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How a resistor works at the atomic level. Why does voltage drop across it?

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PConst167

Member
Hey friends,

this is a question that I've had for a long time. As we know, when a current passes through a resistor, a voltage drop occurs.

What I'd like to know is, what really is this voltage drop? Is it just the electrons bumping into the atomic structure of the resistor? There is something strange about this, because if voltage drop is just electrons bumping inside resistors, losing its energy, then why is it that the voltage drops always happen in exact ratios when you mix resistors in series? If electrons bumping inside a single resistor lose all their energy, then why don't they lose all their energy at the first resistor in a series of resistors?

It seems voltage "knows" how to divide perfectly between the resistors. So it must mean that voltage drop isn't simply electrons bumping against the atoms in a resistor. It seems more like a wave distributing among all the resistors at the same time. Let's say for example we have 2 resistors in series, separated by a distance D made of almost ideal wires. Could it actually be in fact, that after bumping inside the first resistor, the electrons regain their energy in the mean time inside the ideal wire, before entering the second resistor and losing that energy?

If so, then how is the energy gained exactly equal to the energy lost? How can voltage divide itself in perfect ratios ? How does it know!

If I put a multimeter across a resistor and measure the voltage, does the multimeter ever measure voltage itself? Or is it actually measuring the current and calculating the voltage via its internal resistor and using ohms law? Is it actually unaware of the voltage itself? Is it cheating!?


Since all voltages are really relative, then in fact all that must be happening is we are measuring currents and using resistors to find voltages through the abstract mathematical relation called Ohms law. So really the resistors and voltages don't know how to divide, it's just that we measure them in a way that makes it seem like they are.

So when measuring voltages across resistors in series, and the voltages seem to divide in perfect ratios, really what's happening is that by adding a second resistor, the current decreases, and then by attaching a large resistor in parallel with one of them will give a certain current in the large resistor, and then by Ohms law we calculate the voltages. And because Ohms law itself is a perfect mathematical ratio, and we use it, then we get perfect voltage ratios when measuring them.

What do you think?
 
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spec

Well-Known Member
Most Helpful Member
Hi PC,

An electric current is nothing more than a flow of electrons from a negative potential to a positive potential- yes that is correct and is termed, 'electron flow'. How did this incongruous situation arise- it is purely historic. When positive and negative were assigned to a voltage the finer details were not fully understood.

But in most calculations and analysis the opposite direction of current flow is assumed for convenience. This is known as 'conventional flow'.

One amp is 6.2 * 10^18 electrons flowing past a particular point.

Electrons flow easily through conductors, silver, copper, aluminum, gold, graphite... because the outer shell of the atoms of those elements have electrons that are easily dislodged and are thus attracted to a positive voltage (lack of electrons). So, when a voltage is placed across a conductor the electrons from the negative terminal of the battery can flow through the conductor and into the positive terminal of the battery. The formula for the amount of current flow is, I=V/R, where I is current in Amps, V is voltage in Volts, and R is resistance in Ohms.

A conductor will have a resistance of essentially zero Ohms, so if you put any voltage across a conductor, an infinite current would flow- so it is not a good idea to do that.

The next group of materials are semiconductors; germanium, gallium arsenide, and of course the ubiquitous silicon. As their name implies, semiconductors have outer electrons that need more voltage to dislodge them. Semiconductors are also said to have holes (lack of electrons) but don't let that cloud the issue at this stage.

Finally there are insulators: glass, ceramic, mica, aluminum oxide, plastics. The outer electrons of insulators are extremely reluctant to budge from their orbit, so insulators will not conduct any current, even with very high voltages across them. But when insulators do break down under extreme voltage, they break down in a big way and often the atomic structure of the insulator is changed.

And that is all there is to it. (gross oversimplification, of course :))

spec

PS: electrons are not in orbits, rather energy levels, and electrons do not flow- they shuffle from one atom to another. But ignore that for this initial concept stage.
 
