InterviewSolution
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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.
| 151. |
The correct relation between current gains `alpha` and `beta` isA. `beta =(alpha)/(1 -alpha)`B. `beta =(alpha)/(1+alpha)`C. `beta =alpha(1-alpha)`D. `beta = (1-alpha)/(alpha)` |
| Answer» Correct Answer - A | |
| 152. |
Transistors are made of.A. insulatorsB. conductorsC. alloysD. doped semi-conductors |
| Answer» Correct Answer - D | |
| 153. |
In `n-p-n` transistor the arrow head on emitter represents that the convantional current flows fromA. base to emitterB. emitter to baseC. emitter to collectorD. base to collector |
| Answer» Correct Answer - A | |
| 154. |
In transistor the emitter current is.A. slightly more than the collector currentB. slightly less than the collector currentC. equal to the collector currentD. equal to the base current |
| Answer» Correct Answer - A | |
| 155. |
In the use of transistor as an amplifierA. the emitter - base junction is revrese biased and the collector base junction is also revrese biasedB. the emitter - base junction is forward biased and the collector -base junction is reverse biasedC. both the junctions are forward biasedD. any of the two junctions may be forward biased |
| Answer» Correct Answer - B | |
| 156. |
One way in which the operation of an `npn` transistor differ from that of a `pnp` transistor is thatA. the emitter junction is reverse biased in `npn`B. the emitter junction injects minority carriers into the base region of the `pnp`.C. the emitter injects holes into the base of the `pnp` and electrons into the base region of `npn`D. the emitter injects holes into the base of `npn`. |
| Answer» Correct Answer - C | |
| 157. |
The combination of the gates shown below produces .A. `AND` gateB. `XOR` gateC. `NOR` gateD. `NAND` gate |
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Answer» Correct Answer - D The output of `G_(1)` is `overline(A) G_(2)` is `overline(B)`. Hence output of `G_(3)` is `AB :. y = overline(AB)` i.e., `NAND` gate. |
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| 158. |
The combination of gates shown below yields .A. `NAND` gateB. `OR` gateC. `NOT` gatesD. `XOR` gate |
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Answer» Correct Answer - B `X = overline(overline(A). overline(B)) = overline(overline(A))+overline(overline(B)) = A + B`. It is `OR` gate. |
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| 159. |
The expression of `Y` in following circuit is .A. `ABCD`B. `A+BCD`C. `A+B+C+D`D. `AB+CD` |
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Answer» Correct Answer - C `Y_(1)=A+D,Y_(2)=A +D+B`, `Y =A +D+B+C`. |
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| 160. |
The output of the combination of the gates shown in the figure below is .A. `A+A.B`B. `(A+B)A+ overline (B)`C. `(A.B)+(overline(A). overline(B))`D. `(A+B)(overline(A.B))` |
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Answer» Correct Answer - A The input to `OR` gate is `A` and `A.B`. Hence `Y = A + BA`. |
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| 161. |
Assertion (A) : `Si` and `GaAs` are preferred materials for solar cells Reason (R ) : Energy gap of `Si` is `1.1 eV` and that of `GaAs` is `1.53 eV` which gives maximum irradiance where as other materials like `CdS` or `CdSe (E_(g) = 2.4 eV)` and `PbS (E_(g) = 0.4 eV)` given minimum irradiance. |
| Answer» Correct Answer - 1 | |
| 162. |
In the Boolean algebra, the following one is wrongA. `1.0=0`B. `0.1 =0`C. `1.1 = 0`D. `1.1 =1` |
| Answer» Correct Answer - C | |
| 163. |
In Boolean algebra `A.B = Y` implies that :A. product of `A` and `B` is `Y`B. `Y` exists when `A` exists or `B` existsC. `Y` exists when both `A` and `B` exist but not when only `A` or `B` existsD. `Y` exists when `A` or `B` exists but not both `A` and `B` exist. |
| Answer» Correct Answer - C | |
| 164. |
A three terminal device with one terminal common to both the output and input is called.A. rectifierB. transistorC. diodeD. triode |
| Answer» Correct Answer - B | |
| 165. |
Boolean algebra is essentially based onA. symbolsB. logicC. truthD. numbers |
| Answer» Correct Answer - B | |
| 166. |
The gate that has only one input terminalA. `NOT`B. `NOR`C. `NAND`D. `XOR` |
| Answer» Correct Answer - A | |
| 167. |
Among the following one gives output `1` in the `AND` gate.A. `A = 0,B=0`B. `A=1,B=1`C. `A=1,B=0`D. `A =0,B=1` |
| Answer» Correct Answer - B | |
| 168. |
Consider a two-input `AND` gate of figure below. Out of the four entries for the Truth Table given here, the correct ones are. .A. AllB. `1` and `2` onlyC. `1,2` and `3` onlyD. `1,3` and `4` only |
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Answer» Correct Answer - C Cheack the truth table. |
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| 169. |
The output of a 2-input `OR` gate is zero only when itsA. both inputs are `0`B. either input is `1`C. both inputs are `1`D. either input is zero |
| Answer» Correct Answer - A | |
| 170. |
Justify the output wavefrom (y) of the OR gate for input and as gives in Fig. |
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Answer» Note the following : `{:(At,t lt t_(1),,A=0,B=0 ,,HenceY=0),("For",t_(1)"to" t_(2),,A =1,B=0 ,,Hence Y=1),("For",t_(2)"to" t_(3) ,,A =1,B=1,,Hence Y=1),("For",t_(3) "to" t_(4),,A=0,B=1,,Hence Y=1),("For",t_(4)"to" t_(5) ,,A =0,B=0 ,,Hence Y=0),("For",t_(5) "to" t_(6),,A =1,B=0,,Hence Y=1),("For",t gt t_(6) ,,A =0,B=1,,Hence Y=1):}` Therefore the wave from `Y` will be as shown in the Fig. |
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