Classical Electromagnetism: Question on Ampere's Law and Displacement Current

[Defennder]

Hi cabraham,

I don't why you choose to belabor this point repeatedly as you have done so in other threads. The conventional understanding is that charges move under an electric field (they accelerate but effectively move at a constant drift velocity due to electron collisions) which is associated with a potential difference, and it is the rate of net movement of these charges through a given plane (flux) which we call current.

And it's not true you cannot have an electric field without current, simply consider a fully charged capacitor in a circuit. The current flowing through that loop is zero when it is fully charged, yet clearly there is an electric field in the di-electric. I'm not talking about displacement current here because D here is time-invarying. The same thing can be said of any case in which there is a general separation of positive and negative charge. E-field yes, pd yes, current no.

Consider a cell then. What drives the current in the circuit when the cell is connected across a bulb or resistor? Certainly the current isn't already present before you set up the circuit.

 

[cabraham]

Hi cabraham,

I don't why you choose to belabor this point repeatedly as you have done so in other threads. The conventional understanding is that charges move under an electric field (they accelerate but effectively move at a constant drift velocity due to electron collisions) which is associated with a potential difference, and it is the rate of net movement of these charges through a given plane (flux) which we call current.

And it's not true you cannot have an electric field without current, simply consider a fully charged capacitor in a circuit. The current flowing through that loop is zero when it is fully charged, yet clearly there is an electric field in the di-electric. I'm not talking about displacement current here because D here is time-invarying. The same thing can be said of any case in which there is a general separation of positive and negative charge. E-field yes, pd yes, current no.

Consider a cell then. What drives the current in the circuit when the cell is connected across a bulb or resistor? Certainly the current isn't already present before you set up the circuit.

Defennder,

I've explained thoroughly and you just aren't listening. Regarding the charged cap with E field and no I, I've already stated the case. In order to change the E from zero to non-zero, a current was needed. As far as a separation of pos and neg charge, how is that accomplished without current? Also, a current loop with no loss will susain current indefinitely with no E field. Inductance tends to sustain current, but voltage is needed to change current.

The fact that charges move in an E field is not under dispute. You would mislead all to believe that the voltage is solely responsible for the E field, which is incorrect. A static E has no current, but a static charged cap didn't get that way without a current momentarily. ALso, per our good buddy "Eli the ice man", take a cap uncharged, a battery and a switch. At t=0, close the switch. The cap voltage at t=0 is zero but the current is non-zero. The current is present in the cap before the voltage. Eli the ice man has NEVER LIED to us.

Regarding a cell circuit, you don't seem to understand transmission lines at all. Take a cell, a switch, a t-line (2 parallel conductors), and a load resistor. If the switch is closed at time t = 0, the relation between E, I, and V in the load is as follows. At t = 0, the t-line characteristic impedance is Zo, so that the current at t = 0 is Vs/Zo (Vs = cell voltage). This current I = Vs/Zo propogates towards the load. The E field, I, and potential diffreence V are all simultaneous and in unison. At t = 0, all 3 are established at the source end, and travel together. When the E-I-V traveling wave (H is also present) arrives at the load, the relation is Il = Vl/Rl, where Vl = load voltage, Rl = load resistance, Il = load current. If the load resistance Rl matches Zo, then I = Vs/Rl. Otherwise reflections take place and the system eventually settles.

Point is that the load resistance is characterized by Ohm's law, J = sigma*E. The J (current density, A/m^2) and the E arrive in unison.

One does not "drive" the other.

As far as my "bringing up the issue" goes, I didn't start this. The thread was moving along nicely when out of nowhere "V drives I" was brought in. Regarding Ampere's law and displacement current, I would rather stay focused solely on that, as it is the OP's question.

The notion that V drives I is a result of life long use of cells, generators, and electronic bench top power sources that are optimized for constant voltage mode of operation. Some erroneously conclude that constant V and variable I is Mother Nature, when it really is a man made condition. If you were to pedal a bicycle driven ac generator you could monitor the speed, torque, current, and voltage. As the load changes, adjusting the right mechanical quantity allows control of a corresponding electrical quantity. If you force a constant torque, you then adjust the speed so that constant current is maintained as the load varies, resulting in variable voltage. Holding the speed constant, and varying the torque results in constant voltage and variable current, which is very familiar to us all.

