A capacitor of capacitance `C` is charge by a battery of emf `E` and internal resistance r. A resistasnce 2 is also connet in sereis with the capacitor. The amount of heat liberated inside the battery by the time capacitor is `50%` charged is

Options

A. `3/8E^2C`
B. `(E^2C)/6`
C. `(E^2C)/12`
D. `(E^2C)/24`

Answer

D

Explanation

During charging of a capacitor `50%` of the energy supplied by the battery is lost and only `50%` is stored.
`:.` Total energy lost `=1/2 q^2/C=1/2((EC//2)^2)/C=(E^2C)/8`
Now, this total loss is direct ratio `r:2r` or `1:2`
`:.` Energy lost in battery is `1/3rd` of `(E^2C)/8`.