InterviewSolution
Saved Bookmarks
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.
| 101. |
In a transistor configuration `beta` -parameter isA. `(l_(b))/(l_(c))`B. `(l_(c))/(l_(b))`C. `(l_(c))/(l_(a))`D. `(l_(a))/(l_(c))` |
|
Answer» Correct Answer - B `beta=(i_(c))/(i_(b))` |
|
| 102. |
In transistor, a change of `8.0 mA` in the emitter current produces a change of `7.8 mA` in the collector current. What change in the base current is necessary to produce the same change in the collector current ?A. `50 muA`B. `100 muA`C. `150 muA`D. `200 muA` |
|
Answer» Correct Answer - B `Deltai_(e)=Deltai_(c)+Deltai_(b)` `implies 8=7.8+Deltai_(b) implies Deltai_(b)=0.2 mA=200 muA` |
|
| 103. |
For the given circuit of `PN`-junction diode, which of the following statements is correct? A. In forward biasing the voltage across `R` is `V`B. In forward biasing the voltage across `R` is `2V`C. In reverse biasing the voltage across `R` is `V`D. In reverse biasing the voltage across `R` is `2V` |
|
Answer» Correct Answer - A In forward biasing, resistance of `PN` junction diode is zero, so whole voltage appears across the resistance. |
|
| 104. |
In the given figure, which of the diodes are forward biased? A. `1,2,3`B. `2,4,5`C. `1,3,4`D. `2,3,4` |
|
Answer» Correct Answer - B In forward `2,4` and `5`. `P`-crystals are more compared to comapared to `N` -crystals. |
|
| 105. |
For the given circuit of `PN`-junction diode, which of the following statements is correct? A. In forward biasing the voltage across `R` is `V`B. In forward biasing the voltage across `R` is `2V`C. In reverse biasing the voltage across `R` is `V`D. In reverse biasing the voltage across `R` is `2V` |
|
Answer» Correct Answer - A In forward biasing, resistance of `PN` junction diode is zero, so whole voltage appears across the resistance. |
|
| 106. |
In a `PN` -junction diodeA. The current in the reverse biased condition is generally very smallB. The current in the reverse biased condition is small but the forwards biased current is independent of the biase voltageC. The reverse biased current is srongly dependent on the applied biase voltage.D. The forward biased current is very small in comparision to reverse biased current. |
|
Answer» Correct Answer - A In forwards biased `PN`-junction, external voltage decreases the potential barrier, so current is maximum. While in reversed biased `PN`-junction, external voltage incrases the potential barriers, so the current is very small. |
|
| 107. |
The electrical circuit used to get smooth `dc` output from a rectifier circuit is calledA. OscillatorB. FilterC. AmplifierD. Logic gates |
|
Answer» Correct Answer - B Filter circuits are used to get smooth `dc pi`-filter is the best filter. |
|
| 108. |
When npn transistor is used as an ampliflerA. Electrons move from base to emitterB. Electrons move from emitter to baseC. Electrons moves from base to emitterD. Holes moves form base to emitter |
|
Answer» Correct Answer - B In `NPN` transistor when emitter base is forward biased, electrons move from emitter to base. |
|
| 109. |
In the study of transistors as an amplifier, if `alpha=I_(c)//I_(c)` and `beta=I_(c)//I_(b)`, where `I_(c),I_(b)` and `I_(e)` are the collector, base and emitter currents, thenA. `beta=(1-alpha)/(alpha)`B. `beta=(alpha)/(1-alpha)`C. `beta=(alpha)/(1+alpha)`D. `beta=(1+alpha)/(alpha)` |
|
Answer» Correct Answer - B As we know `i_(E)=i_(C)+i_(B)` `implies (i_(e))/(i_(c))=1+(i_(b))/(i_(c)) implies 1/(alpha)=1+1/(beta) implies implies beta=(alpha)/(1-alpha)`. |
|
| 110. |
Thermionic emission from a heated filament varies with its temperature `T` asA. `T^(-1)`B. `T`C. `T^(2)`D. `T^(3//2)` |
