Mentor

## Work done by load in charging a capacitor

 Quote by truesearch It looks like you have taken the path that proves there is no electromagnetic radiation.
It is not a proof, it is an assumption. It is one of the fundamental assumptions of circuit theory, which appears to be the context of the question.

I just see no reason to involve radiation and the full complexity of Maxwell's equations in answering a simple question about circuit theory. It is like a student asking a basic question about a simple pendulum and trying to involve general relativity in the answer. Sure, the more general/complicated theory will give you a correct answer, but why bother with the additional complexity when the more specific/simplified theory also gives you a correct answer.

EDIT: Hmm, I just realized something. The assumption that I was thinking of is that there is no magnetic coupling between different components of the circuit. That is actually not the same as there being no radiation. In fact, a resistor radiates strongly (in the IR range) it just doesn't magnetically couple to other elements. So as long as there is no coupling between elements then it is just resistance, regardless of what frequency is radiated.

The point remains that the energy dissipated in the resistor does not depend on the value of the resistance without any need of anything more than standard circuit theory.

 I read the original context of the question to be about energy. Circuit theory does not determine the forms of energy that are dissipated. Anyway, i think it is now clear that there are 2 means of energy dissipation: Joule heating and electromagnetic radiation. Each can be represented by a 'resistance' term in any equation that is used to analyse the situation. This question is bound to crop up again, it has already been posted at least once and it is to be hoped that due recognition of the means of energy dissipation will be given.
 Recognitions: Gold Member Hello true search.The two means of energy dissipation you refer to are linked together.Electrical energy is converted to heat due to the resistive parts of the circuit(Joule heating)and this heat is transferred to the surroundings. Radiation is just one of the mechanisms of heat transfer but conduction and convection also feature.Any losses due to inductive effects and or sparking depend on the geometry of the circuit and its surroundings.Usually such losses are small.

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 Quote by Dadface Hello true search.The two means of energy dissipation you refer to are linked together.Electrical energy is converted to heat due to the resistive parts of the circuit(Joule heating)and this heat is transferred to the surroundings. Radiation is just one of the mechanisms of heat transfer but conduction and convection also feature.Any losses due to inductive effects and or sparking depend on the geometry of the circuit and its surroundings.Usually such losses are small.
Give him a break.
He introduced the idea of EM radiation -as in Radio Frequency Antennae - and he was perfectly right; I had forgotten all about that, earlier on. There will be plenty of circuit layouts in which that is the main cause of power dissipation.

 Hello Dadface It is wrong to think that radiation from a hot object is the same as electromagnetic radiation from the wire carrying a changing current. They are both electromagnetic radiation but are produced in completely different physical processes. A level students need to realise how electromagnetic radiation from the whole spectrum is produced and this means recognising the different physical mechanisms. I dont think there is much more that can be added to this thread.
 Recognitions: Gold Member Hi truesearch.There has been some misunderstanding here.A current,whether it changes or not,will produce heating which then results in electromagnetic radiation mainly in the infra red region of the spectrum.In addition to this a changing current also produce "electromagnetic radiation" as you referred to it above.THis radiation is mainly in the long wavelength radio band region of the spectrum and involves electromagnetic induction both self and mutual.The point I was making above is that though present these inductive losses are generally very small and for many practical purposes,I suppose, can be considered as negligible when compared to the Joule heating losses. I would like to add that it is not only A level students who need to be familiar with the em spectrum this topic also coming up in AS and GCSE and other courses
 Recognitions: Homework Help truesearch was not simply talking about electromagnetic waves being given off as one of the forms of heat transfer. (But after reading through this thread, it looks like several people have mistakenly thought this is what he meant). What truesearch was actually talking about is the full energy stored in the electromagnetic field. So this takes account of all the energy which is supplied to the circuit. It is all in the pdf link he gave in page 2. What they basically are saying is that half of the energy flows into the capacitor and the other half goes into the rest of the circuit. So it doesn't specify if the other half is lost as ohmic heating or whatever, it shows that half will go to the rest of the circuit, for whatever purpose (heating or otherwise). EDIT: just to summarise, probably the easiest derivation is to just assume there is some resistor in the circuit, and calculate that half the energy is lost due to ohmic heating in that resistor. But the derivation in the pdf link is more general, because it does not mention the resistor. Both derivations come up with the same result.
 The weakest link, the place where most of the energy is dissipated is often the switch. The switch is often ignored and assumed to be ideal, that is instantly transitioning from ∞ to 0 Ohms resistance. In practice there is always transition region where the resistance is finite. In many cases the charging will mostly be done while the switch is still in transit. And yes, the switch can dissipate energy in all sorts of ways, producing heat, sparks and EMI.

