What causes an increase in energy in an electrical system?

[Edi]

.. if we go by the plumbing analogy, ... pressure [difference]
But, if same amount of water (amount of electrons for electric current) flows trough the same sized hose, but with different pressures, in two, or more, different samples, then water in the sample with the highest pressure will flow faster (thus the system would have more power AND total energy ) than water in the sample with the lowest pressure. - can we conclude that electrons/ wave of electrons moves faster trough a given sample if the voltage is higher? (that would account for the increased total energy.. but it, most likely, won't be true, as most of my interpretations tend to. So what EXACTLY does account for the increased energy of the flow and the system?)

 

[rcgldr]

If your using a plumbing as a comparson model, then a better analogy for voltage would be height, since height times the strength of gravity is also a potential.

 

[Drakkith]

The key here is that the voltage determines the current, not the other way around. You say if the same amount of water flows through two hoses, but one has more pressure it will flow faster. But that isn't true, as there is MORE water flowing past a point in the high pressure hose and LESS water flowing past a point in the low pressure hose as a direct result of the increased water pressure.

Similarly, applying more voltage to a circuit directly increases the amount of current flow. This increase results in more charges (electrons) actually flowing past a point in the circuit, not in an increase in speed.

 

[EWH]

Yes, in normal materials (ohmic) the drift velocity of the charges is directly proportional to the voltage. The current is also proportional to the voltage. The signal propagation rate generally isn't, because the charges throughout the circuit move like a conveyor belt - all parts move at the same time. (Delayed by the speed of sound for the conveyor belt material, or light for the electric circuit.

What causes the additional energy for a higher voltage across a given resistance? Well, the unsatisfying answer is that 1Volt=1 Joule/Coulomb. A more satisfying answer is that when there is a higher voltage, there is a greater electromagnetic field outside the wires carrying the current, (or electric field in the dielectric between the plates of a capacitor for the case of a displacement current) and this is where the energy is stored. (Mostly. There is a little inertia of the current itself, but virtually all the energy is in the fields outside the wires.)

See Bill Beatty's detailed explanation of energy flow in a simple circuit http://amasci.com/elect/poynt/poynt.html". His article "What is Voltage? there may also be helpful.

 

[Drakkith]

For some reason that site makes me uneasy. I don't know enough to call it incorrect, but some of the things I've looked at there just don't quite add up to me.

 

[Naty1]

Where does EM energy flow...is just plain ridiculous !

Simply awful...incoherent...

The worst misconceptions I think I have ever seen!

rather funny though...


I only skimmed the first half dozen paragraphs...I would not even bother to go further.

joules "flow"? ...nonsense...

"The field energy flows parallel to the wires, and eventually it dives into the lightbulb filament." silly...swimmers dive not electrical fields.


"Electrons in the metal momentarily speed up before colliding with tungsten atoms."
nonsense.

Whoever wrote this apparently doesn't know that while individual electrons do move at only a few meters per second, the power moves almost instantaneously...analogous to a crowded highway where one car moves and another moves almost simultaneously miles away...but each is crawling along...

 

[RedX]

I looked at "Bill Beaty's" link, and that's actually the way I understand circuits: the wires guide electromagnetic waves produced by a generator or battery to the load, where it gets converted to ohmnic heating. I didn't read the link except for the first couple of paragraphs, but skimming the pictures quickly, they seem all right.

Here's a link to a picture of an electromagnetic wave being guided by two wires (figure 3-14 in the middle of the page):

http://www.tpub.com/neets/book10/41d.htm

One of the currents is going towards you, and another is going away from you. The fields flow down the circuit to the load.

For more information (DC and AC wave being guided along two wires) see this page:

http://www.tpub.com/neets/book10/41f.htm

The e-book I believe is from the Navy, though I'm not sure. They like to draw the direction of the current as the direction of the electrons, which I think is characteristic of the Navy (makes more sense since electrons are really what's flowing)

 

[Drakkith]

It's his terminology and apparent "Everyone else is wrong" attitude that makes me wary. That plus some obvious mistakes I thought I saw. For some reason the site is down right now for me, so I'll go back and take a look at it later.

 

[RedX]

It's his terminology and apparent "Everyone else is wrong" attitude that makes me wary. That plus some obvious mistakes I thought I saw. For some reason the site is down right now for me, so I'll go back and take a look at it later.

I just now saw that attitude. He seems to bash Feynman as if he had some sort of personal grudge. It's ridiculous really.

However, you can calculate the magnetic field produced by a loop of wire, and the electric field, and if you take the Poynting vector ExB, then you'll get that energy is flowing into the resistor from all directions. That's what diagrams 7-10 show: those lines are lines of Poynting flux and they always point towards the resistor. The wires guide the waves to the resistor and prevent the waves from radiating into space.

And this idea is not something that this Bill Beatty guy came up with. It actually comes naturally when you study transmission lines and antennas. When studying these things, there are two viewpoints. One is you can look voltages and currents. The other is you can look at the electric and magnetic fields. Usually you look at the latter for waveguides, and the former for two parallel wires. But you can use either viewpoint for either application.

 

[EWH]

Naty1:"joules "flow"? "

Energy flows, and end energy is measured in Joules. I don't see any misconceptions there, although there are some figures of speech. Bill Beatty has written many meticulously correct articles which attempt to correct the misconceptions fostered by standard presentations of the material. He does try to make his presentation visualizable, but that is a good thing I think.

Naty1: "Whoever wrote this apparently doesn't know that while individual electrons do move at only a few meters per second, the power moves almost instantaneously"
You appear to have totally missed the entire point of the article. That is exactly what he is explaining.

RedX: "He seems to bash Feynman as if he had some sort of personal grudge. "
Er.. well Feynman did mess up there a bit. But B.B calls him "an honest free-thinker", so he clearly has respect for Feynman.

The Poynting field is the ONLY correct description of classical EM field energy and transmission. The power absolutely does NOT flow primarily through the wires. The dielectric constant of the materials near the wires affects the transmission speed substantially, as well as the field strength outside the wires and the voltage of the wires; the dielectric need not be in contact with the wires to do so. The same is true of the magnetic permeability of materials near the wires and the specific geometry of the wires (e.g. coils, plates). Beatty may be belaboring the point a bit with the "speed of light in copper" criticism, since IIRC Feynman did present the drift velocity in the lectures, but Beatty's point was:
"If the common misconception that 'energy flows inside wires' has had such a deleterious effect on an honest free-thinker, imagine the trouble a more conventional mind would have with it."

