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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 transistor, forward bias is always smaller than the reverse bias. The correct reason isA. To avoid excessive heating of transistorB. To maintain a constant base currentC. To produce large voltage gainD. None of these |
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Answer» Correct Answer - A If forward biase is made large, the majority charge carriers would move from the emitter to the collector through the base with high velocity. This would give rise to excessive heat causing damage to transistor. |
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| 52. |
A piece of semiconductors is connected in sereis in an electric circuit. On increasing the temperautre, the current in the circuit willA. DecreasesB. Remain unchangedC. IncreaseD. Stop flowing |
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Answer» Correct Answer - C Because with rise in temperature, resistance of semiconductor decreases, hence overall resistance of the circuit increases, which in turn increases the current in the circuit. |
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| 53. |
Intrinsic semiconductor is electrically neutral. Extrinsic semiconductor having large number of current carriers would beA. Positive chargedB. negetive chargedC. Positively charged or negatively charged depending upon the type of impurity that has been addedD. Electrically charged |
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Answer» Correct Answer - D Extrinsic semiconductor (`N`-type or `P`-type) are neutral |
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| 54. |
Assertion: Two `P-N` junction diodes placed back to back, will work as a `NPN` transistor. Reason: The `P-N` junction of two `PN` junction diodes back to back will form the base of `NPN` transistor.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - D Two `PN`-junction placed back to back cannot work as `NPN` transistor because in transistor the width and concentration of doping of `P`-semiconductor is less as comapred to width doping of `N`-type semiconductor type. |
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| 55. |
An input voltage `V_(i)` of `0.20 V` is applied to an operational amplifier connected as shown in the diagram. What is the output voltage `V_(0)`? A. `0.20 V`B. `1.2 V`C. `0.80 V`D. `8.0 V` |
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Answer» Correct Answer - B The circuit configuration is a non-in-venting amplifier. The gain of the amlifier is given by the equation gain `=(V_(0))/(V_(1))=1+(R_(1))/(R_(2))=1+10/2=6` thus, the output voltage `V_(0)` is `V_(0)=6V_(i)=6(0.20)=1.2 V` |
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| 56. |
When `Ge` crystal are dopped with phosphorus atom, then it becomesA. InsulatorB. `P`-typeC. `N`-typeD. Superconductor |
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Answer» Correct Answer - C Phosphorus is pentavalent. |
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| 57. |
A potential difference of `2V` is applied between the opposite faces of a `Ge` crystal plate of area `1 cm^(2)` and thickness `0.5 mm`. If the concentration of electrons in `Ge` is `2xx10^(19)//m^(3)` and mobilities of electrons and holes are `0.36(m^(2))/(vol t-sec)` and `0.14 (m^(2))/(vol t-sec)` respectively, then the current flowing through the plate will beA. `0.25 A`B. `0.45 A`C. `0.56 A`D. `0.64 A` |
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Answer» Correct Answer - D `sigma=en_(e)mu_(e)` `=2xx10^(19)xx1.6xx10^(-19)(0.36+0.14)=1.6(Omega-m)^(-1)` `R=sigmal/A=l/(sigmaA)=(0.5xx10^(-3))/(1.6xx10^(-4))=25/8 Omega` `:. i=V/R=2/(25//8)=16/25 A =0.64 A` |
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| 58. |
Which is the correct diagram of a half- wave reactifier?A. B. C. D. |
| Answer» Correct Answer - B | |
| 59. |
The correct symbole for Zener diode isA. B. C. D. |
| Answer» Correct Answer - A | |
| 60. |
The value of current gain `alpha` of trasistor is `0.98`. The value of `beta` will beA. `4.9`B. `490`C. `59`D. `49` |
