16 ms·

Two capacitor paradox

438 points by rathel 6y ago

brummm 6y ago
I don't think this should be called a paradox. It's just a case where the limitations of the model (ideal everything) is clear and leads to inconsistent results. Adjusting the model and making it more realistic, quickly clears up the "paradox". To me this seems like something one would use as an example in a physics lecture to show when certain assumptions are necessary and when they aren't.
- JadeNB 6y ago
  Notice the definitions: > The two capacitor paradox or capacitor paradox is a paradox, or counterintuitive thought experiment, in electric circuit theory. > A paradox, also known as an antinomy, is a logically self-contradictory statement or a statement that runs contrary to one's expectation. It is a statement that, despite apparently valid reasoning from true premises, leads to a seemingly self-contradictory or a logically unacceptable conclusion. (emphases mine). It's paradoxical in that it runs counter to expectations or intuition, using apparently valid reasoning. (Of course, maybe it doesn't run counter to your intuition; but that's how correct intuition is developed, by breaking naïve intuition against examples like this!) In fact, the article explicitly acknowledges this: > Unlike some other paradoxes in science, this paradox is not due to the underlying physics, but to the limitations of the 'ideal circuit' conventions used in circuit theory.
  - brummm 6y ago
    Fair enough, wikipedia does include the reference to expectation in its definition of paradox. However, I think my main point is still valid that this should not necessarily be called a paradox exactly because it differs a lot from how this term is commonly used in the naming of scientific thought experiments.
    - mokus 6y ago
      My initial gut response was the same - that this isn’t a paradox, but rather a proof by contradiction that one or more of the assumptions are wrong. But after thinking about it a bit more, I don’t think there’s really that much of a difference. It’s just that some people have already learned not to apply those assumptions in this context (or by chance never stumbled into them in the first place). Perhaps it’s fair to say that all paradoxes are “just” part of a proof by contradiction and the only difference between ones that feel truly paradoxical and those that feel obvious is something subjective about a person’s understanding of the situation. Maybe it’s down to whether someone has enough of an understanding of the conceptual landscape to be able to actually recognize some of the problematic assumptions from the start of the problem statement or not, or maybe it’s something else, but either way I’m suspecting that it’s something subjective related to a person’s state of knowledge and/or approach to organizing their own knowledge.
- techslave 6y ago
  the birthday paradox is also not a paradox. the namers of such things just like to use the word ‘paradox’ for some reason. it probably sounds more puzzling to initiates.
  - mokus 6y ago
    My gut feel is the same. Ironically, the fact that this usage sounds wrong might, in fact, be a paradox :) It’s an apparent contradiction between our understanding of the meaning of the word and the actual usage of the word. The resolution of this paradox could be that there is widespread wrong usage of the word. Alternately it could be that our understanding of the meaning is wrong - which itself could be for multiple reasons. Perhaps our understanding of the definition is incomplete. Or it could be that the definition itself is simply wrong because it doesn’t reflect actual real-world usage. Upon some reflection I think that a core part of the idea of a paradox is that there are many possible solutions to a problem and that some people, but almost certainly not all, will initially fail to even recognize the existence of solutions other than the one they expect, due to unconscious assumptions. Naturally in many cases there will also be people who do not make those problematic assumptions to begin with, but that doesn’t on its own invalidate the use of the word “paradox” in that situation. The birthday paradox probably fits this definition just fine - ask anyone not trained in statistics about the subject and you’ll likely find a lot of people have unconscious assumptions that lead them to wrong answers, and in some cases thy will even be exceptionally confident in their wrong answers. You might object to the “not trained in statistics” part but that objection itself would be circular - modern statistics education was designed with paradoxes like this in mind, specifically to lay down a foundation that doesn’t lead you down the wrong paths in cases like that. The solutions to the basic paradoxes of statistics are built into the way statistics is taught, just like the solutions to Godel’s and Russel’s and many others’ paradoxes are built into the foundations of any modern logical system.
  - cuspycode 6y ago
    Good point. A paradox is something that produces an unexpected contradiction. It must be unexpected, otherwise any trivial false statement such as 2+2=5 would also be a paradox. Russell's paradox is a good example since no one expected a contradiction. The birthday paradox on the other hand does not produce any contradiction, so it's just a case of sloppy labeling.
- korethr 6y ago
  > To me this seems like something one would use as an example in a physics lecture to show when certain assumptions are necessary and when they aren't. IMO, that's the perfect use of this thought experiment. In fact, I dare say if there are more examples like this, where laying out basic clear assumptions and following the model through its rules leads to an answer that is clearly wrong, then this example and those other examples of that type should probably make up a series of lectures in a course (or subsection thereof) on model use. It seems to me that plenty of people pick up on what models are and how to use them. It seems less people internalize, "All models are wrong. Some models are useful."
  - ivan_ah 6y ago
    You can easily get "invalid" circuits when using diodes (the standard assumption of 0.6V or 0.7V drop across didn't make sense). I once got one of those on a final exam and waster 40+ minutes trying to figure out why my solution was "impossible." The teacher later cancelled that question, but I didn't get my 40 minutes back. Many lesson learned: A) don't be stubborn, let go; and B) circuit models ≠ circuit reality.
- kps 6y ago
  Consider a simplified version of the problem, without the second capacitor. When you close the switch, all of the energy in the first capacitor mysteriously disappears! Or not so mysteriously, after the first time you've done it by accident.
- dialamac 6y ago
  Model inconsistencies, unreasonable mathematical idealiazations and false assumptions are what are commonly classified as paradoxes in physics. It’s a common use of the term. https://en.wikipedia.org/wiki/Physical_paradox https://en.wikipedia.org/wiki/Physical_paradox
- anonymfus 6y ago
  There are three tiers of paradoxes in physics: - these which only contradict "common sense" intuitions/expectations but reality conforms to predictions of the theory instead of "common sense", and so these are usually called not real paradoxes like Pascal's hydrostatic paradox; - these which show us limitations of the models we use to construct them, like two capacitor paradox or black hole information paradox; - these true contradictions in the nature of reality which when you discover them the universe disappears in the puff of logic and you wake up as a student in the middle of the physics lecture, and that student was Albert Einstein. A difference between model which we know how to adjust to make more realistic and one we don't know how is about history of physics and not physics itself.
barbegal 6y ago