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Ratchit

Well-Known Member
Hey friends,

this is a question that I've had for a long time. As we know, when a current passes through a resistor, a voltage drop occurs.
Yes, most of us know that. As you will find out, that has to happen.

What I'd like to know is, what really is this voltage drop? Is it just the electrons bumping into the atomic structure of the resistor? There is something strange about this, because if voltage drop is just electrons bumping inside resistors, losing its energy, then why is it that the voltage drops always happen in exact ratios when you mix resistors in series? If electrons bumping inside a single resistor lose all their energy, then why don't they lose all their energy at the first resistor in a series of resistors?
When I see a question like this, it tells me the questioner does not really know what voltage is. Voltage is a density, specifically the energy density of the charge. It is measured in units of energy density, which is joules/coulomb. It takes energy to bring two electrons close together. It takes more energy to bring several electrons close together. It takes still more energy to bring those several electrons even closer together. If one divides the total amount of energy in joule units expended to aggregate the electrons by the amount of electrons in coulombs, the result will be volts. If there is a conduction path to a lower energy density or voltage, the electrons will flow to the lower voltage. During the travel, the charge carriers (electrons) will encounter collisions with the ionic core of the material and electric fields within the conductor material. This will dissipate the energy of the charge carriers by converting it into heat.

Who says the charge carriers lose all their energy? They will lose more energy if the resistance if higher and arrive at the end of the resistance with less energy than before. If the same number of charge carriers arrive with less energy, then naturally, the energy density or voltage will be less. That is called the voltage drop. This explains why the voltage drop in a series circuit is proportional to the resistance.

By the way, the charges do not physically "bump" into each other or the ionic cores of the conductor. They move erratically within the conductor, influenced by the electric fields of the other charge carriers and stationary atoms of the conductor material.

It seems voltage "knows" how to divide perfectly between the resistors. So it must mean that voltage drop isn't simply electrons bumping against the atoms in a resistor. It seems more like a wave distributing among all the resistors at the same time. Let's say for example we have 2 resistors in series, separated by a distance D made of almost ideal wires. Could it actually be in fact, that after bumping inside the first resistor, the electrons regain their energy in the mean time inside the ideal wire, before entering the second resistor and losing that energy?

If so, then how is the energy gained exactly equal to the energy lost? How can voltage divide itself in perfect ratios ? How does it know!
I believe I explained above. The energy loss of the charge carriers is proportional to the resistance through which they travel.

If I put a multimeter across a resistor and measure the voltage, does the multimeter ever measure voltage itself? Or is it actually measuring the current and calculating the voltage via its internal resistor and using ohms law? Is it actually unaware of the voltage itself? Is it cheating!?
No, it is doing both. It is measuring the voltage by determining the current passing through a calibrated resistor.

Since all voltages are really relative, then in fact all that must be happening is we are measuring currents and using resistors to find voltages through the abstract mathematical relation called Ohms law. So really the resistors and voltages don't know how to divide, it's just that we measure them in a way that makes it seem like they are.
Measuring the current through a calibrated resistor is a fool-proof way of determining the voltage.

So when measuring voltages across resistors in series, and the voltages seem to divide in perfect ratios, really what's happening is that by adding a second resistor, the current decreases, and then by attaching a large resistor in parallel with one of them will give a certain current in the large resistor, and then by Ohms law we calculate the voltages. And because Ohms law itself is a perfect mathematical ratio, and we use it, then we get perfect voltage ratios when measuring them.
That statement makes my head explode.

What do you think?
I think you should get to know what voltage is.

Ratch
 

crutschow

Well-Known Member
Most Helpful Member
if voltage drop is just electrons bumping inside resistors, losing its energy, then why is it that the voltage drops always happen in exact ratios when you mix resistors in series? If electrons bumping inside a single resistor lose all their energy, then why don't they lose all their energy at the first resistor in a series of resistors?
The electrons don't lose all their energy in one collision.
They lose a small amount with each collision as they move down the resistor.
Thus two resistor in series are identical to one long resistor of the same total resistance.
The power loss is distributed evenly as it moves down the resistor element or elements in proportion to the resistance of each element.
 

