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.

 

[Martin]

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.

Lumped circuit theory is a self-consistent model that approximates real circuits. There are a number of assumptions inherent in lumped circuit theory that do not hold 100% in “real” circuits. How well the theory models the actual behavior of any “real” circuit will depend upon how closely those assumptions hold for that particular “real” circuit. For example, circuit theory assumes that Kirchhoff’s Laws are valid; that “real” circuit components can be represented by “lumped” parameters (i.e., discrete R, L, and C); and that all associated electric and magnetic fields remain within the confines of the circuit. (Specifically, these assumptions require that electrical effects happen instantaneously throughout the circuit, that there is no accumulation of charges at any point within the circuit, and that there is no magnetic coupling among the various elements of the circuit). These are all approximations that work rather well for many—but not all—applications. Strictly speaking, for example, lumped circuit theory is not valid for time-varying voltages and currents; “real” circuit components are never actually “lumped” quantities; and the electric and magnetic fields associated with circuits never remain fully within the confines of a real circuit.

In a “real” circuit, energy would be lost due to heat dissipation in the finite resistances of the physical wires and components as well as due to radiation resulting from time variations in voltages and current as the circuit responds to the throwing of the switch. These are the only mechanisms that can account for energy loss. But if we assume that we can ignore energy loss due to radiation—which is one of lumped circuit theory’s inherent assumptions—then lumped circuit theory will be “forced” to account for the entire energy loss via resistance (yes, even in the limit of zero resistance!). It does so in a lossless (zero resistance) circuit by requiring a current impulse at the moment that the switch is thrown. It’s only via an “infinite” current that a finite energy dissipation can be created in circuit with zero resistance. Of course, impulses are non-physical—just as zero resistances are non-physical.

In reality (that is, were you close a “real” switch in a “real” circuit consisting of “real” wires and “real” capacitors), the circuit would respond with a large current pulse at the moment that the switch was closed. Some energy would be dissipated in the resistances of the “real” wires, capacitors, and switch. However, there also would be some radiation loss caused by the higher-frequency components of the current pulse. The relative amounts would depend on the dimensions of the circuit and the (physical) properties of the switch, wires, and capacitors.

 

[SGT]

...Strictly speaking, for example, lumped circuit theory is not valid for time-varying voltages and currents; ...

Not true! Lumped circuit theory is valid not only for time-varying voltages and currents but also with time-varying and nonlinear components. The only requirement for a lumped circuit is that it is composed by lumped components and that the overall dimensions of the circuit are small when compared with the shortest wavelength involved. If what you wrote was true, you could not use the theory with an impulsional current, since an impulse is a time-varying waveform.

In a “real” circuit, energy would be lost due to heat dissipation in the finite resistances of the physical wires and components as well as due to radiation resulting from time variations in voltages and current as the circuit responds to the throwing of the switch. These are the only mechanisms that can account for energy loss. But if we assume that we can ignore energy loss due to radiation—which is one of lumped circuit theory’s inherent assumptions—then lumped circuit theory will be “forced” to account for the entire energy loss via resistance (yes, even in the limit of zero resistance!). It does so in a lossless (zero resistance) circuit by requiring a current impulse at the moment that the switch is thrown. It’s only via an “infinite” current that a finite energy dissipation can be created in circuit with zero resistance. Of course, impulses are non-physical—just as zero resistances are non-physical.
In reality (that is, were you close a “real” switch in a “real” circuit consisting of “real” wires and “real” capacitors), the circuit would respond with a large current pulse at the moment that the switch was closed. Some energy would be dissipated in the resistances of the “real” wires, capacitors, and switch. However, there also would be some radiation loss caused by the higher-frequency components of the current pulse. The relative amounts would depend on the dimensions of the circuit and the (physical) properties of the switch, wires, and capacitors.

Do you have a reference where I can read about energy loss without a dissipative element, or are those assumptions a theory of your own?

