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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.
| 101. |
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. 8.4B. 4.8C. 2.4D. 12 |
| Answer» Correct Answer - D | |
| 102. |
A transistor connected in CE mode , has a current gain of 50. If the load resistance is 5 K, input resistance is 1 K and the input peak voltage is 0.4 V, then the peak output voltage will beA. 120B. 150C. 80D. 200 |
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Answer» Correct Answer - B Voltage gain = Current gain x Resistance gain `=30xx5/1`=150 |
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| 103. |
A transistor connected in CE mode , has a current gain of 50. If the load resistance is 5 K, input resistance is 1 K and the input peak voltage is 0.4 V, then the peak output voltage will beA. 25 VB. 50 VC. 75 VD. 100 V |
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Answer» Correct Answer - D `A_V=betaR_o/R_i = 50xx(5xx10^3)/(1xx10^3)`=250 `therefore V_o= 250 xx V_i = 250 xx 0.4` `therefore V_o` = 100 V |
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| 104. |
When the voltage between emitter and the base `V_(EB)` of a transistor is chaged by 5 mV while keeping the collector voltage `V_(CE)` fixed when then its emitter current changes by 0.15 mA. Calculate the input resistance of the transistor. |
| Answer» Correct Answer - 33.33 ohm | |
| 105. |
In a common base transistor amplifier if the input resistance `R_(i)` is 200 `Omega` and load resistance `R_(L)` is `20K Omega` , find voltage gain `(alpha=0.95)`A. 86B. 98C. 94D. 95 |
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Answer» Correct Answer - D `R_(i)=200 Omega, R_(0)=20 K Omega, A_(v)= ? Alpha=0.95` Now `A_(v)=alpha(R_(0))/(R_(i))` `=0.95xx(2000)/(200)=0.95xx200=190` |
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| 106. |
In CE configuration of a transistor, input resistance is 2 `K Omega`, load resistance is `5 k Omega`, current 32 gain is 60. An input signal of 12 mV is applied to the transistor, calculate the voltage gainA. 15B. 150C. 1500D. 1.5 |
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Answer» Correct Answer - B `R_(i)=2K Omega, R_(0)= 5k Omega , V_(i) = 12 m V, , A_(v)=? Beta=60` `A_(V)=beta*(R_(0))/(R_(i))` |
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| 107. |
In the circuit given below, `V(t)` is the sinusiodal voltage source, voltage drop `V_(AB)(t)` across the resistance `R` is A. Is half wave rectifiedB. is full wave rectifiedC. has the same peak value in the positive and negative half cyclesD. has different peak values during positive and negative half cycle |
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Answer» Correct Answer - D In `+ve` half cycle, one diode is in forward bias and other in reverse bias, while in `-ve` half cycle their polarity reverse bias, direction of current is opposite through `R` for `+ve` and `-ve` half cycles so output is not rectified. Since `R_(1)!=R_(2)`, hence the peak values during `+ve` and `-ve` half of input dignal will be different. |
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| 108. |
In a common emitter amplifier, using output reisistance of `5000` ohm and input resistance fo `2000`ohm, if the peak value of input signal voltage is `10 m V` and `beta=50`, then the peak value of output voltage isA. 1.25 VB. 12.5 VC. 125VD. 1250 V |
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Answer» Correct Answer - A `R_(0)=5000 Omega , R_(i)=2000, V_(i)=10 mV, beta=50, V_(0)=?` `A_(v)=(V_(0))/(V_(i))=beta*(R_(0))/(R_(i))` `V_(0)=betaV_(i)(R_(0))/(R_(i))` `=50xx10xx10^(-3)xx(5000)/(2000)` `=50xx10xx2.5xx10^(-3)` `=1250xx10^(-3)V` `=1.25 V` |
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| 109. |