Grossel

Well-Known Member
Hi, I'll try to translate to water-pipe language (just that the water here doesn't have a inertia)

this is a question that I've had for a long time. As we know, when a current passes through a resistor, a voltage drop occurs.
When a flow of water passes through a thin pipe, there will be a difference in pressure from one side to the other.

What I'd like to know is, what really is this voltage drop? Is it just the electrons bumping into the atomic structure of the resistor? There is something strange about this, because if voltage drop is just electrons bumping inside resistors, losing its energy, then why is it that the voltage drops always happen in exact ratios when you mix resistors in series? If electrons bumping inside a single resistor lose all their energy, then why don't they lose all their energy at the first resistor in a series of resistors?
Well, it is the square current times resistance that makes the energy loss.

If so, then how is the energy gained exactly equal to the energy lost? How can voltage divide itself in perfect ratios ? How does it know!
If you send a flow of water through a series of thin pipes, each with different length and diameter (different resistance for water flow) - how could it be that the difference in pressure for each pipe is exactly proportional to the water flow resistance for each pipe?


So with this I hope you see the point, or it will only make more of a mess of what you belies about those small ohm's that is packed inside every resistor ;)
 

cowboybob

Well-Known Member
Most Helpful Member
crutschow is correct.

Although, since nothing is actually crashing into anything (more of a non-contact transference of a small amount of an electron's total energy, which is quite variable) there is, nonetheless, a ton of activity at the atomic level involving valence and conduction bands and the relative distance between them. When electrons are electrostatically moved out of and then fall back into these bands, energy is lost in the form of a photon.

None of this actually stops the movement of the electrons through the conductive material. It does, however, interfere (by deflection) in the electron's maintaining a straight line through the material.

In this case, this is manifested by the appearance of a voltage (potential) across a conductor, especially one designed as a resistor, diode or transistor: hence the voltage drop across them.

Also, in this case, the number of available valence electrons (for instance, copper has one while carbon and lead have 4) also contributes to the conductive "qualities" of an atom. A total lack of valence or conduction electrons or a vast (atomically speaking) distance between them, contributes to the properties of insulating materials.

I can further explain the process, if you wish.
 
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Ratchit

Well-Known Member
crutschow is correct.

Although, since nothing is actually crashing into anything (more of a non-contact transference of a small amount of an electron's total energy, which is quite variable) there is, nonetheless, a ton of activity at the atomic level involving valence and conduction bands and the relative distance between them.
When electrons are electrostatically moved out of and then fall back into these bands, energy is lost in the form of a photon.
There is no visible light involved with simple current within a conductor, unless incandescence occurs. Therefore, no photons are involved. http://whatis.techtarget.com/definition/photon Also, the energy bands overlap in a conductor. No gap exists like it does with a semiconductor.

None of this actually stops the movement of the electrons through the conductive material. It does, however, interfere (by deflection) in the electron's maintaining a straight line through the material.

In this case, this is manifested by the appearance of a voltage (potential) across a conductor, especially one designed as a resistor, diode or transistor: hence the voltage drop across them.
And, what causes this voltage difference to appear?

Also, in this case, the number of available valence electrons (for instance, copper has one while carbon and lead have 4) also contributes to the conductive "qualities" of an atom. A total lack of valence or conduction electrons or a vast (atomically speaking) distance between them, contributes to the properties of insulating materials.
Pure copper (Cu) has two valance electrons, not one. The difference between an insulator and a conductor is the ease or dificulity in which electrons are dislodged and replaced from their atoms, not the number of valance electrons the atom contains. An element that has no valance electrons is a inert gas like radon. The molecules of compounds like sintered quartz hold on to their electrons very tightly, and are considered insulators.

I can further explain the process, if you wish.
Yes, please do.

