Re: [Phys-L] charging a capacitor reversibly (or nearly so)

[brian whatcott <betwys1@sbcglobal.net>]

On 8/6/2014 10:38 AM, John Denker wrote:
> On 08/06/2014 07:36 AM, Bob Sciamanda wrote:
>
> > Perhaps this is a correctly simulated instance of "Mandated Energy
> > Dissipation".   EG., when a battery charges a capacitor, 1/2 of the
> > battery-produced energy must be dissipated.
> More-or-less everybody assumes that, but it's not necessarily
> true!
>
> A whole lot of very smart people have gotten that wrong
> over the years.
>
> If we are clever, we can dissipate a whole lot less than
> half of the energy.  Proof by construction:
>
> A battery, by definition, is a collection of /cells/.
> Assume all N cells are in series.  Assume we have access
> to the individual cells.  Let the cell voltage be u (small
> u) while the battery voltage is V (big V) where V = N u.
>
> Charge the capacitor step by step.  Start by hooking it
> up to just one cell.  This moves a charge (C u) across an
> average voltage drop of u/2, so it dissipates energy on
> the order of (C u^2/2).  This is very small, since u is
> small.  /snip/

A few other people of indeterminate smartitude suggest that in real 
world terms where cells all show some internal resistance, the charging 
voltage waveform is exponential so that the  mean voltage across the 
capacitor would be 0.7 u     :-)

Brian Whatcott     Altus OK