Where is it gone ?

[Dick Rucker]

on 11/8/02 13:48, Andre Kesteloot at andre.kesteloot@verizon.net wrote:

> Assume you have two identical, loss-free 1uF capacitors, one charged
> to 10V, the other is discharged.
> 
> You then connect the two in parallel so the total C is now 2uF -
> charge will flow from the charged capacitor into the discharged
> capacitor.
> 
> Assuming charge Q is conserved, and that Q=CV, the voltage must now be
> 5V.
> 
> But the stored energy in a capacitor =1/2CV^2, so with the single
> charged capacitor, the stored energy is 1/2 x 1uF x 100 = 50uJ, while
> with both capacitors in parallel it is only 1/2 x 2uF x 25 = 25uJ.
> So where has the other 25uJ gone?
> 
> André
> (from the RSGB LF Reflector)

After some head-scratching, here's my analysis:

Initially, C1 has a charge on it of 1uF*10v or 10 Coulombs that represents a
stored potential energy of 1/2*C*V^2 = 1/2*1*100 = 50 uJoules.

When placed in parallel with C2, there is an immediate flow of charge that
reduces the voltage differential between the two Cs to 0.  Initially,
however, the charging voltage seen across C2 is the full 10 volts on C1,
assuming that resistive losses are negligible.  So the average charging
voltage seen by C2 over some very brief time interval is 1/2 [ 10 - 0 ] = 5
volts.

The potential energy stored on C2 at the end of the charging interval is the
charge on C2 times the average charging voltage needed to put it there.  The
charge that C2 ends up with is 1 uF*5 volts since charging ceases when V2
rises to meet the falling V1 in the middle of the range, or 5 volts.  So the
potential energy stored on C2 = 5 coulombs * 5 volts = 25 uJoules.

Since C1 started out with a stored potential energy of 50 uJoules, there
must be another 25 uJoules of energy left on C1 after the charging of C2 has
finished.

Does that work for you?

Dick Rucker, KM4ML