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Answer» Correct Answer - A Current gains are given by `alpha=(Deltai_(C))/(Deltai_(E))` and `beta=(Deltai_(C))/(Deltai_(B))` Where `Deltai_(C),Deltai_(E),Deltai_(B)` are change in collector current emitter current and base current respectively. The relation between `alpha` and `beta` is `beta=(alpha)/(1-alpha)` Given, `beta=0.98`, hence we have `beta=0.98/(1-0.98)=49` |
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| 61. |
If `alpha=0.98` and current through emitter `i_(e)=20 mA`, the value of `beta` isA. `4.9`B. `49`C. `96`D. `9.6` |
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Answer» Correct Answer - B `beta=(alpha)/(1-alpha)=0.98/(1-0.98)=49`. |
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| 62. |
In the circuit shown in the figure, the input voltage `V_(i)` is `20V,V_(BE)=0` and `V_(CE)=0`. The values of `I_(B),I_(C)` and `beta` are given by: A. `I_(B)=40 muA,I_(C)=10mA,beta=250`B. `I_(B)=25 muA,I_(C)=5mA,beta=200`C. `I_(B)=20 muA,I_(C)=5mA,beta=250`D. `I_(B)=40 muA,I_(C)=5mA,beta=125` |
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Answer» Correct Answer - D `V_(i)=I_(B)R_(B)+V_(BE)` `20=I_(B)xx(500xx10^(3))+0` `I_(B)=20/(500xx10^(3))=40 muA` `V_(CC)=I_(C)R_(C)+V_(CE)` `20=I_(C)xx(4xx10^(3))+0` `I_(C)=50xx10^(-3)=5mA` `beta=(I_(C))/(I_(B))=(5xx10^(-3))/(40xx10^(-6))=125` |
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| 63. |
In order to obtain an output `Y=1` form the circuit of fig. the inputs must be A. `A=0,B=1,C=0`B. `A=1,B=0,C=0`C. `A=1,B=0,C=1`D. `A=1,B=1,C=0` |
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Answer» Correct Answer - C The boolean expression for the given combination is output `Y=(A+B)` truth table is `|(A,B,C,Y=(A+B).C),(0,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,1,0),(1,1,0,0),(0,1,1,1),(1,0,1,1),(1,1,1,1)|` Hence `A=1,B=0,C=1` |
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| 64. |
The circuit shown below acts as A. high pass filterB. low pass filterC. tuned filterD. rectifier |
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Answer» Correct Answer - C The circuit is tuned filter circuit. It utilizes one or more tuned circuits (resonant or anti-resonant circuits) to separte signals in relatively narrow bands of frequencies from signal wider frequency spectrums. Most transmitters and recievers incorporate many tuned filters. The give `LCR` tuned filter (also known as `RLC` tuned fillter) is the most flexible of alll frequency or frequency non-linear circuits. It is preferred for its frequency selection property. |
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| 65. |
A triode has a plate resistance of `10 k Omega` and amplification factor `24`. If the input signal voltage is `0.4 V(r.m.s)`, and the load resistance is `10 k` ohm, then, the output voltage is (r.m.s.) isA. `4.8 V`B. `9.6 V`C. `12.0 V`D. None of these |
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Answer» Correct Answer - A Use `V_(0)=AV_(S)` Now `A=(24xx10k)/(10k+10k)=(24xx10)/20=12` Therefore, `V_(0)=12xx0.4=4.8` volt (r.m.s.) |
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| 66. |
In the given circuit as shown the two input waveform `A` and `B` are applied simultaneously. The resultant waveform `Y` is A. B. C. D. |
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Answer» Correct Answer - A (`1`=high,`0`=low) Input to `A` is in the sequence, `1,0,1,0`. Input to `B` is in the sequence, `1,0,0,1`. Sequence is inverted by `NOT`gate. Thus inputs to `OR` gate becomes `0,1,0,1` and ouput `OR` gate becomes `0,1,1,1` Since for `OR` gate `0+1=1`. hence choice `(a)` is correct. |
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| 67. |
When on impurity is doped into an intrinsic semiconductor, the conductivity of the semiconductorA. IncreasedB. DecreasedC. Remains the sameD. Becomes zero |
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Answer» Correct Answer - A There is considerable change in the electrical properties due to the addition of impurity. |
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| 68. |
Which logic gate is represented by the following combination of logic gates ? A. `OR`B. `NAND`C. `AND`D. `NOR` |
