# Old Chestnut

Discussion in 'Physics' started by t_n_k, Apr 29, 2009.

1. ### t_n_k Thread Starter AAC Fanatic!

Mar 6, 2009
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Switching charge between capacitors in a lossless circuit.

See PDF attachment ...

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2. ### mik3 Senior Member

Feb 4, 2008
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Emission of an EMW is the only thing which can consume energy.

3. ### KL7AJ AAC Fanatic!

Nov 4, 2008
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Another source is SOUND energy caused by the arc during make.

Eric

4. ### jpanhalt AAC Fanatic!

Jan 18, 2008
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What about entropy? (Entropy in the uncharged capacitor decreases.)
John

5. ### thingmaker3 Retired Moderator

May 16, 2005
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Dissipated heat is equal to E^2 / R, and is therefore infinite. We can capture the dissipated heat and use it to make HHO gas for running our cars.

6. ### mik3 Senior Member

Feb 4, 2008
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Why is it infinite?

7. ### thingmaker3 Retired Moderator

May 16, 2005
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Because the assignment foolishly specifies R = zero.

8. ### mik3 Senior Member

Feb 4, 2008
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Yes but if you have a 0 ohm resistor then it won't dissipate any power.

This is because the voltage across a 0 ohm resistor is 0V and thus no power is dissipated.

9. ### thingmaker3 Retired Moderator

May 16, 2005
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I apologize for my ambiguity. I will be more plain: any exercise with "zero resistance" or "zero reactance" is pointless. This old problem is nothing more than a slight-of-hand trick.

10. ### studiot AAC Fanatic!

Nov 9, 2007
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I think it's rather more than that and taught in respectable electrical engineering texts. Students should be aware that there is a loss of energy from the system, with or without resistance.
When any switch is closed arcing occurs, minute or larger scale. This dissipates energy. The calculation about the resultant steady state shows the system to be in a lower energy state as TNK observed.

However the system is not lossless as stated.

11. ### b.shahvir Active Member

Jan 6, 2009
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So, indirectly the energy is dissipated by the arc resistance which in turn introduces power loss into the circuit.

12. ### jpanhalt AAC Fanatic!

Jan 18, 2008
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I read this as a thought experiment, not as a real world exercise. From that perspective, switch arcing just doesn't have appeal as a correct answer -- somehow, the switch must know exactly how much to arc to satisfy the equations.

I find the question interesting, perhaps because of my lack of formal education in the subject. By analogy to a physical system involving kinetic energy and momentum (or a corresponding chemical system), one can visualize where the "lost" kinetic energy goes, i.e., the decrease in enthalpy can be explained in terms of entropy, heat, volume, chemical bond strengths, etc.

Does anyone know off hand whether the problem has a name (other than old chestnut), so I can look up the formal answer?

John

13. ### mik3 Senior Member

Feb 4, 2008
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If the switch is an ideal MOSFET with Rdson=0 ohms where is the arc created?

I think there is a better explanation to that.

14. ### steveb Senior Member

Jul 3, 2008
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I agree with mik3 that arcing is not the full explanation.

I agree. This problem, or thought experiment, can teach a couple of good lessons. The interesting thing is that, what was once a thought experiment is no longer unrealistic since the discovery of superconductors.

The solution to the paradox is to realize that circuit theory will break down if very fast and non-arcing discharge occurs. Hence, electromagnetic radiation would need to be considered, as mentioned. The arcing explanation, although a valid consideration, does not fully resolve the paradox since one could implement a fast switch in a vacuum chamber to prevent the arc.

To do a mathematical calculation one could write the electromagnetic equations. In antenna theory, one often talks about a radiation resistance which is an effective resistance seen in the circuit due to radiation loss. For example, assuming sinusoidal signals (which does not apply here), radiation resistance of a circular loop antenna (or circuit) increases as a fourth power of the ratio of loop radius to wavelength (if current is assumed independent of position in the loop, i.e. small loops).

It would be interesting to see if an approximate electromagnetic solution to the above problem can be derived, or whether a numerical solution is needed.

15. ### studiot AAC Fanatic!

