Moderators love me. PLEASE, DON'T CLOSE THIS TOPIC BEFORE I OBTAIN SOME ANSWERS! My last post was closed. It's something i can't understand, i'm censored, and if you want to discuss with me, you're censored also. Alexander, you help me a lot. I see you really understand about all physics phenomena, you have physics quite clear in your mind, so i'm glad to see you take the time to answer me. You said: "There is complete balance of energy in both processes (charging 2C capacitor and discharging C/2 capacitor back to the battery)..." "Parallel transfer of energy: CV^2 (battery) = CV^2/4 (resistance) + 3CV^2/4 (capacitors)" (the 3CV^2/4 on capacitors add to the initial CV^2/4 pre-charge, then capacitors are charged at CV^2) "Series transfer of energy:CV^2 (capacitors) = CV^2/4 (resistance) + CV^2/2 (battery)" "So, no net gain/loss of any energy is detected". You see that the battery gaves CV^2 energy on parallel proccess and takes CV^2/2 on series proccess, But what this means?. The charges on the battery are the same at starting and ending points of the proccess. Will the battery discharge by doing less energy on it? Energy is related to V and I, but finally all this terms are related to charge. Voltage is related to a concentration of charges from one spatial point to another, and current is related with how fast this charges move from one point to another in time. In the parallel proccess, Q charges of the battery goes to the capacitors, because there is a related V from the battery to the capacitors, and the speed of the charges "diffusion" proccess gives the I, so the energy is CV^2. In series proccess, the V is the same than before (from the capacitors to the battery), but the "diffusion" proccess of the Q charges to the battery takes another time or speed to happen, thus giving a different current, so energy is less than before on the battery: CV^2/2. As you said, there is no gain/loss of energy, but the resistor disipated CV^2/4 on both proccess, and also you've got the same Q charges that run out the battery coming in again. Energy on battery is different, but the charges that "produced" this energy are the same as before. Then the battery doesn't discharge, so you can play with the same charges for a long time doing this cycle without waste their concentration (their V), and then creating energy by the charges transfers without finally spending the source of charges. If you have a concentration of charges, and you move them to another place at certain speed, you've got some energy. Now, if you can move them back to the first place without loosing their concentration at another speed, making no work to force them, you obtain another energy, but you're also able to repeat the cycle the times you want, obtaining energy without loose the conditions that lets the energy appear. There is no physical regret to use the same charges once and again to obtain energy, you only need a method to maintain the charges ordered into a certain V at initial and final steps, and the current is given by the proccess you do.