^(-3))/(6xx10^(-2)xx4xx10^(-3))=20A//m^(2)` The drift speed `v_(d)=(j)/(n e)=(20)/(10^(22)xx1.6xx10^(-19))=12.5xx10^(-3)m//sec` Time `t=(l)/(v_(d))=(6xx10^(-2))/(12.5xx10^(-3))=4.8sec` |
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| 57. |
The diode used in the circuit shown in the figure has a constant voltage drop of `0.5 V` at all currents and a maximum power rating fo `100` milliwatts. What should be the value of the resistor `R`, connected in series with the diode for obtaining maximum current? A. `15Omega`B. `5Omega`C. `6.67Omega`D. `200Omega` |
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Answer» Correct Answer - B Current through circuit, `I=(P)/(V)=(100xx10^(-3))/(0.5)=0.2A` Voltage drop across `R=1.5-0.5=1.0V` Hence, `R=1//0.2=5Omega` |
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| 58. |
The diode used in the circuit shown in the figure has a constant voltage drop of `0.5 V` at all currents and a maximum power rating fo `100` milliwatts. What should be the value of the resistor `R`, connected in series with the diode for obtaining maximum current? A. `1.5Omega`B. `5Omega`C. `6.67Omega`D. `200Omega` |
| Answer» Correct Answer - B | |
| 59. |
In the circuit, shown in figure. `R=5000Omega`. If key `K_1` is closed, galvanometer shows a deflection of 30 scale division. On closing key `K_2` and making `S=20Omega`, the deflection of galvanometer reduces to 15 division. The resistance of galvanometer is A. `56Omega`B. `60Omega`C. `64Omega`D. `940Omega` |
| Answer» Correct Answer - C | |
| 60. |
A beam of light travelling along x-axis is described by the electric field, `E_y=(600 Vm^-1) sin omega (t-x//c)` Calculating the maximum electric and magnetic forces on a charge q=2e, moving along y-axis with a speed of `3xx10^7m//s`, where `e=1.6xx10^(-19)C`.A. `19.2xx10^(-17)N`B. `1.92xx10^(-17)N`C. `0.192N`D. none of these |
| Answer» Correct Answer - B | |
| 61. |
An electromagnetic wave in vacuum has the electric and magnetic field `vecE` and `vecB`, which are always perpendicular to each other. The direction of polarization is given by `vecX` and that of wave propagation by `vecK`. ThenA. `vecX||vecB and veck||vecExxvecB`B. `vecX||vecE and veck||vecBxxvecE`C. `vecX||vecB and veck||vecBxxvecE`D. `vecX||vecE and veck||vecExxvecB` |
| Answer» Correct Answer - D | |
| 62. |
A realistic graph depiciting the variation of thhe reciprocal of input resistance in an in- put charcteristics measurement in a com- mon emitter transistor configuration is :A. B. C. D. |
| Answer» Correct Answer - D | |
| 63. |
To get an output of 1 form the circuit shown in figure the input must be :- A. `a=1,b=0,c=1`B. `a=1,b=1,c=0`C. `a=0,b=1,c=0`D. `a=0,b=0,c=1` |
| Answer» Correct Answer - A | |
| 64. |
To get an output `Y=1` in given circuit which of the following input will be correct:A. `A=1,B=0,C=0`B. `A=1,B=0,C=1`C. `A=1,B=1,C=0`D. `A=0,B=1,C=0` |
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Answer» Correct Answer - B When `A=1` `B=0` `C=1` Then `Y=1` |
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| 65. |
A transistor is operated in common emitter configuration at `V_(c)=2 V` such that a change in the base current from `100 muA` to `300 muA` produces a change in the collector current from `10 mA` to `20 mA`. The current gain isA. 50B. 75C. 100D. 25 |
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Answer» Correct Answer - A `beta=(DeltaI_(B))/(DeltaI_(B))=(10mA)/(200muA)=(10xx10^(3))/(200)=50` |
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| 66. |
Why are metals good conductor of electricity? |
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Answer» Metals are good conductors of electricity due to the large number of free electrons (≈ 1028 per m3) present in them. |
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| 67. |
For making p-n junction diode forward biased:A. same potential is appliedB. greater potential is given to n compared to pC. greater potential is given to p compared to nD. unbalanced concentration |
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Answer» Correct Answer - C When p-side of p-n junction is given more positive voltage as compared to n-side, then junction diode becomes forward biased. |
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| 68. |
In a good conductor of electricity the type of bonding that exists is :A. ionicB. van der waalC. covalentD. metallic |
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Answer» Correct Answer - D Presence of large number of free electrons in metallic bond solids, make then good conductor of electricity. |
