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Re: 50%



> >>Subject: Re: 50%
> >
> >   [ snip ]
> >
> >   Well, indeed, thanks to Richard and Malcolm and others commenting
> >   on this thread, I believe I have come to a new realization about
> >   the *resistive* or *capacitive* discharge of capacitors.  Funny I
> >   can't remember this from college.  We must have covered it....
> 
> Fred - 
> 
> Energy is conserved. Mistakes in capacitor circuit theory and algebra are
> the problem. That is why you did not remember it from college. The voltage
> across the two capacitors after the reconnection is not V/2. It is V/sqrt2.
> This can be easily checked by test and correct algebra. 
> 
> The voltage will be a little less than V/sqrt2 because there is some spark
> loss in the reconnection. In the test that I made the voltage across the two
> capacitors was slightly less than V/sqrt2 because of losses but was still
> much more than V/2. Using V/sqrt2 in the equation will give a total of 100%
> energy (50% in each) for the two capacitors if the algebra is done correctly.
> 
> Jack C.

Hold on a minute here, Jack! You really threw me on this one.
If potential energy were to be conserved, I would agree that the
end-point voltage *should be* V/sqrt(2). But I don't see it
that way! Can we validly assume that *charge* is conserved?
If so, then go back to the capicitor equation  Q = C * V.
Doubling the capacitance in a circuit, without adding or
removing charge - just allowing it to "level out" across
the two caps - WILL halve the voltage.

I maintain my stance that it an entropy effect - you are
redistributing potential energy across large "vessels",
at lesser potential. Energy is NOT dissipated, nor destroyed,
but is made less useable to perform work.

I guess it's time to go to the bench and TRY it...

Dave