An uncharged capacitor is connected to a battery. Show that half the energy supplied by the battery is lost as heat while charging the capacitor.

Answer

NA

Explanation

If `V` is e.m.f. of battery and `C` is capacity of the condenser, then chareg given to condenser `Q = CV`.
As work done = charge `xx` potential
`:. W = Q xx V = CV xx V = CV^(2)`
This is the energy supplied by the capacitor `U = (1)/(2) CV^(2)`
`:.` Energy lost in the form of heat
`U' = W - U = CV^(2) - (1)/(2) CV^(2) = (1)/(2) CV^(2)`
`:. (U')/(W) = ((1)/(2) CV^(2))/(CV^(2)) = (1)/(2)`, which was to be proved.