MY PATENTED DEVICE DOES NOT WORK!

[Void]

Indeed resistive losses, no matter how small the resistance between the caps.
See this document clearly illustrating the effect.

Very nice and clear explanations. Thanks poynt99.
All the best...

[Paul-R]

My patent:
https://patentscope.wipo.int/search/en/detail.jsf?docId=WO2011005103

Small translation issue -

Abstract line 1. "comprehending" should be "comprising"

There may be more.

I find this site better:
http://worldwide.espacenet.com/publicationDetails/biblio?DB=EPODOC&II=0&ND=3&adjacent=true&locale=en_EP&FT=D&date=20110113&CC=WO&NR=2011005103A1&KC=A1

but I cannot see what the point of the device is.

[sm0ky2]

The truth is far simpler.  It is almost all just resistive loss.  Charge any capacitor from a voltage source through a resistor and the resistor dissipates the same amount of energy as ends up stored in the capacitor.  The resistor value just contributes to how long the charging process takes.

Hooking two capacitors together does radiate a little bit.  The lion's share of energy loss goes directly to heat dissipated in the wiring and ESR of the two capacitors.

resistance doesn't account for but a fraction of the losses in the system.
we're talking about a small conducting wire, which can be substituted with a superconductor @ 0 Ohms.
E = 1/2 still
This effect was explained in several physics papers 25-30 yrs ago. College professors still like to use it to get their students to think.

If you look at both the kinetic and potential energies of the system, everything is accounted for.
We are only using 1/2 of the potential energy.

[sm0ky2]

@ Tinman,

Nice example with the buckets.

And yes, the Total Charge in the system remains about the same.
This is true. Q = CV, charge = capacitance x voltage

What changes, is the Total Energy.
The equation that defines energy of a capacitor: E = CV^2/2 = QV/2 = Q^2/(2C)
divide the charge in half, and you divide the energy by 4.
there's 2 of them, so total energy is 1/2.

It takes an increasing amount of energy to put more charge into a capacitor.
During discharge, this comes out as momentum, velocity, kinetic energy of a moving charge.

[tinman]

resistance doesn't account for but a fraction of the losses in the system.
we're talking about a small conducting wire, which can be substituted with a superconductor @ 0 Ohms.
E = 1/2 still
This effect was explained in several physics papers 25-30 yrs ago. College professors still like to use it to get their students to think.

If you look at both the kinetic and potential energies of the system, everything is accounted for.
We are only using 1/2 of the potential energy.

That is correct,the other half is still in the cap.