JEE Main 2015PhysicsCapacitanceHardMCQ

JEE Main 2015 — Capacitance Question with Solution

JEE Main 2015 (04 Apr)

Question

In the given circuit, charge Q2 on the 2 μF capacitor changes as C is varied from 1 μF to 3 μF. Q2 as a function of 'C' is given properly by: (figures are drawn schematically and are not to scale)

Choose an option

Show full solutionCorrect option: C
Correct answer
C

Step-by-step explanation



∵1 μF & 2 μF are in parallel.



∴  The equivalent capacitance of the series combination is, Ceq is=3CC+3. So, the total charge supplied by the battery is, 
⇒Q=CeqE=3CEC+3.∴ The potential difference across the parallel combination of 1 μF and 2μF is, ∆V=Q3=CEC+3. So charge on 2 μF capacitor is, Q2=C2∆V=2CEC+3⇒Q22E=CC+3⇒Q22E=C+3-3C+3

⇒Q22E=1-3C+3⇒ So the graph is a hyperbola. With downward curve line. i.e.

 

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About this question

This is a previous-year question from JEE Main 2015, 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.