 

[SGT]

This problem has a mechanical analog. The inelastic collision between two bodies, where momentum is conserved and kinectic energy not.
Before collision the total momentum is
p = mv + 0 = mv
and the kinectic energy is
W = \frac{1}{2}mv^2 + 0 = \frac{1}{2}mv^2
After collision the total momentum is
m\frac{v}{2} + m\frac{v}{2} = mv
while the kinectic energy is
W = \frac{1}{2}m\left(\frac{v}{2}\right)^2 + \frac{1}{2}m\left(\frac{v}{2}\right)^2 = \frac{1}{4}mv^2
We consider that work is done in deforming the bodies and this accounts for the missing energy, that is turned into heat.
If the two bodies are rigid there can be no inelastic collision.
The problem with the two capacitors is that there is no mechanism that can account for the missing energy. The model just does not work, in the same way that the rigid body model is unsuitable to an inelastic collision.

 

[Martin]

Not true! Lumped circuit theory is valid not only for time-varying voltages and currents but also with time-varying and nonlinear components. The only requirement for a lumped circuit is that it is composed by lumped components and that the overall dimensions of the circuit are small when compared with the shortest wavelength involved.

Lumped circuit theory is valid in the sense that it provides useful approximate results whose accuracy depends upon the degree to which its underlying assumptions and restrictions are met in any particular application. When I commented on its validity, I said strictly speaking it was not valid for time-varying voltages and currents—and that is a true statement.

If what you wrote was true, you could not use the theory with an impulsional current, since an impulse is a time-varying waveform.

I didn’t impose a current impulse; the model predicted it.

The model is just that: A model. It is self consistent. It has a set of rules (Kirchoff’s Laws) that must be satisfied at every node and around every loop. Each lumped circuit element is specified by unambiguous terminal relationships. In the circuit presented, the model predicts an impulse of current at t=0, and an instantaneous energy dissipation. The model doesn’t “know” how well its predictions mirror the corresponding “real” circuit that we are using it to represent—it predicts what it predicts, and does so such that all of its “rules” are obeyed. The fact that the model predicts an impulse of current (i.e., we didn’t apply an impulse of current to the circuit) and that an impulse of current (or an impulse of any physical quantity) cannot occur in the “real world,” and that the model predicts an energy loss even though there is no non-zero resistance in the circuit to account for dissipation tells us that the model is attempting to predict the behavior of a real circuit in an extreme situation. We have pushed the use of the model as an accurate representation of a “real” circuit up against its limits.

Specifically, we have ignored the small inductance and resistance that would exist in a configuration of “real” capacitors, a switch, and wires. Why did we do that? We did that because in most “real” configurations that involve “real” circuit elements, ignoring the “incidental” resistances, inductances, and capacitances is legitimate: They are indeed very small quantities compared to the “real” circuit elements themselves. Apparently, then, we can’t directly make such an assumption for this particular configuration. However, we may be able to indirectly do so.

We have to ask: What is this “ideal” circuit trying to tell us? To answer that, we do the simplest thing that we can do: Insert a resistance in the circuit, and do the analysis to determine the relationship between the current and resistance and between the energy dissipated and the resistance; and then we examine how the current and energy dissipation change as we allow the resistance to shrink to zero. In other words, we take as the solution to our resistance-free problem the solution to an analogous problem that includes a resistance, in the limit as that resistance goes to zero. Applying the concept of limits to arrive at solutions that otherwise would escape our grasp is pretty standard stuff.

Do you have a reference where I can read about energy loss without a dissipative element, or are those assumptions a theory of your own?

I have no references, and it is not my “theory.” I simply solved a problem using standard lumped circuit theory, recognizing that sometimes the simplifying assumptions typically used to model a real circuit cannot blindly be applied.

 

[Martin]

... We consider that work is done in deforming the bodies and this accounts for the missing energy, that is turned into heat.
If the two bodies are rigid there can be no inelastic collision.
The problem with the two capacitors is that there is no mechanism that can account for the missing energy. The model just does not work, in the same way that the rigid body model is unsuitable to an inelastic collision.

See my previous post.