 

[Defennder]

Hi cabraham,

I'm a little busy these few days and may not have time to respond to you promptly, so you may have to wait a little longer before getting a response. Thanks for your understanding.

Now, you claim, among other things that you can't get I without E and V and vice versa. This simply isn't true. I've said earlier that E and V can be obtained by mere charge separation. One poster pointed out some time back in another thread that if one considers the triboelectric phenomenon, that of charging insulators by rubbing them together, you get a separation of charges, and hence a potential difference and electric field without the aid of a current. Now of course you may say that any movement of charges would constitute a current and the triboelectric effect requires a movement of charges in order to obtain a separation, but that is a very loose generalisation of the term current.

Consider this then: A neutral body attracts charged particles, by virtue of gravitational attraction; would you consider the subsequent movement of charges to be a current? This would depend on whether you consider movement due to electric field "current" or just any movement of a charged body. By Newton's 1st law, a charged body (with mass) would travel at a constant velocity if no net force acts on it. Would you consider this current as well? Suppose you do, then this is further evidence that current exists indepedent of voltage and E-field; the E-field associated by the moving charge does not affect itself.

And with regards to there being a current without a driving emf, consider the photocurrent from the photoelectric effect. Photocurrent flows in the photoelectric setup without an applied emf or even one in which the applied emf is set up in opposition to the photocurrent.

This shows clearly that one can de-couple E-fields, V and current density. Your point about there being a non-zero current at t=0, but where V=0 across the capacitor in an RC circuit misses the point. The voltage across the potential difference is zero, but there is clearly a potential difference and hence voltage across other circuit elements.

Thirdly, my point earlier about a simple cell circuit was assuming a quasi-static electric setup where the wires are assumed to be ideal, so I don't know why you're bringing in propagation in a transmission line when I did not bring that up in the first place.

Anyway thanks for posting. I have learned more about this and have also clarified certain doubts after I revisited some material I had semi-forgotten.

 

[cabraham]

Defennder,

What part of the following am I not making clear:

I and V can exist independently only under static conditions. A charged cap or energized inductor without loss demonstrates that either can exist on its own. Neither is the cause. By definition a cause comes first, and the effect can never exist in the absence of the cause. I does not "drive" V, and V does not drive I.

I and V cannot exist independently under dynamic, or time-varying conditions. Regarding the charging of a cap, of course there is voltage elsewhere in the circuit. But, wherever there is voltage there is also a current, except for the cap at t = 0. In a resistor they cannot be decoupled, not even at dc (static). But if the source is a lossless inductor, and the switch across it is opened, the inductor will energize the capacitor. Here, the inductive energy is driving the cap, and then when the inductor empties, the cap drives energy back into the inductor. "I drives, V, V drives I, I drives V," etc.

The "driver" is energy, not I or V. Now with the photodiode, I mentioned it to show that I, the photocurrent, is NOT driven by a voltage. You then proceed to lecture me on exactly what I originally said. Read my previous post. When you said that V drives I, I brought in the photodiode to demonstrate otherwise. Now you're telling me that I and V (and E as well) can be "de-coupled", the exact opposite of your original premise. Once again, what "drives" I in a photodiode is *energy* in the form of photons. In a pd, photons incident on the surface transfer energy to the lattice and electrons are elevated from valence to conduction. In an LED, the opposite takes place. The forward current is set up, and electrons dropping from conduction to valence incur an energy decrease. Energy is conserved in the way of photon emission. Once again, the driver is energy.

Changing your position is not a bad thing necessarily. I think it is clear to all reading this, that "apply a voltage and get a current" is just a prejudice based on faulty intuition. I'm glad you don't completely buy into it. You seem to have a good understanding of the photocurrent process. I think we can agree that energy is the key player, or prime mover.

As far as decpupling goes, under dynamic conditions, no one has ever been able to decouple, I, V, H, and/or E. From Ampere to Faraday to Maxwell to Einstein it has been that way. Who are we to argue?