Its easier to think about this using the water pressure analogy. Imagine a large barrel full of water connected to another identical barrel. The first barrel is filled to the top whilst the second is empty. A valve is opened between the two barrels and they equalise with half of the water in each. The initial energy is mgh and the final energy is 2 * (m/2 * g * h/2) = mgh/2 so half of then energy has disappeared. It is clear that work could have been done by the water moving between the two barrels (like in a hydro-electric power station).
- Karto 6y ago
  If I think straight, the initial potential energy in a column of liquid would be the integration of mgh (actually integration of density * g * section * height), therefore mgh/2 at the beginning of the situation. Or said intuitively : the center of gravity of the column at half its height. But obviously, even with this 1/2, your relation after dividing between two tanks still holds. :)
  - thomasahle 6y ago
    I was thinking that too, the initial potential energy should be proportional to h^2/2. However then we still have 2*(h/2)^2/2 = h^2/4.
    - Karto 6y ago
      Yes, one h is hidden inside m... Ending up with energy = something * h^2 actually makes the relationship show the same face as the one for capacitors, where energy = something * v^2.
  - unkulunkulu 6y ago
    * where h = half the water level in the first barrel.
  - sirusdas 6y ago
    Good observation!
- 6y ago
mindslight 6y ago
In the true EE parts reduction fashion, you don't even need two capacitors. One capacitor plus a switch works out similarly - in fact it's an equivalent circuit. Everyone has an intuition for shorting a capacitor (zap!). When there are very few components defining a system, the parasitic components must be significant.
- klyrs 6y ago
  The spring/mass model works really well here, too. Connect two identical springs together between a pair of fixed walls. Pull on it so that one spring is stretched and the other is at its natural length....
  - dguest 6y ago
    Or release a ball bearing on one side of a U-shaped track. The ball ends up on the bottom of the track, where did all the potential energy go?
- cushychicken 6y ago
  When there are very few components defining a system, the parasitic components must be significant. Lee Hill, a popular instructor for EMC compliance trainings, has a saying about that: "There's a lot of money to be made understanding the circuit components that aren't captured in the schematic."
leni536 6y ago
A somewhat analogous effect/paradox in thermodynamics is the Joule expansion: https://en.wikipedia.org/wiki/Joule_expansion https://en.wikipedia.org/wiki/Joule_expansion
hellofunk 6y ago
I once got this question on an interview, and they didn’t know I’d heard it before. So I pretended to figure it out during the interview. I gave him a correct answer after just a few minutes, and they cut the interview short and had me directly proceed to the HR office to proceed with onboarding. Two years later I told them how I already knew the question. I’ve been with the company for 6 years now.
- jvm_ 6y ago
  Hadn't coded much in the 9 months I'd been off. Googled Java interview questions, practiced on the first few results. Interview question was the first result... 7 years later.
  - ComodoHacker 6y ago
    Your interviewer: hadn't coded much in the 7 years, googled Java interview questions...
- lalaland1125 6y ago
  Isn't that quite unethical? You are knowingly misleading a potential employer when you play games like that.
  - yesenadam 6y ago
    I've been sitting here for 5 minutes feeling horrified and sad that your comment is downvoted to near unreadability for suggesting that lying like that in an interview may be unethical, while a sociopathic-sounding child comment extolling the virtues of lying/misleading – they "can actually help things flow smoothly and justly", without lying "the interview might have gone badly" and lecturing you on "radical honesty" being a sign of problems/mild autism like you're a naive idiot who just doesn't understand the world – isn't downvoted at all. I wouldn't call it "radical honesty" not to lie in a situation like this! Sounds like "radical honesty" has a new meaning "not lying in any situation you could gain an advantage by doing so". This reminds me of another time on HN someone told a story about how they'd cheated in a hi-tech way in an exam and a dozen comments supported them, not one person suggesting it might be wrong...until me, then I got virulently attacked for doing so. At least there their comment got [dead]ed. I just wanted to say - you're not alone, that also seems very dodgy to me. Moments like this HN is very disappointing.
TheOtherHobbes 6y ago
Explained here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c4 http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.... For an infinitely small resistor the energy is effectively a spark pulse coupled directly to free space, and half is radiated away in EM waves.
- dvt 6y ago
  Not an electrical engineer, but I wish this was higher. Seeing the integral completely reframes the problem and the paradox vanishes. The intuition is that charge transfer is a linear system (half goes here, half goes there), but seeing the integral form shows otherwise. Thanks for the great source!
- dguest 6y ago
  I'm just excited that hyperphysics is still going strong. It looked outdated in 2010!
amelius 6y ago
The problem with this is that once you close the switch, you get an infinite current (aka short circuit). Bad idea!
- kees99 6y ago
  Yeah, with substantial amount of energy in that capacitor, that is exactly how you weld the contacts in that switch shut.
n-gauge 6y ago
Here's another one - put an oscilloscope across a capacitor and turn off the circuit it's connected too. Quickly short the capacitor with something, and watch the voltage on the oscilloscope as the capacitor discharges, remove the short and watch it start recharging slightly. Where does this charge come from?
- edoceo 6y ago
  It's in the cap
- rathel 6y ago
  Dielectric absorption.
- Cyph0n 6y ago
  1. The short is not an ideal short circuit because whatever you use will have some resistance. 2. The shorted capacitor will discharge some current into the short wire and some into the oscilloscope’s capacitor (probably wire capacitance). 3. Once you remove the short wire, some current will flow back into the capacitor under test to equalize the voltage.
- cushychicken 6y ago
  Potentially the scope cable. Cables have capacitance too. That, or this is the "memory effect" I've heard about in certain dielectrics acted out.
- analog31 6y ago
  It's a non ideal behavior of a cap, and some dielectric types have it worse than others. I once made a sample-and-hold for a data acquisition system, and must have tried every type of cap in the bin until I found one that resulted in the least amount of correlation between subsequent readings. I think it was a polypropylene type. Mica was the worst. I don't think I tested an electrolytic.
dekhn 6y ago
I remember explaining how an NMR machine works to an EE person. They simply refused to believe me that you could inject current into a supercooled superconductive magnet and have it circulate for months at a time. The only way I could convince him it worked was to point out that slowly the system did loose energy and you had to go back and add more current.
cushychicken 6y ago
FYI - this is a popular intro level interview question at semiconductor companies. This mechanism is the whole basis of how DRAM works, as well as the foundation of some cap sense technologies. Likely plenty of applications I don't know about, either!
surewhynat 6y ago
It's an exponential function on the W=CV^2 let's say initially: C = 1 V = 16 W = 1 * 16^2 = 256 When the Voltage is split between the two capacitors, the voltage drops in half from 16 to 8, but because there are two capacitors you count them twice: W = CV^2 + CV^2 C = 1 V = 8 W = (1 * 8^2) + (1 * 8^2) = 128 Where did half the energy go? People are saying it's loss in magnetic radiation during the transfer. But on paper it still seems counter intuitive, so it's called a, "paradox," instead of just how an exponential function works. Something in the universe makes the energy levels exponentially higher as voltage increases. More electric pressure (V) = exponentially more energy. Cool!
