JEE Main 2020PhysicsSemiconductorsEasyNumerical

JEE Main 2020 — Semiconductors Question with Solution

JEE Main 2020 (06 Sep Shift 2)

Question

The output characteristics of a transistor is shown in the figure. When VCE is 10V and IC=4.0 mA, then value of βac is 

   

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Show full solutionCorrect answer: 150
Correct answer
150

Step-by-step explanation

From the formula of current gain, β=ΔiCΔiBVCE=10

ΔiC=4.5-3 mA

⇒ΔiC=1.5 mA

⇒ΔiC=1.5×10-3 A

ΔiB=30-20 μA

⇒ΔiB=10 μA

⇒ΔiB=10×10-6 A

⇒ΔiB=10-5 A

⇒β=ΔiCΔiBVCE=10=1.5×10-310-5

⇒β=1.5×10-3×105

⇒β=150

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About this question

This is a previous-year question from JEE Main 2020, 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.