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 Quote by BruceW truesearch was not simply talking about electromagnetic waves being given off as one of the forms of heat transfer. (But after reading through this thread, it looks like several people have mistakenly thought this is what he meant). What truesearch was actually talking about is the full energy stored in the electromagnetic field. So this takes account of all the energy which is supplied to the circuit. It is all in the pdf link he gave in page 2. What they basically are saying is that half of the energy flows into the capacitor and the other half goes into the rest of the circuit. So it doesn't specify if the other half is lost as ohmic heating or whatever, it shows that half will go to the rest of the circuit, for whatever purpose (heating or otherwise). EDIT: just to summarise, probably the easiest derivation is to just assume there is some resistor in the circuit, and calculate that half the energy is lost due to ohmic heating in that resistor. But the derivation in the pdf link is more general, because it does not mention the resistor. Both derivations come up with the same result.
Other methods such as the one exemplified by DaleSpams post number 31 come up with the same result that "half" the energy is lost.With this method it is shown that the energy dissipated is independant of R.If that is true then it should also be true as R tends to zero.

The method used in the link goes to that limit and assumes one resistance,only namely that the wires are superconducting.If zero resistance is to be assumed then the capacitor plates must be superconducting also and we can have the situation where the wires and capacitor plates are indistinguishable from eachother.We will have an ideal power supply passing an "infinite" current for "zero" time,a pure thought experiment which works because the energy dissipated is independant of R.To my mind methods of the type used by DaleSpam are far more satisfactory and general.

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 Quote by Delta Kilo The weakest link, the place where most of the energy is dissipated is often the switch. The switch is often ignored and assumed to be ideal, that is instantly transitioning from ∞ to 0 Ohms resistance. In practice there is always transition region where the resistance is finite. In many cases the charging will mostly be done while the switch is still in transit. And yes, the switch can dissipate energy in all sorts of ways, producing heat, sparks and EMI.
I would agree that energy is lost due to sparking.To calculate that energy is a rather thorny problem depending ,amongst other things,on the nature of the air/medium between the switch contacts.

Mentor
 Quote by Dadface With this method it is shown that the energy dissipated is independant of R.If that is true then it should also be true as R tends to zero.
Exactly my point. I guess I should have stated it explicitly like you did. Thank you.

 Recognitions: Homework Help yes, the two methods are almost equivalent. It is good that there is more than one way to do this derivation. I'm not sure if it has been said already, but the method which dalespam explained is better for a-level, since I don't think students have been introduced to the Poynting vector that early on? Edit: also, for the derivation using the Poynting vector, the circuit doesn't need to be superconducting. I think they just used that example to show that the energy is not necessarily lost due to a resistor.
 What A level students like to see is experimental evidence of electro magnetic radiation. I demonstrate the radiation coming from the connecting wires when I do this topic. I use an oscilloscope and tomorrow I will set it up and try to take a photograph of the oscilloscope display. If it works I will post the photographs for further discussion.
 Recognitions: Homework Help sweet. onegai shimasu, as the japanese say (not sure if i spelled it right). ps the translation is something like 'if you please' EDIT: the phrase might be yoroshiku onegai shimasu

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 Quote by truesearch What A level students like to see is experimental evidence of electro magnetic radiation. I demonstrate the radiation coming from the connecting wires when I do this topic. I use an oscilloscope and tomorrow I will set it up and try to take a photograph of the oscilloscope display. If it works I will post the photographs for further discussion.
Interestingly that's sort of how Faraday discovered em induction all those years ago.His transmitting circuit was a coil connected to a battery and a switch and his receiving circuit a second coil connected to a galvanometer.Not only was it a demonstration of wireless transmission but it was also the first transformer.

 Recognitions: Gold Member Science Advisor Whilst there is no doubt that some power is radiated due to electromagnetic induction, I might point out that, without some considerable trouble in the circuit layout design, it would be hard to radiate more than a very small fraction of that which is dissipated 'ohmically' (owch - sorry about that word). Your average piece of metal / wire is a very inefficient radiator, which is why they pay antennas engineers vast sums to design efficient antenna systems. Just because you can detect it, doesn't mean there is much of it. Thermal energy loss will be the major one in nearly all circumstances. Actual circuit size and matching would need to be just right for an 'RF' efficiency of more than a few %.
 Sooo confused!