Beatty is also an honest free-thinker, and no sort of crackpot. Though he does have a high affinity for finding out all the potentialities of all sorts of ideas, he also picks apart every little inconsistency in theories. He designs and builds electronic equipment for a university, he used to design and build exhibits for the Boston Museum of Science, and his website has been up since 1994.

 

[ZealScience]

Voltage is energy per unit charge, especially the electric potential energy.Your analogy would be of a pressure potential. So googd analogy, I think.

So you just see where does the potential come from. Pressure poetential in your analogy would be a pump which comsumes fuel. Chemical energy stored would trnsform to pressure. So the battery would be the "pump" for the circuit, which could move charges. Where charge accummulates has higher potential.

 

[cabraham]

The key here is that the voltage determines the current, not the other way around. You say if the same amount of water flows through two hoses, but one has more pressure it will flow faster. But that isn't true, as there is MORE water flowing past a point in the high pressure hose and LESS water flowing past a point in the low pressure hose as a direct result of the increased water pressure.

Similarly, applying more voltage to a circuit directly increases the amount of current flow. This increase results in more charges (electrons) actually flowing past a point in the circuit, not in an increase in speed.

Ref bold quote: not always. If a constant voltage source is connected directly across an impedance, then current is determined by voltage divided by impedance. But if a constant current source is connected to an impedance, then voltage is determined by current and impedance. It can work either way.

A switching power converter is a prime example. Energy is stored in an inductor which tends to behave as a current source. It transfers energy into an output load through a sensing resistor, diode, & recharges the capacitor. The inductor forces a constant current through the sense resistor & diode & the voltage is determined by the impedance of the elements times the current.

The output has a filter capacitor. A cap acts like a voltage source. Any load across the cap sees a constant voltage. So the current in the load is the voltage/load impedance. It can work either way. Voltage does not always determine current. It can, & vice-versa.

Claude

 

[wbeaty]

I just now saw that attitude. He seems to bash Feynman as if he had some sort of personal grudge. It's ridiculous really.

No, the problem is that Feynman is actually wrong in this instance. I point this out. (That's bashing? We're not allowed to criticize?!)

And "personal grudge?" Don't be silly, I'm a huge Feynman worshipper, had "Surely" when it first appeared, and in fact put up one of the internet's early Feynman pages.

Feynman apparently hates the Poynting Vector description of DC circuitry. (Somewhere he has a paper on the details of the difficulty, but I haven't found it yet.) Feynman's apparent mistake was in believing that DC energy sources radiate their energy outwards in all directions to infinity. Nope. The energy is located in the EM fields as always. If fields don't escape, then Poynting-energy doesn't escape either. Dipole fields such as with capacitors, inductors, and wire-pairs all keep their energy mostly local. The strong fields inside a coax cable are the actual location of the moving EM energy. Same thing with the bare wires of a simple flashlight circuit.

IAnd this idea is not something that this Bill Beatty guy came up with.

Obviously not! But I've found only very few physicists ever teach it to students. One was Robert Oppenheimer (of Exploratorium Museum fame.) I've noticed something odd: quite a few physics students become angry when they encounter this discription of circuits, as if it's some kind of heresy. Probably the sticking point is that the description doesn't appear widely in undergrad texts. If it was this simple and useful, why aren't we taught about it in Halliday & Resnik, Sears Zemanski, etc.? It must be WRONG!

:)

Of course double-E students learn all about it in our fields/waves classes. But we're taught to apply it to waveguides. Nobody ever told us that a DC flashlight circuit constituted a valid "waveguide." There is no magical lower cutoff frequency. It works all the way down to DC.

 

[RedX]

The Poynting field is the ONLY correct description of classical EM field energy and transmission. The power absolutely does NOT flow primarily through the wires. The dielectric constant of the materials near the wires affects the transmission speed substantially, as well as the field strength outside the wires and the voltage of the wires; the dielectric need not be in contact with the wires to do so.

Sure, it's easier to see in a coax cable where the dielectric between inner and outer conductors affects the speed of propagation of the voltage/current to the load, but it applies to any two conductors.

But, you can model a long circuit as an LC ladder, and explain the effect of the dielectric as belonging to the dielectric of shunt capacitance between hot and neutral. I believe the terminology is parasitic capacitance. Or in general, having distributed circuit elements rather than lumped elements.

"If the common misconception that 'energy flows inside wires' has had such a deleterious effect on an honest free-thinker, imagine the trouble a more conventional mind would have with it."

I'm visiting the library tomorrow, so I'll look at chapter 27 and see if Feynman made a mistake. I really enjoyed chapters 23 and 24 where Feynman talks about cavity resonators, transmission lines, and waveguides, and Beatty mentions that Feynman doesn't put everything together like Beatty's figure 7 does even though Feynman talks about Poynting vector on chapter 27. Well, maybe that's because chapters 23 and 24 are where he talks about these things, so to go back to chapters 23 and 24 to apply the Poynting vector concept might have been too much of a hassle.

Anyways, here's another picture from a Navy training manual on how a parallel line turns into a waveguide using the fact that a quarter-wave short has infinite impedance, forming a metallic insulator:

http://www.tpub.com/neets/book11/44b.htm

 

[wbeaty]

That plus some obvious mistakes I thought I saw.

PLEASE tell me about any mistakes you find. I'm always updating these pages.

But beware: most of my articles involve common widespread misconceptions. Very often an "obvious mistake" is actually a situation where the entire education community really is wrong, and I'm attempting to debunk the widespread error.

 

[RedX]

Obviously not! But I've found only very few physicists ever teach it to students. One was Robert Oppenheimer (of Exploratorium Museum fame.) I've noticed something odd: quite a few physics students become angry when they encounter this discription of circuits, as if it's some kind of heresy. Probably the sticking point is that the description doesn't appear widely in undergrad texts. If it was this simple and useful, why aren't we taught about it in Halliday & Resnik, Sears Zemanski, etc.? It must be WRONG!

:)

Of course double-E students learn all about it in our fields/waves classes. But we're taught to apply it to waveguides. Nobody ever told us that a DC flashlight circuit constituted a valid "waveguide." There is no magical lower cutoff frequency. It works all the way down to DC.

Perhaps you should consider writing a paper on this for one of the pedagogical physics journals such as The Physics Teacher. You're absolutely correct that it's rarely discussed in physics books. Even looking at the canonical graduate text by John D. Jackson, I can't find a mention of it. A website is good, but I'm not sure if instructors go on the web, so the way to get instructors to adopt and disseminate it to students might be through a journal (although I'm not sure many instructors read journals on how to teach physics, because a lot of them are pretty bad at it!).

edit: actually, I think Jackson briefly mentions it in chapter 6.9, but as is typical of Jackson, it's incomprehensible.

