JEE Main 2022PhysicsCurrent ElectricityEasyMCQ

JEE Main 2022Current Electricity Question with Solution

JEE Main 2022 (24 Jun Shift 1)

Question

Two identical cells each of emf 1.5 V are connected in parallel across a parallel combination of two resistors each of resistance 20 Ω. A voltmeter connected in the circuit measures 1.2 V. The internal resistance of each cell is :

Choose an option

Show full solutionCorrect option: C
Correct answer
C5 Ω

Step-by-step explanation

Equivalent resistance of the parallel 20 Ω branches will be 10 Ω.

Equivalent EMF of the two branches which are in parallel will be, =1.5r+1.5r1r+1r=1.5 V and the equivalent internal resistance for same branches will be, r2.

For loop containing voltmeter and 10 Ω resistor, 1.2-10i=0 i=0.12 A.

Reading of voltmeter should be,

1.2=1.5-ir2 ir2=0.3 r=0.6i=0.60.12= 5 Ω

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About this question

This is a previous-year question from JEE Main 2022, 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.