 

[Defennder]

cabraham,

Perhaps you should reread your OP which spawned my response. Here it is in part:

E field can be expressed as the negative of grad V, but V can be expressed as the line integral of E. Also, J = sigma*E, where J is current density, sigma is conductivity, and E is electric field intensity. There is a functional relationship between J and E, E and V, etc. Also, an electric field cannot be set up unless current and voltage are both present. To change the E field requires work since E fields store energy. Work takes time. The work divided by the time is power. For non-zero power, both I and V must be non-zero. These equations do not imply "causality". Also, E = -grad V only holds for charged particles. For induction in time changing case, E = -dA/dt, where B = curl A. The relation between E, H, I, and V is mutual inclusion. They cannot exist independently, and neither drives the other. Einstein in his 1905 "On The Electrodynamics Of Moving Bodies" stated E and H as having no pecking order. Also, there is no such thing as "applying a voltage". Read up on transmission lines.

I responded initially that V and I are independent (I did not make the distinction between time-varying and time-invariant conditions)

And it's not true you cannot have an electric field without current, simply consider a fully charged capacitor in a circuit. The current flowing through that loop is zero when it is fully charged, yet clearly there is an electric field in the di-electric. I'm not talking about displacement current here because D here is time-invarying. The same thing can be said of any case in which there is a general separation of positive and negative charge. E-field yes, pd yes, current no.

I brought up the photocurrent, as you did, though I think I might have missed your initial mention of it just to show that I and V are independent (as you did).

By definition a cause comes first, and the effect can never exist in the absence of the cause. I does not "drive" V, and V does not drive I.

The point I was trying to make was that saying "V drives I" is not incorrect, but "V always drives I" is incorrect, as I mentioned earlier about separation of charges and photocurrent. Saying that "V drives I" does not preclude "I drives V" is true as well. Just consider this: A force is exerted on a mass and it accelerates by Newton's 2nd law. Suppose someone says that the force causes the mass to accelerate. Would you then argue that the person is indeed sinfully wrong, since the mass accelerates at the same time the force is applied to it? I have argued analogously there is nothing wrong with the notion "V drives I" here. Oddly enough, you have accused me of changing my stand, which is patently untrue.

It's a good thing you finally clarified your point here. It wasn't evident in your original post. I don't think we are in disagreement on this, only the impression that we disagree.

I and V can exist independently only under static conditions...I and V cannot exist independently under dynamic, or time-varying conditions.

 

[cmos]

The "driver" is energy, not I or V. Now with the photodiode, I mentioned it to show that I, the photocurrent, is NOT driven by a voltage. You then proceed to lecture me on exactly what I originally said. Read my previous post. When you said that V drives I, I brought in the photodiode to demonstrate otherwise. Now you're telling me that I and V (and E as well) can be "de-coupled", the exact opposite of your original premise. Once again, what "drives" I in a photodiode is *energy* in the form of photons. In a pd, photons incident on the surface transfer energy to the lattice and electrons are elevated from valence to conduction. In an LED, the opposite takes place. The forward current is set up, and electrons dropping from conduction to valence incur an energy decrease. Energy is conserved in the way of photon emission. Once again, the driver is energy.

This is partly incorrect. First off, an electric potential difference (i.e. voltage) implies an energy difference (energy is directly proportional to electric potential). However, when one speak of currents, it is customary to speak of electric potentials as opposed to energies.

More importantly, your description of light incident on a photodiode is correct up to the point where photons elevate an electron into the conduction band. It is customary to say that you are creating excess minority carriers on either side of the junction. These minority carriers, however, are then swept to the opposite side of the junction by the built-in potential of the junction. So, as you see, it is still a voltage that drives the current.

Your mention of the LED seems to add nothing to the conversation and implies an effort to confuse and draw attention away from the topic at hand.

In reading these last few post, it seems as if you come up with complicated scenarios that are highly convoluted in their explanation. You then go on to mention "On the Electrodynamics of Moving Bodies." Great.

Can you come up with ONE example where current is the ultimate driver? Don't get cute, don't confound the problem. Just present one example where current is the ultimate driver.

 

[cabraham]

This is partly incorrect. First off, an electric potential difference (i.e. voltage) implies an energy difference (energy is directly proportional to electric potential). However, when one speak of currents, it is customary to speak of electric potentials as opposed to energies.

More importantly, your description of light incident on a photodiode is correct up to the point where photons elevate an electron into the conduction band. It is customary to say that you are creating excess minority carriers on either side of the junction. These minority carriers, however, are then swept to the opposite side of the junction by the built-in potential of the junction. So, as you see, it is still a voltage that drives the current.

