JEE Main 2024 · Physics · Capacitors · easy

A capacitor of 4 muF is charged to 100 V and then connected, through a resistor, across an uncharged capacitor of 12 muF. The energy dissipated as…

Numerical answer

A capacitor of

4\ \mu\mathrm{F}

is charged to

100\ \mathrm{V}

and then connected, through a resistor, across an uncharged capacitor of

12\ \mu\mathrm{F}

. The energy dissipated as heat in the resistor (in mJ) is

Show answer and solution

Answer: 15

Initial energy

=\tfrac12(4\times10^{-6})(100)^2=20\ \mathrm{mJ}

. By charge conservation the common potential is

V'=\dfrac{4\times100}{16}=25\ \mathrm{V}

, so the final energy is

\tfrac12(16\times10^{-6})(25)^2=5\ \mathrm{mJ}

. Heat dissipated

=20-5=15\ \mathrm{mJ}

. Equivalent formula:

\dfrac12\dfrac{C_1C_2}{C_1+C_2}V^2=\dfrac12(3\ \mu\mathrm{F})(100)^2=15\ \mathrm{mJ}

. The loss is independent of the resistance value.

Original question written in the JEE Main 2024 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake