JEE Main 2021PhysicsCurrent ElectricityHardMCQ

JEE Main 2021 — Current Electricity Question with Solution

JEE Main 2021 (26 Aug Shift 1)

Question

In the given figure, the emf of the cell is 2.2 V and if internal resistance is 0.6 Ω. Calculate the power dissipated in the whole circuit:

Choose an option

Show full solutionCorrect option: D
Correct answer
D2.2 W

Step-by-step explanation

The above circuit in the question is arranged like that, so all the resistance are connected in parallel and the net resistance of the parallel resistor becomes, 


1RP=1R1+1R2+1R3+1R4=14+16+112+18⇒RP=4830 Ω

Now the net resistance will be, Rnet=4830+0.6 Ω, so the net power, P=V2R=(2.2)22.2=2.2 W

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About this question

This is a previous-year question from JEE Main 2021, 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.