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Answer» Correct Answer - C First two gate are `NOT` gates and the last gate is `NOR` gate Thus, `Y_(1)=bar(A),y_(2)=bar(B)` and `y=bar(y_(1)+y_(2))` The truth table corresponding to this is as follows: `|(A,B,y_(1)=vec(A),y_(2)=vecB,y_(1)+y_(2),y=bar(y_(1)+y_(2)),A.B),(0,0,1,1,1,0,0),(0,1,1,0,1,0,0),(1,0,0,1,1,0,0),(1,1,0,0,0,1,1)|` Thus the combination of gate represents `AND` gate. |
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| 69. |
The diagram of a logic circuit is given below. The output `F` of the circuit is represented by A. `W.(X+Y)`B. `W.(X.Y)`C. `W+(X.Y)`D. `W+(X+Y)` |
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Answer» Correct Answer - C Output of upper `OR` gate=`W+X` Output of lower `OR` gate =`W+Y` Net output `F=(W+X)(W+Y)` `=WW+WY+XW+XY` (Since `WW=W`) `=W(1+Y)+XW+XY` (Since `1+Y=1`) `=W+XW+XY=W(1+W)+XY=W+XY` |
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| 70. |
Which logic gate is represented by following diagram? A. `AND` gateB. `OR` gateC. `NOR` gateD. `XOR` gate |
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Answer» Correct Answer - A 0 |
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| 71. |
In order to obtain an output `Y=1` form the circuit of fig. the inputs must be 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 Here `Y=(A+B).C` `|(A,B,C, Y=(A+B).C),(0,0,0,0),(1,0,0,0),(0,1,0,0),(0,1,0,0),(0,1,1,1),(1,0,1,1),(1,1,1,1)|` `C` should be `1` either `A` and `B` or both `A` and `B` are `1`. therefore, `A=1,B=0,C=1` |
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| 72. |
If `n_(e)` and `n_(h)` are the number of electrons and holes in a semiconductor heavily doped with phosphorus, thenA. `n_(e) gt gt h_(h)`B. `n_(e) lt lt h_(h)`C. `n_(e)leh_(h)`D. `n_(e)=h_(h)` |
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Answer» Correct Answer - A Phosphorus is pentavalent. |
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| 73. |
Zenor diode is used forA. producing oscillations in an oscillatorB. amplificationC. stabilisationD. rectification |
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Answer» Correct Answer - C Zener diode is a silicon crystal diode having an unusual reverse current characteristic which is particularly suitable for voltage regulating purpose. Due to this characterstic, it is used as voltage stabilizer in many application in electronics. |
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| 74. |
In a transistor in `CE` configuration, the ratio of power gain to voltage gain isA. `alpha`B. `beta//alpha`C. `beta alpha`D. `beta` |
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Answer» Correct Answer - D For `CE` configuration voltage gain `=betaxxR_(L)//R_(i)` Power gain `=beta^(2)xxR_(L)//R_(i) implies (Power gai n)/(Vol tag e gai n) =beta` |
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| 75. |
While a collector to emitter voltage is constant in a transistor, the collector current changes by `8.2 mA` when the emitter current changes by `8.3 mA`. The value of forward current ratio `h_(fe)` isA. `82`B. `83`C. `8.2`D. `8.3` |
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Answer» Correct Answer - A `h_(fe)=((Deltai_(c))/(Deltai_(c)))_(V_(ce))=8.2/(8.3-8.2)=82` |
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| 76. |
Which of the following is used to produce radio waves of constant amplitude?A. OscillatorB. `FET`C. RectifierD. Amplification |
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Answer» Correct Answer - A Oscillator can produce radio waves of constant amplitude. |
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| 77. |
In order to obtain an output `Y=1` form the circuit of fig. the inputs must be A. `{:(A,B,C),(0,1,0):}`B. `{:(A,B,C),(1,0,0):}`C. `{:(A,B,C),(1,0,0):}`D. `{:(A,B,C),(1,1,0):}` |
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Answer» Correct Answer - C `A+B=1` `C=1` So, `AND` gate responds. |
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| 78. |
Least doped region in a transistorA. Either emitter or collectorB. BaseC. EmitterD. Collector |
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Answer» Correct Answer - B In transistor, base is least dopped. |