Your mention of the LED seems to add nothing to the conversation and implies an effort to confuse and draw attention away from the topic at hand.

In reading these last few post, it seems as if you come up with complicated scenarios that are highly convoluted in their explanation. You then go on to mention "On the Electrodynamics of Moving Bodies." Great.

Can you come up with ONE example where current is the ultimate driver? Don't get cute, don't confound the problem. Just present one example where current is the ultimate driver.

I never said that current is the ultimate driver. I said that I and V are both driven by energy, or work W. Do you read my posts at all?

Regarding photodiodes, did you ever take and pass a course on semiconductor physics? I took a senior undergrad course from the physics dept, then a grad level course from the EE dept in the latter '70's. Then I took 2 more in 2007-2008 in grad school. You just blind sided me with your theory:

"It is customary to say that you are creating excess minority carriers on either side of the junction. These minority carriers, however, are then swept to the opposite side of the junction by the built-in potential of the junction. So, as you see, it is still a voltage that drives the current."

How can a minority carrier be swept by the built in potential? The built-in potential, herein referred to as Vbi, is a barrier that must be overcome. It doesn't drive carriers anywhere. Recheck your references. In order to change the potential across a junction, the charge distribution must change, i.e. carriers must be increased or decreased. So if charges are increased or decreased, current takes place. The change in current chronologically preceded the change in voltage. The junction voltage does not drive the current, NOR VICE-VERSA. The Vbi is due to minority carrier distribution. Thermally generated hole-electron pairs transit across the junction. When they arrive in the new material, they are minorities, holes in the n material, and electrons in the p material. When new e-h pairs are generated, they cannot be swept across the barrier by Vbi. Vbi consists of the same polarity carriers as the new carriers. I.E. the holes in the n material cannot sweep holes from the p side over to the n side. Like charges repel not attact. Vbi must be overcome. An external source must provide an E field to move the carriers across the barries. This E field must overcome the Vbi since Vbi tends to oppose the migration of carriers. Vbi OPPOSES charge being "swept" across the barrier. Vbi does NO SWEEPING! Take any p-n junction. The barrier Vbi is due to holes in the n side, and electrons in the p side. To move holes from p to n, and electrons from n to p, an E field is needed. This E field points from the p to the n side. But the E field associated with Vbi points from the n to the p side, exactly OPPOSITE that needed.

Regarding an example where current is the ultimate driver, take a lossless inductor shorted. It carriers an non-zero current I, and has inductance L. A switch across it keeps current steady. A cap is placed across the switch. It is uncharged. The energy in the inductor is L*(I^2)/2, and in the cap it is 0.

If the switch is opened, current from the inductor enters the cap and charges it. After the inductor is completely emptied, the cap energy is C*(V^2)/2, and inductor has 0 energy. Then the cap transfers energy back to the inductor, etc. etc.

The reason that the cap was charged, and that the voltage went from non-zero to zero is due to the inductor transferring energy. This energy is in the form of a magnetic field / current. Of course, the instant the switch opened, the inductor began generating an increasing voltage. But at t = 0, V = 0, with non-zero I. Hence the cap voltage is 0 with current entering it. As current charges the cap its voltage increases. Likewise, the cap voltage maxes out, then returns energy to the inductor. The voltage in non-zero while the current is 0.

Thus the current can charge the cap and change the voltage, and the voltage can energize the inductor and change its current. Either I or V can come first. It is energy that is the ultimate driver. If an observer arrives after the switch has been opened, they see the energy bouncing back and forth. If asked, "Which came first, the I or the V?", they couldn't answer. Either can "come first".

This is well known. "Eli the ice man" has been known for nearly 2 centuries. The notion that "voltage drives current" is just a misconception easily acquired since batteries, generators, and electronic power supplies are almost always designed, optimized, and built for *constant-voltage* mode of operation. We can easily lose sight of the fact that the constant voltage and variable current mode of operation is a man-made thing. Mother Nature is neither constant I nor constant V. I hope this explains my position. If anything needs clarified, I'll be glad to do so. Comments are always welcome and appreciated. BR.

Claude

 

[cmos]

Regarding photodiodes, did you ever take and pass a course on semiconductor physics? I took a senior undergrad course from the physics dept, then a grad level course from the EE dept in the latter '70's. Then I took 2 more in 2007-2008 in grad school. You just blind sided me with your theory:

Clearly I have taken such a course. I would question your own claims though as it seems that you simply looked up some definitions but never actually performed any rigorous analysis. As I said once before, EE is not the same is EET.