In a common emitter amplifier, using output reisistance of `5000` ohm and input resistance fo `2000`ohm, if the peak value of input signal voltage is `10 m V` and `beta=50`, then the peak value of output voltage isA. `5xx10^(-6)V`B. `12.50 xx 10^(-6)V`C. 1.25 VD. 125.0 V |
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Answer» Correct Answer - C `A_(v)=(Delta V_(0))/(DeltaV_(i))=beta(R_(0))/(R_(i)) and DeltaV_(0)=DeltaV_(i)xxbeta(R_(0))/(R_(i))` `therefore DeltaV_(0)= 10xx50 xx(500)/(2000)` `DeltaV_(0)=1250mV=1.25V` |
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| 110. |
For a transistor , `alpha_(dc)` and `beta_(dc)` are the current ratios, then the value of `(beta_(dc) - alpha_(dc))/(alpha_(dc) . beta_(dc))`A. 1B. 1.5C. 2D. 2.5 |
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Answer» Correct Answer - A For the transistor `beta_"dc"=alpha_"dc"/(1-alpha_"dc")` `therefore 1-alpha_(dc) = alpha_(dc)/beta_(dc)`…(1) `therefore(beta_"dc"-alpha_"dc")/(alpha_"dc"-beta_"dc")=(beta_"dc"(1-(alpha_"dc")/(beta_"dc")))/(alpha_"dc" beta_"dc")` `=(1-alpha_"dc"/beta_"dc")/alpha_"dc"` `=(1-(1-alpha_"dc"))/alpha_"dc"` ...from (1) `=alpha_"dc"/alpha_"dc"=1` ...Option (a) |
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| 111. |
For a transitor the current ratio `alpha_(DC)` is 69/70 the current gain `beta_(DC)` isA. 66B. 67C. 69D. 71 |
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Answer» Correct Answer - C `beta=(alpha)/(1-alpha)=((69)/(70))/(1-(69)/(70))=((69)/(70))/((1)/(70))` `beta=69`. |
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| 112. |
Consider the MB shown in the diagram, let the resistance X have temperature coefficient `alpha_(1)` and the resistance from the RB have the temperature coefficient `alpha_(2)`. Let the reading of the meter scale be 10 cm from the LHS. if the temperature of the two resistance increase by small temperature `DeltaT` then what is the shift in the position of the null point ? Neglect all the other changes in the bridge due to temperature rise.A. `9(alpha_(1)-alpha_(2))DeltaT`B. `9(alpha_(1)+alpha_(2))DeltaT`C. `(1)/(9)(alpha_(1)+alpha_(2))DeltaT`D. `(1)/(9)(alpha_(1)-alpha_(2))DeltaT` |
| Answer» Correct Answer - A | |
| 113. |
What is the output X of the following logic gate circuit? A. ABCDB. AB+C+DC. A+B+C+DD. AB+CD |
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Answer» Correct Answer - C All are OR gates. `therefore Y_1=A+B` `Y_2=A+B+C, X=A+B+C+D` |
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| 114. |
What is the output X of the following logic gate circuit? A. (A+B)+(A+C)B. (A+`barB`).(`barA`+C)C. (A.B) + (A.C)D. (A+B) . (A+C) |
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Answer» Correct Answer - D The outputs of the OR gates A + B and A+ C, become the inputs for the AND gate. `:.` X = (A+ B) · (A+ C) |
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| 115. |
A glass tube filled with colored water, sealed at both the ends is bent into an arc. There is a small air bubble inside. The tube is held with its plane vertical. When the tube moves with constant acceleration either of left or right the bubble shifts and settles at some plane either to the left or right of the highest point. For the situation shown, what can you conclude about acceleration vector of the tube?A. right towards experimental wireB. towards compensating wire towards either of themC. dows not shiftD. |
| Answer» Correct Answer - B | |
| 116. |
The transfer ration of a transistor is `50`. The input resistance of the transistor when used in the common -emitter configuration is `1 kOmega`. The peak value for an `A.C.` input voltage of `0.01 V` peak isA. `100muA`B. `0.01mA`C. `0.25mA`D. `500muA` |
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Answer» Correct Answer - D `beta=50,R_(i)=1000 Omega,v_(i)=0.01V` `v_(i)=DeltaI_(B)R_(i)impliesDeltaI_(B)=(0.01)/(1000)=10^(-5)A` `beta=(DeltaI_(C ))/(DeltaI_(B))impliesDeltaI_(C )=betaDeltaI_(B)=50xx10^(-5)A=500muA` |
