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If we use an ideal capacitor to charge another ideal capacitor, my intuition tells me no heat is generated since capacitors are just storage elements. It shouldn't consume energy.

Original question

But in order to solve this question, I used two equations (conservation of charge and equal voltage for both capacitors at equilibrium) to find that energy had indeed been lost.

My diagram

My solution

What's the mechanism by which heat is lost in this case? Is it the energy required to push the charges closer together on C1? Is it energy spent to accelerate charges, to make it move? Am I right in claiming that no "heat" is generated?

I noticed that the energy lost equals that stored in the "equivalent" series capacitance if it was charged to \$V_0\$. Is there any reasoning why it is so?

Parallel capacitance

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    \$\begingroup\$ Have you read: en.wikipedia.org/wiki/Two_capacitor_paradox. In my personal opinion the correct answer isn't listed. In my opinion the correct answer is "0" (zero) as there are no elements in the circuit which can dissipate power. So yes, I agree with your intuition. I also think it is a stupid idea to make a (study) question out of this controversial paradox. Basically you only need to know what answer the teacher expects and choose that. No one learns anything from that. \$\endgroup\$ – Bimpelrekkie May 21 at 9:53
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    \$\begingroup\$ @Bimpelrekkie thanks! That link will really help. I agree with you too. \$\endgroup\$ – Aditya P May 21 at 10:04
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    \$\begingroup\$ As @Huisman correctly points out, this is a nonsense question. The circuit you drew violates our definitions of ideal circuit elements because of a built-in contradiction: parallel elements must have the same voltage but the voltage across a capacitor cannot change instantaneously. So, connecting two capacitors in parallel with different voltages is an invalid circuit and cannot be analyzed by normal circuit techniques. Get a different book. \$\endgroup\$ – Elliot Alderson May 21 at 12:05
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    \$\begingroup\$ @BenVoigt A schematic is an ideal drawing tool that has basic elements, one of which is the ideal wire. To indicate parasitics like wire resistance, it must be indicated with an ideal resistor. Anything else is an egregious and imprecise abuse of notation that leads to ambiguities. Huisman gives the correct answer. \$\endgroup\$ – Shamtam May 21 at 20:42
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    \$\begingroup\$ @BenVoigt Students learning circuit analysis always assume that components are ideal...you can not mathematically analyze the circuit otherwise. This question was clearly about a homework problem and needs to be answered from the student's perspective. \$\endgroup\$ – Elliot Alderson May 22 at 12:49
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The problem with these theoretical examples lies in the fact the current is assumed infinite for 0 seconds. Crudely substituting this in the conservation law:

$$ \frac {\partial \rho }{\partial t} +\nabla \cdot \mathbf {J} = 0 $$

$$ \frac { \rho }{ 0 }+ \infty \neq 0 $$

Since charge is conserved, the assumption of infinite current in zero time is wrong.

How much power is dissipated \$P_{diss}=VI\$ cannot be defined, since the definition of the current is false.

So, the answer is: cannot be defined

EDIT
Note that the dissipation neither is 0 W because R = 0 \$ \Omega\$. For the same reason as above: \$ P = I^2R = \infty^2 \cdot 0 \$, which is not defined.

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    \$\begingroup\$ Yes. This is the only correct answer. \$\endgroup\$ – Elliot Alderson May 21 at 14:36
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    \$\begingroup\$ The power lost cannot be calculated, but the loss of energy can. \$\endgroup\$ – Ben Voigt May 21 at 19:31
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    \$\begingroup\$ You can make the conservation law work with Dirac’s delta. You cannot add infinity to the real / complex set and expect calculus to keep working. It makes the set not partially ordered. If it is not partially ordered, no Zorn’s lemma, which means no axiom of choice. \$\endgroup\$ – user110971 May 21 at 21:29
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When masses collide in an inelastic manner, momentum is conserved but energy has to be lost. It's the same with the two-capacitor paradox; charge is always conserved but, energy is lost in heat and EM waves. Our schematic model of the simple circuit isn't sufficient to show the subtler mechanisms at play such as interconnection resistance.