Anyway, if your claims are true, then I will encourage you to draw the band diagram of a pn-junction under sufficient illumination. Then you can see how the built-in potential sweeps photogenerated minority carriers thus inducing current.

Regarding an example where current is the ultimate driver, take a lossless inductor shorted. It carriers an non-zero current I, and has inductance L. A switch across it keeps current steady. A cap is placed across the switch. It is uncharged. The energy in the inductor is L*(I^2)/2, and in the cap it is 0.

If the switch is opened, current from the inductor enters the cap and charges it. After the inductor is completely emptied, the cap energy is C*(V^2)/2, and inductor has 0 energy. Then the cap transfers energy back to the inductor, etc. etc.

Yes, but you still need to have a source present. The source will be fundamentally a voltage driver.

The reason that the cap was charged, and that the voltage went from non-zero to zero is due to the inductor transferring energy. This energy is in the form of a magnetic field / current. Of course, the instant the switch opened, the inductor began generating an increasing voltage. But at t = 0, V = 0, with non-zero I. Hence the cap voltage is 0 with current entering it. As current charges the cap its voltage increases. Likewise, the cap voltage maxes out, then returns energy to the inductor. The voltage in non-zero while the current is 0.

Charge builds up on the capacitor because, initially, the plate is at a different potential than the end of the inductor. So when you make a connection, current flows because of that potential difference.

This is well known. "Eli the ice man" has been known for nearly 2 centuries.

This is a phase relation and not a causality relation.

The notion that "voltage drives current" is just a misconception easily acquired since batteries, generators, and electronic power supplies are almost always designed, optimized, and built for *constant-voltage* mode of operation.

So then how would one make a "true" current driven source? (as opposed to a current source that is fundamentally driven by a voltage.)

 

[cabraham]

cmos, this is absurd. First of all, why do you keep referring to me as "EET" and not "EE". My profile explicitly states my EE status, as well as PhD candidacy. I know that EET is not the same as EE.

Regarding the inductor, NO, a source (voltage or current) need NOT be present. An energized inductor will charge up a cap without any other source(s) present. Elementary circuit theory that is. The cap is initially uncharged w/ zero voltage. Yet current immediately exists at t=0. There is zero voltage and non-zero current. Current flows because the inductor is transferring energy to the cap. No voltage is driving said current. The plate is NOT at a different potential than the inductor.

Regarding "Eli the ice man", I never said it was about causality. I've been inferring that there is no causality between I and V and Eli is proof. Since I can lead V, or V can lead I, that is prima facie evidence that neither causes the other. I brought Eli in just to demonstrate the non-causal relation between I and V. You are now arguing my case.

How to make a true current driven source. Run a generator (ac) at constant torque. Imagine a bicycle type ac generator. If you pedal with constant torque, and the load resistance can change, the current remaons fixed while the voltage and speed vary with load.

A photodiode is a current source. I just took an advanced semiconductor physics course spring 08. The Vbi built in potential in the energy band diagram is a barrier that must be overcome. I covered that previously. If the built in potential could sweep minority carriers, light wouldn't be needed. Minority carriers are constantly being generated by thermal energy. The Vbi opposes charge migration across the junction.

Nothing personal, but you're batting 0. BR.

 

[cmos]

cmos, this is absurd. First of all, why do you keep referring to me as "EET" and not "EE". My profile explicitly states my EE status, as well as PhD candidacy. I know that EET is not the same as EE.

Because your tactless side comments, absurd claims, and misuse of fundamental concepts leads to me suspect.

Regarding the inductor, NO, a source (voltage or current) need NOT be present. An energized inductor will charge up a cap without any other source(s) present.

So the inductor spontaneously became charged? Splendid, you've just overthrown conservation of energy. In less sarcastic words; the source will be necessary to charge the inductor in the first place.

Elementary circuit theory that is. The cap is initially uncharged w/ zero voltage. Yet current immediately exists at t=0. There is zero voltage and non-zero current. Current flows because the inductor is transferring energy to the cap. No voltage is driving said current. The plate is NOT at a different potential than the inductor.