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| 117. |
A semiconductor is known to have an electron concentration of `6xx10^(12)` per cubic centimeter and a hole concentration of `8xx10^(13)` per cubic centimeter. Is this semiconductor N-type or P-type ?A. a p-type semiconductorB. an n-t ype semiconductorC. an intrinsic semiconductorD. either a p-type or an n-type semiconductor |
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Answer» Correct Answer - A Since the hole density is more than the electron density, it is p-type semiconductor. |
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| 118. |
Zener diode is used asA. RectificationB. StabilisationC. AmplificationD. producing oscillations in an oscillator |
| Answer» Correct Answer - B | |
| 119. |
Application of a forward biase to a `p-n` junction:A. widens the depletion zoneB. increase the number of donors on the n sideC. increses the potential difference across the depletion zoneD. increases the electric field in the depletion zone |
| Answer» Correct Answer - B | |
| 120. |
Carbon , silicon and germanium have four valence elcectrons each . These are characterised by valence and conduction bands separated by energy band - gap respectively equal to ` (E_g)_(c) (E_g)_(si) ` and ` (E_g)_(Ge) `. Which of the following statements ture ?A. `(E_(g))_(C)lt(E_(g))_(ge)`B. `(E_(g))_(c)gt(E_(g))_(si)`C. `(E_(g))_(c)=(E_(g))_(si)`D. `(E_(g))_(c)lt(E_(g))_(si)` |
| Answer» Correct Answer - B | |
| 121. |
Carbon , silicon and germanium have four valence elcectrons each . These are characterised by valence and conduction bands separated by energy band - gap respectively equal to ` (E_g)_(c) (E_g)_(si) ` and ` (E_g)_(Ge) `. Which of the following statements ture ?A. `(E_(g))_(Si) lt (E_(g))_(Ge) lt (E_(g))_(C)`B. `(E_(g))_(C) lt (E_(g))_(Ge) lt (E_(g))_(Si`C. `(E_(g))_(C) gt (E_(g))_(Si) gt (E_(g))_(C)`D. `(E_(g))_(C) = (E_(g))_(Si) lt (E_(g))_(Ge)` |
| Answer» Correct Answer - C | |
| 122. |
In a....... baised pn junction , the net flow of holes is from the `n` region to the `p` region .A. forwards biasB. reverse biasC. noD. both 1 and 2 |
| Answer» Correct Answer - B | |
| 123. |
Three energy levels `L_1,L_2` and `L_3` of a hydrogen atom correspond to increasing values of energy i.e., `E_(L_1) lt E_(L_2) lt E_(L_3)` . If the wavelength corresponding to the transitions `L_3` to `L_2, L_2` to `L_1` and `L_3` to `L_1` are `lambda_3, lambda_2` and `lambda_1` respectively thenA. `l_1 gt l_2 gt l_3`B. `l_3 lt l_2 lt l_1`C. `l_3 gt l_2 gt l_1`D. `l_1 = l_2 = l_3` |
| Answer» Correct Answer - C | |
| 124. |
A piece of copper and another of germanium are cooled from room temperature to `80 K`. The resistance ofA. each of these decreasesB. copper strip increases and that of germanium decreasesC. copper strip decreases and that of germanium increasesD. each of the above increases |
| Answer» Correct Answer - C | |
| 125. |
Cu and Ge are cooled to 70 K then :-A. the resistance of Cu will decrease and that of Ge will decreaseB. the resistance of Cu will decrease and that of Ge will increaseC. the resistance of both Cu and Ge decreaseD. the resistance of both Cu and Ge increase |
| Answer» Correct Answer - B | |
| 126. |
In semiconductor the valence band at 0 K isA. completely filledB. completely emptyC. partially filledD. nothing can be said |
| Answer» Correct Answer - A | |
| 127. |
When there is current during only one-half of the A.C. input cycle in a ciruit, then it is calledA. an amplifierB. an oscillatorC. full-wave rectifiedD. half-wave rectified |
| Answer» Correct Answer - D | |
| 128. |
Avalanche breakdown in a semi conductor diode occurs whenA. the forward current exceeds a certain valueB. forward bias exceeds a certain valueC. reverse bias exceeds a certain valueD. the depletion region is reduced to zero |