An elastic collision can be said to be equivalent to adding series inductors in the wires. Somewhere between the two is reality - the connections are composed of resistors and inductors; the fact that our schematic may not show them is just a weakness of our imagination.

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    \$\begingroup\$ I noticed it too, in the other answer you wrote. Maybe you should try contacting stackexchange, they can find the user who is targeting you. You should really report this. \$\endgroup\$ – Aditya P May 21 at 18:04
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    \$\begingroup\$ Have an upvote :) \$\endgroup\$ – Sombrero Chicken May 21 at 18:26
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    \$\begingroup\$ I down-voted this answer because I didn't feel that it addressed the original question. It seemed to me that you wandered off into a discussion of particle and wave physics that was of no help to the OP. And I think there is a reason that anonymous downvotes are allowed. Now, you have a lot more reputation than I do so go ahead, do your worst. I've up-voted a lot of your other answers in the past but I won't bother anymore. Report me as necessary. \$\endgroup\$ – Elliot Alderson May 21 at 22:48
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    \$\begingroup\$ @ElliotAlderson I don't report anything I just observe and comment. I never mentioned particle or wave physics. I made a comparison with masses in the newtonian way i.e. conservation of momentum is very similar to conservation of charge. \$\endgroup\$ – Andy aka May 22 at 7:58
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    \$\begingroup\$ the fact that our schematic may not show them is just a weakness of our imagination. Um, I think it's either sloppy question drafting, or an attempt to illustrate the gulf between ideal circuits and real circuits. The collisions analogy is good physics, the units and the mechanisms are right, especially the total energy before minus after leaves a deficit that is independent of the means of dissipation, for instance the undrawn component may have been a transformer primary with an antenna and a radiation resistance on it. As drawn, the circuit is a paradox, wrong, SPICE would choke on it \$\endgroup\$ – Neil_UK May 22 at 15:34
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What's the mechanism by which heat is lost in this case?

Normally, the wires and the switches have some resistance. Because current flows through the wires, heat is produced.

I noticed that the energy lost equals that stored in the "equivalent" series capacitance if it was charged to V0. Is there any reasoning why it is so?

If you charge an "ideal" capacitor where charge and voltage are proportional, 50% of the energy will be converted to heat.

However, if you have "real" capacitors where charge and voltage are not exactly proportional (as far as I know this is the case for DLCs) the percentage of energy which is converted to heat is NOT exactly 50%.

This means that the key to your observation lies in the equation of the capacitors (q ~ v) and there is no "intuitive" explanation that is independent of that equation.

(If there was an explanation that is independent of the equation, the percentage would also be 50% for "real" capacitors.)

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I have to go with "The question is invalid".

It looks like the problem was edited from a previous one to a different question.

The "answers" all have units of Q^2 * C / C^2 or Q/C.

It's been 40 years for me since I had that EE class, but isn't that Voltage? How do you answer a "heat dissapated" question with units of voltage?

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    \$\begingroup\$ where'd the 2 in $$Q^2$$ go in your dimensional analysis? The answers have units of $$\frac{Q^2}{C} = Q \Delta V$$ which is energy \$\endgroup\$ – Ben Voigt May 21 at 22:01
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    \$\begingroup\$ Apparently lost in my brain. Right so the units are q^2/C. What the heck is that unit? And the winner is Joules. So I probably need to downvote my own answer. \$\endgroup\$ – pbm May 21 at 22:07
  • \$\begingroup\$ @pbm \$ Q^2/C \$ has units of \$ \text{C}^2/\text{F} = \text{C}^2/(\text{C}/\text{V}) = \text{C} \, \text{V} = \text{J} \$, i.e. of joules, which is a unit of energy. You can derive yourself some of the equivalent units I used, or read en.wikipedia.org/wiki/Farad#Equalities and en.wikipedia.org/wiki/Joule \$\endgroup\$ – Alejandro Nava Aug 24 at 15:31
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EDIT: To those of you uncomfortable with me proclaiming that \$R = 0\$ in the end, it is analogous to taking the resistance of air to be infinite. And if you are still uncomfortable, read "infinite" as "really really large", and "zero" as "really really small".