Now let your clock run. A voltage will be dropped across the capacitor; this will reach a steady-state at which point no more charge builds up on the capacitor plates. This means no more current is running from inductor to capacitor. So the two adjacent end nodes must be at the same potential. This implies that this were not originally the case (i.e. the plate was at a different potential than the inductor initially - this drove the current to charge the capacitor).

Regarding "Eli the ice man", I never said it was about causality. I've been inferring that there is no causality between I and V and Eli is proof. Since I can lead V, or V can lead I, that is prima facie evidence that neither causes the other. I brought Eli in just to demonstrate the non-causal relation between I and V. You are now arguing my case.

This does not follow. This relation says nothing for or against your argument. It speaks oly of the phases and nothing else.

How to make a true current driven source. Run a generator (ac) at constant torque. Imagine a bicycle type ac generator. If you pedal with constant torque, and the load resistance can change, the current remaons fixed while the voltage and speed vary with load.

I don't quite follow this. Maybe you can cite some source?

Really, what it comes down to is Maxwell's equations. None of the four equations say that current may be induced but that it is simply an inducer. The fact that current (i.e. the movement of charge) may occur is given by Coulomb's law. This explicitally demands that an electric field (voltage) be present to move the charge (drive the current).

A photodiode is a current source. I just took an advanced semiconductor physics course spring 08. The Vbi built in potential in the energy band diagram is a barrier that must be overcome. I covered that previously. If the built in potential could sweep minority carriers, light wouldn't be needed. Minority carriers are constantly being generated by thermal energy. The Vbi opposes charge migration across the junction.

What you are describing is a non-illuminated diode or an LED in forward operation. An illuminated photodiode as a current source (sometimes commonly called a solar cell) runs a current that is negative with respect normal diode operation. This is because the built-in potential in the photodiode drives the current the opposite way with respect to normal diode operation. Pay better attention in class.

Nothing personal, but you're batting 0. BR.

Nothing personal, but you are still batting 0.

 

[cabraham]

cmos, you've raised *getting it wrong* to an art form.

You twist everything I say. First, of course the inductor is not spontaneously charged. It has been charged by another source, but that source is decoupled, leaving an energized inductor terminated in a short. Then a switch opens with a cap across the inductor. Current from the inductor charges the cap at t=0. The voltage at t=0 is *zero*, with non-zero current. Any EE prof will explain this to you. The voltage builds up until the current is zero. t this point the voltage maxes out. Then energy is returned from the cap to the inductor. Pretty straightforward. Of course, a charged cap can be the original source of energy, and the inductor gets switched in. The voltage is non-zero at t=0, and the current is zero. The current builds up, and when the voltage is zero, the current maxes out. Then the inductor returns energy to the cap.

Energy is transferred between the 2, with no pecking order.

Eli the ice man demonstrates the futility of trying to establish a pecking order. Since I can lead V, V is not the cause of I. Since V can lead I, I is not the cause of V. Eli disproves causality.

Regarding ac generators, you ask for a reference. Any machines book will confirm that torque and current are strongly related, as are speed and voltage. The reverse holds w/ motors. Anyone experienced with motor drives will affirm that to control torques you use current drive, and to control speed you use voltage drive. Any text will affirm.

I'll see if I can scan the photodiode diagrams and post them.

Regarding Maxwell's equations, you keep equating E field with voltage, which is not equivalent. Of course, charges are moved by an E field. But the voltage is not what moves charges. F = qE. The Coulomb force is observed and the finite velocity of field propogation is observed. Thus when the field arrives, the force is felt. Charges move due to the presence of other charges. Potential is derived from force distance and charge.

To establish an E field, you need volts and amp-seconds both. E fields exist when charges are separated, but separating requires moving them. Moving charges are what is called current. So, charges move, i.e. current, in an E field, but it takes current (and voltage) to establish an E field.

There is no pecking order.

 

[cmos]

cmos, you've raised *getting it wrong* to an art form. You twist everything I say.

Well this just isn't true. I believe all my statement thus far have been correct. And my recent style has been to quote exactly what you have said and respond accordingly; no twisting anything you say in that.

First, of course the inductor is not spontaneously charged. It has been charged by another source, but that source is decoupled, leaving an energized inductor terminated in a short.

I feel like any attempt to do this would result in a shorted inductor and no current; similar to taking a piece of wire and making a ring out of it.