 

[Drakkith]

PLEASE tell me about any mistakes you find. I'm always updating these pages.

But beware: most of my articles involve common widespread misconceptions. Very often an "obvious mistake" is actually a situation where the entire education community really is wrong, and I'm attempting to debunk the widespread error.

Actually I think I just understood what you were talking about in one of your articles that I wasn't sure was correct. Once I understood what you were saying I was like "Oh, yeah, i thought that was obvious" lol.

 

[wbeaty]

Naty1:"joules "flow"? "

Energy flows, and end energy is measured in Joules. I don't see any misconceptions there

If radio waves, light, as well as b-fields/e-fields all contain "electromagnetic energy," then we could be more nitpickingly precise, :) and instead say that "EM energy propagates" across a circuit.

When the beam from a laser pointer zips across a room, that's a flow or propagation of EM energy. When RF waves are gathered in by an antenna and sent along a 2-wire cable to your 1950s antique television set, EM energy is flowing from the antenna to the TV receiver. And when a D-cell sends some joules to a flashlight bulb, EM energy is flowing from battery to bulb in the space around the connecting wires.

If our descriptions must apply to all frequencies, then we don't want to say "light is flowing" or "RF energy is flowing." Instead, the generalized moving entity is "EM energy."

But too bad we have no single word which means light, microwave, rf, and DC circuit energy. Grammar-engineering sometimes can greatly clarify physics. Speaking of which...

I suggest "Wakalixes."

Submitted for your approval: From this point in time forward, Wakalixes means "a certain quantity of EM energy occupying a certain volume of space."

The word comes from "Judging Books By Their Covers:"

"I turned the page. The answer was, for the wind-up toy, "Energy makes it go." And for the boy on the bicycle, "Energy makes it go." For everything, "Energy makes it go."

"Now that doesn't mean anything. Suppose it's "Wakalixes." That's the general principle: "Wakalixes makes it go." There's no knowledge coming in. The child doesn't learn anything; it's just a word!
http://www.textbookleague.org/103feyn.htm"
​

So Susie, what comes out of the front of this flashlight? "Wakalixes of optical wavelength, sir!" Correct. And Johnnie, describe the flashlight operation from the EM fields perspective. "The electrochemical cell emits Wakalixes which are guided by the pair of conductors. Some of it glides inwards to warm the conductors, but the major portion arrives at the bulb and dives into the surface of the tungsten filament."

:)

Oooh, I may regret this. Crap, what if the above meme catches fire!

 

[RedX]

No, the problem is that Feynman is actually wrong in this instance. I point this out. (That's bashing? We're not allowed to criticize?!)

And "personal grudge?" Don't be silly, I'm a huge Feynman worshipper, had "Surely" when it first appeared, and in fact put up one of the internet's early Feynman pages.

Feynman apparently hates the Poynting Vector description of DC circuitry. (Somewhere he has a paper on the details of the difficulty, but I haven't found it yet.) Feynman's apparent mistake was in believing that DC energy sources radiate their energy outwards in all directions to infinity.

Yeah, I was wrong about that. You have a a large Feynman page on your website that has some footage I've never seen, including Gell-Mann on Feynman:

http://amasci.com/feynman.html

So obviously you weren't bashing, and I misinterpreted your words

 

[Drakkith]

Ref bold quote: not always. If a constant voltage source is connected directly across an impedance, then current is determined by voltage divided by impedance. But if a constant current source is connected to an impedance, then voltage is determined by current and impedance. It can work either way.

Are we talking about an AC or DC circuit here? If it is DC then the impedance is simply resistance, correct? Then it just becomes Ohms law correct?

A switching power converter is a prime example. Energy is stored in an inductor which tends to behave as a current source. It transfers energy into an output load through a sensing resistor, diode, & recharges the capacitor. The inductor forces a constant current through the sense resistor & diode & the voltage is determined by the impedance of the elements times the current.

What? The inductor causes a voltage that resists the change in current through it. This causes the circuit to have a steady current, but only because the voltage in the inductor is changing in response to the current, either resisting the increase or compensating for the decrease. Either way the current in the circuit is directly caused by voltage sources.

 

[Drakkith]

Can someone answer this for me? On Beaty's site he has the following:

Everything connected to one battery terminal acquires the same electrical potential (voltage.) The circuit acts like two separate conductors, one with a positive charge imbalance and one with negative.

I don't understand that. If one were to chop the circuit up into a million little pieces wouldn't each piece also behave in the same manner? IE the end towards the positive side would act positive and the side on the negative end would act negative. In his diagram he has E fields between the 2 sides of the circuit. I don't see how that is possible.

Also, is that even accurate to what's really happening? For example, if one applied a positive voltage to one end of a wire, and a negative voltage to the other, the wire would develop an imbalance of charges on each end with a greater difference between the ends and a gradual fall off towards the middle correct?

Now, for a circuit, there isn't a chance for the charges to build up. The electrons flow through the conductor without gathering up so no potential builds up, right? Hence no E field between two different spots in the circuit.

Does that make sense at all, or am I horribly confused?

 

[Edi]

wow.. I honestly was expecting a simple answer. Few comments at best.
Who would have thought that electricity (im not even sure how to call it now! :O ?? )could be so .. SO confusing and complicated, at the fundamental level..
Now I actually know that I understand less than I thought I do [about electricity]. Fascinating.

 

[EWH]

Can someone answer this for me? On Beaty's site he has the following:

Everything connected to one battery terminal acquires the same electrical potential (voltage.) The circuit acts like two separate conductors, one with a positive charge imbalance and one with negative.

I don't understand that. If one were to chop the circuit up into a million little pieces wouldn't each piece also behave in the same manner? IE the end towards the positive side would act positive and the side on the negative end would act negative. In his diagram he has E fields between the 2 sides of the circuit. I don't see how that is possible.

Well, it is a slight simplification, but assuming perfect conductors (which copper almost is compared to a usual resistor - several orders of magnitude difference) one can treat all parts of a given conductor as being at the same voltage (assuming DC or frequencies low enough that the wavelength is much larger than the length of the conductor - say less than 3MHz for each 1 meter of length to be within 4% of constant V all along the conductor at any given instant).

If the wires are 0.005 ohms each and the resistor is 1KOhm, then ~99.999% of the voltage will be dropped across the resistor, and the voltage in each wire will be constant along its length to about 5 parts per million. So one wire will essentially be at the potential of the + battery terminal, the other at the potential of the - battery terminal.