There is infinite current flowing through zero resistance, and this results in a finite energy being dissipated in the wire. To make sense of this, we need to do a bit of calculus. Suppose there is also a resistance \$R\$ in the circuit, which we shall set to zero at the end.

Let \$V_0 = q_0 / C_1\$. Doing the usual Laplace transform for circuits, the transformed current \$I(s)\$ is given by $$ \begin{align} \frac{V_0}{s} &= I(s) \left[ R + \frac{1}{s C_1} + \frac{1}{s C_2} \right] \\ &= I(s) \left[ R + \frac{1}{s C} \right] \\ \end{align} $$ where \$1/C = 1/C_1 + 1/C_2\$. Thus $$ \begin{align} I(s) &= \frac{V_0 / s}{R + 1 / (s C)} \\ &= \frac{V_0 / R}{s + 1 / (R C)} \\ i(t) &= \frac{V_0}{R} \cdot \mathrm{e}^{-t / (R C)}. \end{align} $$ The instantaneous power dissipated is $$ \begin{align} P(t) &= i(t)^2 \cdot R \\ &= \frac{{V_0}^2}{R} \cdot \mathrm{e}^{-2t / (R C)} \end{align}, $$ and so the total energy dissipated is $$ \int_0^\infty \frac{{V_0}^2}{R} \cdot \mathrm{e}^{-2t / (R C)} \,\mathrm{d}t = \frac{1}{2} C {V_0}^2 = \frac{{q_0}^2 C_2}{2 C_1 (C_1 + C_2)}. $$ Note that this is independent of \$R\$, and I would argue it even holds for \$R = 0\$.

Indeed setting \$R\$ to zero in the context of generalised functions, we have that $$ \begin{align} i(t) &= C V_0 \cdot \delta(t) \\ P(t) &= \frac{1}{2} C {V_0}^2 \cdot \delta(t), \end{align} $$ where \$\delta(t)\$ is the Dirac delta (or unit impulse) in time, which has dimensions \$1/\text{time}\$. Thus all of the energy is dissipated in the instant \$t = 0\$.

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  • \$\begingroup\$ If R=0 then where does the dissipated energy go? Specifically, how is it converted to heat as the question asks? How can you derive equations assuming nonzero R and then set R to zero? \$\endgroup\$ – Elliot Alderson May 22 at 12:55
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    \$\begingroup\$ @ElliotAlderson: The actual case of R = 0 is a red herring. Even in "real circuits", we don't assume that R = 0 in wires. We assume that R is non-zero but "negligible", which is not the same thing (and it's an assumption that can get us into trouble sometimes). What this derivation shows is that no matter how small R is, so long as it's non-zero, the power dissipated is always the same. \$\endgroup\$ – Michael Seifert May 22 at 14:09
  • \$\begingroup\$ @MichaelSeifert Yes, what you said! so long as it's non-zero That was my point exactly. \$\endgroup\$ – Elliot Alderson May 22 at 14:17
  • \$\begingroup\$ @ElliotAlderson The energy is dissipated as heat in the wire. Although \$R = 0\$, \$i^2 = \infty\$ at \$t = 0\$ is large enough that the indeterminate expression \$i^2 R\$ yields a finite energy when integrated over time. Assuming things are nonzero and setting them to zero is, well, calculus. For an analogy, a mass \$m \ne 0\$ in a gravitational field of strength \$g\$ has weight \$m g\$ and thus acceleration \$a = m g / m = g\$. But even massless objects (\$m = 0\$) fall with acceleration \$g\$. \$\endgroup\$ – u54112 May 22 at 16:15
  • \$\begingroup\$ @lastresort From what I read, within the newtonian framework massless particles do not experience g. It is due to the how gravity bends space that massless objects experience g. \$\endgroup\$ – Aditya P May 23 at 11:33

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