vegetablepotpie 6y ago
This example violates one of the core assumptions of circuit theory, which is that you cannot have a node with two voltages. As soon as the switch closes, is the voltage at the switch Vi or 0v? It would be both, which is impossible. If you had some kind of component between the two then there would be a voltage drop across it, and you would get realistic results.
- aliceryhl 6y ago
  I mean, that is a simplification, and in the real world it really follows some sort of differential equation that just happens to be very fast. And the description on Wikipedia does seem to use an explanation based on this differential equation, e.g. see "When a steady state is reached".
  - jackhalford 6y ago
    The differential equation is still a model and not what's happening in "the real world", remember, all models are false.
- zoomablemind 6y ago
  The system in question is not closed, as the state change requires application of external energy. Thus expecting a conservation of the initial energy only may be unwarranted. My intuition is that there's a missing counter-action energy which would be required in order to maintain the initial state in such a circuit. Once the state change is being effectuated (switch closed), it would change the balance of action-counter-action which would lead to dissipation in the real world. Not sure how to quantify/formalize it.
- wrycoder 6y ago
  It’s insufficiently specified. There is actually a small resistance in series with an inductor between the nodes. There’s a time varying voltage across those. Edit: as a follow-on, the paradox is posed as a problem in electrostatics, when it’s actually a problem in electrodynamics.
mdturnerphys 6y ago
Adding the second capacitor is equivalent doubling the area of a single capacitor. If you had a plate capacitor where the plates were extendable you could charge it and decrease the energy by extending the plates, which tells you that there's a force that is pushing to extend the plates. Of course there is--the charges on one plate repel each other and "want" to increase their spacing. If you do extend the plates there is a force times a distance, which equals work done by the charges, decreasing the capacitive energy.
- compumike 6y ago
  This is a beautiful explanation, thank you! It's also compatible with barbegal's hydraulic pressure analogy elsewhere in the discussion. A liquid- or gas-filled tank feels pressure (i.e. force per area) on the walls as the fluid "wants" to either (1) flow to a lower gravitational potential energy location, for a liquid tank in a gravitational field, or (2) expand in volume, in the case of a pressurized gas tank.
irrational 6y ago
I knew nothing about electricity until I started studying for my technicians license recently. It was actually very interesting and I learned enough that I could actually understand this wiki article. It made me interested in studying for the general and extra licenses since they have more advanced electrical knowledge required.
dheera 6y ago
I wouldn't really consider this a paradox. It is a failure mode of the idealization you made in saying that wires and capacitors have zero resistance and inductance. Even if you work by your idealization, saying wires have zero resistance means that every piece of connected conductor in the circuit is at the same potential (in the absence of a magnetic field) so saying you have two connected capacitors charged to different potentials is already violating your assumption. You can find similar edge cases in almost every situation and field of study where you try to simplify things with an approximation, and most of them aren't called paradoxes.
moultano 6y ago
Does anyone have a sense of how much the idealization of circuit design limits people's imagination when designing real circuits? Are there fruitful possibilities that could be explored but aren't because the abstraction doesn't contain them?
- pps43 6y ago
  This question tests understanding of limitations of idealized circuits. Anyone with hands-on experience of shorting a capacitor knows about sparks and arcing that accompanies it.
  - fdavison 6y ago
    "While in theory there is no difference between theory and practice, in practice there is."
- zokier 6y ago
  The effect is smaller than you might imagine, because many of parasitics can be modeled as adding additional basic elements and are understood fairly well. For example here is just some random model you could use for capacitor http://www.iequalscdvdt.com/cap_model.html http://www.iequalscdvdt.com/cap_model.html If simple lumped element model really fails to model your circuit, there is distributed element model which can analyze all sorts of RF voodoo https://en.m.wikipedia.org/wiki/Distributed-element_model https://en.m.wikipedia.org/wiki/Distributed-element_model and https://en.m.wikipedia.org/wiki/Distributed-element_circuit https://en.m.wikipedia.org/wiki/Distributed-element_circuit
ummonk 6y ago
Without reading the solution: wouldn’t this impossible zero resistance zero inductance ideal setup result in infinite frequency infinite magnitude oscillations? Add in some resistance and the missing energy goes into heat generated by the resistance on the way to equilibrium. Edit: also when I think about it there is a little bit of additional energy in the open switch which is itself a capacitor. Edit 2: for this circuit to stay the way it is in the initial state, wouldn’t the open switch need to have equal capacitance to the capacitors? Or some kind of voltage generating field applied across it that is removed when the switch is closed?
- pps43 6y ago
  If inductance is zero, there would be no oscillations.
  - ummonk 6y ago
    That would be true if resistance were non-zero.
    - pps43 6y ago
      To get oscillations, you need to shift the phase by 180 degrees. Capacitance alone (without some inductance) can only do 90.
rzimmerman 6y ago
This is effectively the same paradox some friends and I discussed in college. Another way of looking at it is: If you connect an ideal voltage source through a resistor R to a capacitor C, the amount of energy required to charge the capacitor is CV^2 while the amount of energy that winds up in the charged capacitor is 1/2 CV^2. The other 1/2 CV^s is dissipated across R. This is unaffected by the value of R, even as R -> 0. The only thing that changes is the charge time. R = 0 is impossible for real circuits, but no matter how close you get you still lose half of that energy in the resistor.
Animats 6y ago
It's simpler than that. When you short a capacitor, where does the energy go? An ideal capacitor has an infinite current in that situation. Inductors have a comparable situation - when you open-circuit an ideal inductor, you get an infinite voltage. In practice, you can get hundreds or even thousands of volts from inductors that way, which is how auto ignitions and boost-type switching power supplies work. Similar problems come up in the idealized physics of impulse/constraint physics engines. Getting rid of the energy in collisions requires hacks to prevent things from flying off into space, a problem with early physics engines.
- pps43 6y ago
  That's one way to solve it. More generally, thinking of edge cases or simplifying the problem is a good problem-solving strategy.
aj7 6y ago
Try it in Spice with realistic small resistances and inductance. Every electrical engineer is taught, in basic theory: The voltage on a capacitor cannot change instantaneously. The current in an inductor cannot change instantaneously. These become much clearer when you write down the INTEGRAL form of the V-I relations.
pps43 6y ago
A real-world analogy is a brick standing upright on the table. If you topple it and it falls, some of the potential energy is gone.