There will thus be an electric field between the wires, and the intensity of this field will depend on the distance between the wires and what sort of dielectric was separating them. (Also, since all parts of a conductor are at essentially the same potential, there is essentially no electric field inside the conductor at all. )

Also, is that even accurate to what's really happening? For example, if one applied a positive voltage to one end of a wire, and a negative voltage to the other, the wire would develop an imbalance of charges on each end with a greater difference between the ends and a gradual fall off towards the middle correct?

That would be a different case - the wire would be the only resistance in the circuit and it would be absorbing the whole voltage drop. The voltage would vary continuously down the wire. (In real circuits the internal resistance of the voltage source would usually become dominant, making the voltage across the wire less.)

Now, for a circuit, there isn't a chance for the charges to build up. The electrons flow through the conductor without gathering up so no potential builds up, right? Hence no E field between two different spots in the circuit.

If the resistance of the conductors is much less than that of the resistor, the potential is dropped across the resistor.

Current is a completely orthogonal concept to voltage - depending on resistance there can be tiny currents with high voltage (e.g. "static electricity"), huge currents with low voltage, or huge currents with high voltage. Only the voltage across two points affects the electric field between those points, not the current. (Well, changes in current affect the electric field, but we're talking about DC. And distance and dielectric affect field strength in volts/meter. But the absolute value of the current doesn't enter into it at all.) So you can have high electric fields in high- or low-current circuits so long as there is a large voltage difference.

 

[Drakkith]

So between the negative side and the resistor, is there an excess of negative charges? And vice versa on the positive side with positive charges? If so then that makes complete sense.

 

[RedX]

No, the problem is that Feynman is actually wrong in this instance. I point this out. (That's bashing? We're not allowed to criticize?!)

I just read chapter 27 and I'm typing this post in the library with the book in front of me. Feynman didn't get any of the physics wrong. Nor does he say that the idea that energy flows in from the field outside (rather than along the wires) is wrong. Quite the contrary, his last sentence on the subject is:

"It is not a vital detail, but it is clear that our ordinary intuitions [of energy flowing down the wire] are quite wrong."

All Feyman does is recommend that you don't get too wrapped up in explaining circuits using Poynting flux. My opinion is that viewing circuits this way is beautiful and counter-intuitive (but more importantly the only correct view), and every physics student should see it once, but Feynman does have a point when he says:

"You might believe that you must revamp all your intuitions, and therefore have a lot of things to study here...it seems to be only rarely of value, when using the idea of energy conservation, to notice in detail what path the energy is taking."

Feynman did make sure that the people who used his textbook would see it once. He gave the example of the capacitor and resistor and the paradox of a bar magnet and charge.

Also, when Feynman uses the word "nutty" and "absurd" and "crazy" to describe Poynting flux, I didn't read it as disparaging words. I just took it to mean his way of saying "counterintuitive."

But utimately Feynman makes it clear that the Poynting view is the correct view with his last sentence.

I think you misinterpreted Feynman, just as I had misinterpreted your comments about Feynman. It happens.

I've told three of my fellow physics students to view your webpage on this correct view of circuits. You explanation is very lucid. Great page.

 

[wbeaty]

So obviously you weren't bashing, and I misinterpreted your words

If you want some Feynman Bashing, ask a (very minor) Feynman expert from a university chemistry dept. where his weak vulnerabilities have long been hidden. Like so.

In the book "Surely You're Joking Mr. Feynman," the title refers to an incident where a Proper Upstanding faculty wife hosting a Proper Upstanding faculty party, asked the young Dick Feynman if he'd like lemon or cream in his tea. He asked for both. The hostess was aghast.

Feynman assumed that he'd violated some weird secret social rule. But what happens if you ACTUALLY add lemon and cream to your tea? It curdles. Usually it turns into cottage cheesevomit-flavored tea filth.

RP Feynman: not a chemist.

:)

And this chemistry faux pas ended up on the front cover of his first book! (If I was Feynman's spirit, I'd be laughing and laughing right now. Oh, the humanity.)

 

[wbeaty]

Can someone answer this for me? On Beaty's site he has the following:

Everything connected to one battery terminal acquires the same electrical potential (voltage.) The circuit acts like two separate conductors, one with a positive charge imbalance and one with negative.
​

I don't understand that. If one were to chop the circuit up into a million little pieces wouldn't each piece also behave in the same manner?

Nope. That sort of polarization would only happen during a current in the wire. If you just connect two wires to a D-cell, there is no direct current. The wires have a surface charge imbalance, and if you chop them up, the charge remains unchanged.

But about the battery wires: one entire wire will be positive, and the other negative. The two wires behave like two capacitor plates, with 1.5V across that capacitor, and some e-field flux lines between them. (Then connect a light bulb, and the bulb "discharges" that capacitor.)

If you want the full-blown physics description, see this paper by Chabay and Sherwood, the authors of the college physics text "Matter & Interactions." I hope my version is simpler!

A Unified Treatment of Electrostatics and Circuits (PDF)
"[URL
[/URL]

Also, is that even accurate to what's really happening? For example, if one applied a positive voltage to one end of a wire, and a negative voltage to the other, the wire would develop an imbalance of charges on each end with a greater difference between the ends and a gradual fall off towards the middle correct?

Exactly. In that case the wire is being used as a resistor. There is a current, and a smoothly-varying voltage-drop along the wire. Use that wire as a voltage divider circuit, or as a volume control pot.

Now, for a circuit, there isn't a chance for the charges to build up.

Wherever there are voltages, there are buildups of charge. If a circuit is being run by a 9v battery, then the battery + and battery - wires will have a significant charge. (In fact, this charge is the cause of the wires' voltage. The wires are like capacitor plates, and capacitor plates can only have a voltage if they first have a surface charge.)

 

[Drakkith]

Nope. That sort of polarization would only happen during a current in the wire. If you just connect two wires to a D-cell, there is no direct current. The wires have a surface charge imbalance, and if you chop them up, the charge remains unchanged.

I didn't actually mean you physically chop them up if that's what you were thinking. Sorry for the confusion.

But about the battery wires: one entire wire will be positive, and the other negative. The two wires behave like two capacitor plates, with 1.5V across that capacitor, and some e-field flux lines between them. (Then connect a light bulb, and the bulb "discharges" that capacitor.)

I can understand if they aren't connected, but with a flow of charges how does that happen?
Edit: Hrmm. Looking up some more stuff now and thinking it all over. I'll come back to this.

Exactly. In that case the wire is being used as a resistor. There is a current, and a smoothly-varying voltage-drop along the wire. Use that wire as a voltage divider circuit, or as a volume control pot.

