Work done by load in charging a capacitor

[jsmith613]

The energy stored on a charged capacitor is 0.5*Q*V or 0.5*C*V²
BUT
what is the electrical work done by the supply as it transfers the charge to the capacitor?

I have heard two answers:
W=QV and W=0.5*QV
the first answer implies, i think, that energy is required to move charges against the already present charges on the capacitor plate. is this correct?

 

[mfb]

That depends a bit on your setup. If you connect both without any resistance in between, you get a short circuit and waste a lot of energy in cables/power supply/whatever.

In general, if your power supply has the same voltage at all time, you need W=QV and waste 50% of this energy somewhere.

 

[Emilyjoint]

What do you mean by 'waste energy somewhere'
Is it exactly 50% that gets wasted every time

 

[mfb]

The power supply supplies a charge of Q, with a voltage of V (unless it is a bad power supply or has a variable voltage or something else), which means that you take an energy of Q*V from it.
The capacitor stores an energy of 1/2 Q*V.

Both are simple equations, and if you divide both you get the result that only 50% of the energy is stored in the capacitor. You can use the other 50% for something else (heat a resistor, light a room, charge some other energy storage or whatever), but they won't get into your capacitor in this charging process.

 

[sophiecentaur]

What do you mean by 'waste energy somewhere'
Is it exactly 50% that gets wasted every time

He wasn't being unspecific about the 50% loss. He was implying that the energy is lost "somewhere" and that can be in the connecting wires, the power supply internal resistance- in addition to the explicit resistance that is introduced.
This may not apply where there is an Inductor involved in the circuit.

 

[Emilyjoint]

OK, I have seen how the energy stored on a capacitor is 0.5QV because the graph of Q against V is a straight line and the energy is the area under the graph. (this is like the equation for energy stored in a stretched spring = 0.5Fx)
Also I know that the charge coming from the battery is always at the voltage V and so the energy from the battery is QV. So 0.5QV is missing.
I don't think the battery internal resistance comes into this because the V is at the terminals of the battery. If energy is missing it must appear as another kind of energy somewhere . What kind of energy is the missing energy?

 

[sambristol]

This is an old problem folks but it would appear the answer is not well known !

Consider an idealised situation of a 'stiff' voltage source (a power supply which maintains voltage whatever the current drawn) connected via wires with no resistance to two parallel metal plates with a slab of dielectric between them.

A charge Q is transferred to one plate and -Q to the other plate.

But where is the energy? Remove dielectric slab and put it into identical set of plates connected not to a power supply but to a voltmeter. The voltmeter will resister a voltage and if replaced by say a lamp the lamp will glow.

The energy you are missing was stored in the polarisation of the molecules in the dielectric.

Regards

Sam

 

[Emilyjoint]

"The energy you are missing was stored in the polarisation of the molecules in the dielectric"
Surely that can't be right !Isnt the dieelctric part of the operation of the capacitor, just like the plates? Isnt that where the 0.5QV is stored?
And it is possible to have no dielectric in a capacitor, just 2 metal plates.

 

[sambristol]

In the original capacitor disconect the power supply and text the voltage between them - it will surprise you !

 

[Emilyjoint]

we have done measurements on capacitors and the voltage is equal to the battery voltage when it is disconnected. We measured the voltage falling in an exam practical to measure time constant.
It is the energy explanation that sounds wrong. I cannot see anything in my textbook along those lines.

 

[sophiecentaur]

Surely it would be better to discuss what happens in a vacuum capacitor first before getting bogged down with what happens inside a dielectric when it is introduced. The Energy Situation is the same and there is one less factor to consider.

The fact is that supplying energy to a Capacitor, in a practical situation, always involves series resistance. If you really want to ignore any series resistance, you still can't ignore the inductance of connecting leads. If the series inductance dominates then you just get an oscillation, in which the energy is shared (over time) between the L and the C.
One needs always to avoid the 'irresistable force and immoveable object' type of scenario.

 

[Emilyjoint]

There is something wrong here...when the caoacitor is charged ther is no current so I do not see what part inductance would play

 

[sophiecentaur]

Not wrong. Whilst charging (through the conductor) , energy is stored in the magnetic field associated with the Inductor. As the Capacitor volts increase, the current decreases, and the voltage on the inductor changes (change producing volts). Eventually, the volts at the capacitor get higher than battery volts and current is zero. Then the capacitor discharges through the L and volts drop. This process goes on for ever or until the resistive elements have dissipated the 50% of energy supplied.

 

[mfb]

I don't think the battery internal resistance comes into this because the V is at the terminals of the battery.

This voltage is the voltage at zero load - while you are charging the capacitor, you get a current and therefore a lower voltage in a realistic power supply.


@sambristol: Your post explains how the 1/2 QV in the capacitor are stored, but that is the part which is not discussed here.

 

[truesearch]

There is an awful ot of confusing and misleading physics in these responses.
There has been no mention of the fact that a changing electric current produces electromagnetic radiation and this amounts to an energy 'loss'.
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.
It is also possible that a spark is produced when the switch is closeds which is also an energy loss mechanism.
The same processes also apply when charged capacitors are connected in parallel to uncharged capacitors.