pholos 6y ago
Hearing about this problem while I was taking a stats mech course (online from Leo Susskind) this reminded me more of energy / entropy relationship than a paradox. For example if we have hot and cold reservoirs separated by a divider in a bath, when we remove the divider entropy of the system increases when it reaches equilibrium. After reaching equilibrium, moving the divider back will not cause the two reservoirs to return to the initial hot cold state. The system with 2 capacitors seems like a good analogy to a heat bath; when the switch is flipped the charge is divided equally between the capacitors when it reaches equilibrium. The entropy of the system has increased, and the potential to do work has decreased. The number of possible states the system can be identified in has decreased by half (it is not possible to know which side had the initial charge after the switch has been flipped). Flipping the switch back will not bring the charge back to only one side. While this line of thinking doesn't explain the physics of what is happening, there is clearly a (statistically) irreversible change going on which seems like the natural language is energy, entropy, and temperature.
CoffeeDregs 6y ago
Got this question in an interview at National Semiconductor (out of MSEE school). I didn't think it was too paradox-y: we didn't work the math through fully but my answer of "start with the circuit with a resistor with value R and the energy will burn up in R; now recalculate as R→0 and the energy will still burn up in R (even though R is zero-y...". That seemed to satisfy the interviewer (who had not heard that answer before).
- gnramires 6y ago
  I think an interesting question following it up is, how do you transfer energy between capacitors then, without losing so much energy? (someone else asked if this is a fundamental limitation) There are many alternatives. A good start is noting that the inefficiency is actually lower the lower the starting voltage difference: V1 = V, V2 = V-dV V' = (V + (V-dV))/2 = V+dV/2 We can define efficiency as the ratio of energy lost by the first to energy transferred to the other. dU1 = U1-U1' = CV^2/2-CV'^2/2 = C/2 (V dV-dV^2/4) dU2 = U2'-U2 = CV'^2/2-C(V-dV)^2/2 = C/2 (-V dV+dV^2/4+2VdV-dV^2) = C/2 (V dV - 3dV^2/4 ) n = (dU2/dU1) = ( V - 3dV/4 ) / ( V - dV/4) ~ 1 - 3/4 dV/V (for small dV) as dV → 0, n → 1. You lose 3/4 of the fractional difference in efficiency, so for 10% difference your efficiency is ~92.5%, pretty great. Now, there are indeed devices that change voltages without energy losses! (transformers, for example). So if you plug in a variable transformer that keeps the voltage close to the target, your (dis)charging efficiency can be arbitrarily high. Of course, if your second voltage is 0, the efficiency must start at 1/3 no matter what (which can seem to imply this cannot be changed) -- but as soon as you have a small voltage you can start tracking it and keep efficiency high. Challenge to the reader: Use quantum mechanics and thermodynamics to derive a fundamental limit of efficiency (which must be less than 1 at positive temperatures) :)
  - mindslight 6y ago
    Inductor in series with a diode. The voltage differential stores energy in the inductor's magnetic field, which comes back out when the destination capacitor is at a higher voltage than the source capacitor. The diode keeps the process from repeating in reverse (oscillation).
Stierlitz 6y ago
My brain hurts, is there a simpler version?
- Dylan16807 6y ago
  Yes there is, and it's listed at the end. You have an ideal capacitor, charged to some voltage. You connect both ends, which empties it. Where did all the energy go? The answer is that there's no such thing as an ideal capacitor or an ideal wire. There are a few effects you can't avoid, but most importantly there are tiny bits of internal resistance that usually go ignored. In a short circuit, they become important, and will turn all the energy into heat.
Stierlitz 6y ago
@surewhynat Why did someone mod you down dead? - quote - It's an exponential function on the W=CV^2 let's say initially: C = 1 V = 16 W = 1 * 16^2 = 256 When the Voltage is split between the two capacitors, the voltage drops in half from 16 to 8, but because there are two capacitors you count them twice: W = CV^2 + CV^2 C = 1 V = 8 W = (1 * 8^2) + (1 * 8^2) = 128 Where did half the energy go? People are saying it's loss in magnetic radiation during the transfer. But on paper it still seems counter intuitive, so it's called a, "paradox," instead of just how an exponential function works. Something in the universe makes the energy levels exponentially higher as voltage increases. More electric pressure (V) = exponentially more energy. Cool! - unquote -
- gnramires 6y ago
  Correction: It's not exponential, it's a quadratic (or polynomial). (I think the difficult part isn't to accept that some of the energy vanished from the system, it just contradicts our expectation from having supposedly no dissipation elements)
wolfgke 6y ago
What experiments were done to settle the question which property of the "non-idealized" models is the dominant one?
kees99 6y ago
Funny thing is - this is exactly what happens many millions times every single clock cycle inside pretty much any modern(-ish) digital CMOS CPU/ASIC, with the right-hand-side capacitor being parasitic gate capacitance in the driven gate.
- jacquesm 6y ago
  Which in turn is why your CPU needs a heatsink. And why reversible computing can be done without using energy ;)
simonblack 6y ago
The 'paradox' in this experiment lies in believing that you can have a voltage across one of the capacitors and zero volts across the other before you hit the switch. In other words, you can't have two connected locations resting at different voltages. --------| |---------------| |---------- --- +6 -6 ------------ 0 0 ----- ^^^^^^^^^^^^^^^^ NOT POSSIBLE! Two capacitors in series are equivalent to one (combined) capacitor, just like a bunch of batteries can be considered to be one (combined) battery.
- Dylan16807 6y ago
  You can definitely have a charged capacitor in series with an uncharged capacitor. They only act as an evenly-split combined capacitor if they start at the same charge. Your diagram is wrong. Let's mark out some contact points: ---| |-------| |--- A B C D Voltages only exist between points. To measure against ground, part of the circuit has to be connected to ground. Ground is a point like any other. So when you measure A at 12 volts above B, it doesn't mean B is at any absolute voltage. If only one capacitor is charged, then A can be 12 volts above B while B=C=D. It works out fine. If you attach ground to B or C, then A is 12 volts relative to ground, and B and C and D are all 0 volts relative to ground. If you attach ground to A, then A is 0 volts relative to ground, B and C and D are all -12 volts relative to ground. If you somehow attach ground to the midpoint of A and B, then A is 6 volts relative to ground, and B and C and D are all -6 volts relative to ground. None of these are impossible. Nothing goes wrong until you try to close the switch.
userbinator 6y ago
This is another instance of those situations where "the abstraction is leaky". Relatedly, I've seen (unfortunately) a few textbooks which attempt to teach basics of computing and digital logic by assuming that gates have no propagation delay, or at least that's what their timing diagrams seem to show. It's very puzzling because a lot of sequential circuit elements rely on that in order to work.
ars 6y ago
I learned that this is also why you should never connect batteries in parallel - the power will oscillate in much the same manner, draining away the energy and wearing out the batteries.
phendrenad2 6y ago
This is an easy one. The assumptions are flawed. W = (0.5) x(C)x(V) is a back-of-the-envelope shorthand for a whole realm of formulas in the study of capacitors, and it doesn't work in edge cases like this.