Hold on, if the wire isn't connected to itself how is there a current? (Past the initial movement that happens until the charges are balanced)

 

[Evil Bunny]

Wherever there are voltages, there are buildups of charge.

If you connected a wire to one side of a battery (the other terminal of the battery not hooked to anything), you're saying that there would be a buildup of charge on that wire?

Wouldn't that buildup of charge be at a different potential than... say the dirt under our feet? Wouldn't that also mean that this charge would want to flow between the dirt and the wire if we were to touch this wire to the ground?

Sounds like you are saying that we could drain a battery by connecting only one of its terminals to the ground. This is not true is it? What am I missing here?

 

[Evil Bunny]

Exactly. In that case the wire is being used as a resistor. There is a current, and a smoothly-varying voltage-drop along the wire. Use that wire as a voltage divider circuit, or as a volume control pot.

Hold on, if the wire isn't connected to itself how is there a current? (Past the initial movement that happens until the charges are balanced)

In your example, you made it sound as if you shorted the + and - terminals of a battery with a wire. In this case, you would have tremendous current flow... Maybe that's not what you meant, but it's what it sounded like. To me anyway...

 

[RedX]

Also, is that even accurate to what's really happening? For example, if one applied a positive voltage to one end of a wire, and a negative voltage to the other, the wire would develop an imbalance of charges on each end with a greater difference between the ends and a gradual fall off towards the middle correct?

This is correct, but for DC current, the difference between the imbalance at the ends and the imbalance at the middles will approach zero.

See this picture, figure 4-4 (c):

http://www.tpub.com/neets/book10/42.htm

In this picture you see a dramatic difference between the charge imbalance at the ends and the imbalance between the middles, but that is because the length of the wires are comparable to the wavelength of the frequency (the picture is of your typical center-fed half-wave antenna).

The difference in DC is less dramatic, and you'll just get something like a step function for your charge distribution.

addendum:

Note that saying the two wires are capacitors is perfectly reasonable. It's an unintended capacitance, which is why it's sometimes called parasitic capacitance. When studying transmission lines you usually give the wires a capacitance per unit length. In this case, the charge distribution on the wire follows the voltage distribution, since Q is proportional to V, i.e., Q=CV. Check out figure 3-20 here:

http://www.tpub.com/neets/book10/41f.htm

You can understand the physics of a real circuit using lumped elements: the capacitor represents the charges on the wire, and the inductance represents the magnetic fields of the wires. This is one of the reasons I believe that the Poynting view is not critical. The reason the voltage propagates slower if there is a dielectric is because it takes longer to fill each capacitor since the capacitance increases. Of course the Poynting view is correct and more aesthetically pleasing in my opinion, but for actual calculations, it's not necessary one has the Poynting view in mind.

 

[Drakkith]

In your example, you made it sound as if you shorted the + and - terminals of a battery with a wire. In this case, you would have tremendous current flow... Maybe that's not what you meant, but it's what it sounded like. To me anyway...

I guess it does look like that from a different perspective. I meant that the wire isn't connected to anything, but merely has a charged object near each end.

This is correct, but for DC current, the difference between the imbalance at the ends and the imbalance at the middles will approach zero.

See this picture, figure 4-4 (c):

http://www.tpub.com/neets/book10/42.htm

In this picture you see a dramatic difference between the charge imbalance at the ends and the imbalance between the middles, but that is because the length of the wires are comparable to the wavelength of the frequency (the picture is of your typical center-fed half-wave antenna).

The difference in DC is less dramatic, and you'll just get something like a step function for your charge distribution.

Ok. That is exactly what I thought was happening in that regard.

You can understand the physics of a real circuit using lumped elements: the capacitor represents the charges on the wire, and the inductance represents the magnetic fields of the wires. This is one of the reasons I believe that the Poynting view is not critical. The reason the voltage propagates slower if there is a dielectric is because it takes longer to fill each capacitor since the capacitance increases. Of course the Poynting view is correct and more aesthetically pleasing in my opinion, but for actual calculations, it's not necessary one has the Poynting view in mind.

Alright. I guess what I need to look into is what exactly is happening with the way the emf works in the conductor.

 

[RedX]

Ok. That is exactly what I thought was happening in that regard.

Reading over the other answers, I think I might have misinterpreted your question. I thought you were talking about the charge distribution on two wires in an open circuit. But everyone else thought you were talking about a short circuit, except giving the wires a tiny bit of resistance. So their answers would apply to the latter case, and mine would to the former case.

Alright. I guess what I need to look into is what exactly is happening with the way the emf works in the conductor.

Usually we make idealizations in circuits, such as there is no resistance in a circuit except at a resistor, that there is no capacitance at a circuit except at a capacitor, and the same with inductance/inductor. Even if you connect a battery to only a resistor, your circuit has some inductance because you have a loop of wire with current flowing in it. And the same goes with capacitance: there is capacitance between the two wires coming out of the battery even though you didn't put a capacitor in there.

If you focus on what a real circuit is like, with capacitance and inductance all along the wires and not just at places where you purposely insert capacitors and inductors, then all these ideas will seem perfectly natural. You don't have to know in detail how a battery works or an AC generator - you just need to know that they output some voltage. Charge in wires automatically follows from Q=CV, so that once you have voltage you have charge.

 

[Drakkith]

I understand RedX, it's just that I'm trying to understand what's happening in each piece. I think the biggest problem I have is understanding what happens to the charges in the circuit and how.

 

[Naty1]

If I were a moderator, I'd remove this entire thread due to so many inaccuracies...

This is my last comment here as this entire thread is not worth anyone's time.

I'll just pick a few incorrect posts on which to comment:

beaty:

But about the battery wires: one entire wire will be positive, and the other negative. The two wires behave like two capacitor plates, with 1.5V across that capacitor, and some e-field flux lines between them. (Then connect a light bulb, and the bulb "discharges" that capacitor.

No, the bulb remains lit as long as there is battery power.

Drakkith: So between the negative side and the resistor, is there an excess of negative charges? And vice versa on the positive side with positive charges?

You would be led to think that from the inaccurate comments previosuly posted, but that is NOT correct. There is a difference in potential (voltage)...the same number of charges (electrons) enters one end of the resistor as leaves the other end.

Drakkith
Also, is that even accurate to what's really happening? For example, if one applied a positive voltage to one end of a wire, and a negative voltage to the other, the wire would develop an imbalance of charges on each end with a greater difference between the ends and a gradual fall off towards the middle correct?

