Why doesn't C1 get hot?

[steveB]

In post #13 of this thread you said, "Displacement current is a valid form of current and it is the current the exists in the capacitor dielectric." Since "displacement current" is an equivalency and not a real physical current, and doesn't exist in the dielectric, I can be forgiven for assuming you are confusing the real current in a wire with the assumed current in the capacitor.

I did say that, which is not the same thing as saying displacement current is drift current. But, I see that what I said is not quite right, so I need to correct it. The displacement current is really better described as something due to fields in space, rather than in the dielectric. The dielectric is better said to have polarization current, which is what gives the dielectric its higher effective permittivity.

As I said above, even the drift current does not go through the wire with AC, the charges on average move back and forth. At 60 Hz, the drift can only be about 10 μm in either direction, and 60 Hz is a relatively low frequency. That's hardy describable as "going through" but we say it anyway because that's the terminology that exists. In the same way polarization current does not represent charges actually going through the dielectric, but we say it anyway, and we know what it really means. It means charge going back and forth, but in this case the charge is bound to an atom and the electron cloud distorts. With polarization current the charge movement is in Angstroms rather than μm.

Is language always perfectly descriptive? No, of course not, yet it communicates effectively when we rely on the common knowledge base rather than a literal interpretation of words.

The author says in the 5th paragraph "The current through the capacitor (red) will 'lead' the mains voltage (green) by 90º. " Since he says the current passes through the capacitor, he is also saying that current is passing through the dielectric, which it does not. He should have said something to the effect of (the branch containing the capacitor). Drift current is the only type current in a wire, so the author can only mean drift current.

No, this is your own interpretation. I would interpret what he said differently. You seems to like to enforce your own definitions and try to impose them on the rest of the world. The term current is general and can include drift current, relative charge movement of any type, polarization current and displacement current. That's just terminology. It seems they are all grouped together for good reason, since one gives rise to the others and all are consistent under Maxwell's generalized version of Ampere's law which includes displacement current. Yes, there are some people that don't like these definitions, but too bad, because they are already defined and existing in the history and literature. You are free to include the adjective if you want to say "drift current" but don't require that the rest of the world should mean "drift current" when they say "current". We all understand the difference (or at least we should) between the types of current that have been classified, and when someone talks about current going through a wire and through a capacitor, we understand (or at least should understand) what is happening physically, and we should not need to every time write several sentences of clarification to say such a well understood thing.

 

[Ratchit]

steveB,

The displacement current is really better described as something due to fields in space, rather than in the dielectric.

I will agree to that.

The dielectric is better said to have polarization current, which is what gives the dielectric its higher effective permittivity.

I have heard of polarization density, but not polarization current. Converting molecules of dielectric material into dipoles does not generate a current, because the charges are bound to one location.

As I said above, even the drift current does not go through the wire with AC, the charges on average move back and forth. At 60 Hz, the drift can only be about 10 μm in either direction, and 60 Hz is a relatively low frequency. That's hardy describable as "going through" but we say it anyway because that's the terminology that exists.

As you probably know, I reject terminology that is not accurately descriptive.

In the same way polarization current does not represent charges actually going through the dielectric, but we say it anyway, and we know what it really means. It means charge going back and forth, but in this case the charge is bound to an atom and the electron cloud distorts. With polarization current the charge movement is in Angstroms rather than μm.

As I said before, polarization current is a nonstarter for me.

Is language always perfectly descriptive? No, of course not, yet it communicates effectively when we rely on the common knowledge base rather than a literal interpretation of words.

They don't operate that way in law, so why should scientific personnel do so?

No, this is your own interpretation. I would interpret what he said differently. You seems to like to enforce your own definitions and try to impose them on the rest of the world.

I did not define the meaning of the words. If the terminology defined by the words is incorrect, then it should be called out, even if everyone thinks they know what it really means.

The term current is general and can include drift current

OK, that is charge movement.

relative charge movement of any type

OK, charge movement such as a beam of electrons in a CRT.

polarization current

What's that?

displacement current

Not real, a mathematical artifice.

You are free to include the adjective if you want to say "drift current" but don't require that the rest of the world should mean "drift current" when they say "current". We all understand the difference (or at least we should) between the types of current that have been classified, and when someone talks about current going through a wire and through a capacitor, we understand (or at least should understand) what is happening physically, and we should not need to every time write several sentences of clarification to say such a well understood thing.