438 points by rathel 6y ago

I don't think this should be called a paradox. It's just a case where the limitations of the model (ideal everything) is clear and leads to inconsistent results. Adjusting the model and making it more realistic, quickly clears up the "paradox". To me this seems like something one would use as an example in a physics lecture to show when certain assumptions are necessary and when they aren't.

JadeNB 6y ago
Notice the definitions: > The two capacitor paradox or capacitor paradox is a paradox, or counterintuitive thought experiment, in electric circuit theory. > A paradox, also known as an antinomy, is a logically self-contradictory statement or a statement that runs contrary to one's expectation. It is a statement that, despite apparently valid reasoning from true premises, leads to a seemingly self-contradictory or a logically unacceptable conclusion. (emphases mine). It's paradoxical in that it runs counter to expectations or intuition, using apparently valid reasoning. (Of course, maybe it doesn't run counter to your intuition; but that's how correct intuition is developed, by breaking naïve intuition against examples like this!) In fact, the article explicitly acknowledges this: > Unlike some other paradoxes in science, this paradox is not due to the underlying physics, but to the limitations of the 'ideal circuit' conventions used in circuit theory.
- brummm 6y ago
  Fair enough, wikipedia does include the reference to expectation in its definition of paradox. However, I think my main point is still valid that this should not necessarily be called a paradox exactly because it differs a lot from how this term is commonly used in the naming of scientific thought experiments.
  - mokus 6y ago
    My initial gut response was the same - that this isn’t a paradox, but rather a proof by contradiction that one or more of the assumptions are wrong. But after thinking about it a bit more, I don’t think there’s really that much of a difference. It’s just that some people have already learned not to apply those assumptions in this context (or by chance never stumbled into them in the first place). Perhaps it’s fair to say that all paradoxes are “just” part of a proof by contradiction and the only difference between ones that feel truly paradoxical and those that feel obvious is something subjective about a person’s understanding of the situation. Maybe it’s down to whether someone has enough of an understanding of the conceptual landscape to be able to actually recognize some of the problematic assumptions from the start of the problem statement or not, or maybe it’s something else, but either way I’m suspecting that it’s something subjective related to a person’s state of knowledge and/or approach to organizing their own knowledge.
techslave 6y ago
the birthday paradox is also not a paradox. the namers of such things just like to use the word ‘paradox’ for some reason. it probably sounds more puzzling to initiates.
- mokus 6y ago
  My gut feel is the same. Ironically, the fact that this usage sounds wrong might, in fact, be a paradox :) It’s an apparent contradiction between our understanding of the meaning of the word and the actual usage of the word. The resolution of this paradox could be that there is widespread wrong usage of the word. Alternately it could be that our understanding of the meaning is wrong - which itself could be for multiple reasons. Perhaps our understanding of the definition is incomplete. Or it could be that the definition itself is simply wrong because it doesn’t reflect actual real-world usage. Upon some reflection I think that a core part of the idea of a paradox is that there are many possible solutions to a problem and that some people, but almost certainly not all, will initially fail to even recognize the existence of solutions other than the one they expect, due to unconscious assumptions. Naturally in many cases there will also be people who do not make those problematic assumptions to begin with, but that doesn’t on its own invalidate the use of the word “paradox” in that situation. The birthday paradox probably fits this definition just fine - ask anyone not trained in statistics about the subject and you’ll likely find a lot of people have unconscious assumptions that lead them to wrong answers, and in some cases thy will even be exceptionally confident in their wrong answers. You might object to the “not trained in statistics” part but that objection itself would be circular - modern statistics education was designed with paradoxes like this in mind, specifically to lay down a foundation that doesn’t lead you down the wrong paths in cases like that. The solutions to the basic paradoxes of statistics are built into the way statistics is taught, just like the solutions to Godel’s and Russel’s and many others’ paradoxes are built into the foundations of any modern logical system.
- cuspycode 6y ago
  Good point. A paradox is something that produces an unexpected contradiction. It must be unexpected, otherwise any trivial false statement such as 2+2=5 would also be a paradox. Russell's paradox is a good example since no one expected a contradiction. The birthday paradox on the other hand does not produce any contradiction, so it's just a case of sloppy labeling.
korethr 6y ago
> To me this seems like something one would use as an example in a physics lecture to show when certain assumptions are necessary and when they aren't. IMO, that's the perfect use of this thought experiment. In fact, I dare say if there are more examples like this, where laying out basic clear assumptions and following the model through its rules leads to an answer that is clearly wrong, then this example and those other examples of that type should probably make up a series of lectures in a course (or subsection thereof) on model use. It seems to me that plenty of people pick up on what models are and how to use them. It seems less people internalize, "All models are wrong. Some models are useful."
- ivan_ah 6y ago
  You can easily get "invalid" circuits when using diodes (the standard assumption of 0.6V or 0.7V drop across didn't make sense). I once got one of those on a final exam and waster 40+ minutes trying to figure out why my solution was "impossible." The teacher later cancelled that question, but I didn't get my 40 minutes back. Many lesson learned: A) don't be stubborn, let go; and B) circuit models ≠ circuit reality.
kps 6y ago
Consider a simplified version of the problem, without the second capacitor. When you close the switch, all of the energy in the first capacitor mysteriously disappears! Or not so mysteriously, after the first time you've done it by accident.
dialamac 6y ago
Model inconsistencies, unreasonable mathematical idealiazations and false assumptions are what are commonly classified as paradoxes in physics. It’s a common use of the term. https://en.wikipedia.org/wiki/Physical_paradox https://en.wikipedia.org/wiki/Physical_paradox
anonymfus 6y ago
There are three tiers of paradoxes in physics: - these which only contradict "common sense" intuitions/expectations but reality conforms to predictions of the theory instead of "common sense", and so these are usually called not real paradoxes like Pascal's hydrostatic paradox; - these which show us limitations of the models we use to construct them, like two capacitor paradox or black hole information paradox; - these true contradictions in the nature of reality which when you discover them the universe disappears in the puff of logic and you wake up as a student in the middle of the physics lecture, and that student was Albert Einstein. A difference between model which we know how to adjust to make more realistic and one we don't know how is about history of physics and not physics itself.

barbegal 6y ago

Its easier to think about this using the water pressure analogy. Imagine a large barrel full of water connected to another identical barrel. The first barrel is filled to the top whilst the second is empty. A valve is opened between the two barrels and they equalise with half of the water in each. The initial energy is mgh and the final energy is 2 * (m/2 * g * h/2) = mgh/2 so half of then energy has disappeared. It is clear that work could have been done by the water moving between the two barrels (like in a hydro-electric power station).