(answer 1)

Beaty: Exactly.

no. There is an imbalance of potential (voltage)...there is conservation of charge within any current carrying conductor...electrons do not appear and disappear along a conductor as if there were sources and sinks along it's length...if that were true it would invalidate Kirchoffs current lawe.

(answer 2)

EWH: That would be a different case - the wire would be the only resistance in the circuit and it would be absorbing the whole voltage drop. The voltage would vary continuously down the wire. (In real circuits the internal resistance of the voltage source would usually become dominant, making the voltage across the wire less.)

no it is NOT a different case...in either case there is a gradually varying voltage along the wire. In this this case, with lower resistance, the current will be much larger...but the rest of the quote is accurate.

 

[RedX]

You would be led to think that from the inaccurate comments previosuly posted, but that is NOT correct. There is a difference in potential (voltage)...the same number of charges (electrons) enters one end of the resistor as leaves the other end.

no. There is an imbalance of potential (voltage)...there is conservation of charge within any current carrying conductor...electrons do not appear and disappear along a conductor as if there were sources and sinks along it's length...if that were true it would invalidate Kirchoffs current lawe.

I think you're confusing an idealized resistor with a real resistor. Real resistors have capacitance. So any resistor you buy, you could model as an idealized resistor and (idealized) capacitor in parallel. So using this picture, that's where you sink is: onto the plates of the capacitor. So charge is conserved, and the sink are the plates of the capacitor.

 

[Drakkith]

You would be led to think that from the inaccurate comments previosuly posted, but that is NOT correct. There is a difference in potential (voltage)...the same number of charges (electrons) enters one end of the resistor as leaves the other end.

Initially there is a positive potential applied to one side, and a negative to the other. When these are applied do electrons "pile up" at the resistor since they can't get through as quickly compared to the conductor? Assuming that is correct then this would also result in a reduction in electrons on the other side of the resistor since there is no resistance for the electrons to get from the end of the resistor to the positive potential. As soon as the voltage is applied and these effects equalize, the current through the whole circuit would remain steady.

Now, what I would like to know is if any of that is correct or not. I really don't know at all.

 

[RedX]

Initially there is a positive potential applied to one side, and a negative to the other. When these are applied do electrons "pile up" at the resistor since they can't get through as quickly compared to the conductor? Assuming that is correct then this would also result in a reduction in electrons on the other side of the resistor since there is no resistance for the electrons to get from the end of the resistor to the positive potential. As soon as the voltage is applied and these effects equalize, the current through the whole circuit would remain steady.

Now, what I would like to know is if any of that is correct or not. I really don't know at all.

That's correct. But how would you model that?

Assume you have a battery connected to a resistor R1, and in series with this resistor is another resistor R2 that is in parallel with a capacitor C2. Now close the switch. Initially there is no voltage drop across the capacitor since there is no charge on it initially. So the current should be V/R1, and no current flows through R2. Instead, the current flowing through R1 hops across the capacitor C2, thus charging C2. Once C2 is completely charged, you get a steady state current V/(R1+R2) through R2. But until C2 is charged, you won't get the maximum current possible passing through R2.

So charges have to pile up on C2 before they can flow through R2.

So in this way you can model a real resistor as containing R2 and C2, which kind of mimics the process of charge piling up at the leads.

At least that's how I understand it.

 

[wbeaty]

If I were a moderator, I'd remove this entire thread due to so many inaccuracies...

This is my last comment here as this entire thread is not worth anyone's time.

I'll just pick a few incorrect posts on which to comment:

beaty:
But about the battery wires: one entire wire will be positive, and the other negative. The two wires behave like two capacitor plates, with 1.5V across that capacitor, and some e-field flux lines between them. (Then connect a light bulb, and the bulb "discharges" that capacitor.

No, the bulb remains lit as long as there is battery power.

Obviously it does. But from a basic physics perspective, the battery "charges" the two-wire capacitor, and the bulb simultaneously discharges it. That's how the EM energy is able to travel from the battery to the bulb.

In addition, the circuit loop acts as a 1-turn inductor, and the battery injects energy into this inductor and produces a current (and the energy is stored in the surrounding b-field.) Simultaneously the bulb presents a resistive load on this inductor, and thus extracts the energy.

In a flashlight circuit, the battery puts EM energy into the pair of wires, and the bulb extracts the energy again. Electrons move slowly in a circuit, while the EM energy propagates across the circuit at ~c light speed.

This is the simple basic physics of transmission lines from your engineering course on EM/Waves. Go look it up in your old double-E textbook if you still have it.

Drakkith: So between the negative side and the resistor, is there an excess of negative charges? And vice versa on the positive side with positive charges?

You would be led to think that from the inaccurate comments previosuly posted, but that is NOT correct. There is a difference in potential (voltage)...the same number of charges (electrons) enters one end of the resistor as leaves the other end.

Yes, charge-conservation is the rule for current.

The physics rule for voltage is: circuit voltages are caused by surface charges on the conductors. To produce a voltage drop across a pair of wires, the battery produces a tiny positive charge on one wire, and a tiny negative charge on the other. If voltage is constant, these surface charges don't change. For a complete description of the physics see this excellent paper: http://www.matterandinteractions.org/Content/Articles/circuit.pdf" (1999 Chabay and Sherwood)

 

[wbeaty]

Also, is that even accurate to what's really happening? For example, if one applied a positive voltage to one end of a wire, and a negative voltage to the other, the wire would develop an imbalance of charges on each end with a greater difference between the ends and a gradual fall off towards the middle correct?

Beaty: Exactly. In that case the wire is being used as a resistor. There is a current, and a smoothly-varying voltage-drop along the wire. Use that wire as a voltage divider circuit, or as a volume control pot.

Hold on, if the wire isn't connected to itself how is there a current? (Past the initial movement that happens until the charges are balanced)

If you apply a positive voltage to one end of a wire, and a negative voltage to the other end, then it must be connected (connected to a battery for example,) and there will be a large current. Visualize a piece of nichrome wire connected to a D-cell. One end of the wire has the same positive surface-charge as the (+) battery terminal. The other end has the (-). And the surface-charge in the exact center of the wire is zero. And finally whenever a conductor has varying voltage along its length, because of ohm's law a large current must exist.

 

[cabraham]

Are we talking about an AC or DC circuit here? If it is DC then the impedance is simply resistance, correct? Then it just becomes Ohms law correct?



What? The inductor causes a voltage that resists the change in current through it. This causes the circuit to have a steady current, but only because the voltage in the inductor is changing in response to the current, either resisting the increase or compensating for the decrease. Either way the current in the circuit is directly caused by voltage sources.