|
Answer» Correct Answer - C According to Richardson-Dushman equation, number of thermions emitted per sec unit area `J=AT^(2) e^(-W0//kT) implies J propT^(2)` |
|
| 111. |
In the following circuit, a voltmeter `V` is connected across a lamp `L`. What change would occure in voltmeter reading if the resistance `R` is reduced in value? A. IncreasesB. DecreasesC. Remains sameD. None of these |
|
Answer» Correct Answer - A Hence the emitter base junction of `N-P-N` transistor is forward biased with battery `V_(BB)` through resistance `R`. When the value of `R` is reduced, then the emitter current `i_(e)` will increases. As a result the collector current will also increases `(i_(c)=i_(e)-i_(b))`. Due to increase in `i_(e)`, the potential difference across `L` increases and hence the reading of voltmeter will increases. |
|
| 112. |
For the transistor circuit shown below, if `beta=100`, voltage drop between emitter and base is `0.7V` then value of `V_(CE)` will be A. `10 V`B. `5 V`C. `13 V`D. `0 V` |
|
Answer» Correct Answer - C `i_(b)=(5-0.7)/8.6=0.5 mA implies I_(c)=betaI_(b)=100xx0.5 mA` By using `V_(CE)=V_(C C)-I_(C)R_(L)=18-50xx10^(-3)xx100=13V` |
|
| 113. |
A transistor is used as an amplifier in `CB` mode with a load resistance of `5 k Omega` the current gain of amplifier is `0.98` and the input resistance is `70 Omega`, the voltage gain and power gain respectively areA. `70,68.6`B. `80,75.6`C. `60,66.6`D. `90,96.6` |
|
Answer» Correct Answer - A In common base mode `alpha=0.98, R=5 kOmega,R_(i n)=70 Omega` `:.` Voltage gain `A_(v)=alphaxxR/(R_(i n)) =0.98xx(5xx10^(3))/70=70` power gain =Current gain xVoltage gain `=0.98xx70=68.6` |
|
| 114. |
In the following common emitter configuration an `NPN` transistor with current gain `beta=100` is used. The output voltage of the amlifier will be A. `10 mV`B. `0.1V`C. `1.0V`D. `10V` |
|
Answer» Correct Answer - C `"Voltage gain" ("Output voltage")/("Input voltage")` `implies V_(out)=V_("in")xx"Voltage gain"` `implies V_(out)=V_("in")xx"Current gain"xx"Resistance gain"` `=V_("in")xxbetaxx(R_(L))/(R_(BE))=10^(-3)xx100xx10/1=1V`. |
|
| 115. |
A transistor is used in common-emitter mode in an amplifier circuit. When a signal 20mV is added to the base-emmiter voltage,the base current changes by `20(mu)A `and the collector current changes by 2mA.The load resistance is `5k(Omega)`.Calculate (a)the factor`(beta)`(b)the input resistance `R_(BE),`(c) the transconductance and (d) the voltage gain .A. `200`B. `300`C. `400`D. `500` |
|
Answer» Correct Answer - D Voltage gain `=(Output Vol tag e)/(Input vol tag e)` `=(2mAxx5 k Omega)/(20 mV)=(1xx5xx10^(3))/10=500` |
|
| 116. |
Assertion : In transistor common emitter mode as an amplifier is prefered over common base mode. Reason: In common emitter mode the input signal is connected in series with the voltage applied to the base emitter function.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. |
|
Answer» Correct Answer - B Common emitter is prepared over common base because all the current, voltage and power gain of common emitter amplifier is much more than the gains of common base amplifier. |
|
| 117. |
Which of the following gates will have an output of `1`? A. `4`B. `1`C. `2`D. `3` |
|
Answer» Correct Answer - C For `NAND` gate, `bar(0.1)=bar(0)=1` |
|
| 118. |
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. `OR`C. `AND`D. `NAND` |
|
Answer» Correct Answer - D We know that the output of a `NAND` gate is zero if both the inputs are `1`. Hence the given truth table is for `NAND` gate. |
|
| 119. |