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| 69. |
you have three identical pn junction, junction 1 is unbiased, junction 2 is reverse biased and junction 3 is forward biased. From largest to smallest, rank the three junction according to the magnitude of their diffusion currents.A. 1,2,3B. 3,1,2C. 3,2,1D. 2,3,1 |
| Answer» Correct Answer - B | |
| 70. |
The intrinsic carrier density in germanium crystal at 300 K is `2.5xx10^(13)` per `cm^(3)` if the electron density in an N-type germanium crystal at 300 K be `0.5xx10^(17)` per `cm^(3)` the hle density (per `cm^(3)`) in this N-type crystal at 300 K would be expected around-A. `2.5xx10^(13)`B. `5xx10^(6)`C. `1.25xx10^(10)`D. `0.2xx10^(4)` |
| Answer» Correct Answer - C | |
| 71. |
(i) Name the type of a diode whose characteristic are shown figure . (ii) What does the point P in Figure represent? |
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Answer» Correct Answer - A::B::C::D (i) Diode is a rectifier diode. (ii) Point P represents zener breakdown potential. |
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| 72. |
The physical property (ies) which (is) are same along all the directions in an isotropic solid-A. Refractive indexB. Electrical conductivityC. Thermal conductivityD. None of the above. |
| Answer» Correct Answer - A::B::C | |
| 73. |
There is a small energy gap between the conduction and valence bands ofA. copperB. silverC. sodiumD. aluminium |
| Answer» Correct Answer - C | |
| 74. |
In a semiconductor, the energy gap between the valence and conduction bands is 1. 1 eV. It is expressed in joules asA. `1.2xx10^(-19)` JB. `1.76xx10^(-19)` JC. `1.6xx10^(-19)` JD. `3.2xx10^(-19)` J |
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Answer» Correct Answer - B Energy gap =1.1 eV = `1.1 xx 1.6 xx 10^(-19)=1.76xx10^(-19) ` J |
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| 75. |
In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance u and the image distance v, from the lens, is plitted using the same scale for the two axes. A straight line passing through the origin and making an angle of `45^circ` with x-axis meets the experimental curve at P. The coordinates of P will be.A. (2f,2f)B. `((f)/(2),(f)/(2))`C. `(f,f)`D. `(4f,4f)` |
| Answer» Correct Answer - A | |
| 76. |
If the frequency of input alternating potential in n, then the ripple frequency of output potential of full wave rectifier will be-A. 2nB. nC. `(n)/(2)`D. `(n)/(4)` |
| Answer» Correct Answer - A | |
| 77. |
A student uses a simple pendulum of exactly `1m` length to determine `g`, the acceleration due ti gravity. He uses a stop watch with the least count of `1sec` for this and record `40 seconds` for `20` oscillations for this observation, which of the following statement `(s) is (are)` true?A. Error `DeltaT` in measuring T, the time period, is 0.05 secondsB. Error `DeltaT` in measuring T, the time period, is 1 secondC. Percentage error in the determination of g is 5%D. Percentage error in the determination of g is 2.5% |
| Answer» Correct Answer - A::C | |
| 78. |
In a half wave rectifier output is taken across a `90 ohm` load resistor. If the resistance of diode in forward biased condition is `10 ohm`, the efficiency of rectification of `ac` power into `dc` power is.A. 0.406B. 0.812C. 0.7308D. 0.3654 |
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Answer» Correct Answer - D `eta=0.406xx(6)/(R_(l)+r_(f))=(0.406xx90)/(90+10)` `=36.54%` |
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| 79. |
In a circuit a diode was used and the output voltage across the diode was always 50 volts, even if the input voltage fluctuated between 110 V to 90 V. The diode used in the circuit wasA. a junction diodeB. photodiodeC. zener diodeD. light emitting diode |
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Answer» Correct Answer - C Zener diode was used as a voltage regulator |
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| 80. |
A`p-n-p` transistor is used in common-emitter mode in an amplifier circuuit. A change of `40muA` in the base current brings a change of `2mA` in collector current and `0.04V` in base-emitter volatage. Find the (a) input resistance `r_(i)` and (b) the base current amplification factor `(beta)`. (c ) if a load of `6kOmega` is used, then also find the voltage gain of the amplifier. |
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Answer» Given `DeltaI_(B)=40muA=40xx10^(-6)A` `DeltaI_(C)=2mA=2xx10^(-3)A` `DeltaV_(BE)=0.04"volt",R_(L)=6kOmega=6xx10^(3)Omega` (1) input resistance `R_("input")=(DeltaV_(BE))/(DeltaI_(B))=(0.04)/(40xx10^(-6))=10^(3)Omega=1kOmega` (2) Current amplification factor, `beta=(DeltaI_(C))/(DeltaI_(B))=(2xx10^(-3))/(40xx10^(-6))=50` (3). Voltage gain in common-emitter configuration, `A_(upsilon)=beta(R_(L))/(R_("inp."))=50xx(6xx10^(3))/(1xx10^(3))=300` |