The inductor causes a voltage?! Just how does the inductor "know" which voltage to output across an unknown resistance to maintain a steady current? An inductor has 1.0 amp w/ the FET switch transitioning from on to off. This stored energy in the inductor now discharges into a diode, current sense resistor, & output load resistance parallelled by an output filter capacitor.

The inductor current continues through the csr (current sense resistor). The electrons have energy. When the input power is cut off from the inductor when the FET switch opens, the energy stored in the inductor will transfer to the diode, csr, output cap & load. The inductor does not know what is the impedance of the elements receiving its energy.

The 1.0 amp current transits through the csr, & incurs collisions with the lattice ions of the resistor's material. In doing so, the energy given up by electrons colliding results in photon emission in the infrared region. The resistor heats up due to I^2*R loss. This results in a buildup of a barrier. Charges accumulate in the csr resulting in an electric field in the opposite direction. The integral of this E field wrt the csr path is the voltage drop.

If the csr value is 0.10 ohm, then the 1.0 amp into the 0.10 ohm csr determines the 0.10 volt voltage drop across the csr. The voltage drop is literally determined by the csr value & the inductor current value.

If we construct an identical supply w/ the same elements except that the csr is 0.15 ohm, then the 1.0 amp through the 0.15 ohm csr results in a voltage drop of 0.15 volts. When the source providing emergy/power to the passive elements is a constant current source, which an inductor is, the current times the resistance determines the voltage.

Common power sources like batteries & generators are intentionally designed for constant voltage operation. Hence the voltage divided by resistance determines current. But it could be designed to work the other way as well.

Either one can determine the other. Neither is independent unless made that way.

Claude

 

[Drakkith]

The inductor causes a voltage?! Just how does the inductor "know" which voltage to output across an unknown resistance to maintain a steady current?

Per my knowledge:

Electric current through the conductor creates a magnetic flux proportional to the current. A change in this current creates a corresponding change in magnetic flux which, in turn, by Faraday's Law generates an electromotive force (EMF) that opposes this change in current.

Looks to me like the change in current results in the creation of EMF that opposes that change. EMF is voltage right? I believe they produce a set amount of voltage depending on their construction and how much current runs through them.

Common power sources like batteries & generators are intentionally designed for constant voltage operation. Hence the voltage divided by resistance determines current. But it could be designed to work the other way as well.

Either one can determine the other. Neither is independent unless made that way.

Batteries and generators produce a voltage which causes current to flow in a circuit. They produce a constant voltage because they are designed that way. The only way that they would produce a constant current in the circuit would be to vary the voltage as the resistance of the circuit increases and decreases. Current is the flow of charges, and the only way to get them to flow is with EMF.

That is correct to my knowledge, but if you know a case where that is incorrect then please let me know.

 

[cabraham]

Per my knowledge:



Looks to me like the change in current results in the creation of EMF that opposes that change. EMF is voltage right? I believe they produce a set amount of voltage depending on their construction and how much current runs through them.



Batteries and generators produce a voltage which causes current to flow in a circuit. They produce a constant voltage because they are designed that way. The only way that they would produce a constant current in the circuit would be to vary the voltage as the resistance of the circuit increases and decreases. Current is the flow of charges, and the only way to get them to flow is with EMF.

That is correct to my knowledge, but if you know a case where that is incorrect then please let me know.

But the current did not change yet. The instant the FET switch opens, the 1.0 amp inductor current continues throught the csr. No change in current, no induced emf. The energy is dissipated in the resistance. Electrons transfer energy by colliding with lattice ions in the resistor. Heat is radiated. The inductor energy decreases & the voltage drop across the csr, V, is determined by R & I. Otherwise what determines the initial value of V?

Before the current has had time to change it is present in the csr. The V value is simply I*R. As the current decreases so does the voltage since Ohm's law is always in effect. I & R literally determine V.

Again if 2 identical inductors w/ 1.0 amp were terminated in 1 ohm & 100 ohm, the voltages at the instant of conduction beginning are 1.0V & 100V. They both decay but the 100 ohm circuit decays 100 times faster. The voltage at time 0 is simply I*R. Current determines voltage. I've developed dozens of switching power supplies, motor drivers, custom transformers & inductors, so I need to ask you what experience & education you have on this topic.

If you're looking to Wikipedia for your reference, may I remind you that Wiki is not peer-reviewed. Even AP high school students are told not to use Wiki as reference. I speak from 33 yrs. of experience, & near completion on an EE doctorate. What makes you think I'm wrong?

Claude

 

[Drakkith]

But the current did not change yet. The instant the FET switch opens, the 1.0 amp inductor current continues throught the csr. No change in current, no induced emf. The energy is dissipated in the resistance. Electrons transfer energy by colliding with lattice ions in the resistor. Heat is radiated. The inductor energy decreases & the voltage drop across the csr, V, is determined by R & I. Otherwise what determines the initial value of V?

When the FET switch opens, the input is cut correct? At that instant the current begins to stop as voltage declines, causing the inductor to apply a voltage using it's stored energy to keep it going. The voltage applied from the inductor to the circuit is proportional to the change in current right?

Before the current has had time to change it is present in the csr. The V value is simply I*R. As the current decreases so does the voltage since Ohm's law is always in effect. I & R literally determine V.

I think this means that if you measure the current and you know the resistance you can determine what the voltage is, not that the current and resistance determine voltage. Current can cause effects which then cause a voltage in another part of the circuit, but in that circuit the current is still determined by the voltage and resistance. Is that what you mean by current causing voltage?

Again if 2 identical inductors w/ 1.0 amp were terminated in 1 ohm & 100 ohm, the voltages at the instant of conduction beginning are 1.0V & 100V. They both decay but the 100 ohm circuit decays 100 times faster. The voltage at time 0 is simply I*R. Current determines voltage. I've developed dozens of switching power supplies, motor drivers, custom transformers & inductors, so I need to ask you what experience & education you have on this topic.

The voltage at time 0, before the switch is opened or right as it opens is 100v and 1v, correct? That is because the resistance of the circuits are 1 ohm and 100 ohms. I don't follow how this means that the voltage is determined by the current. You have to apply 100v and 1v to GET a current of 1 amp in the first place. If both inductors are identical then the one on the 100 ohm circuit would have to apply a voltage of up to 100v to maintain the 1 amp current, hence running out of power 100x faster correct?