Well, the author said current existed through the capacitor. The point is that none of the currents you named will light up the leds except drift current, and that does not exist through the capacitor. I am sure a lot of folks including the author still think it does. Threrefore, it is not well understood as you aver it is. So it needs to be corrected.

Ratch

 

[steveB]

I have heard of polarization density, but not polarization current. ... As I said before, polarization current is a nonstarter for me. ... polarization current? - What's that?

The fact that you don't know what it is says a lot.

Polarization density you know, so let's start there and call it P. Take the time rate of change of that to get dP/dt and you have polarization current density. Then integrate that over a cross sectional area and you have total polarization current.

Some on-line references are as follows, but of course text books are better.

**broken link removed**

http://en.wikipedia.org/wiki/Polarization_density

https://www.electro-tech-online.com/custompdfs/2013/08/06.pdf

Displacement Current? - Not real, a mathematical artifice.

It's called a "current", even if it is inspired as a mathematical artifice or correction term. Certainly, it represents something real or we wouldn't deal with it in science and engineering. It has many of the properties of a current which is why it's called a current. It doesn't have all properties of charge motion, but that's exactly why it's such an amazing concept and discovery.

 

[ronsimpson]

Capacitors (and inductors) are not like resistors. A 'perfect' cap does not heat.

Yes, that is well known.
Ratch

It was not known in post #1. Here we are in post #24. By #3 to #5 those who don't know have moved on with a answer, or are very lost. I should have moved on.
I am not that invested in language usage. I just want to say caps don't get hot. (with out 20 posts on the fact that they do) I don't know how to say current (does something) down a wire with out causing a complicated argument. There is more than one thread on this.
Why did I say anything? I know better.

 

[tvtech]

Oh boy

Not again. Relatively simple, accurate explanation becomes complicated.

It's like one person is trying to be cleverer than the next one now. Friends start arguing etc. And Ratch starts the ball on the roll every time.

Unnecessary stuff. Don't be baited guys. Really tired of this. You know what you know. And leave it at that.

Leave the Pedantic Troll alone. Don't fall for his cleverness. All he is looking for are people to prove wrong.

Really tired of this

Ratchit, I am sure many Forums out there are maybe hungering for your knowledge. Please visit them and leave ETO alone.

So far Ratch, you have taught me nothing. Just complicated Basic Stuff.

Expecting once again your Pedantic pathetic reply quoting everything I typed here. Dingbat.

Please quote "Dingbat". And then look at yourself in a mirror......Very important..

tvtech

 

[Ratchit]

steveB,

The fact that you don't know what it is says a lot.

Yes, it does.

Polarization density you know, so let's start there and call it P. Take the time rate of change of that to get dP/dt and you have polarization current density. Then integrate that over a cross sectional area and you have total polarization current.

Stop right there. You have a polarization charge density P, which is the number of dipoles per unit volume. Then you differentiate that with respect to time to get the rate at which the dipoles are either forming or disappearing per unit volume. Each of those stationary bound molecular dipoles has a net charge of zero. Next you integrate these dipoles over some area in a direction to get what? Current? How can that be? The dipoles are stationary and have no net charge. OK, so the dipoles appear/disappear, and rotate. That causes an energy transaction that affects the Maxwell electric and magnetic field equations, but they don't have lateral motion to constitute flowing charges to light up those LEDs.

Some on-line references are as follows, but of course text books are better.

The first reference does the math like you explained, but does not show how current comes from dipoles. Nowhere does it say the words "polarization current".

The second reference presents the material better, but still does not contain the words "polarization current".

The third is a catalog of courses given by MIT OpenCourseWork. I don't know what I am supposed to do with that.

It's called a "current", even if it is inspired as a mathematical artifice or correction term.

Yes, another misnomer.

Certainly, it represents something real or we wouldn't deal with it in science and engineering. It has many of the properties of a current which is why it's called a current. It doesn't have all properties of charge motion, but that's exactly why it's such an amazing concept and discovery

It is related to the energy that passes through a capacitor. It has the same dimensions as current. The energy through the capacitor is transformed into charge motion when it arrives at the opposite plate of the capacitor.

Ratch

 

[Ratchit]

ronsimpson,

It was not known in post #1. Here we are in post #24. By #3 to #5 those who don't know have moved on with a answer, or are very lost. I should have moved on.

The question that started the thread was answered by MrAl in post #3. Then the thread moved on. Why mention the same thing in post #18 which was contained in post #3? Nobody had to stick with the thread after the posted question was answered in post #3. Post #25 is about polarization current. Stay or leave according to your preference.