Ratch
 

crutschow

Well-Known Member
Most Helpful Member
When electrons are electrostatically moved out of and then fall back into these bands, energy is lost in the form of a photon.
I think that's actually a phonon inside the material.
 

cowboybob

Well-Known Member
Most Helpful Member
There is no visible light involved with simple current within a conductor, unless incandescence occurs. Therefore, no photons are involved. http://whatis.techtarget.com/definition/photon Also, the energy bands overlap in a conductor. No gap exists like it does with a semiconductor.
Photons are not limited to only visible light. A photon carries energy proportional to the radiation frequency, for instance, infra red (heat). Also, energy bands do, indeed, overlap but not always. Sometimes theu merely exchange electrons. Depend on the elements involved.
Pure copper (Cu) has two valance electrons, not one.
My error (a typo).
The difference between an insulator and a conductor is the ease or dificulity in which electrons are dislodged and replaced from their atoms, not the number of valance electrons the atom contains.
My emphasis.
That also is true, but the greater the number of available valence electrons, the greater the number of possible interactions between them and the introduced electrons of the current. Thus the greater the reduction in current electrons' energy levels and the greater the level of photon production per given number of elemental atoms, thus, the greater reduction in the available current, resulting in a greater voltage drop.
I think that's actually a phonon inside the material.
A phonon is just another concept based on the photon which is, of course, still not all that well defined (as in, "wave particle" as described in quantum physics).
 

PConst167

Member
Who says the charge carriers lose all their energy? They will lose more energy if the resistance if higher and arrive at the end of the resistance with less energy than before. If the same number of charge carriers arrive with less energy, then naturally, the energy density or voltage will be less. That is called the voltage drop. This explains why the voltage drop in a series circuit is proportional to the resistance.


Ratch

That still doesn't explain why the voltage drops in exact ratios. Let's have a 6V battery powering a circuit with 2 series resistors, R and 2R, the voltage drop at 2R will be 4V, and the voltage drop at R will be 2V. However, if you use a single resistor R, the voltage drop will be 6V at that single resistor. My question is, if the energy lost at the single circuit resistor is 6 Joules per Coulomb, why doesn't the same energy drops across the first resistor in the series arrangement? Why does voltage know ahead that it has to give only 4 Joules per Coulomb at the first resistor and only 2 J/C at the second?

This seems just strange. If you use 2 resistors, then the energy lost at each one is less than if you use a single one. It seems voltage is planning ahead.

Can anyone explain this? I know it has to be this way because this is the logical way. Of course it would make no sense if the voltage dropped all across the first resistor and none at the second. But it's still strange that it can divide perfectly among them both.

I also understand that 2 resistors are equivalent to one large resistor. So in this case, if we use one single resistor then the voltage drop at each point in the resistor will be different. Why is this?

Let's say we have an electric field inside the resistor, which comes from the negative end of the battery (Assuming positive current). Then at each point in the resistor will the electric field be variable or constant? If voltage is the line integral of the electric field with respect to distance, then we multiply the electric field at each point by an infinitesimal distance dr, and add the small pieces of voltage together to find the total voltage drop. This can of course be divided into a sum of integrals or voltage drops across the resistor. If the electric field is constant, then the integral will simply be E*r where r is a single dimensional length (assuming 1 dimension). This is proportional to the distance travelled. Is the electric field constant ?

Who can explain this in terms of the electric fields created by the voltage source and through the resistor?
 

crutschow

Well-Known Member
Most Helpful Member
Suppose you have water pressure applied to one end of two pipes in series (the other end open), with one being smaller and having twice the resistance to water flow as the other.
So how does the water know to drop 2/3rds of the pressure across the one with twice the resistance and 1/3rd across the other?
That's the same question your are asking in a different form.
Seems like you are looking for some sort of metaphysical answer to your question. :rolleyes:
 

