# Lost energy when capacitor discharges

1. Feb 15, 2015

### Tekk

1. The problem statement, all variables and given/known data

In state (a), I charged A with 1 Coulomb of charges, as depicted in figure (a).
In state (b), I paralleled A with B, by doing this I discharged A, as depicted in figure (b).

I try to find the total energy of state (a) and state (b). From what I have calculated, the system lost 90% of its energy when it goes from (a) to (b). Where does the energy go?

2. Relevant equations

$Q=C\phi$
$U=(1/2)C\phi^2$

3. The attempt at a solution

Please refer to the image below.

Last edited: Feb 15, 2015
2. Feb 15, 2015

### Staff: Mentor

You heat the wires and the capacitors from the resistances in the setup. A tiny bit is also lost to electromagnetic radiation.

3. Feb 15, 2015

### Tekk

I think my approach to find energy assumes there is no resistance on wires. So I suspect that energy would not turn to heat on these wires.

What I see during the discharge process is a flow of electrons. Can you explain how electromagnetic radiation is produced?

4. Feb 15, 2015

### Staff: Mentor

Your approach is independent of the resistance.
Connecting capacitors like that without any resistors is a very special case, it will give you an extremely good oscillator (as you always have a small inductance) and all the lost energy will be emitted as radiation over time.
You have a variable current flow = accelerated charges. With small resistances, current will oscillate back and forth for a while.