I am not that invested in language usage. I just want to say caps don't get hot. (with out 20 posts on the fact that they do) I don't know how to say current (does something) down a wire with out causing a complicated argument. There is more than one thread on this.

Caps get hot or stay cool according to the operating circumstances.

Why did I say anything? I know better.

Only you can answer that question.

Ratch

 

[Ratchit]

tvtech,

I thought you were going to ignore me.

Not again. Relatively simple, accurate explanation becomes complicated.

It's like one person is trying to be cleverer than the next one now. Friends start arguing etc. And Ratch starts the ball on the roll every time.

Unnecessary stuff. Don't be baited guys. Really tired of this. You know what you know. And leave it at that.

Leave the Pedantic Troll alone. Don't fall for his cleverness. All he is looking for are people to prove wrong.

Really tired of this

Then why do you keep reading what I write? I cannot prove anyone wrong unless they are, in fact, wrong. Folks are supposed to argue their point of view about a subject. If you are correct, then you can defend your position.

Ratchit, I am sure many Forums out there are maybe hungering for your knowledge. Please visit them and leave ETO alone.

I have and do.

So far Ratch, you have taught me nothing. Just complicated Basic Stuff.

Sometimes, basic things are complicated.

Expecting once again your Pedantic pathetic reply quoting everything I typed here. Dingbat.

Please quote "Dingbat". Very important..

As you wish.

Ratch

 

[tvtech]

Dear Ratchit

I am moving on. I am no longer prepared to put time and trouble into this thread. Teach at will.

Good luck.

tvtech

 

[steveB]

OK,

First, I've expressed my opinion enough on the debate issue so that any readers can get a perspective from different points of view. I won't say more on that part because I'll just be repeating, which won't help anything and may make tvtech, moderators or others mad. Enough said on my part for that.

However, I'll answer a couple of questions from your post, since that is the appropriate thing to do in a discussion. And, I do feel we are discussing, not arguing. So, I hope it does not seem like arguing to others aside from tvtech.

Stop right there. You have a polarization charge density P, which is the number of dipoles per unit volume. Then you differentiate that with respect to time to get the rate at which the dipoles are either forming or disappearing per unit volume. Each of those stationary bound molecular dipoles has a net charge of zero. Next you integrate these dipoles over some area in a direction to get what? Current? How can that be?

What you described can't be, but that's the wrong description.

Polarization is density (per unit volume) of dipole-moments, not dipoles. Dipole moment is the charge times the separation distance (call it Qd) which has SI units of Cm (charge in Coulombs times distance in meters). If you add up all dipoles in a volume and then divide by the volume you get P, which has SI units of C/m^2. But, P is a point-property meaning that we take the limit of density as volume gets small. However, not so small as to get to the molecular level, because P is intended to be a macroscopic property usually. In other words, we want the overall averaged effect at perhaps μm scale rather than at Angstrom scale.

If the P is changing in time, you can visualize it as if the distance of charge-separation of the dipoles are changing. In some sense, the positive and negative charges are swapping places with + going one way and - going the other way. This represents a current because the directions are opposite and the charges are opposite. Hence, they represent current in the same direction. This is just a crude description to get the mental picture.

But, this is per unit volume, so we end up with dP/dt having units of A/m^2, which represents a current density, which is a vector because P was a vector. So, Jp=dP/dt and any current density can be vector integrated over an area to get an effective current Ip, which ( as much as you disagree with this also, it seems) is a scalar quantity.

The first reference does the math like you explained, but does not show how current comes from dipoles. Nowhere does it say the words "polarization current".

The second reference presents the material better, but still does not contain the words "polarization current".

OK, so between the two reference we have a decent presentation, but no mention of the words "polarization current". However, they both mention polarization current density.

But, charge density is the thing we integrate over area to get current. So, once we have the former, we automatically have the latter according to a long established way of doing things. The first reference has Ip as part of the sum that creates total current I. What else would you call Ip except polarization current, given that Jp is called polarization current density?

The third is a catalog of courses given by MIT OpenCourseWork. I don't know what I am supposed to do with that.

Link seems fine to me? Did you scroll down? It's your choice what you do with it. You don't have to do anything with it, but I would recommend you use it (or another textbook) to get the proper understanding of what polarization, polarization current density and polarization current are conceptually. I just decided to include it because I think textbooks are better than websites and wiki, and that link seemed more "textbook-like" and might carry more weight for some readers. It might also seem unnecessarily complicated, which is why I included the other two references. The first reference is baby-simple, the second is sufficient, and the third reference is comprehensive.