Nov 9, 2007
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First of all arcing dissapates energy as sound, light, heat evaporation of metal ions , local temperature rises in the contacts and temperaturerise/ionisation/breakdown of any medium between the contacts.

It is impossible to avoid arcing as one brings two charged contacts (even mercury wetted ones) closer and closer. When they are microscopically close, but not touching, the electric field is intense.

All of this is independant of bulk wiring resistance.

All FETs have some on resistance in the channel so will dissipate the energy resistively.

I think this is one of those cases of sudden change where the best solution is to calculate the situation before and after the change and subtract. Attempts to partition the energy loss between various mechanisms are doomed to failure because of the uncertain and random nature of each switch closing.

Here is a standard textbook answer.

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16. ### steveb Senior Member

Jul 3, 2008
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Why do you say impossible? What if we do the experiment in vacuum and rapidly close a mechanical switch?

Surely a sufficiently evacuated chamber and a fast enough switch would prevent all energy loss from dissipating in an arc. How do we then explain the loss in energy. EM radiation loss seems the only answer. Yes/No?

EDIT: (assuming perfect conductors of course)

17. ### studiot AAC Fanatic!

Nov 9, 2007
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Imagine two conductors, one connected to the charged capacitor, one connected to the uncharged capacitor. They are widedly separated, but connected at their other ends to the common connection between the two cpacitors.

There is a potential difference between the conductors, by virtue of the charge on one cap.

Now let one conductor approach the other. As they approach the field intensity, given by the PD / separation increases as the separation decreases, without limit.

At some non zero separation the field intensity will exceed the breakdown voltage of any intervening matter.

If there is no intervening matter, at some closer, but still non zero, separation the field will be strong enough to cause particles to leave the surface of a conductor, either as electrons or ions, thus establishing the current that forms the mcro arc.

In real world conductors the process is further enhanced by the grain structure of the metal. Not all atoms in the imperfect metal crystals are equally placed and impurities will set up side currents as well.

Do you have no comment on prosecution exhibit 'A' ?

18. ### jpanhalt AAC Fanatic!

Jan 18, 2008
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Here's another textbook answer ("The two-capacitor problem with radiation." TB Boykin, D Hite, and N Singh. Am J. Phys. 70(4)415-420 (2002)):

I did a little library research this morning and kept coming up with two solutions: 1) as described above by Boykin et al.; and 2) solutions that introduce a little bit of resistance. Radiation loss seems to be the favored answer today.

Unfortunately, I did not find any analyses that considered entropy. Many physics books use an example of the spontaneous adiabatic expansion of 1 mole of ideal gas into an evacuated space to introduce the subject of entropy, which seems to me is at least superficially similar to the capacitor problem. I will keep looking. I did find one reference that had entropy as a keyword, but it is not available to me online without cost. I will have to go visit my local library that has it in print. The citation is: Am J Physics 54:742-744(1986).

John

19. ### studiot AAC Fanatic!

Nov 9, 2007
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Would this analysis be along the lines (presented without maths)

Following closure of the switch charge moves to redistribute itself in the system between the capacitors.

This charge has a finite velocity in the conductors, but moving charge generates EM radiation which dissipates.

Once the charge has equilibrated it stops moving so stops generating EM.

20. ### steveb Senior Member

Jul 3, 2008
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Yes. This is why my suggested thought experiment said to work in vaccum and bring the switch closed very radidly. If the switch closes faster than a significant number of ions can leave the surface, then not all of the missing energy can be dissipated in a microarc.

Note that I'm not saying that the arc concept is not valid. I'm just saying that it doesn't completely resolve the paradox. If an arc does happen, then there is resistance and a finite discharge time - so no problem. However, if there is no arc, or if it's duration is too small, we need another answer to account for the missing energy. EM radiation will always be some component to the resolution, even if it is a small part in a particular experiment.

By the way, I wasn't sure what you mean by exhibit A. The attachment? I didn't have an issue with that. Was there something I missed there? I'll go back and look at it carefully.

EDIT: Ah OK, I reread this. The last sentence says that the EM radiation will be from oscillatory current flow. This strikes me as being misworded. A very rapid current impuse can generate radiation too. All that is required is accellerating charges to generate radiation loss.

Last edited: May 11, 2009