JEE Main 2019PhysicsCapacitanceMediumMCQ

JEE Main 2019 — Capacitance Question with Solution

JEE Main 2019 (10 Apr Shift 1)

Question

Figure shows charge (q) versus voltage (V) graph for series and parallel combination of two given capacitors. The capacitances are:

Choose an option

Show full solutionCorrect option: D
Correct answer
D40 μF  and  10 μF

Step-by-step explanation

Q=CV
The slope of graph will give the equivalent capacitance of the combination.
From graph A: C1+C2=50010=50 …(i)
From graph B: C1C2C1+C2=8010=8 …(ii)
From equation (i) & (ii)
C1C2=400 …(iii)
C1+400C1=50 ⇒C12-50C1+400=0
C1=50±2500-16002=50±302
=10 μF,  40 μF

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About this question

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