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Re: non-dissipative circuitry



Sorry. I conflated the current discussion (charging a cap.) with
discharging one slowly or rapidly (through a large value or zero
resistance).

The answer must be yes to conserve energy, yes? As you answer,
indirectly, in part B.

I haven't followed closely the latter part of this thread, but ...
would connecting a cap to a chemical cell at low temp. and raising
veerrry slowly the temp. be a method of charging with low dissipation?

bc who heard liquid air was used to disarm Jerry UXBs during the blitz.

"John S. Denker" wrote:

Bernard Cleyet wrote:

Is this confirmation or .... of my claim that the total energy
dissipation (radiation + I^2*R) is constant.

If the answer concerns constancy of the dissipation,
we need to be rather explicit about what the
question is.

A) If the question is what's the minimum dissipation
for bicharging a capacitor C to a voltage V, then
the answer is _not_ .5 C V^2 or any other constant.

The dissipation could be very small, using methods
discussed in previous notes.

B) If you really want the answer to be .5 C V^2,
then you can cobble up a suitable question,
perhaps along the lines
-- voltage source V, switch, resistor R, capacitor C
-- V = constant
-- no other circuit elements
-- no variable capacitances
-- et cetera.
and then you can make it possible to prove, using
conservation of charge, conservation of energy, etc.,
that there will be dissipation of .5 C V^2, independent
of R.

It is a question of taste whether you choose to
restrict the question as in (B). Choose what
you like, but remember that others may choose
differently. Remember that it is a choice, not
a law of nature.

I've had the experience of being yelled at by
an entire roomful of "experts" who "knew" that
.5 C V^2 dissipation was a law of nature. And
they had a stack of textbooks to "prove" it.
... But all their yelling didn't change the
facts. There really are low-dissipation ways of
charging a capacitor.