Karto 6y ago
If I think straight, the initial potential energy in a column of liquid would be the integration of mgh (actually integration of density * g * section * height), therefore mgh/2 at the beginning of the situation. Or said intuitively : the center of gravity of the column at half its height. But obviously, even with this 1/2, your relation after dividing between two tanks still holds. :)
- thomasahle 6y ago
  I was thinking that too, the initial potential energy should be proportional to h^2/2. However then we still have 2*(h/2)^2/2 = h^2/4.
  - Karto 6y ago
    Yes, one h is hidden inside m... Ending up with energy = something * h^2 actually makes the relationship show the same face as the one for capacitors, where energy = something * v^2.
- unkulunkulu 6y ago
  * where h = half the water level in the first barrel.
- sirusdas 6y ago
  Good observation!
6y ago

mindslight 6y ago

In the true EE parts reduction fashion, you don't even need two capacitors. One capacitor plus a switch works out similarly - in fact it's an equivalent circuit. Everyone has an intuition for shorting a capacitor (zap!). When there are very few components defining a system, the parasitic components must be significant.

klyrs 6y ago
The spring/mass model works really well here, too. Connect two identical springs together between a pair of fixed walls. Pull on it so that one spring is stretched and the other is at its natural length....
- dguest 6y ago
  Or release a ball bearing on one side of a U-shaped track. The ball ends up on the bottom of the track, where did all the potential energy go?
cushychicken 6y ago
When there are very few components defining a system, the parasitic components must be significant. Lee Hill, a popular instructor for EMC compliance trainings, has a saying about that: "There's a lot of money to be made understanding the circuit components that aren't captured in the schematic."

leni536 6y ago

A somewhat analogous effect/paradox in thermodynamics is the Joule expansion: https://en.wikipedia.org/wiki/Joule_expansion https://en.wikipedia.org/wiki/Joule_expansion

hellofunk 6y ago

I once got this question on an interview, and they didn’t know I’d heard it before. So I pretended to figure it out during the interview. I gave him a correct answer after just a few minutes, and they cut the interview short and had me directly proceed to the HR office to proceed with onboarding. Two years later I told them how I already knew the question. I’ve been with the company for 6 years now.

jvm_ 6y ago
Hadn't coded much in the 9 months I'd been off. Googled Java interview questions, practiced on the first few results. Interview question was the first result... 7 years later.
- ComodoHacker 6y ago
  Your interviewer: hadn't coded much in the 7 years, googled Java interview questions...
lalaland1125 6y ago
Isn't that quite unethical? You are knowingly misleading a potential employer when you play games like that.
- yesenadam 6y ago
  I've been sitting here for 5 minutes feeling horrified and sad that your comment is downvoted to near unreadability for suggesting that lying like that in an interview may be unethical, while a sociopathic-sounding child comment extolling the virtues of lying/misleading – they "can actually help things flow smoothly and justly", without lying "the interview might have gone badly" and lecturing you on "radical honesty" being a sign of problems/mild autism like you're a naive idiot who just doesn't understand the world – isn't downvoted at all. I wouldn't call it "radical honesty" not to lie in a situation like this! Sounds like "radical honesty" has a new meaning "not lying in any situation you could gain an advantage by doing so". This reminds me of another time on HN someone told a story about how they'd cheated in a hi-tech way in an exam and a dozen comments supported them, not one person suggesting it might be wrong...until me, then I got virulently attacked for doing so. At least there their comment got [dead]ed. I just wanted to say - you're not alone, that also seems very dodgy to me. Moments like this HN is very disappointing.

TheOtherHobbes 6y ago

Explained here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c4 http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.... For an infinitely small resistor the energy is effectively a spark pulse coupled directly to free space, and half is radiated away in EM waves.

dvt 6y ago
Not an electrical engineer, but I wish this was higher. Seeing the integral completely reframes the problem and the paradox vanishes. The intuition is that charge transfer is a linear system (half goes here, half goes there), but seeing the integral form shows otherwise. Thanks for the great source!
dguest 6y ago
I'm just excited that hyperphysics is still going strong. It looked outdated in 2010!

amelius 6y ago

The problem with this is that once you close the switch, you get an infinite current (aka short circuit). Bad idea!

kees99 6y ago
Yeah, with substantial amount of energy in that capacitor, that is exactly how you weld the contacts in that switch shut.

n-gauge 6y ago

Here's another one - put an oscilloscope across a capacitor and turn off the circuit it's connected too. Quickly short the capacitor with something, and watch the voltage on the oscilloscope as the capacitor discharges, remove the short and watch it start recharging slightly. Where does this charge come from?

edoceo 6y ago
It's in the cap
rathel 6y ago
Dielectric absorption.
Cyph0n 6y ago
1. The short is not an ideal short circuit because whatever you use will have some resistance. 2. The shorted capacitor will discharge some current into the short wire and some into the oscilloscope’s capacitor (probably wire capacitance). 3. Once you remove the short wire, some current will flow back into the capacitor under test to equalize the voltage.
cushychicken 6y ago
Potentially the scope cable. Cables have capacitance too. That, or this is the "memory effect" I've heard about in certain dielectrics acted out.
analog31 6y ago
It's a non ideal behavior of a cap, and some dielectric types have it worse than others. I once made a sample-and-hold for a data acquisition system, and must have tried every type of cap in the bin until I found one that resulted in the least amount of correlation between subsequent readings. I think it was a polypropylene type. Mica was the worst. I don't think I tested an electrolytic.

dekhn 6y ago

I remember explaining how an NMR machine works to an EE person. They simply refused to believe me that you could inject current into a supercooled superconductive magnet and have it circulate for months at a time. The only way I could convince him it worked was to point out that slowly the system did loose energy and you had to go back and add more current.

cushychicken 6y ago

FYI - this is a popular intro level interview question at semiconductor companies. This mechanism is the whole basis of how DRAM works, as well as the foundation of some cap sense technologies. Likely plenty of applications I don't know about, either!

surewhynat 6y ago

It's an exponential function on the W=CV^2 let's say initially: C = 1 V = 16 W = 1 * 16^2 = 256 When the Voltage is split between the two capacitors, the voltage drops in half from 16 to 8, but because there are two capacitors you count them twice: W = CV^2 + CV^2 C = 1 V = 8 W = (1 * 8^2) + (1 * 8^2) = 128 Where did half the energy go? People are saying it's loss in magnetic radiation during the transfer. But on paper it still seems counter intuitive, so it's called a, "paradox," instead of just how an exponential function works. Something in the universe makes the energy levels exponentially higher as voltage increases. More electric pressure (V) = exponentially more energy. Cool!

