JEE Advanced 2025 Paper 1 · Physics · Electromagnetic Induction and Alternating Current · hard

A capacitor C1=2 muF is charged to 100 V and then, at t=0, connected through a 2 kΩ resistor to an uncharged capacitor C2=3 muF. The time (in ms) at…

Numerical answer

A capacitor

C_1=2\ \mu\mathrm{F}

is charged to

100\ \mathrm{V}

and then, at

t=0

, connected through a

2\ \mathrm{k\Omega}

resistor to an uncharged capacitor

C_2=3\ \mu\mathrm{F}

. The time (in ms) at which the heat dissipated in the resistor becomes exactly half of the total heat that is eventually dissipated is ______. Give the answer correct to two decimal places.

Show answer and solution

Answer: 0.83

The two capacitors and the resistor form a single loop with an effective capacitance

C_{eq}=\dfrac{C_1C_2}{C_1+C_2}=1.2\ \mu\mathrm{F}

, so the time constant is

\tau=RC_{eq}=2000\times1.2\times10^{-6}=2.4\ \mathrm{ms}

. The current decays as

I=I_0e^{-t/\tau}

with

I_0=V/R

. The heat dissipated up to time

t

H(t)=\int_0^tI^2R\,dt'=\tfrac12I_0^2R\tau\left(1-e^{-2t/\tau}\right)=H_\infty\left(1-e^{-2t/\tau}\right)

, with

H_\infty=\tfrac12C_{eq}V^2=6\ \mathrm{mJ}

. Setting

H=H_\infty/2

gives

e^{-2t/\tau}=\tfrac12

, so

t=\dfrac{\tau\ln2}{2}=\dfrac{2.4\ln2}{2}=0.832\ \mathrm{ms}\approx0.83\ \mathrm{ms}

. Note that it is not

\tau\ln2=1.66\ \mathrm{ms}

(the half-life of the current).

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