Capacitors, charge and energy: a question

[Skywave]

I know I'm missing something fundamental here, but I can't quite put my (metaphorical) finger on it. Any clues would be welcome.

The charge, Q, on a capacitor of capacitance C1, when a voltage, V1, is applied to it is given by: Q = C1.V1
Consequently, if that capacitor is now connected to another, initially discharged capacitor of capacitance C2, the voltage, V2, appearing across its terminals is determined by the relation: C1.V1 = C2.V2, since all the charge is transferred. (This is the ideal case: assume resistive losses are zero).
Hence, V2/V1 = C1/C2 . . . . equation (i)
so V2 = V1.(C1/C2)

The energy stored in C1 = ½.C1.(V1)²;
the energy stored in C2 = ½.C2.(V2)².
Since zero energy is lost in the transfer, that means that
C1.(V1)² = C2.(V2)², so: C1/C2 = (V2/V1)², which is different to equation (i).

So, where is the error in my thinking, guys?

Cheers,
Al. / Skywave.

[Nick]

Basically, you need to think of this in terms of conservation of energy (Joules)- you are mixing charge in Coulombs and energy in Joules and assume all the energy is transferred to the second capacitor (it isn't). The final energy will be the same as the initial stored energy in C1, i.e. 0.5 * C1 * sqr(V1) (#1) where V1 is the initial voltage.

When you connect C2 in parallel with C1, some energy is transferred to C2, and consequently, as the total energy (assuming a lossless system) remains constant, the voltage will be V2 across both capacitors, so:

0.5 * C1 * sqr(V1) = 0.5 * (C1 + C2) * sqr(V2) (#2)

Dropping the 0.5 and dividing both sides by (C1 + C2) and sqr(V1) we get

1 + C1/C2 = sqr(V2/V1), i.e. V2/V1 = sqroot(1 + C1/C2) (#3)

So, the ratio of the new voltage, V2, to the original voltage, V1, is given by the above equation (#3) and the energy stored remains constant.

For safety, we can now work backwards to check our result...

Using equation #3, V2 = V1*sqroot(1+C1/C2)

Substituting in #2 and using #1, 0.5 * (C1 + C2) * sqr(V1*sqroot(1 + C1/C2)) = 0.5 * C1 * sqr(V1)

...which if you cancel through you will find 1 = 1 !! i.e. energy gets conserved.

HTH

[Skywave]

Thanks Nick.
In other words, I made the following oversight.
When I connected the first, charged capacitor to the second, initially discharged capacitor, my argument was that all the energy from the first capacitor would be transferred to the second one, with zero stored energy remaining in the first capacitor. And that, of course, is simply and obviously incorrect: the energy is shared between the two capacitors - and the mathematics needs to be 'adjusted' to reflect that - as you have done.
I just knew I was missing something totally obvious: 'woods and trees' spring to mind.

Al. / March 27. //

[Kalee20]

Sorry folks, that's not right!

If you have a charged capacitor, and you connect it in parallel with a discharged capacitor, charge will balance, and energy HAS to be removed. You can't do it 'losslessly' and end up with a new equilibrium.

Charging 1µF to 10V, you have Q = CV = 1µF x 10V = 10µC of charge. Energy = ½CV² = 50µJ

If you connect this to a discharged 1µF capacitor, some of the electrons will flow around. The first capacitor will discharge, the second will charge. It'll balance when the two capacitors share the same voltage. None of the electrons can 'disappear', so charge conservation applies.

The new total capacitance will be 2µF; the new voltage will be 5V. Charge will be Q = CV = 10µC exactly as before. However, energy = ½CV² = 25µJ so half has been lost!

If you parallel-up the capacitors by a resistor and wait for several time constants, equilibrium will have been nearly achieved, and it's easy to see that heat will be generated (any current flowing through a resistor creates heat). It turns out that the total heat generated, if you wait to infinity, is independent of resistor value so even doing the parallelling with a knife switch will give the same energy loss from the capacitors. It's unavoidable.

If you parallel-up using an inductor ("lossless"), and wait a bit, the same applies - the capacitor voltages will equalise at 5V, the energy in the capacitors will be 25µJ, and you'll have lost some. But in this case, there will be significant current flowing in the inductor and the energy 'lost' from the capacitors will exist in the inductor. If you wait a bit longer, the first capacitor voltage will fall below 5V, the second will rise above 5V, and not long after, you'll find that the first has fallen to 0V, the second now is charged to 10V, there is no current in the inductor, charge has been conserved, and all the energy is back in the capacitors only the other way round! At all times during this oscillation, charge will be conserved (C1V1 + C2V2 = constant), but energy in the capacitors alone won't be constant.

[Nick]

Yep - I can see that - silly of me - the number of electrons constituting the charge is fixed so the total charge not the total energy will be constant.

[Skywave]

If you have a charged capacitor, and you connect it in parallel with a discharged capacitor, charge will balance, and energy HAS to be removed. You can't do it 'losslessly' and end up with a new equilibrium.

Thanks for your contribution Kalee: it is valued.

I understand your arithmetic and I certainly don't want to open up a can of worms here, but can you explain to me where that lost energy has gone and why, bearing in mind that we are talking about an energy transfer here using conductors with zero resistance?

Thank you.

Al . / Skywave.

[pwdrive]

Maybe it can not be answered if we are talking about hyperthetical absolute zero resisitance, In practice there will be some resistance, I guess you already know that, wether that resistance is high or low is of no real consequence as the energy changed to heat over the time periods will be the same, but don't ask me to prove the maths.....

http://www.edaboard.com/thread168490.html

Lawrence.

[Refugee]

In order to calculate the loss you would need to know the resistance of the link placed between the capacitors although this could be made very small indeed.
The most important thing that needs to be taken into account is the ESR of the capacitors.
The ESRs of both would have to be added together and ohms law used to calculate the loss at the moment they are connected. Then you need a couple of screens of mathematics to calculate the change in the rate of loss as the voltages equalise.

[Skywave]

Interesting thoughts, gentlemen, but resistance of conductors and ESRs do not feature in my question, which is the idealised case: zero resistance; perfect capacitors; hence, zero energy losses. And I say "zero energy losses" on account of the Principle of the Conservation of Energy.
My puzzlement remains unabated.

Al.

[pwdrive]

Hi Al, what about the inductance of the capacitors and thus the effect of radiation resistance due to oscillation:

http://puhep1.princeton.edu/~kirkmcd/exa...wocaps.pdf

Lawrence.