# Energy dissipated in RC circuit without Resistance [duplicate]

This question already has an answer here:

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• Link of high relevance: hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c4 – DepressedDaniel Nov 27 '16 at 1:00
• I think that the question "Energy Loss in Capacitors" (physics.stackexchange.com/questions/209215/…) and the answer I supplied there also provides some insight into the answer to the current question. – Samuel Weir Nov 27 '16 at 1:03
• This has been asked and answered here already; the short version is that you 'break' ideal circuit theory in this case because, in reality, there is an inescapable radiation resistance that is ignored in ideal circuit theory. Essentially, the energy is radiated away. – Alfred Centauri Nov 27 '16 at 2:23
• – Alfred Centauri Nov 27 '16 at 2:31
• 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