 

[SGT]

Lumped circuit theory is valid in the sense that it provides useful approximate results whose accuracy depends upon the degree to which its underlying assumptions and restrictions are met in any particular application. When I commented on its validity, I said strictly speaking it was not valid for time-varying voltages and currents—and that is a true statement.

So, in your opinion, lumped circuit theory is valid only for constant voltages and currents?

I didn’t impose a current impulse; the model predicted it.

You have imposed a step voltage , that is a time varying waveform. The impulse appeared because it is the time derivative of the step.

The model is just that: A model. It is self consistent. It has a set of rules (Kirchoff’s Laws) that must be satisfied at every node and around every loop. Each lumped circuit element is specified by unambiguous terminal relationships. In the circuit presented, the model predicts an impulse of current at t=0, and an instantaneous energy dissipation. The model doesn’t “know” how well its predictions mirror the corresponding “real” circuit that we are using it to represent—it predicts what it predicts, and does so such that all of its “rules” are obeyed. The fact that the model predicts an impulse of current (i.e., we didn’t apply an impulse of current to the circuit) and that an impulse of current (or an impulse of any physical quantity) cannot occur in the “real world,” and that the model predicts an energy loss even though there is no non-zero resistance in the circuit to account for dissipation tells us that the model is attempting to predict the behavior of a real circuit in an extreme situation. We have pushed the use of the model as an accurate representation of a “real” circuit up against its limits.
Specifically, we have ignored the small inductance and resistance that would exist in a configuration of “real” capacitors, a switch, and wires. Why did we do that? We did that because in most “real” configurations that involve “real” circuit elements, ignoring the “incidental” resistances, inductances, and capacitances is legitimate: They are indeed very small quantities compared to the “real” circuit elements themselves. Apparently, then, we can’t directly make such an assumption for this particular configuration. However, we may be able to indirectly do so.
We have to ask: What is this “ideal” circuit trying to tell us? To answer that, we do the simplest thing that we can do: Insert a resistance in the circuit, and do the analysis to determine the relationship between the current and resistance and between the energy dissipated and the resistance; and then we examine how the current and energy dissipation change as we allow the resistance to shrink to zero. In other words, we take as the solution to our resistance-free problem the solution to an analogous problem that includes a resistance, in the limit as that resistance goes to zero. Applying the concept of limits to arrive at solutions that otherwise would escape our grasp is pretty standard stuff.
I have no references, and it is not my “theory.” I simply solved a problem using standard lumped circuit theory, recognizing that sometimes the simplifying assumptions typically used to model a real circuit cannot blindly be applied.

The affirmations you made about modelling are true. The mistake is to use a model in a situation where it does not apply. In my mechanical example, using a rigid body model to study the behavior of an inelastic collision. In the electrical example, using ideal capacitors and switch to study the charging of one capacitor by another.
Back to the mechanical example, the collision makes appear a force f = m\frac{dv}{dt} in the same way that the charge and discharge of the capacitors give origin to a current i = C\frac{dV}{dt}. In both cases, if dt tends to zero, the derivative becomes an impulse. The difference is that in the mechanical case dt is only zero in an elastic collision. In this case both momentum and kinectic energy are conserved. Lumped circuit theory has no correspondent element to a rigid body. Is this a failure of the theory? No, the failure is to use it to model things it is not supposed to do.

 

[Martin]

So, in your opinion, lumped circuit theory is valid only for constant voltages and currents?

I said:

“Lumped circuit theory is valid in the sense that it provides useful approximate results whose accuracy depends upon the degree to which its underlying assumptions and restrictions are met in any particular application.”

Inherent to the theory’s applicability is the assumption that there is no radiation—that the only way energy can be lost from the circuit is via heat dissipation in resistances. The moment that you allow time varying fields, you necessarily allow for energy loss via radiation. Of course, we can apply the theory when the voltages and currents vary in time “slowly” enough (the “quasi-static” approximation—that is, when the circuit dimensions are “small” enough compared to the wavelength) that radiation can be neglected. Nevertheless, strictly speaking, time-varying voltages and currents do give rise to radiation, and there is no mechanism to account for radiation in lumped circuit theory. Consequently, strictly speaking, the theory is not valid for time-varying voltages and currents.

