A capacitor of capacitance C, which is initially uncharged , is connected with a battery of emf `epsilon`. Find the heat dissipated in the circuit during the process of charging .

Explanation

Initial charge in the capacitor is `Q=0`. When the capacitor is connected across battery, the final potential difference across the capacitor will be `E`. Final charge in the capacitor is `Q_(f)=C_(epsilon)`
Initial potential energy stored is `U_(i)-0`. Final potential energy stored is `U_(f)=(1//2)Cepsilon^(2)`. So the charge suppplied by the battery is
`DeltaQ=Q_(f)-Q_(i)=(Cepsilon-0)=Cepsilon`.
Work done by the battery in charging the capacitor is
`W_("battery")=(DeltaQ)epsilon=(Cepsilon)epsilon=Cepsilon^(2)`
We know that `W_("battery")-DeltaU +` heat developed. So
`H = H_("battery") - DeltaU = Cepsilon^(2) - (1)/(2)Cepsi^(2)`
`:.` Heat developed `= (1)/(2) Cepsi^(2)`