Assertion : The crystalline solids have a sharp melting point. Reason: All the bonds between the atoms or molecules fo a crystalline solids are equally strong, that they get broken at the same temperature.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. |
|
Answer» Correct Answer - A At a particular temperature all the bonds of crystalline solids breaks and show sharp melting point. |
|
| 120. |
In semiconductor at a room temperatureA. Valence band is partially empty and the conduction band is partially filledB. Valence band is completely filled and the conduction band is partially emptyC. Valence band is completely filledD. Conduction band is completely empty |
| Answer» Valence band is partially empty and the conduction band is partially filled | |
| 121. |
In semiconductor at a room temperatureA. The velence band is partially empty and the conduction band is partially filledB. The velence band is completely filled and the conduction band is partially filledC. The velence band is completely filledD. The conduction band is completely empty |
|
Answer» Correct Answer - A Copper, Aluminium, Iron are conductors, while `Ge` is semiconductor. |
|
| 122. |
In case of a semiconductor, which of the following statement is wrong?A. Dopping increases conductivityB. Temperature coefficient of resistance is negativeC. Resistivity is in between that of a conductor and insulatorD. At absolute zero temperature, it behaves like a conductor |
|
Answer» Correct Answer - D At absolute zero temperature, semiconductor. |
|
| 123. |
The circuit shown in following figure contanis two diode `D_(1)` and `D_(2)` each with a forward resistance of `50 ohm` and with infinite backward resistance. If the battery voltage is `6V`, the current through the `100` ohm resistance (in amperes) is A. zeroB. `0.02`C. `0.03`D. `0.036` |
|
Answer» Correct Answer - B Accroding to the given polarity, diode `D_(1)` is forward biased while `D_(2)` is reverse biased. Hence current will pass through `D_(1)` only. so current `i=6/((150+50+100))=0.02 A` |
|
| 124. |
When a battery is connected to a `P`-type semiconductor with a metallic wire, the current in the semiconductor (predominantly), inside the metallic wire and that inside the bettery respectively due toA. Holes,electrons,ionsB. Holes, ions, electronsC. Electrons, ions,holesD. Ions,electron,holes |
|
Answer» Correct Answer - A Charge carriers inside the `P`-type semiconductor are holes (mainly). Inside the conductor charge carriers are electrons and for cell ions are the charge carriers. |
|
| 125. |
`P`-type semiconductors are made by adding impurity elementA. `As`B. `P`C. `B`D. `Bi` |
|
Answer» Correct Answer - C Because boron is a trivelent impurity. |
|
| 126. |
Which of the following materials can be used for making solar cell ?A. IronB. copperC. Lead ziconateD. Gallium arsenide |
| Answer» Correct Answer - D | |
| 127. |
Which of the following diodes is forward-biased ?A. B. C. D. |
| Answer» Correct Answer - 4 | |
| 128. |
Which of the following diodes is used in unbiased condition ?A. Zener diodeB. Photo diodeC. Solar cellD. LED |
| Answer» Correct Answer - C | |
| 129. |
In a p-n junction , the barrier potential offers resistance toA. Free electrons in n-side and holes in p-sideB. free electrons in p-side and holes in n-sideC. Only free electrons in n-sideD. Only holes in p-side |
| Answer» Correct Answer - A | |
| 130. |
Which of the following is not equal to 1 in Boolean algebra ?A. A+1B. `A + bar(A)`C. `bar(A.bar(A))`D. `A.bar(A)` |
| Answer» Correct Answer - D | |
| 131. |
With reference to figure, which of the following is possible? A. `A=0,B=0,X=1`B. `A=0,B=1,X=0`C. `A=0,B=0,X=0`D. `A=1,B=1,X=0` |
|
Answer» Correct Answer - D The output of both `AND` gates will be zero. So, `OR`gate does not respond. |
