# Energy dissipated in RC circuit without Resistance [duplicate]

The good old RC circuit consists of a battery, a resistor ($R$) and a capacitor ($C$).

Once the capacitor has been charged with charge $Q$, the battery will have done work $W = QV$.

But the energy stored in the capacitor is only $\frac{QV}{2}$. Personaly I interpret this as being due to the fact that charging the capacitor becomes more and more difficult as its charge builds up, since we need to do work against the repulsion.

This energy must go somewhere though, I am assuming that it is dissipated in the resistor. However, $\frac{QV}{2}$ is independent of resistance.

So, in the ideal situation of NO resistance, where would the other half of the energy actually go?

## marked as duplicate by Floris, Alfred Centauri, Kyle Kanos, Community♦Nov 27 '16 at 14:36

• If there is zero resistance loss, then the circuit will "ring", with energy oscillating between the capacitor and whatever inductance is present, falling to $E=QV/2$ only after loss. physics.stackexchange.com/questions/187774/… – BowlOfRed Nov 27 '16 at 7:22