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| 81. |
A`p-n-p` transistor is used in common-emitter mode in an amplifier circuuit. A change of `40muA` in the base current brings a change of `2mA` in collector current and `0.04V` in base-emitter volatage. Find the (a) input resistance `r_(i)` and (b) the base current amplification factor `(beta)`. (c ) if a load of `6kOmega` is used, then also find the voltage gain of the amplifier.A. `1000 Omega, 50`B. `2000 Omega, 40`C. `5000 Omega, 20`D. none of these |
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Answer» Correct Answer - A `R_(i)=(V_("be"))/(I_(b))` `=(0.04)/(40xx10^(-6))=1000 Omega` `beta=(I_(c))/(I_(b))=(2xx10^(-3))/(40xx10^(-6))=50` |
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| 82. |
If a small amount of antimony is added to germanium crystalA. the antimony becomes an acceptor atomB. there will be more free electrons than holes in the semiconductorC. its resistance is increasedD. it becomes a p-type semiconductor |
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Answer» Correct Answer - B When a small amount antimony is added to germanium crystal becomes n-type semiconductor because antimony is a pentavalent substrate. It excess free electrons. |
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| 83. |
The input resistance of a silicon transistor is `100Omega` base current is changed by `40muA` which results in a change in collector current by `2mA`. This transistor is used as a common emitter amplifier with a load resistance of `4kOmega`. The voltage gain of the amplifier isA. 2000B. 3000C. 4000D. 1000 |
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Answer» Correct Answer - A We know voltage gain is, `A_(V)=beta(R_(L))/(R_(i))` On putting given values, `A_(V)=(2xx10^(-3))/(40xx10^(-6))xx(4xx10^(3))/(100) " " [because beta=(DeltaI_(C))/(DeltaI_(B))]` `A_(V)=2000` |
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| 84. |
The input resistance of a silicon transistor is `665Omega`. Its base current is changed by `15muA` which results in the change in collector current by `2mA`. This transistor is used as a common emitter amplifier with a load resistance of `5kOmega`. What is the voltage gain of the amplifier.A. 1002B. 1004C. 1006D. 1008 |
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Answer» Correct Answer - A `R_(i)=665 Omega, delta I_(b)=15 muA , deltaI_(c)=2mA` `R_(0)=5 K Omega, A_(v)= ?` `A_(v)=beta(R_(0))/(R_(i))=(deltaI_(c))/(deltaI_(b))*(R_(0))/(R_(i))` `=(2xx10^(-3)xx5xx10^(3))/(15xx10^(-6)xx665)` `=(10xx10^(6))/(15xx665)` `=(10000xx10^(3))/(9975)` `=1.002xx10^(3)` =1002 |
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| 85. |
The symbolic representation of four logic gates are given in Fig.The logic symbol for OR, NOT and NAND gates are respectively A. (iii),(iv),(ii)B. (iv),(i),(iii)C. (iv),(ii),(i)D. (i),(iii),(iv) |
| Answer» Correct Answer - C | |
| 86. |
For a common emitter configuration , the base current is `60 muA` and the collector current is 6 mA . The current gain of transistor isA. 30B. 60C. 100D. 200 |
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Answer» Correct Answer - C `I_(b)=60muA, I_(c)=6mA" "beta=?` `beta=(I_(c))/(I_(b))=(6xx10^(-3))/(60xx10^(-6))=100` |
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| 87. |
In a silicon transistor, a change of `7.89 mA` in the emitter current produce a change of 7.8 mA in the collector current. What change in the base current is necessary to produce an equivalent change in the collector current?A. 0.9 `mum`B. 900 `mum`C. 90 `mum`D. 9 `mum` |
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Answer» Correct Answer - C `DeltaI_b=DeltaI_E-DeltaI_C = (7.89-7.8) mA=0.09` mA `=9xx10^(-2)` mA = `9xx10^(-2) xx 10^(3) muA= 90 muA` |
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| 88. |
In a common base mode of transistor, collector current is `5.488 mA` for an emitter current of `5.60mA`. The value of the base current amplification factor `(beta)` will be :A. 45B. 50C. 55D. 60 |
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Answer» Correct Answer - C `beta=I_C/I_B=I_C/(I_E-I_B) = 5.5/(5.6-5.5)`=55 |
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| 89. |
In a silicon transistor, a change of `7.89 mA` in the emitter current produce a change of 7.8 mA in the collector current. What change in the base current is necessary to produce an equivalent change in the collector current?A. 0.9 `muA`B. `9 muA`C. `90 muA`D. 900 `muA` |
