Where Did Half the Energy of Two Capacitors Go?

[Martin]

Two ideal capacitors (i.e., purely capacitance—no resistance or inductance), each with a capacitance of C, are connected together through ideal wires (zero resistance), and an ideal switch (i.e., when open, the switch offers infinite resistance; when closed, it offers zero resistance), which initially (say, for t<0) is open.

A charge Q is placed on one of the capacitors; the other has zero charge. Since the energy stored in an ideal capacitor is given by

Energy = (1/2) (1/capacitance) X (charge)^2

we see that the energy stored in the two capacitors is

E(1st Capacitor, t<0) = (1/2) (1/C) X Q^2

and

E(2nd Capacitor, t<0) = 0.

Hence the total energy stored in the circuit while the switch is open is

E(total, t<0) = (1/2) (1/C) X Q^2.

https://www.physicsforums.com/attachment.php?attachmentid=5476&stc=1&d=1131215321

Suppose (say, at t=0) the switch is then closed. The charge Q will flow from the 1st (charged) capacitor onto the 2nd (uncharged) capacitor until it’s evenly distributed between the two capacitors (because each has the same capacity to store charge). Hence, each will end up storing a charge of Q/2.

The energy stored in the two capacitors after the switch is closed is, therefore,

E(1st Capacitor, t>0) = (1/8) (1/C) X Q^2

and

E(2nd Capacitor, t>0) = (1/8) (1/C) X Q^2.

Hence the total energy stored in the circuit after the switch is closed is

E(total, t>0) = (1/4) (1/C) X Q^2 .

https://www.physicsforums.com/attachment.php?attachmentid=5477&stc=1&d=1131215312

Clearly, half of the initial (before the switch is closed) energy is lost after the switch is closed.


Where did it go?

(The answer is not at all obvious.)

 

[Reality_Patrol]

This is actually quite a practical problem, this loss places limits on the peformance of charge-pumps (DC-DC converters). The answer is entropy, but as far as I know the details of the energy transfer mechanism are unknown.
Don't get me wrong, it clearly produces heat in charge-pump circuits, but whether that heat comes from "transient" conduction losses, or "IR" type radiation between the plates is uncertain. Do you know?

 

[Martin]

We are dealing with an “ideal” circuit (i.e., lumped circuit theory), which means that you must analyze the circuit’s behavior in terms of idealized circuit elements—that is, electrical properties (resistance, capacitance, and inductance) as opposed to actual (physical) electrical devices (resistor, capacitor, and inductor). In this instance, the circuit specifies that the only circuit elements present are capacitances. The property “capacitance” accounts only for electrical energy storage. Nevertheless, if you carefully apply Kirchoff’s Laws, you can arrive at the answer.

 

[SGT]

Your first capacitor has a voltage $$V=\frac{Q}{C}$$, while the second one has a voltage of zero. Connecting both of them in parallel violates Kirchoff's Voltage Law, so you are not allowed to do it.
Of course, with real components, there will be a resistence associated, the charge of the second capacitor will not be instantaneous and there will be energy dissipated as heat in the resistive part of the circuit, so no paradox exists.

 

[Martin]

... Connecting both of them in parallel violates Kirchoff's Voltage Law, so you are not allowed to do it. ... Of course, with real components, ... no paradox exists.

The “violation” is only apparent, and no “paradox” exists.

 

[SGT]

The “violation” is only apparent, and no “paradox” exists.

Is this an opinion, or do you have arguments to support your affirmation?

 

[Martin]

Lumped circuit theory is self consistent: By insisting that Kirchoff’s Laws hold, there will be no violations, and the circuit will be forced to respond in such a manner that avoids any apparent paradox.

 

[SGT]

Lumped circuit theory is self consistent: By insisting that Kirchoff’s Laws hold, there will be no violations, and the circuit will be forced to respond in such a manner that avoids any apparent paradox.

According to KVL, two elements in parallel must have the same voltage. You cannot link together two capacitors with different voltages, because it violates that law. You can find this in any circuits theory book. In Basic circuits Theory by Charles Desoer and Ernest Kuh it is specifically written. My edition is in Portuguese, so I don't know if the page reference would be valid in the American edition, but it is in item 5.2, page 89.

 

[Martin]

According to KVL, two elements in parallel must have the same voltage.

The capacitors are simultaneously in parallel as well as in series.

You cannot link together two capacitors with different voltages, because it violates that law.

By insisting that Kirchoff’s Voltage Law holds, the circuit will respond by forcing the charge to redistribute itself such that the capacitors attain the same voltage.

