JEE Main 2019PhysicsCurrent ElectricityMediumMCQ

JEE Main 2019Current Electricity Question with Solution

JEE Main 2019 (12 Apr Shift 1)

Question

The resistive network shown below is connected to a D.C. source of 16 V. The power consumed by the network is 4 Watt. The value of R is:

Choose an option

Show full solutionCorrect option: B
Correct answer
B8 Ω

Step-by-step explanation

Req for mixed series and parallel combination of

  given circuit Req=114R+14R+R+116R+112R+R

Req=8R
Power dissipated by circuit P=V2Req
4=1628R
R=25632=8Ω

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About this question

This is a previous-year question from JEE Main 2019, covering the Current Electricity 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.