|
| 132. |
Name the Gate represented by the following circuit. A. `OR` gateB. `XOR` gateC. `AND` gateD. `HAND` gate |
|
Answer» Correct Answer - B Output of `OR` gate is `A+B`. Output of `NAND` gate is `bar(A).bar(B)`. New, `Y=(A+B).bar(A.B)=(A+B).(bar(A)+bar(B))` If `A=1` and `B=1` then `A+B=1` and `bar(A)+bar(B)=0` So, `Y=0` If `A=0` and `B=0`, then `Y=0` If `A=1` and `B=0`, then `A+B` gives `1 bar(A)+bar(B)=1` gives So, `Y=1` If `A=1` and `B=1`, then `Y=1` So, the given combination is `XOR` gate |
|
| 133. |
Which gate is represented by this figure? A. `NAND` gateB. `AND` gateC. `NOT`gateD. `OR` gate |
|
Answer» Correct Answer - A The given symbole is of `NAND` gate. |
|
| 134. |
Identify the gate from the following A. `NOT`gateB. `AND` gateC. `OR` gateD. None of these |
|
Answer» Correct Answer - D "Single input" not acceptable. |
|
| 135. |
In the Boolean algebra `bar((bar(A).bar(B))). A` equal toA. `bar(A+B)`B. `A`C. `bar(A.B)`D. `A+B` |
|
Answer» Correct Answer - B `bar(bar(A)+bar(B)).A=(bar(bar(A))+bar(bar(B))).A=(A+B).A` `=A.A+AB=A+AB=A(1+B)=A` |
|
| 136. |
The following configuration of gates is equivalent toA. `HAND`B. `XOR`C. `OR`D. None of these |
|
Answer» Correct Answer - A Output of `G_(1)+A+B` Output of `G_(1)=bar(A.B)` Output of `G_(3)` `=(A+B).bar(A.B)=(A+B).(bar(A)+bar(B))` If `A=1` and `B=1`, then `A+B=1` `A+B=0` So, output is zero, If `A=0` and `B=0`, then `A+B=0` `bar(A)+bar(B)=1+1=1` So, output is zero. Clearly, the given combination is `XOR` gate. |
|
| 137. |
A diode is connected to `220V (rms)` `ac` in series with a capacitor as shown in fig. The voltage across the capacitor is A. `220 V`B. `110 V`C. `311.1 V`D. `220/(sqrt(2))V` |
|
Answer» Correct Answer - D The diode `D` will conductor for positive half cycle of a.c. supply because this is forward biased. For negative half cycle of `a.c.` supply, this is reverse biased and does not conduct. So output would be half wave rectified output `V_(rms)=(V_(0))/2=(200sqrt(2))/2=200/(sqrt(2))` |
|
| 138. |
A transistor cannot be used as anA. AmplifierB. OscillatorC. ModulatorD. Rectifier |
| Answer» Correct Answer - D | |
| 139. |
What is the voltage gain in a common emitter amplifier, where output resistance is `3 Omega` and load resistance is `24 Omega (beta=0.6)` ?A. 8.4B. 4.8C. 2.4D. 480 |
| Answer» Correct Answer - B | |
| 140. |
The P-N junction is-A. Ohmic - resistanceB. Non-ohmic resistanceC. Negative resistanceD. Positive resistance |
| Answer» Correct Answer - B | |
| 141. |
What is the value of output voltage `V_(0)` in the circuit shown in the figure ? A. 6VB. 14VC. 20 VD. 26 V |
| Answer» Correct Answer - A | |
| 142. |
Iron and silicon wires are heated form `30^(@)C` to `50^(@)C`. The correct statement is thatA. Resistance of both wires IncreasesB. Resistance of both wires decreasesC. Resistance of iron wire increases and that of silicon wire decreasesD. Resistance of iron wire decreases and that of silicon wire increases |
|
Answer» Correct Answer - C Temperature coeffiecient of resistance of metals is positive while that of semiconductor is negative |
|
| 143. |
The energy gap of silicon is `1.14 eV`. The maximum wavelength at which silicon will begin absorbing energy isA. `10888Å`B. `1088.8Å`C. `108.88Å`D. `10.888Å` |
|
Answer» Correct Answer - A `lambda_(max)=(hc)/E=(6.6xx10^(-34)xx3xx10^(8))/(1.14xx1.6xx10^(-19))=10888Å` |
|
| 144. |
Consider an `n-p-n` transistor amplifer in common-emitter configuration. The current gain of the transistor is `100`. If the collector current changes by `1mA`, what will be the change in emitter current?A. `1.01 mA`B. `1.1 mA`C. `0.01 mA`D. `10 mA` |
|