You can find this in any circuits theory book. In Basic circuits Theory by Charles Desoer and Ernest Kuh it is specifically written. My edition is in Portuguese, so I don't know if the page reference would be valid in the American edition, but it is in item 5.2, page 89.

I am not familiar with that particular textbook. I can assure you, however, that there is nothing inherent in lumped circuit theory that prohibits “link[ing] together two capacitors with different voltages.”

In lumped circuit theory, there are no “disallowed” circuit configurations: As long as you properly apply the voltage/current relationships for each circuit element and enforce Kirchoff’s Laws throughout the circuit, you will arrive at self-consistent results.

 

[SGT]

In lumped circuit theory, there are no “disallowed” circuit configurations: As long as you properly apply the voltage/current relationships for each circuit element and enforce Kirchoff’s Laws throughout the circuit, you will arrive at self-consistent results.

can you explain how lumped circuit theory allows two different branches of a parallel circuit to have different voltages?

 

[Martin]

can you explain how lumped circuit theory allows two different branches of a parallel circuit to have different voltages?

Lumped circuit theory does not allow parallel branches to have different voltages. Quite the contrary, the theory insists—via Kirchoff’s Voltage Law—that they have the same voltage. This is the key to analyzing the above circuit and answering the question that I posed.

 

[Reality_Patrol]

I'm not exactly sure where you're going with all of this. I'd say the problem formulation given in your initial description contained the result of applying circuit laws: there is a loss of stored energy.
So, you've got me, why not just share your answer? I think you might have found something interesting; something mathematically correct but physically inaccurate since heat production is the experimentally found answer. Models are just that, the real thing is what it is. Still, I'd like to see it. Come on, don't be scared.

 

[SGT]

Lumped circuit theory does not allow parallel branches to have different voltages. Quite the contrary, the theory insists—via Kirchoff’s Voltage Law—that they have the same voltage. This is the key to analyzing the above circuit and answering the question that I posed.

Are capacitors allowed to change voltage instantaneously in that theory?

 

[Martin]

I'm not exactly sure where you're going with all of this. I'd say the problem formulation given in your initial description contained the result of applying circuit laws: there is a loss of stored energy.
So, you've got me, why not just share your answer? I think you might have found something interesting; something mathematically correct but physically inaccurate since heat production is the experimentally found answer. Models are just that, the real thing is what it is. Still, I'd like to see it. Come on, don't be scared.

Why not at least attempt to find a solution analytically, the way you would go about determining the solution to any circuit problem? By doing so, you might get some insight into what makes this such an interesting problem.

 

[Martin]

Are capacitors allowed to change voltage instantaneously in that theory?

Why shouldn’t a capacitor’s voltage be able to change instantaneously?

 

[Corneo]

Why shouldn’t a capacitor’s voltage be able to change instantaneously?

I believe you would need infinite energy. One of the conditions used in transient analysis is $$v_c({t_0}^{-}) = v_c(t_0^+)$$

 

[SGT]

Why shouldn’t a capacitor’s voltage be able to change instantaneously?

Because the current in a capacitor is $$i=C\frac{dV}{dt}$$. If $$dt$$ is zero, $$i$$ is infinite.

 

[Martin]

I believe you would need infinite energy.

How so? We begin with a finite amount of energy ((1/2) (1/C) X Q^2), and we end up with a finite amount of energy ((1/4) (1/C) X Q^2).

One of the conditions used in transient analysis is $$v_c({t_0}^{-}) = v_c(t_0^+)$$

Where does this come from? Are there any situations where this is not true?

 

[SGT]

How so? We begin with a finite amount of energy ((1/2) (1/C) X Q^2), and we end up with a finite amount of energy ((1/4) (1/C) X Q^2).
Where does this come from? Are there any situations where this is not true?

Yes, if you have an impulse of current. Infinite amplitude and zero time, so the energy is finite.

 

[Martin]

Because the current in a capacitor is $$i=C\frac{dV}{dt}$$. If $$dt$$ is zero, $$i$$ is infinite.

What is wrong with infinite current?

 

[Martin]

Yes, if you have an impulse of current. Infinite amplitude and zero time, so the energy is finite.

You are correct, and on the right track to solving the problem.

 

[Reality_Patrol]

Why not at least attempt to find a solution analytically, the way you would go about determining the solution to any circuit problem? By doing so, you might get some insight into what makes this such an interesting problem.

Well you're just being difficult. Even so I spent a few minutes finding an analytical solution to a more general circuit: your circuit plus a resistor. I calculated one thing, the energy dissipated in the resistor. The answer is:
Ediss = (1/(4C)) Q^2
This is exactly what's expected of course. Notice it doesn't depend at all on the value of the resistor, the energy dissipated always equals the amount of stored energy that was "lost".
Now I'm tired from all that work, so I think I'll simply wait to hear how your "resistor-free" circuit accounts for the same loss. I hope I won't be dissappointed!

