Re: Energy loss in a capacitor
- From: ianparker2@xxxxxxxxx
- Date: Sun, 15 Oct 2006 01:45:11 +0000 (UTC)
jumanicus wrote:
This is a typical problem in undergraduate physics: When a charged
capacitor in connected to an equal uncharged capacitor through a
resistor, the final energy in the capacitors is less than the initial
energy. This loss in energy is dissipated in the resistor as heat
(Joule Heating). Now if the capacitors are connected with
resistanceless wires (superconducting, for example), then the deficite
in energy has to be explained as energy radiated as electromagnetic
waves (due rapid change in current).
My question is, why do we consider the deficite energy to be dissipated
only as heat when a resistor is connected? Because even in this case,
due to non-linear variation of current, some energy is bound to get
radiated away. Is this because its amount is negligible? Do the laws of
Electromagnetism entail preference for Joule Heating over radiation?
Can anyone shed light on this ??
It is a matter of frequency. You will recall that the early experiments
of Hertz on radio waves did just that. Radio waves were generated by a
tuned circuit activated by a spark. This seems quite similar to what
you are saying. Your superconducting wire has INDUCTANCE and will form
a tuned circuit with the capacitor. Of course all the energy will
eventually be dissipated as radio waves. If you have a coil (not
necessarily superconducting) you will again get radio waves - as Hertz
did.
Frequency and size of attenae gives the coupling to free space.
.
- References:
- Energy loss in a capacitor
- From: jumanicus
- Energy loss in a capacitor
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