| Answer» Correct Answer - C | |
| 129. |
Avalanche breakdown is due toA. heavy dopingB. thermal ionisationC. impact ionisationD. combination of holes and electrons |
| Answer» Correct Answer - C | |
| 130. |
For the circuit shown in the figure:A. current through zener diode is 4 mA.B. current through zener diode is 9 mAC. the output voltage is 50 VD. the output voltage is 40 V |
| Answer» Correct Answer - B::C | |
| 131. |
Avalanche breakdown in a `PN` junction diode is toA. shift of fermi levelB. widening of forbidden gapC. high impurity concentrationD. commulative effect of conduction band electrons collision |
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Answer» Correct Answer - D At a particular value of reverse bias voltage, the covalent bonds near the junction break, so electron-hole pairs get liberated, this process rapidly multiplies and an avalanche of electron-hole pairs is produced. |
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| 132. |
Avalanche breakdown is obtained in a p-n junction when there isA. forward biasB. reverse biasC. zero biasD. very high bias |
| Answer» Correct Answer - B | |
| 133. |
Avalanche breakdown is due toA. collision of minority charge carrierB. depletion layer thickness increaseC. DL thickness decreaseD. none of the above |
| Answer» Correct Answer - A | |
| 134. |
The avalanche break down in p-n junction is not caused due toA. shift of fermi energy levelB. widening of forbidden bandC. cumulative effect of conduction band electron collisionsD. High impurity concentration. |
| Answer» Correct Answer - A::B::D | |
| 135. |
In a vemier callipers the main scale and the vernier scale are made up different materials. When the room temperature increases by `Delta T^@ C` , it is found the reading of the instrument remains the same. Earlier it was observed that the front edge of the wooden rod placed for measurment crossed the `N^(th)` main scale division and (N + 2) msd coincided with the 2nd vsd. Initially, 10 vsd coincided with 9 msd. If coefficient of linear expansion of the main scale is `alpha_1` and that of the vermier scale is `alpha_2` then what is the value of `alpha_1//alpha_2` ? (Ignore the expansion of the rod on heating)A. `1.8//(N)`B. `1.8//(N+2)`C. `1.8//(N-2)`D. none |
| Answer» Correct Answer - B | |
| 136. |
In the junction transistor voltage amplifier circuit of figure, if `R_(1)=100kOmega,R_(2)=1kOmega,V_(C C)=6.0V` and `V_(BE)=0.6V` current gain=60A. `I_(B)=54muA`B. `I_(C)=3.24mA`C. the voltage across `R_(2)=3.24V`D. the voltage across the collector-emitter=3.24V |
| Answer» Correct Answer - A::B::C | |
| 137. |
When N-P-N transistor is used as an amplifier-A. electrons move from base to amitterB. electrons move from emitter to baseC. electrons move from collector to baseD. holes move from base to emitter |
| Answer» Correct Answer - B | |
| 138. |
A common emitter circuit is used as an amplifier, its current gain is 50. if input resistance is `1kOmega` and input voltage is 5 volt then output current will beA. 250 mAB. 30 mAC. 50 mAD. 100 mA |
| Answer» Correct Answer - A | |
| 139. |
Which of the following is true?A. common base transistor is commonly used bacause current gain is maximum.B. common-emitter is commonly used because current gain is maximum.C. Common collector is commonly used because current gain is maximum.D. Common emitter is the least used transistor. |
| Answer» Correct Answer - B | |
| 140. |
In an n-p-n transistor, the collector current is 10 mA. If 90% of the electrons emitted reach the collector, then the emitter current will beA. 9 mAB. 1 mAC. 2 mAD. 8 mA |
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Answer» Correct Answer - B `I_E=10 ` mA `I_C=90/100xx10`= 9 mA `therefore I_B=I_E-I_C` 1 mA |
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| 141. |