 

[Martin]

Well you're just being difficult. Even so I spent a few minutes finding an analytical solution to a more general circuit: your circuit plus a resistor. I calculated one thing, the energy dissipated in the resistor. The answer is:
Ediss = (1/(4C)) Q^2
This is exactly what's expected of course.

If the result of my “just being difficult” was to get you to expend a little bit of effort thinking through the problem, then perhaps my “just being difficult” isn’t such a bad thing.

Notice it doesn't depend at all on the value of the resistor, the energy dissipated always equals the amount of stored energy that was "lost".

That is (seemingly) counter intuitive, don’t you think?

Now I'm tired from all that work, ...

There you go, complaining again!

... so I think I'll simply wait to hear how your "resistor-free" circuit accounts for the same loss. I hope I won't be dissappointed!

You already have the answer. Take a look at your own analysis: The energy dissipated is independent of the resistance—even in the limit of zero resistance. But although the energy dissipated is independent of the resistance, the power dissipated depends very much upon the resistance. (SGT was on the right track when he noted that the zero-resistance circuit responds by producing an impulse of current when the switch is closed at t=0.)

 

[SGT]

If the result of my “just being difficult” was to get you to expend a little bit of effort thinking through the problem, then perhaps my “just being difficult” isn’t such a bad thing.
That is (seemingly) counter intuitive, don’t you think?
There you go, complaining again!
You already have the answer. Take a look at your own analysis: The energy dissipated is independent of the resistance—even in the limit of zero resistance. But although the energy dissipated is independent of the resistance, the power dissipated depends very much upon the resistance. (SGT was on the right track when he noted that the zero-resistance circuit responds by producing an impulse of current when the switch is closed at t=0.)

And where does the missing energy go in a lossless circuit? The impulsional current cannot account for it.

 

[Martin]

And where does the missing energy go in a lossless circuit? The impulsional current cannot account for it.

Why not? “Infinity times zero” can very well be finite—you yourself realized that when you noted that:

... if you have an impulse of current. Infinite amplitude and zero time, so the energy is finite.

Go through the analysis that Reality_Patrol performed: Insert a series resistance R in the circuit. Calculate the current, power, and, finally, the energy dissipated by the resistance from t=0 (when the switch is closed) till hell freezes over (“infinity”). Examine the equations as a function of t, R, and C. When this is done, you will see that the current and power depend upon t, R, and C, peaking at t = 0, and decaying exponentially thereafter at a rate determined by R and C. As R goes to 0, the current peak becomes infinite at t = 0, as does the rate of its decay for t = 0. But the total energy dissipated remains constant, independent of R.

Consequently, the answer to the question “Where did it [the (1/4) (1/C) X Q^2 energy] go?” is: “It was instantly dissipated in the zero resistance of the ideal wires by an instantaneous current of infinite amplitude.”

 

[SGT]

Why not? “Infinity times zero” can very well be finite—you yourself realized that when you noted that:
Go through the analysis that Reality_Patrol performed: Insert a series resistance R in the circuit. Calculate the current, power, and, finally, the energy dissipated by the resistance from t=0 (when the switch is closed) till hell freezes over (“infinity”). Examine the equations as a function of t, R, and C. When this is done, you will see that the current and power depend upon t, R, and C, peaking at t = 0, and decaying exponentially thereafter at a rate determined by R and C. As R goes to 0, the current peak becomes infinite at t = 0, as does the rate of its decay for t = 0. But the total energy dissipated remains constant, independent of R.
Consequently, the answer to the question “Where did it [the (1/4) (1/C) X Q^2 energy] go?” is: “It was instantly dissipated in the zero resistance of the ideal wires by an instantaneous current of infinite amplitude.”

This is metaphysics, not physics. Where did the energy go? Turned into heat? Into EM radiation? Gone into the 5th dimension?

 

[Cliff_J]

Its been a while for me and I thought this was already debated once, but I thought it ended up at 70.7% V on each cap and not 50% V on each cap after they are connected in parallel.

That basically the premise of the question is the problem, because the charge does not represent the columbs of electrons but instead the pressure under which they are stored. So Q/2 is incorrect because it never represented the quantity but instead indirectly the pressure.

No metaphysics needed there.