Then a switch opens with a cap across the inductor. Current from the inductor charges the cap at t=0. The voltage at t=0 is *zero*, with non-zero current. Any EE prof will explain this to you. The voltage builds up until the current is zero. t this point the voltage maxes out. Then energy is returned from the cap to the inductor. Pretty straightforward.

Assuming that your "decoupled inductor" works: well the voltage is actually zero at t=0- and then steps to some value at t=0+. The point that I have been driving is that the current tends to zero as the adjacent ends of the inductor and the capacitor equilibrate in potential. This is clearly a voltage-driven phenomenon.

Of course, a charged cap can be the original source of energy, and the inductor gets switched in. The voltage is non-zero at t=0, and the current is zero. The current builds up, and when the voltage is zero, the current maxes out. Then the inductor returns energy to the cap.

Similar arguments to the two I just made.

Eli the ice man demonstrates the futility of trying to establish a pecking order. Since I can lead V, V is not the cause of I. Since V can lead I, I is not the cause of V. Eli disproves causality.

As I keep saying, the term "lead" is in reference to the phase and not cause and effect.

Regarding ac generators, you ask for a reference. Any machines book will confirm that torque and current are strongly related, as are speed and voltage. The reverse holds w/ motors. Anyone experienced with motor drives will affirm that to control torques you use current drive, and to control speed you use voltage drive. Any text will affirm.

You speak of Faraday induction. In this, it is the electric field that is induced, not the current. When we speak of the electric field in circuits, it is convenient to speak of voltages. As a result, voltage again drives the current.

I'll see if I can scan the photodiode diagrams and post them.

If you need help drawing them out, just ask.

Regarding Maxwell's equations, you keep equating E field with voltage, which is not equivalent.

The negative gradient of voltage is the electric field. Whichever one is used is more of a matter of convenience. This is similar to formulating mechanics in terms of energy or in terms of forces.

Of course, charges are moved by an E field. But the voltage is not what moves charges. F = qE. The Coulomb force is observed and the finite velocity of field propogation is observed. Thus when the field arrives, the force is felt. Charges move due to the presence of other charges. Potential is derived from force distance and charge.

But the potential energy is U=qV. The movement of charge can thus be analyzed by the mechanics of Hamilton and Lagrange.

To establish an E field, you need volts and amp-seconds both. E fields exist when charges are separated, but separating requires moving them. Moving charges are what is called current. So, charges move, i.e. current, in an E field, but it takes current (and voltage) to establish an E field.

All it takes is a single charge to create an electric field. You may then place a test

Sadly, I think that at this point we are going in circles and more or less repeating ourselves. Unless you have a specific question or something genuinely new to add, I will let you have the last word on this one and leave it at that.

 

[atyy]

I'm getting the impression from some of the sources that I've read (Wikipedia and others) that this results in an electromagnetic wave. Does this explain why wires heat up or something? I guess not, since clearly Joules heating occurs in DC current flowing in wires. But if not, what happens to the EM waves?

I'm not certain on this, but I believe that the radiative loss (heating of a resistor) is correctly taken into account by circuit theory. The main place where circuit theory should fail with respect to energy conservation is radiative loss from a capacitor or an inductor. A related thing is "stray capacitance" - there is no such thing as a pure resistor or pure inductor or pure capacitor. Any "pure resistor" that you can buy is only an approximation, and if it has significant capacitance that you don't take into account, then your circuit may not work as planned.

In the most general terms, there is no such thing as capacitance. Capacitance as a property of a conductor, as opposed to being a property of the conductor and field configuration, is defined only if no charge is moving, so that we can predict everything using a potential (a scalar field) instead of an electric field (a vector field). cabraham made the same point above:

I and V cannot exist independently under dynamic, or time-varying conditions.

For induction in time changing case, E = -dA/dt, where B = curl A.

As he points out (I think there may be a minor typo in his equation, but the sense is correct), under general conditions, if we insist on using a scalar electric potential, we must supplement it with a vector magnetic potential. The scalar and vector potential together form the electromagentic four-potential. In classical field theory, the E and B fields are primary, and the potentials secondary (a physical state can be labelled by anyone of many four-potential configurations). But in quantum field theory the four-potential is primary, and the multiple labels for the same physical state can cause lots of confusion in complex equations.