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Re: non-dissipative circuitry



For me the main point is that there are several channels through which the
dissipation of energy can occur. I^2 r losses only being only one of them.
There are many books and lab manuals that give the impression that the
dissipation is always I^2 r in nature and omit any discussion of radiative
(sparks?) losses. Hence, the discusstion of superconducter wiring in order
to bring to the fore the radiative channel.

Joel R.

-----Original Message-----
From: Chuck Britton [mailto:britton@NCSSM.EDU]
Sent: Tuesday, February 04, 2003 8:06 AM
To: PHYS-L@lists.nau.edu
Subject: Re: non-dissipative circuitry


At 12:07 AM -0500 2/4/03, John S. Denker wrote:

In the context of non-dissipative switching,
with a nonzero voltage across the switch,
Ludwik Kowalski wrote:

Since we are "gedankening" let me observe that there should be no
spaaaakking in a "perfect vacuuuuum," i.e. in space without atoms.

I don't think you can avoid dissipation just
by switching in a vacuum.

Vacuum, shamacuum. spaaaks, sparks

It don't make no nevermind.
My gedanken exp was to connect two capacitors that are charged to
different voltages.

Dissipation is of COURSE a factor - 'cause that is 'where' the lost
electrostatic energy 'goes'.

It seems to ME that the METHOD of connecting the capacitors is
absolutely NO effect on the final state of the two capacitors.

Connect them with and inductor, connect them with a resistor. Once
things calm down, the charge will have redistributed itself in a
totally predictable way.
Energy has been 'dissipated'. Exactly the SAME amount of energy,
regardless of the connection method. nicht wahr???

(Unless John's spaaaaking is such that it carries net charge from the
circuit on the molten drops of metal)