Bozidar Lisac

[TinselKoala]

Energy (Joules) on a capacitor is given by E = (CV²)/2.

Charge two 1 F capacitors in parallel to 1 V.  Energy contained in the system is E = (2*1²)/2 = 1 Joule.

Now connect the two caps in series. Voltage of the system is now 2 V, but capacitance has decreased to 0.5 F.
So energy contained in the system is E = (0.5*2²)/2 = 1 Joule.

So no cheezburger today, sorry.   

[citfta]

What you said TK is true but consider this. The 1 Joule of energy would have gone to ground if not used to charge the caps.  So we now have 1 Joule of energy that can be returned back to the source.  Well not quite because the 2 caps in series can only discharge to close to the same level as the battery voltage so the caps never really return all that was put into them.  But by returning almost half of the energy put into them they do make the system much more efficient.  We are powering a load and charging some caps at the same time with the current leaving the load.  Then depending on the circuit we can power the load again as we send half of the cap energy back to the source or just send it directly back to the source.  Anyone that understands the charge and discharge actions of caps will see there is no extra energy here, just a simple way to reuse some of the energy and make a more efficient circuit.

Carroll

[Low-Q]

What you said TK is true but consider this. The 1 Joule of energy would have gone to ground if not used to charge the caps.  So we now have 1 Joule of energy that can be returned back to the source.  Well not quite because the 2 caps in series can only discharge to close to the same level as the battery voltage so the caps never really return all that was put into them.  But by returning almost half of the energy put into them they do make the system much more efficient.  We are powering a load and charging some caps at the same time with the current leaving the load.  Then depending on the circuit we can power the load again as we send half of the cap energy back to the source or just send it directly back to the source.  Anyone that understands the charge and discharge actions of caps will see there is no extra energy here, just a simple way to reuse some of the energy and make a more efficient circuit.

Carroll

The capacitors can't discharge 100%, but not 100% charged either. The energy you put in is the excact energy you can get out - which make sense.