Capacitors and the Missing Energy

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spsnyder

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I have a question that is stumping me...

Imagine you have a capacitor charged to V that is initially sitting all alone. The Energy of the system is 1/2CV^2. Now if you connect an identical capacitor in parallel, the voltage across each capacitor becomes V/2. The total energy of the system is now 2*(1/2 * C * (V/2)^2) = 1/4CV^2. Where did the other 1/4CV^2 go? My thoughts are that the energy went to heat through I^2R losses in the conductor to the second cap. The resistance in the wire is very low. Therefore the initial current is extremely high and falls due to the RC circuit (time constant). Am I on the right path?
 
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jim dungar

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Re: Capacitors and the Missing Energy

If the capacitors are in parallel then there is no voltage divider circuit, so each will still see the full voltage.
 
charlie b

charlie b

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Re: Capacitors and the Missing Energy

I am "shooting from the hip" on this answer, and I may be all wrong. But here goes:

The two capacitors are in parallel, so they will see the same voltage. But that voltage need not be the same as it was when the charge appeared on only one capacitor. You will note that the capacitance of two identical capacitors in parallel is twice that of a single capacitor alone. Let us use "Vi" as the "initial voltage" (one capacitor), and use "Vf" as the "final voltage" (two capacitors in parallel).

The energy stored by the system cannot have changed. Therefore, since you had a stored energy of (1/2) C (Vi)^2 at the beginning of the experiment, then you will have a stored energy of (1/2) (2C) (Vf)^2 at the end.

</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">(1/2) (2C) (Vf)^2 = (1/2) C (Vi)^2</font>
<font size="2" face="Verdana, Helvetica, sans-serif">Dividing both sides by (1/2) C,
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">(2) (Vf)^2 = (Vi)^2</font>
<font size="2" face="Verdana, Helvetica, sans-serif">Solving for Vf,
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">(Vf)^2 = (1/2) (Vi)^2</font>
<font size="2" face="Verdana, Helvetica, sans-serif"></font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">(Vf) = sqrt (1/2) (Vi)</font>
<font size="2" face="Verdana, Helvetica, sans-serif">or as an approximation,
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">(Vf) = .707 (Vi)</font>
<font size="2" face="Verdana, Helvetica, sans-serif">So current will flow from the first capacitor to the second, until the voltage on the two equalizes at about 70.7% of the initial voltage.
 
charlie b

charlie b

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Re: Capacitors and the Missing Energy

I've thought through my previous answer and have concluded that it must be right. What occurred to me is that the current flowing into the second capacitor from the first would be governed by the formula I = C (dV/dt). That is, current is proportional to the rate at which voltage is changing. Therefore, the voltage in the first capacitor must be changing, for there to be any current flow. And we know that current is flowing (until charge becomes equalized), since the two capacitors will have an equal stored charge when everything settles. From this, it becomes clear that the "final voltage" cannot be the same as the "initial voltage."
 
steve66

steve66

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Re: Capacitors and the Missing Energy

Posted by Charlie:


The energy stored by the system cannot have changed.
Click to expand...
That was my first thought also, and everything else you did looks right Charlie. But I see one other problem. The equation Q=CV says if the charge stays the same, and the capacitance is doubled, the voltage has to be halved.

Either some energy left, or some of the charge left, or I am missing something.

Steve
 
Mike03a3

Mike03a3

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Re: Capacitors and the Missing Energy

Or you could consider that E=(1/2)QV

Since the energy is constant, the charge must vary as the voltage changes.
 
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spsnyder

Senior Member
Re: Capacitors and the Missing Energy

Steve, you're correct that the voltage is halved because of Q=CV. If the Cap is doubled the voltage has to be halved as we have no Q available to add and it has nowhere to go.

Jim, the charged cap is not connected to a voltage source at any point so the both caps at the end would not be V, but V/2. Do you agree?

Charlie, Wouldn't the Q be split evenly between the two Caps? And therefore Vf = Vi/2?

Thanks.

[ February 24, 2006, 01:42 PM: Message edited by: spsnyder ]
 
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jim dungar

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Re: Capacitors and the Missing Energy

Ooops, my bad.

Only skimmed the OP before answering.
 
charlie b

charlie b

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Re: Capacitors and the Missing Energy

Originally posted by steve66: Either some energy left, or some of the charge left, or I am missing something.
Click to expand...
I knew the voltage could not have remained the same. But I had presumed, as the natural course of nature, that it was energy that had to be conserved. That much is true.

But I further presumed that there would be no energy lost to the system, as the second capacitor was connected. That part is not true. The thing that must be conserved in this system is charge.

The article provided by spsnyder explains where the "lost energy" goes. It is converted into heat in the arc that takes place as the switch is closed.
Originally posted by spsnyder: Charlie, Wouldn't the Q be split evenly between the two Caps? And therefore Vf = Vi/2?
Click to expand...
Yes, that is correct.
Originally posted by charlie b: I am "shooting from the hip" on this answer, and I may be all wrong.
Click to expand...
At least I got that statement right. :D
 
steve66

steve66

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Re: Capacitors and the Missing Energy

I thought for sure you would have something to say about the "harmonics" topic, Rattus.

As for the energy being lost in the arc or the wire resistance, I'm not quite sold. Switches usually arc when they open a circuit, not when they close a circuit. Anyhow, we should be able to imagine a perfect switch, and zero resistance between the capacitors. Then we should still be able to say where the energy went.

Textbooks always said the energy in a capacitor is stored in the dielectric. The charge on the plates distorts the orbits of the electrons in the dielectric. I'm thinking mabye the energy is lost in the dielectric, maybe as heat or something like that.

Steve
 
steve66

steve66

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Re: Capacitors and the Missing Energy

Here is additional food for thought on this topic:

What happens if we do put a resistor between the capactiors? (i.e. connnect the caps with a resistor instead of a wire). Do we get less than 1/2 the voltage after everything has settled? The same charge still has to be present, right? But don't we have to loose more energy? If so, then it seems like the voltage should be less.

If someone can show that the final voltage and final energy is the same if we connect the caps with a 1000 ohm resistor or a short, then I might be convinced the energy is lost in the connection between the caps.

Steve

[ February 24, 2006, 06:23 PM: Message edited by: steve66 ]
 
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rattus

Senior Member
Re: Capacitors and the Missing Energy

There is no question that charge is conserved, therefore voltage is halved. Then, assuming a resistor connecting the two caps, it is easily shown that half the energy is lost in the charge transfer. This fact is independent of the value of the resistor. Just take the limit of the expression as R approaches zero to show this for the limiting case.

Completion of just about any circuit causes an arc--caps included. So some energy will be lost in the arc and some in radiation.
 
R

rattus

Senior Member
Re: Capacitors and the Missing Energy

Gents, it is like this:

Assume that the initial voltages on the caps are 2Vo on C1 and 0 on C2.

When we connect the two identical caps with a resistor of any value, the midpoint of the resistor acquires a voltage of Vo.

Since the two caps charge and discharge at the same rate, this voltage stays at Vo--forever in the ideal case. We can treat this node as a battery of voltage Vo.

Now we have a two simple RC problems where the current equations may be determined by inspection.
e.g.,

Ic2 = (Vo/R)*e^-t/RC = -Ic1

We know that charging and discarging a cap through a resistor from a voltage source causes half the energy to be dissipated in the resistor. If you don't believe that, you can work it out with some simple differential equations.

[ February 26, 2006, 12:02 AM: Message edited by: rattus ]
 
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