 

[Reality_Patrol]

Martin,
Well that was worth waiting for, and I agree with your answer. Of course it's completely accurate physically in cases with finite resistances. The big question is how accurate it is for zero-R cases. I think physically there will always be a finite current, even for superconductors. In that case there would be an "R" that could be inferred from the experimentally measured current decay profile. This R would however vanish under steady-state conduction, so it's more of a non-equilibirium resistance. That was one of my early guesses, what I referred to in my first post as a transient conduction loss. What this problem brings out cleary is that all materials share this unique property, and that is something that's not well understood. Cool.
And you might want to remind those people who don't think a simple wire can support a voltage about a transformer.
Cliff,
I have no idea what you're referring to. Care to clarify it?

 

[Martin]

This is metaphysics, not physics. Where did the energy go? Turned into heat? Into EM radiation? Gone into the 5th dimension?

As I suggested above, go through the analysis that Reality_Patrol performed: Insert a series resistance R in the circuit, and do the math. You will find, as Reality_Patrol found, that the total energy dissipated by R is (1/4) (1/C) X Q^2, independent of the value of R. The resistance does affect the amplitude and duration of both the current i(t) in the circuit and the power p(t) consumed by the resistance (the smaller the R, the larger the amplitudes and shorter the durations of i(t) and p(t)), but not the total amount of charge transferred from the 1st capacitor to the 2nd capacitor (which is the total area under the i(t) curve), nor the total energy dissipated by the resistance (which is the total area under the p(t) curve). In the limit as R approaches zero, i(t) and p(t) both become impulse functions.

 

[Martin]

Its been a while for me and I thought this was already debated once, but I thought it ended up at 70.7% V on each cap and not 50% V on each cap after they are connected in parallel.

That basically the premise of the question is the problem, because the charge does not represent the columbs of electrons but instead the pressure under which they are stored. So Q/2 is incorrect because it never represented the quantity but instead indirectly the pressure.

No metaphysics needed there.

After the switch is closed, both capacitors must have the same voltage. Since the relationship between charge and voltage for an ideal capacitor is given by charge = capacitance X voltage, and since the capacitors are the same size and have the same voltage, they must have the same charge. And since total charge is conserved, they must also add up to the total initial charge Q.

Q/2 is indeed correct.

 

[Cliff_J]

Nevermind, got my terms confused. The energy is lost just performing the work charging the 2nd cap, correct?

 

[SGT]

As I suggested above, go through the analysis that Reality_Patrol performed: Insert a series resistance R in the circuit, and do the math. You will find, as Reality_Patrol found, that the total energy dissipated by R is (1/4) (1/C) X Q^2, independent of the value of R. The resistance does affect the amplitude and duration of both the current i(t) in the circuit and the power p(t) consumed by the resistance (the smaller the R, the larger the amplitudes and shorter the durations of i(t) and p(t)), but not the total amount of charge transferred from the 1st capacitor to the 2nd capacitor (which is the total area under the i(t) curve), nor the total energy dissipated by the resistance (which is the total area under the p(t) curve). In the limit as R approaches zero, i(t) and p(t) both become impulse functions.

With a finite R, energy is dissipated as heat, so we can account for the difference between initial and final energy. What happens with the missing energy in a lossless circuit? I understand the maths, but I want a physical answer. Energy cannot be destroyed nor created.

 

[SGT]

Nevermind, got my terms confused. The energy is lost just performing the work charging the 2nd cap, correct?

Wrong! The work performed by charging the capacitor is exactly the stored energy.

 

[Martin]

Nevermind, got my terms confused. The energy is lost just performing the work charging the 2nd cap, correct?

No. The energy is dissipated in the zero resistance of the wires: The circuit responds to the closing of the switch by producing an impulse of current—which is a current of infinite magnitude lasting zero time, such that the charge transferred from the 1st capacitor to the second capacitor (the area under the impulse) is finite. (This is one of those cases where you are multiplying an infinite amount by a zero amount, and getting a finite result.)

For comparison, imagine replacing the second ideal capacitor with an ideal inductor (pure inductance). In such a circuit, the charge would leave the capacitor, and then return to the capacitor, in a sinusoidal manner, ad infinitum. Essentially, the electric potential energy stored in the capacitor is converted into the kinetic energy of the moving charge, which is then converted into magnetic potential energy stored in the inductor, ...and so on. (As an analogy, imagine dropping a mass from some height onto an ideal spring: The potential energy of the mass due to the gravitational field is converted into kinetic energy as the mass gains speed, and then into potential energy as it compresses the spring, then back again into kinetic energy as it bounces upward, ...and so on.) Bottom line: The performance of work does not, per se, imply any loss of energy.

 

[Martin]

... I have no idea what you're referring to. Care to clarify it?

I would, if only I knew what you were referring to.