🚀 go-pugleaf

vaso@mips.COM (Vaso Bovan) writes:

>A Paradox of Capacitor Energy Storage

>I've heard several competing answers to this paradox. None is entirely
>satisfactory:

>Consider an ideal 2uF (for computational ease) capacitor charged by a 10 volt 
>source. Eventually, the energy stored is (1/2)*CV^2=100 joules.
                                          ^^^^^^^^^^^^^^^^^^^^^
You forgot that the 'u' in 2 uF stands for 'micro', that is 2*10^-6 F.
So the energy stored is 100 uJ (100*10^-6 J). But this does not affect 
your 'Paradox of Capacitor Energy Storage'.

>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 ?

The current in a capacitor is defined as: I = - dV/dt * C (that
is: change in voltage pr. sec. times the capacity, forget the sign). 

This means that you cannot change the voltage across a capacitor
instantaneously : this would inply infinite current.

It also means that when you short your two capacitors across each other,
then the current will rise to an level where internal resistance and wire
resistance is not neglectable.

When this happens then you loose energy in form of heat.

Also: when you start shorting the capacitors, you will get a spark -
this is also a waste of enery.

If you use an inductor when connecting the capacitors then you 
would not loose energy (ok, ok a little if the components is not ideal).

But then the resulting voltage would not be 5 V - it would be 7.07 V 
(for energy preservation).

By the way: I'm not so sure that the resulting voltage would be 5 V
in your example.
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