PConst167

Member
Suppose you have water pressure applied to one end of two pipes in series (the other end open), with one being smaller and having twice the resistance to water flow as the other.
So how does the water know to drop 2/3rds of the pressure across the one with twice the resistance and 1/3rd across the other?
That's the same question your are asking in a different form.
Seems like you are looking for some sort of metaphysical answer to your question. :rolleyes:

Not metaphysical, just physical :p It seems to me that if you have one thin pipe followed by a large pipe, only as much water as the thin pipe can provide will pass through the second large pipe ? That's if the second pipe is at the end of the water flow. If things are reversed then yes. But In electricity it doesn't make a difference

It seems the large pipe would be almost empty, wouldn't it ? I can't actually visualize it being full! Whereas a second large resistor would be full of voltage!
 

spec

Well-Known Member
Most Helpful Member
It seems to me that if you have one thin pipe followed by a large pipe, only as much water as the thin pipe can provide will pass through the second large pipe ? That's if the second pipe is at the end of the water flow.
That is correct and the water flow analogy holds true for electricity.

Water pressure is analogous to voltage (potential difference) and water flow is analogous to electron flow.

The large diameter pipe is a low resistance to water flow and is analogous to a conductor. In the water situation there will be no pressure difference across the large cross section pipe just like there would be no voltage across a conductor

If you have a resistor (thin pipe) in series with a conductor (big pipe) the current will be defined solely by the resistor (V/R). Also, the resistor will have all the input voltage across it.
If things are reversed then yes. But In electricity it doesn't make a difference.
This statement is too vague to be comprehensible- sorry
It seems the large pipe would be almost empty, wouldn't it?
Absolutely- that is fundamentally correct. But the big pipe could also be full of water. But much of the water in the big pipe would be static and would not be flowing through the circuit.
I can't actually visualize it being full!
There is no law that says that a pipe must be full of water, although in the case of the two pipes you mention, the big pipe could well be full of water in a sealed system (circuit).

You have introduced another force- gravity which may apply to a water circuit, but has no significant effect on electrons (for the present discussion that is).

Also, the water situation that you describe is not a circuit, or at least if it is a circuit, it has an airlock in it.
Whereas a second large resistor would be full of voltage!
When you say large resistor, do you mean high value resistor?

No component can be full of voltage. Components can only have a voltage across their terminals. Similarly a pipe cannot be full of water pressure, but a pipe can have a pressure across it, and it can be full of water.

The small cross section pipe equates to a high value resistor and the large cross section area pipe equates to an electrical conductor which, by definition cannot have any voltage across it.

You mention the big pipe not being full of water. Likewise a resistor, or conductor, does not have every valance electron stripped from every atom comprising the resistor or conductor.

The thing to bear in mind is that the voltage (pressure) is the force that causes the electrons (water) to flow. Or, in the parlance, voltage is the electromotive force (EMF)

spec
 
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PConst167

Member
That is correct and the water flow analogy holds true for electricity.

Water pressure is analogous to voltage (potential difference) and water flow is analogous to electron flow.

The large diameter pipe is a low resistance to water flow and is analogous to a conductor. In the water situation there will be no pressure difference across the large cross section pipe just like there would be no voltage across a conductor

If you have a resistor (thin pipe) in series with a conductor (big pipe) the current will be defined solely by the resistor (V/R). Also, the resistor will have all the input voltage across it.

This statement is too vague to be comprehensible- sorry

Absolutely- that is fundamentally correct. But the big pipe could also be full of water. But much of the water in the big pipe would be static and would not be flowing through the circuit.

There is no law that says that a pipe must be full of water, although in the case of the two pipes you mention, the big pipe could well be full of water in a sealed system (circuit).

You have introduced another force- gravity which may apply to a water circuit, but has no significant effect on electrons (for the present discussion that is).

Also, the water situation that you describe is not a circuit, or at least if it is a circuit, it has an airlock in it.

When you say large resistor, do you mean high value resistor?

No component can be full of voltage. Components can only have a voltage across their terminals. Similarly a pipe cannot be full of water pressure, but a pipe can have a pressure across it, and it can be full of water.

