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Capacitor charge transfer

-August 18, 2012

The oldie-but-goodie question is posed as follows. (The numbers are of course, arbitrary.)

I have two 1 µF capacitors, one of which is charged to 10 volts and the other of which is at zero volts. The stored charge in the first capacitor is Q = C*V = 1X10-6 farad * 10 volts = 1X10-5 coulombs. The energy in that one capacitor is ½*C*V² = 0.5 * 1X10-6 * 10² = 5X10-5 joules = 50 µ-joules. The second capacitor has no charge and therefore has zero stored energy. This means that the two capacitors taken together have that 50 µ-joules of stored energy.

Now we connect the two capacitors in parallel with each other to form a 2 µF capacitor. That stored charge will redistribute itself so that now, since Q=C*V, we may write V = Q/C = 1X10-5 coulombs / 2X10-6 farads which comes to 5 volts. The stored energy in each capacitor is again ½*C*V² which now comes to 0.5 * 1X10-6 * 5² = 1.25X10-5 joules each for a sum of 2.5X10-5 joules = 25 µ-joules.

Hey! What happened to the stored energy? It's only half of what we had started out with. Why???

When I was young and impressionable, someone gave me an answer that the missing energy had been taken away by "radiation". I was more inclined to think that elves might have had something to do with it.

With age however, I gained insight as to what had happened. Consider these two analytic approaches to the issue:

 

The energy loss is that energy consumed by the resistor during charge transfer. If you wait long enough for everything to settle out, the energy loss does not vary versus the resistor's ohmic value. Only the timing changes. Even if you let the ohms go to zero so that the energy transfer time goes to zero too, the total energy loss itself is determined only by the capacitor values.

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