What causes an increase in energy in an electrical system?

[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 through 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 through 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.