JEE Main 2019PhysicsSemiconductorsMediumMCQ

JEE Main 2019 — Semiconductors Question with Solution

JEE Main 2019 (08 Apr Shift 2)

Question

A common emitter amplifier circuit, built using an NPN transistor, is shown in the figure. Its dc current gain is 250, RC=1 kΩ and VCC=10 V . The minimum base current for VCE to reach saturation is

Choose an option

Show full solutionCorrect option: B
Correct answer
B40 μA

Step-by-step explanation



Given, β=250

RC=1000 Ω

VCC=10 V


At saturation, VCE=0

⇒ie=10 V1000 Ω=10 mA

Current gain factor, β=ieiB

⇒iB=ieβ=10 mA250

=40 μA

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Semiconductors chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

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.