If an uncharged capacitor is connected to a battery, then show that half of the energy supplied by the battery to charge the capacitor is dissipated as heat.

Explanation

Let a capacitor of capacitance C be connected to a battery of emf V.
The final charge on the capacitor, Q= CV
The work done by the battery to fully charge the capacitor, W= VQ
So energy supplied by the battery = VQ
The energy stored in the fully charged capacitor `= (1)/(2)CV^2 = (1)/(2)QV`
The remaining amount of energy dissipated as heat
`= VQ - (1)/(2)QV = (1)/(2) QV`
Hence half of the energy supplied by the battery is dissipated as heat.