Work done by load in charging a capacitor

[Dale]

When the connection to the battery is made a current flows to charge the capacitor. This current will produce heat energy in any resistance and electromagnetic radiation from the wires. If the resistance is small then the current will be large and the radiation could be the major means of energy loss. If the resistance is large the current will be small and resistive heating could be the major energy loss mechanism.

Actually, the energy dissipated in the resistance for a charging series RC circuit is independent of the actual value of the resistance. For a series RC circuit with initial voltage 0 across the capacitor and with a constant supply voltage v_s after t=0 we have the differential equation:
C \frac{dv_c}{dt}+\frac{v_c-v_s}{R}=0
Which has the solution:
v_c=v_s(1-e^{-\frac{t}{RC}})

So the energy dissipated in the resistor is given by:
\int_0^{\infty} i_r v_r \, dt = \int_0^{\infty} \frac{v_c-v_s}{R} (v_c-v_s) \, dt = C \frac{v_s^2}{2}
which is not a function of R, and is also exactly equal to the "missing" energy.

As to whether or not resistance or radiation is the dominant source, if you are using circuit theory or doing circuit analysis then one of the basic assumptions/approximations that you are making is that the radiation is 0. Even making the assumption of no radiation, you can still show that energy is always conserved, so it is not necessary to invoke radiation to explain the missing energy.

 

[sophiecentaur]

which is not a function of R, and is also exactly equal to the "missing" energy.

.

You are so right.
And it amazes me that people don't believe the two values of energy involved - QV and QV/2 are enough to prove the point. You shouldn't have to go as far as you have done in order to prove it.

 

[truesearch]

VrxIr tells you how much energy (power) is dissipated by the wire. It does not tell you what form of energy is dissipated.
You could assume there was no Joule heating just as easily as you could assume there is no radiation. You are quite right to recognise that circuit analysis does not reveal the forms of energy involved.
It looks like you have taken the path that proves there is no electromagnetic radiation.
Just for the moment I like the assumption that there is no Joule heating.
Have you considered 'radiation resistance' in your analysis?

 

[sophiecentaur]

I mentioned Radiation Resistance way back in the thread. There is no detectable difference as far as the source is concerned although a quick frequency sweep could allow R(ohmic) and R(radiation) to be identified.

 

[Dale]

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.

 

[truesearch]

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.

 

[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.

 

[sophiecentaur]

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.

 

[truesearch]

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 realize how electromagnetic radiation from the whole spectrum is produced and this means recognising the different physical mechanisms.
I don't think there is much more that can be added to this thread.

 

[Dadface]

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

 

[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.

 

[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.

 

[Dadface]

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 independent 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 each other.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 independent of R.To my mind methods of the type used by DaleSpam are far more satisfactory and general.

 

[Dadface]

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.

 

[Dale]

With this method it is shown that the energy dissipated is independent 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.

 

[BruceW]

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.

 

[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.

 

[BruceW]

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

 

[Dadface]

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.

 

[sophiecentaur]

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 %.

 

[PriceMike]

Sooo confused!

 

[truesearch]

I have attached photographs of the demonstration of em radiation from the connecting wires in the capacitor circuit.
The first picture shows the components, the powersupply is a square wave signal generator to charge and discharge the capacitor. The variable resistor is 0-15Ω.
The black wire is one of the current carrying leads in the capacitor circuit and the yellow wire is the 'pickup' coil.
The second photo is the square wave supply and the third photo is what is picked up by the yellow coil from the black wire. I passed the black wire through the coil several times to obtain a trace that could be photographed. There is a trace with the coil just sitting on the straight wire.
When the resistance is decreased the pickup increases (current is greater)
There is 'something' passing from the current carrying wire to the coil and this is energy being lost from the capacitor circuit.
There are essentially 2 means of dissipating the 0.5QV of energy...(1) Joule heating due to resistance (2) electromagnetic (radio) radiation from the wire as a result of the changing current. The 2 together account for the missing energy.

 

[sophiecentaur]

Very interesting and not at all surprising that you can 'detect' some radiated RF. We have established that there is some. But do you actually know (how can you?) how much Power is being radiated by this mechanism and how do you know how much Power is being dissipated by resistive elements in the rest of the circuit? Do you actually know the resistance in the circuit? What about the output impedance of the sig genny at your operating frequency?
Introducing the idea of superconductivity is a complete red herring because there are NO power sources with zero internal resistance - even if there are some superconducting wires involved in the rest of the circuit.
If all the 'extra' power is being coupled out into space then you have achieved a remarkable feat.

 

[jsmith613]

Very interesting and not at all surprising that you can 'detect' some radiated RF. We have established that there is some. But do you actually know (how can you?) how much Power is being radiated by this mechanism and how do you know how much Power is being dissipated by resistive elements in the rest of the circuit? Do you actually know the resistance in the circuit? What about the output impedance of the sig genny at your operating frequency?
Introducing the idea of superconductivity is a complete red herring because there are NO power sources with zero internal resistance - even if there are some superconducting wires involved in the rest of the circuit.
If all the 'extra' power is being coupled out into space then you have achieved a remarkable feat.

I did not realize anyone answered the q...seems like lots of people had fun but I just want to check what I asked first time
anyway to summarise MY initial question:
- work done by power source = QV
- energy stored = 1/2*QV
- missing energy is transferred to wires

right?

 

[BruceW]

I did not realize anyone answered the q...seems like lots of people had fun but I just want to check what I asked first time
anyway to summarise MY initial question:
- work done by power source = QV
- energy stored = 1/2*QV
- missing energy is transferred to wires

right?

Exactly. And if there are other components in the circuit, some of the missing energy will also be transferred to them.

 

[truesearch]

the missing energy is dissipated (lost) from the wires
There are 2 clear means of dissipating energy
1) resistance of wires (joule heating)...could be as near zero as you want to get
2) electromagnetic radiation from the wires...can never be zero if the current is changing.