JEE Main 2019PhysicsSemiconductorsMediumMCQ

JEE Main 2019 — Semiconductors Question with Solution

JEE Main 2019 (10 Apr Shift 1)

Question

An NPN transistor operates as a common emitter amplifier, with a power gain of  60 dB . The input circuit resistance is  100 Ω and the output load resistance is  10 kΩ . The common emitter current gain β is:

Choose an option

Show full solutionCorrect option: B
Correct answer
B102

Step-by-step explanation

Av×β=Pgain 

60=10log10PP0
P=106=β2×Rout Rin =β2×104100
β2=104
β=100=102

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About this question

This is a previous-year question from JEE Main 2019, covering the Semiconductors chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.