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Re: 50%
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Hi Malcolm!
> Wow! I just have to respond to this post on the 2 cap problem.....
>
> > 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.
>
> So we now have Q = 2C x V/sqrt2 when we started with Q = CV. I think
> your result begs an explanation. If no energy is expended in
> connecting the two caps, how did the extra charge separation come
> about?
> I have just this very minute done this experiment with two high
> quality capacitors and the voltage comes to V/2 near as.
>
> Malcolm
Well Malcolm can always be relied on to do a good experiment. The two
capacitor problem is a classic physics "paradox". Of course, it isn't
reallly a paradox at all if you are careful in thinking about the problem.
Since charge is not destroyed ( this only happens when electrons meet
positrons) then this is the only conserved quantity we know of for sure in
this problem. I think Ed Phillips already addressed thhis problem by the
way.
So starting with two caps of capacticance C , one charged to initial votage
Vi and total initial charge Qi=CVi, and the other with zero charge or
voltage. When the two are connected in parallel, the voltage must
equilibrate after some time which depends on the resistance. So the final
voltages for each capacitors is Vf and therefore the final charges on each
of must be equal and are Qf=CVf. Now since the two final charges must equal
the total initial charge Qf+Qf=Qi, we get Qf=0.5Qi and
Vf=Qf/C=0.5Qi/C=0.5Vi.
The initial total energy is Ei=0.5CVi*Vi
The final energy is then (split between two caps):
Ef=0.5CVf*Vf+0.5CVf*Vf=CVf*Vf
then substituting Vf=0.5Vi we get Ef=0.25CVi*Vi
Ei=0.5CVi*Vi
Ef=0.25CVi*Vi=0.5Ei !!!
Half of the energy is lost. Physcally, energy is dissipated in the
connecting wires and the in the spark which occurs when you connect the two
caps together. There is really no way around it in this situation. In fact,
you can ( as Ed said) do the problem using differnetial equations and it
just doesn't matter how big the dissipating resistance is, the energy lost
is always half of the original energy.
See yall later,
-Ed