You have imposed a step voltage , that is a time varying waveform. The impulse appeared because it is the time derivative of the step.

I was responding to your statement:

“If what you wrote was true, you could not use the theory with an impulsional current, since an impulse is a time-varying waveform.”

I pointed out that I didn’t impose a current impulse; the model predicted it.

While I agree that closing the switch leads to the appearance of a step voltage across the ideal capacitor (across both ideal capacitors, actually), the appearance of that step voltage, like the appearance of the current impulse, is predicted by the model (via KVL and the voltage/current relationship required by the ideal capacitors), not imposed by me.

The affirmations you made about modelling are true. The mistake is to use a model in a situation where it does not apply. ... In the electrical example, using ideal capacitors and switch to study the charging of one capacitor by another. ...Is this a failure of the theory? No, the failure is to use it to model things it is not supposed to do.

I doubt that that you would’ve found it objectionable had I proposed a “more-complete” model—one that incorporated a “small” series resistance. Doing so leads to the results I described previously.

If using the resistance-free circuit to model an actual circuit truly were a “failure,” we would not get results consistent with such a more-complete model. Both models predict the same total loss of energy. The “more-complete” model enables us to see how the charge, voltage, current, power, and energy dissipated depend on the size of the resistance. It shows that, regardless of the size of the resistance, both the total charge transferred and the total energy dissipated does not change. And the “more-complete” model and its results lead naturally to the “resistance-free” model and its results as limits.

But, if you prefer, place a small—infinitesimal even—series resistance in the circuit to account for the inevitable resistance that must exist in a “real” circuit. Then, instead of the answer that I offered (“It [the (1/4) (1/C) X Q^2 energy] was instantly dissipated in the zero resistance of the ideal wires by an instantaneous current of infinite amplitude”), you can more-comfortably respond with “It [the (1/4) (1/C) X Q^2 energy] was very rapidly dissipated in the very small resistance of the very low-resistance wires by a very short-duration current of very large amplitude.”

 

[SGT]

I said:
“Lumped circuit theory is valid in the sense that it provides useful approximate results whose accuracy depends upon the degree to which its underlying assumptions and restrictions are met in any particular application.”
Inherent to the theory’s applicability is the assumption that there is no radiation—that the only way energy can be lost from the circuit is via heat dissipation in resistances. The moment that you allow time varying fields, you necessarily allow for energy loss via radiation. Of course, we can apply the theory when the voltages and currents vary in time “slowly” enough (the “quasi-static” approximation—that is, when the circuit dimensions are “small” enough compared to the wavelength) that radiation can be neglected. Nevertheless, strictly speaking, time-varying voltages and currents do give rise to radiation, and there is no mechanism to account for radiation in lumped circuit theory. Consequently, strictly speaking, the theory is not valid for time-varying voltages and currents.

The motive lumped circuit theory demands that the dimensions of the circuit are small compared to the wavelengths involved has nothing to do with radiation. The limitation arises from the applicability of Kirchoff's laws. You can't predict voltages and currents along a distributed circuit with them, you must use Maxwell's laws. This has nothing to do with radiation, since waveforms in a waveguide or in a coaxial cable don't emit radiation, but they vary along the length of the component.
There is not in lumped circuit theory that forbids steps, ramps or sinusoids tobe analysed.

