JEE Main 2019PhysicsSemiconductorsMediumMCQ

JEE Main 2019 — Semiconductors Question with Solution

JEE Main 2019 (09 Apr Shift 1)

Question

An NPN transistor is used in common emitter configuration as an amplifier with 1 kΩ load resistance. Signal voltage of 10 mV is applied across the base-emitter. This produces a 3 mA change in the collector current and 15 μA  change in the base current of the amplifier. The input resistance and voltage gain are:

Choose an option

Show full solutionCorrect option: B
Correct answer
B0.67 kΩ, 300

Step-by-step explanation

load resistace=output resistaceRL=1 kΩ=1000 Ω
 input voltage=ΔVi=10 mV=10×10-3 V
input current=base current=ΔiB=15 μA=15×10-6 A
output current=collector current=ΔiC=3 mA=3×10-3 A
ΔiBRi=ΔVi=0.67 kΩ
Ri=10×10-315×10-6=0.67 kΩ
Voltage gain =ΔICRLΔVi=3×10-3×100010×10-3=300

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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.