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| 79. |
In a common emitter transistor, the current gain is `80`. What is the change in collector current, when the change in base current is `250 muA`?A. `80xx250 muA`B. `(250-80) muA`C. `(250+80) muA`D. `250//80 muA` |
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Answer» Correct Answer - A Current gain `beta=(Deltai_(c))/(Deltai_(b)) implies betaxx Deltai_(b)=80xx250 muA`. |
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| 80. |
Assertion: The current gain in common base circuit is always less than one. Reason: At constant collector votalge the change in collector current is more than the change in emitter current.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - C The current gain in common base circuit `alpha=((DeltaI_(C))/(DeltaI_(E)))_(V_(C))`. The change in collector current is always less than the change in the emitter current. `DeltaI_(C)ltDeltaI_(E)`. Therefore, `alphalt1`. |
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| 81. |
In the circuit, if the forward voltage drop for the diode is `0.5 V`, the current will be A. `3.4 mA`B. `2 mA`C. `2.5 mA`D. `3 mA` |
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Answer» Correct Answer - A The voltage drop across resistance `=8-0.5=7.5 V` `:.` Current `i=7.5/(2.2xx10^(3))=3.4 mA` |
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| 82. |
A `2V` battery is connected across the points `A` and `B` as shown in the figure given below. Assuming that the resistance of each diode is zero in forward bias and infinity in reverese biase, the current supplied by the battery when its positive terminal is connected to `A` is A. `0.2 A`B. `0.4 A`C. zeroD. `0.1 A` |
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Answer» Correct Answer - A Since diode in upper branch is forward biased and in lower branch is reversed biased. So current through circuit `I=V/(R+r_(d))`, here `r_(d)`=diode resistance in forward biasing `=0` `implies V/R=2/10=0.2A` |
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| 83. |
If the two ends of a p-n junction are joined by a wire ,A. There will not be a steady current in the circuitB. There will be a steady current form `N` side to `P` sideC. There will be a steady current from `P` side to `N` sideD. There may not a be a current depending upon the resistance of the connecting wire |
| Answer» Correct Answer - A | |
| 84. |
In a `PN`-junctionA. `P` and `N` both are at same potentialB. High potential at `N` side and low potential at `P` sideC. High potential `P` side and low potential at `N` sideD. Low potential at `N` side and zero potential at `P` side |
| Answer» Correct Answer - B | |
| 85. |
The current through an ideal `PN`-junction shown in the following circuit diagram will be A. zeroB. `1mA`C. `10 mA`D. `30 mA` |
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Answer» Correct Answer - A The diode is in reverse biasing so current through it is zero. |
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| 86. |
In the diagram, the input is across the terminals `A` and `C` and the output is across the terminals `B` and `D`, then the outputs is A. zeroB. same as inputC. Full wave rectifierD. Half wave rectifier |
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Answer» Correct Answer - C The given circuit is full wave rectifier. |
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| 87. |
In `P`-type semiconductor, there isA. An excess of one electronB. Absence of one electronC. A missing atomD. A donor level |
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Answer» Correct Answer - B Absence of one electron, creates the positive charge of magnitude equl to that of electronic charge. |
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| 88. |
What is the current through an ideal p-n junction diode shown in figure below ? |
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Answer» 20 mA Join the grounded points |
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| 89. |
Symbolic representation of four logic gates are shown as A. (iii),(ii) and (i)B. (iii),(ii) and (iv)C. (ii),(iv) and (iii)D. (ii),(iii) and (iv) |
