🚀 go-pugleaf

< >Consider the capacitor to be isolated from the voltage source, and then
< >directly shorted across an identical (ideal) capacitor. Eventually, the
< >voltage across each capacitor will be 5V. Now, there are two equally
< >charged capacitors, each storing (1/2)*CV^2= 25 joules, for a total of
< >of 50 joules.  What happened to the other 50 joules ?
< 
< 
<	[Much correct stuff deleted for brevity]
<
< 	The point is that the *details* of how the capacitors are
< 	connected together determines what the final energy of the
< 	system will be.  Modelling a series resistance shows that 
< 	half of the original energy is dissipated in the resistor.
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The final energy of the system is independent of the details of the system,
assuming only that electric charge is conserved.  Half of the original
energy must be lost as the system decays to the final state.  The time
dependence of the voltage and current do of course depend critically on
the details of the system.
It is not necessary to assume any ohmic resistance is present.  Even if
all material present is superconducting, radiative losses will cause
energy dissipation, although the time scale may be long.
-- 
Bill Tanenbaum - AT&T Bell Labs - Naperville IL  att!ihlpb!tan