 

[sophiecentaur]

"Confusing"? I should say, rather, that we're probably being a bit too simplistic.
Your EM idea is very fair comment and, as we try to get more and more 'ideal' in our thought experiment, it must be the limiting factor for energy loss, as it can't be eliminated, once you've chosen the dimensions of the set up. (Unless you do it all in a silver lined box, of course)

Wouldn't the radiation resistance of a small structure be pretty low (arm waving values) compared with the likely internal resistance of power supply and wires?
Would you really expect an arc when a switch closes? Breaking a circuit when current is flowing through an inductor would certainly be an issue but I think you are going another step deeper if you want to introduce sparks here.

We'd have to be a bit more specific about the actual setup before taking this much further, I think.

 

[truesearch]

I typed 'energy loss charging capacitors' into Google and got a reference to physicsforums :Nov4-06 by jumanicus who also states energy loss by EM radiation...it has already been discussed here!
I hope (I am not certain how to do it) that I am attaching an anaysis of EM radiation loss in capacitors.
In my humble opinion this is the FIRST loss mechanism that should be discussed when dealing with this topic...it is always there. It is a pure physics, fundamental explanation that does not rely on resistance, inductance, dieelectric constant etc.
How to get sparks...Touch a charged capacitor onto the terminals of an uncharged capacitor (almost zero resistance)
PS: I am a physics teacher and passionate that students get sensible explanations to their questions.

 

[mfb]

There has been no mention of the fact that a changing electric current produces electromagnetic radiation and this amounts to an energy 'loss'.

And this is reasonable, as the energy released EM radiation is really small, unless you add some fancy oscillators to the setup, charge it with another capacitor and superconducting wires or do something else unusual I don't think of at the moment.

Resistance is always there, too, unless you use superconductors everywhere.

It is also possible that a spark is produced when the switch is closeds which is also an energy loss mechanism.

Right.

 

[truesearch]

Would anyone agree that the answer to the original post:
"the first answer implies, i think, that energy is required to move charges against the already present charges on the capacitor plate. is this correct?"
is incorrect and a good answer would be :
The energy lost is due to
a) Heat produced in any resistance
b) electromagnetic radiation from the connecting wires
c) sparks (if there are any)

 

[sophiecentaur]

I typed 'energy loss charging capacitors' into Google and got a reference to physicsforums :Nov4-06 by jumanicus who also states energy loss by EM radiation...it has already been discussed here!
I hope (I am not certain how to do it) that I am attaching an anaysis of EM radiation loss in capacitors.
In my humble opinion this is the FIRST loss mechanism that should be discussed when dealing with this topic...it is always there. It is a pure physics, fundamental explanation that does not rely on resistance, inductance, dieelectric constant etc.
How to get sparks...Touch a charged capacitor onto the terminals of an uncharged capacitor (almost zero resistance)
PS: I am a physics teacher and passionate that students get sensible explanations to their questions.

It's certainly always there but it can hardly be thought of as the major loss in most 'real' circumstances. In many ways, it can be looked upon as a bit of a distraction for a student who is probably struggling with the basics of circuits and Maths. It appears as a resistance in series with the Ohmic resistances in the circuit and can be included in the little rectangle that's always drawn in the 'charging a capacitor' circuit.
'Sensible' explanations don't usually bring in yet another can of worms for a poor student to deal with. A level doesn't actually deal with launching em waves into space. What level do you teach?

 

[sambristol]

folks check out this

http://www.physics.princeton.edu/~mcdonald/examples/twocaps.pdf

Regards

Sam

 

[sweet springs]

Hello.

The More charges gather to the plates, The More work we need to mount additional charge against repulsion forces. Thus the whole work are integration of V dQ = Q/C dQ. Thus 1/2 appears from formula of calculus, as well as it appears in Mechanics the distance is expressed 1/2 a t^2 by time t and acceleration a or in Mathematics of area of triangle.

Regards.

 

[sophiecentaur]

That's all fine and accurate but the power supply has delivered QV of energy - twice as much as that. That's where people find the difficulty.

 

[truesearch]

I teach A level physics and electronics.
Sensible explanations do not avoid covering physics in detail.
A level does deal with 'launching em waves into space' Students need to know about the whole range of electromagnetic waves, how they are produced and how they are transmitted. We need to cover electromagnetic induction, fields due to current carrying conductors etc. It is easy to demonstrate transmission of electromagnetic waves using 2 coils, a signal generater and an oscilloscope.
The most basic work on capacitors requires an understanding of the energy stored and why it is 0.5QV. When a student asks (and they do ask!) what happens when resistance is zero (they have heard of superconductors at A level) they deserve an explanation.
The explanation is easy...it has been covered in this forum before.

 

[mfb]

Sensible explanations do not avoid covering physics in detail.