vegetablepotpie 6y ago

This example violates one of the core assumptions of circuit theory, which is that you cannot have a node with two voltages. As soon as the switch closes, is the voltage at the switch Vi or 0v? It would be both, which is impossible. If you had some kind of component between the two then there would be a voltage drop across it, and you would get realistic results.

aliceryhl 6y ago
I mean, that is a simplification, and in the real world it really follows some sort of differential equation that just happens to be very fast. And the description on Wikipedia does seem to use an explanation based on this differential equation, e.g. see "When a steady state is reached".
- jackhalford 6y ago
  The differential equation is still a model and not what's happening in "the real world", remember, all models are false.
zoomablemind 6y ago
The system in question is not closed, as the state change requires application of external energy. Thus expecting a conservation of the initial energy only may be unwarranted. My intuition is that there's a missing counter-action energy which would be required in order to maintain the initial state in such a circuit. Once the state change is being effectuated (switch closed), it would change the balance of action-counter-action which would lead to dissipation in the real world. Not sure how to quantify/formalize it.
wrycoder 6y ago
It’s insufficiently specified. There is actually a small resistance in series with an inductor between the nodes. There’s a time varying voltage across those. Edit: as a follow-on, the paradox is posed as a problem in electrostatics, when it’s actually a problem in electrodynamics.

mdturnerphys 6y ago

Adding the second capacitor is equivalent doubling the area of a single capacitor. If you had a plate capacitor where the plates were extendable you could charge it and decrease the energy by extending the plates, which tells you that there's a force that is pushing to extend the plates. Of course there is--the charges on one plate repel each other and "want" to increase their spacing. If you do extend the plates there is a force times a distance, which equals work done by the charges, decreasing the capacitive energy.

compumike 6y ago
This is a beautiful explanation, thank you! It's also compatible with barbegal's hydraulic pressure analogy elsewhere in the discussion. A liquid- or gas-filled tank feels pressure (i.e. force per area) on the walls as the fluid "wants" to either (1) flow to a lower gravitational potential energy location, for a liquid tank in a gravitational field, or (2) expand in volume, in the case of a pressurized gas tank.

irrational 6y ago

I knew nothing about electricity until I started studying for my technicians license recently. It was actually very interesting and I learned enough that I could actually understand this wiki article. It made me interested in studying for the general and extra licenses since they have more advanced electrical knowledge required.

dheera 6y ago

I wouldn't really consider this a paradox. It is a failure mode of the idealization you made in saying that wires and capacitors have zero resistance and inductance. Even if you work by your idealization, saying wires have zero resistance means that every piece of connected conductor in the circuit is at the same potential (in the absence of a magnetic field) so saying you have two connected capacitors charged to different potentials is already violating your assumption. You can find similar edge cases in almost every situation and field of study where you try to simplify things with an approximation, and most of them aren't called paradoxes.

moultano 6y ago

Does anyone have a sense of how much the idealization of circuit design limits people's imagination when designing real circuits? Are there fruitful possibilities that could be explored but aren't because the abstraction doesn't contain them?

pps43 6y ago
This question tests understanding of limitations of idealized circuits. Anyone with hands-on experience of shorting a capacitor knows about sparks and arcing that accompanies it.
- fdavison 6y ago
  "While in theory there is no difference between theory and practice, in practice there is."
zokier 6y ago
The effect is smaller than you might imagine, because many of parasitics can be modeled as adding additional basic elements and are understood fairly well. For example here is just some random model you could use for capacitor http://www.iequalscdvdt.com/cap_model.html http://www.iequalscdvdt.com/cap_model.html If simple lumped element model really fails to model your circuit, there is distributed element model which can analyze all sorts of RF voodoo https://en.m.wikipedia.org/wiki/Distributed-element_model https://en.m.wikipedia.org/wiki/Distributed-element_model and https://en.m.wikipedia.org/wiki/Distributed-element_circuit https://en.m.wikipedia.org/wiki/Distributed-element_circuit

ummonk 6y ago

Without reading the solution: wouldn’t this impossible zero resistance zero inductance ideal setup result in infinite frequency infinite magnitude oscillations? Add in some resistance and the missing energy goes into heat generated by the resistance on the way to equilibrium. Edit: also when I think about it there is a little bit of additional energy in the open switch which is itself a capacitor. Edit 2: for this circuit to stay the way it is in the initial state, wouldn’t the open switch need to have equal capacitance to the capacitors? Or some kind of voltage generating field applied across it that is removed when the switch is closed?

pps43 6y ago
If inductance is zero, there would be no oscillations.
- ummonk 6y ago
  That would be true if resistance were non-zero.
  - pps43 6y ago
    To get oscillations, you need to shift the phase by 180 degrees. Capacitance alone (without some inductance) can only do 90.

rzimmerman 6y ago

This is effectively the same paradox some friends and I discussed in college. Another way of looking at it is: If you connect an ideal voltage source through a resistor R to a capacitor C, the amount of energy required to charge the capacitor is CV^2 while the amount of energy that winds up in the charged capacitor is 1/2 CV^2. The other 1/2 CV^s is dissipated across R. This is unaffected by the value of R, even as R -> 0. The only thing that changes is the charge time. R = 0 is impossible for real circuits, but no matter how close you get you still lose half of that energy in the resistor.

Animats 6y ago

It's simpler than that. When you short a capacitor, where does the energy go? An ideal capacitor has an infinite current in that situation. Inductors have a comparable situation - when you open-circuit an ideal inductor, you get an infinite voltage. In practice, you can get hundreds or even thousands of volts from inductors that way, which is how auto ignitions and boost-type switching power supplies work. Similar problems come up in the idealized physics of impulse/constraint physics engines. Getting rid of the energy in collisions requires hacks to prevent things from flying off into space, a problem with early physics engines.

pps43 6y ago
That's one way to solve it. More generally, thinking of edge cases or simplifying the problem is a good problem-solving strategy.

aj7 6y ago

Try it in Spice with realistic small resistances and inductance. Every electrical engineer is taught, in basic theory: The voltage on a capacitor cannot change instantaneously. The current in an inductor cannot change instantaneously. These become much clearer when you write down the INTEGRAL form of the V-I relations.

pps43 6y ago

A real-world analogy is a brick standing upright on the table. If you topple it and it falls, some of the potential energy is gone.

pholos 6y ago

Hearing about this problem while I was taking a stats mech course (online from Leo Susskind) this reminded me more of energy / entropy relationship than a paradox. For example if we have hot and cold reservoirs separated by a divider in a bath, when we remove the divider entropy of the system increases when it reaches equilibrium. After reaching equilibrium, moving the divider back will not cause the two reservoirs to return to the initial hot cold state. The system with 2 capacitors seems like a good analogy to a heat bath; when the switch is flipped the charge is divided equally between the capacitors when it reaches equilibrium. The entropy of the system has increased, and the potential to do work has decreased. The number of possible states the system can be identified in has decreased by half (it is not possible to know which side had the initial charge after the switch has been flipped). Flipping the switch back will not bring the charge back to only one side. While this line of thinking doesn't explain the physics of what is happening, there is clearly a (statistically) irreversible change going on which seems like the natural language is energy, entropy, and temperature.

