Re: There's work, and then there's work

[Bernard Cleyet <anngeorg@PACBELL.NET>]

Can one insert a PM into a super conducting coil to obtain the energy to
charge the cap.?   The plates of the cap. must also be super conducting and
the dielectric should be a vacuum.   The sum of the radiative and I^2*R
dissipations is constant?

bc

p.s. I suspect there will be losses associated with inserting the PM.  One
one has charged the cap. it discharges through the coil and ejects the PM?

Hugh Haskell wrote:

> At 10:28 -0800 2/3/03, David Rutherford wrote:
> > I don't know much about superconductivity, but shouldn't you be able
> > charge the capacitor in the absence of resistance (superconducting
> > circuit). Then, since the energy loss doesn't depend on R, you would
> > still have to get 1/2 CV^2 for the energy stored on the capacitor. But
> > where does the other 1/2 CV^2 energy go, in this case? Or maybe the
> > energy stored on the capacitor in the first place is actually CV^2, not
> > 1/2 CV^2, even in the presence of nonzero R.
> Even if you have superconducting wires and no resistor in the
> circuit, we still, as far as I know, don't have superconducting
> batteries. Every battery has an internal resistance, and in the
> absence of resistance anywhere else in the circuit, the (CV^2)/2 that
> is lost to the resistor, will be lost to the resistor in the battery.
>
> Sorry, there ain't no free lunch.
>
> Hugh
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>
> Hugh Haskell
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