I was responding to your statement:
“If what you wrote was true, you could not use the theory with an impulsional current, since an impulse is a time-varying waveform.”
I pointed out that I didn’t impose a current impulse; the model predicted it.
While I agree that closing the switch leads to the appearance of a step voltage across the ideal capacitor (across both ideal capacitors, actually), the appearance of that step voltage, like the appearance of the current impulse, is predicted by the model (via KVL and the voltage/current relationship required by the ideal capacitors), not imposed by me.
I doubt that that you would’ve found it objectionable had I proposed a “more-complete” model—one that incorporated a “small” series resistance. Doing so leads to the results I described previously.
If using the resistance-free circuit to model an actual circuit truly were a “failure,” we would not get results consistent with such a more-complete model. Both models predict the same total loss of energy. The “more-complete” model enables us to see how the charge, voltage, current, power, and energy dissipated depend on the size of the resistance. It shows that, regardless of the size of the resistance, both the total charge transferred and the total energy dissipated does not change. And the “more-complete” model and its results lead naturally to the “resistance-free” model and its results as limits.
But, if you prefer, place a small—infinitesimal even—series resistance in the circuit to account for the inevitable resistance that must exist in a “real” circuit. Then, instead of the answer that I offered (“It [the (1/4) (1/C) X Q^2 energy] was instantly dissipated in the zero resistance of the ideal wires by an instantaneous current of infinite amplitude”), you can more-comfortably respond with “It [the (1/4) (1/C) X Q^2 energy] was very rapidly dissipated in the very small resistance of the very low-resistance wires by a very short-duration current of very large amplitude.”

I think you did a very good analysis of the behaviour of the circuit. My objection is that your conclusions are not consistent with the accepted laws of Physics. Capacitors and inductors store energy in their electric and magnectic fields respectively and resistors dissipate energy as heat. You are postulating an extension to that theory allowing a zero resistence element to dissipate energy if you have an impulsional power waveform. I agree with you that this extension is useful in the sense that it eliminates the paradox of the missing energy.

 

[Martin]

The motive lumped circuit theory demands that the dimensions of the circuit are small compared to the wavelengths involved has nothing to do with radiation.

Lumped circuit theory assumes that all electrical effects happen instantaneously throughout the circuit, which is approximately realized if the dimensions of the circuit are small compared to the wavelength. This assumption is equivalent to ignoring the finite velocity of electromagnetic field propagation (i.e., equivalent to assuming “instantaneous propagation”), which amounts to ignoring radiation effects.

The limitation arises from the applicability of Kirchoff's laws. You can't predict voltages and currents along a distributed circuit with them, you must use Maxwell's laws.

When the dimensions of the circuit are not small compared to the wavelength, the “lumped-parameter” representation of a physical circuit can be replaced with a “distributed-parameter” representation, and the circuit analyzed by applying Kirchoff’s Laws to incremental sections of the circuit. The analysis results in predicting voltage and current waves that travel along (propagate in the direction of) the wires. This is the model used in Transmission Line Theory.

This has nothing to do with radiation, since waveforms in a waveguide or in a coaxial cable don't emit radiation, but they vary along the length
of the component.

At cross-sectional points of discontinuity (either within the circuit, or in the region exterior to the circuit) the electromagnetic fields associated with these “guided waves” of voltage and current can be scattered into electromagnetic waves that propagate away (i.e., radiate) from the circuit. Radiation does not occur with closed waveguides and coaxial cables because those structures provide an inherent shielding, trapping all of the electromagnetic energy within themselves.

I think you did a very good analysis of the behaviour of the circuit. My objection is that your conclusions are not consistent with the accepted laws of Physics. Capacitors and inductors store energy in their electric and magnectic fields respectively and resistors dissipate energy as heat. You are postulating an extension to that theory allowing a zero resistence element to dissipate energy if you have an impulsional power waveform. I agree with you that this extension is useful in the sense that it eliminates the paradox of the missing energy.

I am postulating nothing; I am presenting the results of the analysis.

The analysis tells us that finite energy is dissipated by a zero resistance. This seemingly paradoxical result is a consequence of the model’s insistence that conservation laws (charge and energy) not be violated (i.e., by requiring that Kirchoff’s Laws be obeyed), regardless of the size of the resistance. The conservation of charge and the conservation of energy are the accepted laws of physics; the mechanisms by which they are satisfied are not.

The only way that energy dissipation can occur when the resistance is zero (a physical impossibility) is if an infinite current (another physical impossibility) flows through it. Infinity times zero is meaningless, unless a finite result is arrived at (as it is in this instance) as a limiting process.