 

[Ratchit]

steveB,

However, I'll answer a couple of questions from your post, since that is the appropriate thing to do in a discussion. And, I do feel we are discussing, not arguing. So, I hope it does not seem like arguing to others aside from tvtech.

Thank you, I appreciate your answers. I am afraid, however, that we are having an argument. An argument is defined as a discussion in which disagreement is expressed. But who cares, as long as interesting points are made.

Polarization is density (per unit volume) of dipole-moments, not dipoles. Dipole moment is the charge times the separation distance (call it Qd) which has SI units of Cm (charge in Coulombs times distance in meters). If you add up all dipoles in a volume and then divide by the volume you get P, which has SI units of C/m^2. But, P is a point-property meaning that we take the limit of density as volume gets small. However, not so small as to get to the molecular level, because P is intended to be a macroscopic property usually. In other words, we want the overall averaged effect at perhaps μm scale rather than at Angstrom scale.

Yes, dipoles are measured as dipole-moments and defined as you state above. I knew that, but I got careless is describing it.

If the P is changing in time, you can visualize it as if the distance of charge-separation of the dipoles are changing. In some sense, the positive and negative charges are swapping places with + going one way and - going the other way. This represents a current because the directions are opposite and the charges are opposite. Hence, they represent current in the same direction. This is just a crude description to get the mental picture.

Yes, I see what you are saying, and I agree that a current will exist in one direction for a limited time. The limit of charge movement is determined by how far the dielectric molecule can be distorted. It is not a current that exists in a circuit.

But, this is per unit volume, so we end up with dP/dt having units of A/m^2, which represents a current density, which is a vector because P was a vector. So, Jp=dP/dt and any current density can be vector integrated over an area to get an effective current Ip, which ( as much as you disagree with this also, it seems) is a scalar quantity.

You are correct about my disagreement. A current has magnitude and direction, so it is a vector. However, the current you describe is a transient, because of the charge constituting the current is bound by the molecular dimensions of the dielectric.

OK, so between the two reference we have a decent presentation, but no mention of the words "polarization current". However, they both mention polarization current density.

That is correct.

But, charge density is the thing we integrate over area to get current. So, once we have the former, we automatically have the latter according to a long established way of doing things. The first reference has Ip as part of the sum that creates total current I. What else would you call Ip except polarization current, given that Jp is called polarization current density?

OK, now that you explained it, it seems logical to me. However, I still say is a short-lived current that is very localized.

Link seems fine to me? Did you scroll down? It's your choice what you do with it. You don't have to do anything with it, but I would recommend you use it (or another textbook) to get the proper understanding of what polarization, polarization current density and polarization current are conceptually. I just decided to include it because I think textbooks are better than websites and wiki, and that link seemed more "textbook-like" and might carry more weight for some readers.

I did scroll down, but I could not see any reference to textbooks. No matter, I do have textbooks on electromagnetics. None of my textbooks say anything about polarization current or equate current to the dipoles changing their dimensions. Interesting.

Ratch

 

[steveB]

Hmmm, strange that the link doesn't work for you. I checked it several times and it works for me. It's supposed to be Chapter 6 on Polarization from

https://www.amazon.com/Electromagne...&qid=1375924380&sr=8-1&keywords=9780132490207

I downloaded it and attached it here, but it may be a reader compatibility issue, so it may not help. The amazon cite says this is out of print, but someone mentions it's offered free electronically from MIT, which I assume is why this chap. 6 was available to me.

 

[Ratchit]

steveB,

Thanks I got it now from your link. Now to digest it. Did you see what The Amazon wants for the book? $1330.22 USD for a new book and 295.00 USD a used book. Are those pages embossed with gold leaf? Anyway, thanks again.

Ratch

 

[steveB]

Ha, yes I saw that. This kind of thing happens sometimes with the out of print books. I don't know who actually pays those amounts. Haus is well known, but even if he signed the book, I can't see that amount as reasonable.

There was a book I wanted that came up listed at about $500 for a new hardcover, but I waited a few weeks and a $40 copy came up from a second hand book store. When I got it, it was a like new library copy, with better than standard binding.

 

[hanhan]

Hi,
I have a PDF copy of the book. If anyone needs it, please private me your email.