Answer» Correct Answer - A Current gain `=(chang e i n coll ect or current)/(chang e i n base current)` `beta=(Deltai_(C))/(Deltai_(B))` Also `i_(E)=i_(B)+i_(C) implies Deltai_(E)=Deltai_(B)+Deltai_(C)` `beta=(Deltai_(C))/(Deltai_(E)-Deltai_(C))` Given, `beta=100,Deltai_(C)=1mA` `:. 100=1/(Deltai_(E)-1), Deltai_(E)-1=1/100=0.01` `Deltai_(E)=1+0.01=1.01 mA` |
|
| 145. |
Assertion: Base in transistor is made very thin as compared to collector and emitter regions. Reason: Due to thin base power gain and voltage gain is obtained by a 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. |
|
Answer» Correct Answer - A In a transistor, the base is made extermely thin to reduce the combinations of holes and electrons. Under this conditions, most of the holes (or electrons) arriving from the emitter diffuses across base and reach the collector. hence, the collector current, is almost equal to base the emitter current, the base current being comparatively much smaller. This is the main reason that power gain and voltage gain are obtained by a transistor. if the base region was made quite thick, then majority of carriers form emitter will combine with the carriers in the base and only small number of carriers will reach the collector, so there would be little collector current and the purpose of transistor would be defeated. |
|
| 146. |
Which of the following energy band diagrams shows the `N`-type semiconductor?A. B. C. D. |
|
Answer» Correct Answer - B In `N`- type semiconductor impurity energy level lies just below the conduction band. |
|
| 147. |
If `n_(e)` and `n_(h)` be the number of electrons and drift velocity in a semiconductor. When the temperature is increasedA. `n_(e)` increases and `v_(d)` decreasesB. `n_(e)` decreases and `v_(d)` increasesC. Both `n_(e)` and `v_(d)` increasesD. Both `n_(e)` and `v_(d)` decreases |
|
Answer» Correct Answer - A Because `v_(d)=i/((n_(e))eA)` |
|
| 148. |
The contribution in the total current flowing through a semiconductor due to electrons and holes are `3/4` and `1/4` respectively. If the drift velocity of electrons is `5/2` times that of holes at this temperature, then the ratio of concentration of electrons and holes isA. `6:5`B. `5:6`C. `3:2`D. `2:3` |
|
Answer» Correct Answer - A As we know current density `J=nqv` `implies J_(e)=n_(e)qv_(e)` and `J_(h)=n_(h)qv_(h)` `implies (J_(e))/(J_(h))=(n_(e))/(n_(h))xx(v_(e))/(v_(h)) implies (3//4)/(1//4)=(n_(e))/(n_(h))xx5/20 implies (n_(e))/(n_(h))=6/5` |
|
| 149. |
A peice of copper and the other of germanium are cooled from the room temperature to `80K`, then which of the following would be a correct statement?A. Resistance of each increasesB. Resistance of each decreasesC. Resistance of copper increases whil that of germenium decreasesD. Resistance of copper decreases while that of germenium increases |
|
Answer» Correct Answer - B Resistance of conductors `(Cu)` decreases with decreas in temperature while that of semi-conductors `(Ge)` increases with decreases in temperature. |
|
| 150. |
In forward bias, the width of potential barrier in a `P-N` junction diodeA. the positive terminal of the battery is connected to `n`-side and the depletion region becomes thinB. the positive terminal of the battery is connected to `n`-side and the depletion region becomes thickC. the positive terminal of the battery is connected to `p`-side and the depletion region becomes thinD. the positive terminal of the battery is connected to `p`-side and the depletion region becomes thick |
|
Answer» Correct Answer - C In forward biasing of `p-n`junction the positive terminal of the battery is connected to `p`-side and the depletion region becomes thin. |
|