🚀 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.
>
>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 ?
>----------

Not a paradox at all.  If you put a resistor (pick _any_ value) between
the 10-volt-charged and the uncharged caps, it's easy to figure out
that you loose half the energy in the resistor.  That assumes "ideal"
components.  If you put an ideal inductor between the two (instead of
a resistor), you end up with a resonant circuit that will "ring" forever.
In practice, you will have both inductance and resistance, and if you
didn't go out of your way to put in inductance, the circuit will most
likely be overdamped and get rid of its energy in a non-oscillatory way.

That's a fairly simplistic view.  In a practical sense, some of the
energy will be radiated as electromagnetic waves.  If you do the
experiment, you would probably be able to hear a "click" on a nearby
properly setup AM radio.

There's a practical class of circuit that's closely related to this
problem.  Consider a circuit wherein a capacitor is discharged
quickly into a load (like in a flashlamp/strobe, or a pulsed
magnetron).  You want to efficiently recharge the capacitor to get
ready for the next pulse.  You know if you charge it through a
resistance that power will be lost in that resistance -- total
energy equal to the energy change in the capacitor, unless you do
something special.  So -- you put an inductor in series with the
charging supply, and a diode so the current can't come back through
the inductor.  And - viola - you get a much reduced energy loss and
a voltage step-up to boot.  (This circuit is deceptively simple; it
makes a good interview problem, because it tests the interviewee's
knowledge of some very basic components in a way he likely hasn't
thought of them.)  This is also closely related to many switching
power supply circuits.