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Answer» Correct Answer - C `I_(b)=I_(e)-I_(c)` `=7.89 - 7.8` `=0.09 ` mA `=90 muA` |
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| 90. |
In an oscillator circuit `L=10^(-3)` H, and C =`2 muF` . The frequency of oscillation isA. 3.5 KHzB. 2.5 KHzC. 10 KHzD. 15 KHz |
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Answer» Correct Answer - A `f=(1)/(2pisqrt(LC))` `=(1)/(2xx3.14sqrt(10^(-3)xx2xx10^(-6)))` `=3.5` KHz |
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| 91. |
Given below are four logic gates symbol (figure). Those for OR, NOR and NAND are respectivelyA. 1,4,3B. 4,1,2C. 1,3,4D. 4,2,1 |
| Answer» Correct Answer - C | |
| 92. |
For a transistor, the current amplification factor is 4. when the transistor is connected in common emitter configuration, the change in collector current, when the base current changes by 6mA, isA. 2.4 mAB. 3.6 mAC. 24 mAD. 36 mA |
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Answer» Correct Answer - C `beta=4, I_(b)=6 mA , I_(c)=?` `b=(I_(c))/(I_(b))` `therefore I_(c)=betaI_(b)=4xx6=24 mA` |
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| 93. |
In a common base transistor circuit, the current gain is 0.98. On changing the emitter current by 5.00 mA, the change in collector current isA. 0.196 mAB. 2.45 mAC. 4.9 mAD. 3.1 mA |
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Answer» Correct Answer - C `alpha=0.98,I_(e)=5 mA, I_(c)= ?` `alpha=(I_(c))/(I_(e))` `therefore I_(c)=alphaI_(e)=0.98xx5=4.9` mA |
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| 94. |
A transistor, connected in common emitter configuration, has input resistance `R_(i)=2kOmega` and load resistance `6kOmega`. If `beta=60` and an input signal `10mV` is applied, calculate the (i) resistance gain, (ii) voltage gain (iii) the power gainA. 9000B. 4000C. 6000D. 8000 |
| Answer» Correct Answer - A | |
| 95. |
What is the voltage gain in a common emitter amplifier where input resistance is `3 Omega` and load resistance is `24 Omega` :- `(beta = 6)` ?A. `2.2`B. `1.2`C. `4.8`D. `48` |
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Answer» Correct Answer - D Voltage gain `=beta(R_(0))/(R_("in"))=6xx(24)/(3)=48` |
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| 96. |
For a common emitter amplifier, input resistance `(R_i) = 500Omega` and load resistance `R_L = 5000Omega`. If `beta` = 60, then the voltage gain isA. 60B. 600C. 6D. 100 |
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Answer» Correct Answer - B Voltage gain = `betaR_L/R_i=60xx5000/500`=600 |
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| 97. |
In a common emitter amplifier , the input resistance is 1 `Omega` and load resistance is 10 `K Omega` . If the current gain is 100, the voltage gasin of the amplifier isA. 1000B. 100C. 10000D. 10 |
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Answer» Correct Answer - A `A_(v)=beta(R_(o))/(R_(i))=(100xx10xx10^(3))/(1xx10^(3))=1000` |
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| 98. |
A common emitter amplifier has current gain 70. Its load resistance is 5 `k Omega` and input resistance is 500 `Omega` . The voltage gain isA. 500B. 700C. 1000D. 1400 |
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Answer» Correct Answer - B `beta=70, R_(1)=5K Omega, R_(i)=500 Omega, A_(v)= ?` `A_(v)=beta*(R_(0))/(R_(i))=(70xx5xx10^(3))/(500)=700` |
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| 99. |
The current gain in the common emitter mode of a transistor is 10. The input impedance is `20kOmega` and load of resistance is `100k Omega`. The power gain isA. 300B. 500C. 200D. 100 |
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Answer» Correct Answer - B The power gain is defined as the ratio of charge in output power to the charge in input power. Since, `P=Vi` therefore Power gain = current gain `xx` voltage gain `=beta xx beta ((R_("out"))/(R_("in")))=beta^(2)((R_("out"))/(R_("in")))` Given, `beta=10, R_("in") =20kOmega and R_("out")=100kOmega` `therefore" Power gain"=(10)^(2)((100)/(20))=100xx5=500` |
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| 100. |
What is the voltage gain in a common emitter amplifier when the input resistance is 200 `Omega` and the load resistance is 1`K Omega` ? (`beta` = 50 )A. 100B. 150C. 200D. 250 |
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Answer» Correct Answer - D Voltage gain `A_V=betaR_o/R_i=50xx1000/200`=250 |
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