In an `NPN` transistor the collector current is `10 mA`. If `90%` of electrons reach collector, the emitter current `(i_(E))` and base current `(i_(B))` are given byA. `I_(E)=1mA,I_(B)=11mA`B. `I_(E)=11mA,I_(B)=1mA`C. `I_(E)=-1mA,I_(B)=9mA`D. `I_(E)=9mA,I_(B)=-1mA` |
| Answer» Correct Answer - B | |
| 142. |
In an `NPN` transistor the collector current is `10 mA`. If `90%` of electrons reach collector, the emitter current `(i_(E))` and base current `(i_(B))` are given byA. `I_(E) = 1mA , I_(B) = 11 mA`B. `I_(E) = 11 mA l I_(B) = 1mA`C. `I_(E) = -1mA , I_(B) = 9mA`D. `I_(E) = 9mA , I_(B) = -1 mA` |
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Answer» Correct Answer - B `I_(C)=90%I_(E)=(90)/(100)I_(E)rArr I_(E)=(100)/(90)xxI_(C)=11mA` `I_(B)=I_(E)-I_(C)=11mA-10mA=1mA`. |
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| 143. |
In an NPN transistor the values of base current and collector current are `100muA` and `9mA` respectively the emitter current will be-A. 9.1 mAB. 18.2 mAC. `9.1muA`D. `18.2muA` |
| Answer» Correct Answer - A | |
| 144. |
In case of `NPN`-transistor the collector current is always less than the emitter current becauseA. collector side is reverse biased and the emitter side is forward biasedB. a few electrons are lost in the base and only remaining ones reach the collectorC. collector being reverse biased, attracts less electronsD. collector side is forward biased and emitter side is reverse biased |
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Answer» Correct Answer - D Due to forward bias at the emitter-base junction, the majority charge carrier electrons of emitter get repelled from the negative terminal and move towards base. Some of these electrons combine with the majority charge carrier holes present in the base and most of the electrons reach the collector, crossing the collector-base junction. |
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| 145. |
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 coltage gainD. None of the above |
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Answer» Correct Answer - A If forward bias 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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| 146. |
If a full wave reactifier circuit is operating from `50 Hz` mains, the fundamental frequency in the ripple will beA. 50 HzB. 25 HzC. 100 HzD. 70.7 Hz |
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Answer» Correct Answer - C Cu is conductor and Ge is semiconductor. Increase of temperature increases resistance of conductor but decreases resistance of semiconductor. |
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| 147. |
If a full wave reactifier circuit is operating from `50 Hz` mains, the fundamental frequency in the ripple will beA. `50Hz`B. `70.7Hz`C. `100Hz`D. `25Hz` |
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Answer» Correct Answer - C `2xx50=100HZ` |
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| 148. |
The maximum effeciency of full wave rectifier isA. `100%`B. `25.20%`C. `40.2%`D. `81.2%` |
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Answer» Correct Answer - D For full wave rectifier `eta=(81.2)/(1+r_(f)//R_(L))` `eta_(max)=81.2%(r_(f)lt lt R_(L)` |
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| 149. |
A p-type semoconductor has acceptor levels `57meV `above the valence band. Find the maximum wavelength of light which can create a hole. |
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Answer» `(hc)/(lambda)ge57xx10^(-3)xx1.6xx10^(-19)` `lambdale(6.6xx10^(-34)xx3xx10^(8))/(57xx1.6xx10^(-22))` `=2.18xx10^(-5)m` |
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| 150. |
Photodiode is always operated inA. forward biasB. reverse biasC. unbiasedD. none of these |
| Answer» Correct Answer - B | |