The question is "how detailed?"
Would you skip Newtonian mechanics, as GR is more precise? ;)
Or skip an introduction to classical EM waves, as quantum electrodynamics is more precise?
Of course not - both are beyond the scope of your classes.

While it is useful to know that there is something more behind this, it is a good idea to use the easiest explanation which gives good results. For an object falling in the lab, this is Newtonian gravity. For charging a capacitor, this is energy losses in a resistor. Most regular power supplys have something similar to an internal resistance, even if you don't use an external one.
It might be interesting to consider the setup with two capacitors and superconductors, but it is certainly not the main point the students should learn about charging capacitors.

 

[sweet springs]

Hi.

but the power supply has delivered QV of energy - twice as much as that.

Why do you have such an idea? I suppose you believe that voltage between plates is constant during charging immediately after the moment the condenser is connected to battery. When battery is connected to the condenser, V=0 because no charge at the plates mean no voltage between. V would increase as charge is accumulated on the plates.

For example tall glass of water and short glass of water is connected by tube so that water can move between. Is water level equal at the moment of the connection?

Regards.

 

[sophiecentaur]

Hi.



Why do you have such an idea? I suppose you believe that voltage between plates is constant during charging immediately after the moment the condenser is connected to battery. When battery is connected to the condenser, V=0 because no charge at the plates mean no voltage between. V would increase as charge is accumulated on the plates.

For example tall glass of water and short glass of water is connected by tube so that water can move between. Is water level equal at the moment of the connection?

Regards.

Of course not. Your connecting pipe requires a pressure drop along it in order to allow a flow of water. The Maths correspond to an RC circuit and not just a pair of Capacitors. The water transfer takes TIME, during which, energy is dissipated.
If we are discussing a real situation in which there is some series resistance and a source of emf, then, of course, there will be a finite time during which the capacitor becomes charged. The source of emf will have been producing (by definition) V volts all the time. The PD across the capacitor will have started at zero and eventually reached V. During this time, current is flowing through the resistive element (which always has a real, radiative component). QV is the energy supplied by the source. QV/2 is the energy stored in the capacitor. The rest has been dissipated.
If you want to do the analysis more thoroughly then you could introduce the Inductive properties of any connecting wires. If the L is high enough, then the reactance is high enough to treat the system as a damped oscillator then you will get a different path towards the equilibrium condition but the same end result applies.

 

[sophiecentaur]

I teach A level physics and electronics.
Sensible explanations do not avoid covering physics in detail.
A level does deal with 'launching em waves into space' Students need to know about the whole range of electromagnetic waves, how they are produced and how they are transmitted. We need to cover electromagnetic induction, fields due to current carrying conductors etc. It is easy to demonstrate transmission of electromagnetic waves using 2 coils, a signal generater and an oscilloscope.
The most basic work on capacitors requires an understanding of the energy stored and why it is 0.5QV. When a student asks (and they do ask!) what happens when resistance is zero (they have heard of superconductors at A level) they deserve an explanation.
The explanation is easy...it has been covered in this forum before.

The explanation of how EM waves are launched into space is far from easy. Why do you think it is not dealt with (i.e. analysed) at A level? Electromagnetic induction is dealt with very superficially (appropriately so). The fact that em waves can be radiated from objects is fairly easy to take on board but how many 'informed' adults ever make the connection between the good old H atom radiates (involving photons) and the way a wire dipole radiates 10MHZ signals? Would you also recommend using QM to discuss the way a capacitor charges?
I would also take issue when you state that they have "heard of superconductivity", implying that they have not confused it with 'very high conductivity'. Students have a very limited scientific life experience (not helped by the TV these days) and they very easily confuse Science with Science Fiction. Down - to - Earth explanations get my vote every time.

As a teacher, I would have assumed that you would take into account the limitations of your students and that you'd avoid confusing their superficial fascination with actual understanding. Or perhaps you have a particularly bright set of students, in which case you are very lucky.

 

[sweet springs]

Hi. Now I think I get your point.

Of course not. Your connecting pipe requires a pressure drop along it in order to allow a flow of water.

In the example of connecting pipe, you will see see-saw of water levels in both the glasses and it continues eternally if the liquid is perfect fluid.

Similarly the increasing electric field between plates of capacitor by rushing charges cause magnetic field. Disappear of magnetic field cause excess current to the capacitor. In the case battery is also charged capacitor of the same type, see-saw of charges begin. If all the dynamic parts of energy dissipate, the remaining static part of energy is half the initial one.

Regards.

 

[sophiecentaur]

Hi. Now I think I get your point.



In the example of connecting pipe, you will see see-saw of water levels in both the glasses and it continues eternally if the liquid is perfect fluid.

Similarly the increasing electric field between plates of capacitor by rushing charges cause magnetic field that restore energy. In the case battery is also charged capacitor of the same type, see-saw of charges begin. If all the dynamic parts of energy dissipate, the remaining static part of energy is half the initial one.

Regards.

That's it.
The argument continues as to the relative amounts of dissipation by the various mechanisms.

 

[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 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 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 independant 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 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.

 

[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 %.