Edit: Also, I missed this part of your other thread:

If we construct an identical supply w/ the same elements except that the csr is 0.15 ohm, then the 1.0 amp through the 0.15 ohm csr results in a voltage drop of 0.15 volts. When the source providing emergy/power to the passive elements is a constant current source, which an inductor is, the current times the resistance determines the voltage

Are you saying that voltage drop is = to voltage? It kind of looks that way to me. Also, any source of current is also a source of voltage, otherwise how would any current flow in the first place? Let's look at a constant current power supply for example. To my knowledge these keep a constant current flowing as the resistance of the circuit increases and decreases. How does it do this? It varies the voltage applied to keep the current the same. Is that not the same effect or at least similar to an inductor?

 

[cabraham]

When the FET switch opens, the input is cut correct? At that instant the current begins to stop as voltage declines, causing the inductor to apply a voltage using it's stored energy to keep it going. The voltage applied from the inductor to the circuit is proportional to the change in current right?



I think this means that if you measure the current and you know the resistance you can determine what the voltage is, not that the current and resistance determine voltage. Current can cause effects which then cause a voltage in another part of the circuit, but in that circuit the current is still determined by the voltage and resistance. Is that what you mean by current causing voltage?



The voltage at time 0, before the switch is opened or right as it opens is 100v and 1v, correct? That is because the resistance of the circuits are 1 ohm and 100 ohms. I don't follow how this means that the voltage is determined by the current. You have to apply 100v and 1v to GET a current of 1 amp in the first place. If both inductors are identical then the one on the 100 ohm circuit would have to apply a voltage of up to 100v to maintain the 1 amp current, hence running out of power 100x faster correct?Edit: Also, I missed this part of your other thread:


Are you saying that voltage drop is = to voltage? It kind of looks that way to me. Also, any source of current is also a source of voltage, otherwise how would any current flow in the first place? Let's look at a constant current power supply for example. To my knowledge these keep a constant current flowing as the resistance of the circuit increases and decreases. How does it do this? It varies the voltage applied to keep the current the same. Is that not the same effect or at least similar to an inductor?

Ref bold. The inductor does not "apply" anything. You claim that the inductor apllies 100V to keep the 1 amp going in the 100 ohm circuit, 1V for the 1 ohm circuit. But the inductor does not sense the 100 ohm resistor value. How can an inductor "apply a specific voltage"? What happens is the following. If the resistor was a 0 ohm superconductor, SC, the current would continue forever. No collisions occur, so no energy is lost. No voltage is present because the electrons have initial energy that never dissipates. The current stays forever w/o voltage.

Now same problem w/ 1 ohm R & 1 A current. When the time is 0, the current is still 1 amp. But it takes a short time, in the picosecond range before the electrons reach the resistor. Up until then, the current is 1 A. When the electrons reach the resistor, collisions occur, & energy is converted into thermal via photon emission due to lattice collisions. The inductor cannot generate a voltage because the current has not yet changed. But once the collisions occur, a charge buildup takes place at the boundaries between the wire & resistor. This results in an E field which counters the direction of current. Energy is lost & current decreases. Collisions continue until zero is reached.

By the way, a constant current supply can be made by simply spinning a generator at constant torque. Spinning at constant speed results in constant voltage (& frequency). Losses w/ constant current are much greater so constant voltage is used.

I know that current sources also produce voltage. But the inductor is not actively outputting a voltage based on resistance. If the resistor is a foot away, it takes a nanosecond before the current reaches the resistor. The 1 amp at t = 0 continues at 1 amp not knowing what resistance value lies ahead. How fast the energy is dissipated depends on the R value. The inductor is not actively regulating its output voltage in response to the resistance encountered. How would the inductor adjust its own voltage based on a resistance value a nanosecond or more away from it?

Again, are you an EE? Do you practice power, analog, networks, what is your level of education. I just want to know your sources & references. This subject takes years of intense study to master. To instruct others requires at least an MSEE or MS-phy. Do you have such? No offense, just curious. BR.

Claude

 

[Drakkith]

Ref bold. The inductor does not "apply" anything. You claim that the inductor apllies 100V to keep the 1 amp going in the 100 ohm circuit, 1V for the 1 ohm circuit. But the inductor does not sense the 100 ohm resistor value. How can an inductor "apply a specific voltage"? What happens is the following. If the resistor was a 0 ohm superconductor, SC, the current would continue forever. No collisions occur, so no energy is lost. No voltage is present because the electrons have initial energy that never dissipates. The current stays forever w/o voltage.

Of course! The current doesn't change so there is no change in magnetic flux and hence no induced voltage!

Now same problem w/ 1 ohm R & 1 A current. When the time is 0, the current is still 1 amp. But it takes a short time, in the picosecond range before the electrons reach the resistor. Up until then, the current is 1 A. When the electrons reach the resistor, collisions occur, & energy is converted into thermal via photon emission due to lattice collisions. The inductor cannot generate a voltage because the current has not yet changed. But once the collisions occur, a charge buildup takes place at the boundaries between the wire & resistor. This results in an E field which counters the direction of current. Energy is lost & current decreases. Collisions continue until zero is reached.

Then what is the inductor doing? In this case it looks like it isn't doing anything. When the current starts to decay shouldn't the inductor counteract it with an EMF?

By the way, a constant current supply can be made by simply spinning a generator at constant torque. Spinning at constant speed results in constant voltage (& frequency). Losses w/ constant current are much greater so constant voltage is used.

Yes, the generator is producing voltage in the circuit which causes current to flow. To keep a steady current as the resistance increases or decreases you would need to increase or reduce the voltage in the circuit as necessary, or design the power supply to increase or decrease its own resistance to keep the total resistance the same as the load changes.

I know that current sources also produce voltage. But the inductor is not actively outputting a voltage based on resistance. If the resistor is a foot away, it takes a nanosecond before the current reaches the resistor. The 1 amp at t = 0 continues at 1 amp not knowing what resistance value lies ahead. How fast the energy is dissipated depends on the R value. The inductor is not actively regulating its output voltage in response to the resistance encountered. How would the inductor adjust its own voltage based on a resistance value a nanosecond or more away from it?

For that nanosecond the inductor applies a very low voltage to keep the current going due to the very low resistance. As soon as the current starts to fall quickly because of the resistor, the inductor has to apply more voltage to resist the accelerated change in current. At that point it IS basing its voltage off of the resistor because the resistor is causing the reduction in current.

Again, are you an EE? Do you practice power, analog, networks, what is your level of education. I just want to know your sources & references. This subject takes years of intense study to master. To instruct others requires at least an MSEE or MS-phy. Do you have such? No offense, just curious. BR.

Claude

I'll give my sources and experience once we come to some kind of agreement. Either one way or the other, or agree to disagree.

And honestly, has anything I've said been wrong so far? Ignoring our discussion over voltage and current.