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Answer» Correct Answer - C Symbols give in problem are `(i) OR` gate `(ii) AND` gate `(iii) NOT` gate `(iv) NAND` gate |
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| 90. |
The nature of binding for a crystal with alternate and evenly spaced positive and negative ions isA. CovalentB. MetallicC. DipolarD. Ionic |
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Answer» Correct Answer - D Ionic bonds cone being when atoms that have low ionization energies, and hence lose electrons rapidly, interact with other atoms that and to acquire excess electrons. The former atoms give up electrons to the latter and they there upon becomes positive and negative ions respectively. |
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| 91. |
In a common base ampifier , the phase difference between the input signal and output voltage isA. `0`B. `pi//4`C. `pi//2`D. `pi` |
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Answer» Correct Answer - A In `CB` amplifer Inputs and output voltage signal are in same phase. |
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| 92. |
An n-type semi- conductor isA. Positively chargedB. Negatively chargedC. Either positively charged or negatively chargedD. Electrically neutral |
| Answer» Correct Answer - D | |
| 93. |
The electrical resistance of the following decreases with rise in temperatureA. MetalsB. SemiconductorC. GoldD. Constantan |
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Answer» Correct Answer - B Temperature coeffiecient of resistance of metals is positive while that of semiconductor is negative |
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| 94. |
The temperature coefficient of resistance of a conductor isA. Positive alwaysB. Negative alwaysC. zeroD. Infinite |
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Answer» Correct Answer - A The temperature coeffiecient of conductor is positive. |
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| 95. |
For a transistor, in a common emitter arragement, the alternating current gain `beta` is given byA. `beta=((DeltaI_(C))/(I_(B)))_(V_(C))`B. `beta=((DeltaI_(B))/(DeltaI_(C)))_(V_(C))`C. `beta=((DeltaI_(C))/(DeltaI_(E)))_(V_(C))`D. `beta=((DeltaI_(E))/(DeltaI_(C)))_(V_(C))` |
| Answer» Correct Answer - A | |
| 96. |
The truth table given below is for which gate? `|(A,B,C),(0,0,1),(0,1,1),(1,0,1),(1,1,0)|`A. `XOR`B. `NOT`C. `NAND`D. `AND` |
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Answer» Correct Answer - C `Y=bar(A.B)` `|(A,B,Y),(0,0,1),(0,1,1),(1,0,1),(1,1,0)|` Which is the truth of `NAND` gate. |
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| 97. |
Which of these is unipolar transistor?A. Point contanct transistorB. Field effect transistorC. `PNP` transistorD. None of these |
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Answer» Correct Answer - B `FET` is unipolar. |
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| 98. |
Current in the circuit will be A. `5/40 A`B. `5/30 A`C. `5/10 A`D. `5/20 A` |
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Answer» Correct Answer - B The diode in lower branch is forward biased and diode in upper branch is reverse biased `:. i=5/(20+30)=5/20 A` |
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| 99. |
In a `P`-type semi-conductorA. Current is mainly carried by holesB. Current is mainly carried by electronsC. The material is always positively chargedD. Dopping is done by pentavalent material |
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Answer» Correct Answer - A In `P`-type semicondutor, holes are majority charge carriers. |
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| 100. |
In a `p-n` junctionA. `P` and `N` both are at same potentialB. High potential at `N` side and low potential at `P` sideC. High potential `P` side and low potential at `N` sideD. Low potential at `N` side and zero potential at `P` side |
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Answer» Correct Answer - B For conduction, `p-n` junction must be forward biased. For this `p`-side should be connected to higher potential and `n`-side to lower potential. |
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