CoffeeDregs 6y ago

Got this question in an interview at National Semiconductor (out of MSEE school). I didn't think it was too paradox-y: we didn't work the math through fully but my answer of "start with the circuit with a resistor with value R and the energy will burn up in R; now recalculate as R→0 and the energy will still burn up in R (even though R is zero-y...". That seemed to satisfy the interviewer (who had not heard that answer before).

gnramires 6y ago
I think an interesting question following it up is, how do you transfer energy between capacitors then, without losing so much energy? (someone else asked if this is a fundamental limitation) There are many alternatives. A good start is noting that the inefficiency is actually lower the lower the starting voltage difference: V1 = V, V2 = V-dV V' = (V + (V-dV))/2 = V+dV/2 We can define efficiency as the ratio of energy lost by the first to energy transferred to the other. dU1 = U1-U1' = CV^2/2-CV'^2/2 = C/2 (V dV-dV^2/4) dU2 = U2'-U2 = CV'^2/2-C(V-dV)^2/2 = C/2 (-V dV+dV^2/4+2VdV-dV^2) = C/2 (V dV - 3dV^2/4 ) n = (dU2/dU1) = ( V - 3dV/4 ) / ( V - dV/4) ~ 1 - 3/4 dV/V (for small dV) as dV → 0, n → 1. You lose 3/4 of the fractional difference in efficiency, so for 10% difference your efficiency is ~92.5%, pretty great. Now, there are indeed devices that change voltages without energy losses! (transformers, for example). So if you plug in a variable transformer that keeps the voltage close to the target, your (dis)charging efficiency can be arbitrarily high. Of course, if your second voltage is 0, the efficiency must start at 1/3 no matter what (which can seem to imply this cannot be changed) -- but as soon as you have a small voltage you can start tracking it and keep efficiency high. Challenge to the reader: Use quantum mechanics and thermodynamics to derive a fundamental limit of efficiency (which must be less than 1 at positive temperatures) :)
- mindslight 6y ago
  Inductor in series with a diode. The voltage differential stores energy in the inductor's magnetic field, which comes back out when the destination capacitor is at a higher voltage than the source capacitor. The diode keeps the process from repeating in reverse (oscillation).

Stierlitz 6y ago

My brain hurts, is there a simpler version?

Dylan16807 6y ago
Yes there is, and it's listed at the end. You have an ideal capacitor, charged to some voltage. You connect both ends, which empties it. Where did all the energy go? The answer is that there's no such thing as an ideal capacitor or an ideal wire. There are a few effects you can't avoid, but most importantly there are tiny bits of internal resistance that usually go ignored. In a short circuit, they become important, and will turn all the energy into heat.

Stierlitz 6y ago

@surewhynat Why did someone mod you down dead? - quote - It's an exponential function on the W=CV^2 let's say initially: C = 1 V = 16 W = 1 * 16^2 = 256 When the Voltage is split between the two capacitors, the voltage drops in half from 16 to 8, but because there are two capacitors you count them twice: W = CV^2 + CV^2 C = 1 V = 8 W = (1 * 8^2) + (1 * 8^2) = 128 Where did half the energy go? People are saying it's loss in magnetic radiation during the transfer. But on paper it still seems counter intuitive, so it's called a, "paradox," instead of just how an exponential function works. Something in the universe makes the energy levels exponentially higher as voltage increases. More electric pressure (V) = exponentially more energy. Cool! - unquote -

gnramires 6y ago
Correction: It's not exponential, it's a quadratic (or polynomial). (I think the difficult part isn't to accept that some of the energy vanished from the system, it just contradicts our expectation from having supposedly no dissipation elements)

wolfgke 6y ago

What experiments were done to settle the question which property of the "non-idealized" models is the dominant one?

kees99 6y ago

Funny thing is - this is exactly what happens many millions times every single clock cycle inside pretty much any modern(-ish) digital CMOS CPU/ASIC, with the right-hand-side capacitor being parasitic gate capacitance in the driven gate.

jacquesm 6y ago
Which in turn is why your CPU needs a heatsink. And why reversible computing can be done without using energy ;)

simonblack 6y ago

The 'paradox' in this experiment lies in believing that you can have a voltage across one of the capacitors and zero volts across the other before you hit the switch. In other words, you can't have two connected locations resting at different voltages. --------| |---------------| |---------- --- +6 -6 ------------ 0 0 ----- ^^^^^^^^^^^^^^^^ NOT POSSIBLE! Two capacitors in series are equivalent to one (combined) capacitor, just like a bunch of batteries can be considered to be one (combined) battery.

Dylan16807 6y ago
You can definitely have a charged capacitor in series with an uncharged capacitor. They only act as an evenly-split combined capacitor if they start at the same charge. Your diagram is wrong. Let's mark out some contact points: ---| |-------| |--- A B C D Voltages only exist between points. To measure against ground, part of the circuit has to be connected to ground. Ground is a point like any other. So when you measure A at 12 volts above B, it doesn't mean B is at any absolute voltage. If only one capacitor is charged, then A can be 12 volts above B while B=C=D. It works out fine. If you attach ground to B or C, then A is 12 volts relative to ground, and B and C and D are all 0 volts relative to ground. If you attach ground to A, then A is 0 volts relative to ground, B and C and D are all -12 volts relative to ground. If you somehow attach ground to the midpoint of A and B, then A is 6 volts relative to ground, and B and C and D are all -6 volts relative to ground. None of these are impossible. Nothing goes wrong until you try to close the switch.

userbinator 6y ago

This is another instance of those situations where "the abstraction is leaky". Relatedly, I've seen (unfortunately) a few textbooks which attempt to teach basics of computing and digital logic by assuming that gates have no propagation delay, or at least that's what their timing diagrams seem to show. It's very puzzling because a lot of sequential circuit elements rely on that in order to work.

ars 6y ago

I learned that this is also why you should never connect batteries in parallel - the power will oscillate in much the same manner, draining away the energy and wearing out the batteries.

phendrenad2 6y ago

This is an easy one. The assumptions are flawed. W = (0.5) x(C)x(V) is a back-of-the-envelope shorthand for a whole realm of formulas in the study of capacitors, and it doesn't work in edge cases like this.