JEE Main 2019PhysicsCommunication SystemEasyMCQ

JEE Main 2019 — Communication System Question with Solution

JEE Main 2019 (12 Apr Shift 2)

Question

In a CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2 V, if the base resistance is 1 kΩ and the current amplification of the transistor is 100 then the input signal voltage is

Choose an option

Show full solutionCorrect option: A
Correct answer
A10 mV

Step-by-step explanation

Current amplification factor,

β=ΔICΔIB

Collector current,

ΔIC=2V2×103Ω=1×10-3 A

Base current,

ΔIB=VBRB=VB1×103=VB×10-3

Given, β=100

Now, 100=10-3VB×10-3

⇒VB=1100 V=10 mV

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Communication System 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 Communication System 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.