Energy Loss

[BWRX]

You're assuming the voltage will become halved by connecting an equal capacitor in parallel with the first. What you must assume is that the energy will be shared between the caps. From that you can determine what the voltage across the caps will be.

First situation with one cap charged:
U1=0.5*C*V1*V1

Second situation where an uncharged capacitor of equal value is added in parallel to the first:
U2=0.5*2*C*V2*V2

Assuming U1=U2 (conservation of energy), it follows that V2=V1*sqrt(0.5)

So if you had 2V across a single cap and added another uncharged cap of equal value in parallel the new voltage across those caps would be 1.414V.

[poobah]

Sorry Brian,

V = Q / C

[BWRX]

Yes, and your point?

Let me rewrite the equation.

0.5*C*(Q1/C)^2 = 0.5*(2C)*(Q2/(2C))^2

What does that boil down to?

[poobah]

Well... if the charge is halved... will not the voltage be halved?

Conservation of charge of charge and energy must both hold.

[BWRX]

Here's what it boils down to.

(Q1^2)/C = (Q2^2)/(2*C)

which can be simplified to

Q2 = Q1*sqrt(2)

[BWRX]

So now we have the following equations:

Q2 = Q1*sqrt(2)

and

V2=V1*sqrt(0.5)

C=Q1/V1, right?

from the equations 2C = Q2/V2 = (Q1*sqrt(2))/(V1*sqrt(0.5)) = 2*(Q1/V1)

Is that not right, or am I missing something?

[[Ro]Sharky]

The energy is lost moving the electrical charge from the source (the charged cap.) to destination (the fully discharged cap)

[BWRX]

How is that possible if the capacitors are ideal? In reality you are correct, some energy will be lost to resistances and some to heat and light because there will be a nice spark if you connect a discharged cap across a charged cap

[poobah]

Hey BWRX,

No one is screwing up the math here... what has to screwed up is the assumptions being made.

Consider this: a lake and a dock at exactly the same height. Pull a bucket of water from the lake and set it on the dock. Now... we have 1 bucket-mass (B) moved an average distance of 1/2 the bucket's height (H)... ie work or potential energy... B x 1/2 H amount. Now... do it again... twice the energy (this is the final state in the cap quandry... we have 2B x 1/2H. Then... put one bucket atop the other the other. The end result is we have 2B x H (average height of mass) worth of potential energy.

Now it is easy to see why we had the add energy to put the 2nd bucket on top... and how the total energy doubled when we did so. Now return the top bucket to the dock... it gives up energy when doing so.

So in the cap world we lost half the energy... makes perfect sense. Nothing can be violated here... we just aren't seeing where the energy in our cap went.

[sawreyrw]

I don't think anyone has this exactly correct. It can be shown that the charge shared between the 2 capacitors is conserved, but energy is lost because the circuit has some finite resistance. Assume initially C1 is charged to Uinit and C2 is charged to 0. After the switch connects the two caps in parallel, the voltage across them is Ufinal.

Before the switch is closed, Q1=C1*Uinit.
After the switch is closed, Qfinal=(C1+C2)*Ufinal.

From these equations: Ufinal=Uinit*C1/(C1+C2).

The initial energy is Einit=.5*C1*Uinit^2
The final energy is Efinal=.5*(C1+C2)*Ufinal^2

The energy lost is Elost=.5*C1*Uinit^2-.5(C1+C2)*Ufinal^2.

The energy is lost in the resistance; although low it will never be zero.