The small cross section pipe equates to a high value resistor and the large cross section area pipe equates to an electrical conductor which, by definition cannot have any voltage across it.

You mention the big pipe not being full of water. Likewise a resistor, or conductor, does not have every valance electron stripped from every atom comprising the resistor or conductor.

The thing to bear in mind is that the voltage (pressure) is the force that causes the current (electrons) to flow. Or, in the parlance, voltage is the electromotive force (EMF)

spec

Wow I am impressed at how good the water analogy is.

So if we have a certain pressure, 2 pipes, the first one having a larger cross area, and the second having a small one, then the second pipe will have only as much pressure available as the first one can deliver. Amazing!

So in voltage, after the voltage drop on the first resistor, only a certain voltage drop will be left for the second resistor. The amazing thing is that even for the water analogy, the pressure magically distributes proportionally to the resistances.

So if the second pipe/resistor offers much more resistance, then at its entry there will be a larger proportion of the total pressure/voltage than at the first pipe/resistor!

Hence why the voltage/pressure magically distributes proportionally.

Woooooow. I'm mind blown.

Since at the end of the pipes there is zero relative pressure then the pressure must divide among the pipes/resistors before reaching the end.

Very interesting.

The pressure divides magically because all the molecules are close together and exerting forces on each other almost instantanously.

Amazing stuff.
 
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spec

Well-Known Member
Most Helpful Member
I had it the other way around when cruchow mentioned the big pipe.

The water analogy seems perfect, are there any cases where it doesn't apply ?
No. But the water analogy gets more difficult with capacitance and inductance. But, by that time, you will be thinking about electrons flowing, not water.:) I have been involved in electronics for so long that I use electron flow as an analogy for water flow.:joyful: I also use electronic theory for analyzing mechanical systems, like automobile car suspensions, or the pendlum on a clock.

The next hurdle is semiconductors: diodes, bipolar junction transistors, where the human comprehensible models are rather fanciful, but still give a good working model so that you can understand and deign with semiconductors.

But don't forget, that all models designed for human comprehension are not true at the elementary particle level, and even the most rigorous laws fall apart at extremes anyway.:D

spec
 

PConst167

Member
No. But it gets more difficult with capacitance and inductance. But, by that time, you will be thinking about electrons flowing, not water.:)

The next hurdle is semiconductors: diodes, bipolar junction transistors, where the human comprehensible models are rather fanciful, but still give a good working model so that you can understand and deign with semiconductors.

But don't forget, that all models designed for human comprehension are not true at the elementary particle level, and even the most rigorous laws fall apart at extremes anyway.:D

spec

Thank you Spec :)

Yes I'm familiar with mosfets, bjt and diodes. I have done some studying on them before. Electronics is great :D
 

spec

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Most Helpful Member
Thank you Spec :)

Yes I'm familiar with mosfets, bjt and diodes. I have done some studying on them before. Electronics is great :D
No sweat PC,

once you get Ohm's law and Kirchhoff's two laws under your belt (all three extremely simple) , the sky is the limit, especially if you are with mathematics.:)

But, electronics is such a vast subject that you never stop learning, and new devices and technology are introduced daily.

Not only is electronics fascinating, but it is very empowering- you can do almost anything with it.:cool:

spec
 

Ratchit

Well-Known Member
Photons are not limited to only visible light. A photon carries energy proportional to the radiation frequency, for instance, infra red (heat). Also, energy bands do, indeed, overlap but not always. Sometimes theu merely exchange electrons. Depend on the elements involved.
Although all electromagnetic (EM) radiation including heat occurs in discrete packets, most of the literature defines photons as energy packets of visible light. But, I believe the OP was wondering more about the amount of energy lost by the charge carriers traveling through a resistance, not the form the lost energy took or the way it was dissipated.

My emphasis.
That also is true, but the greater the number of available valence electrons, the greater the number of possible interactions between them and the introduced electrons of the current. Thus the greater the reduction in current electrons' energy levels and the greater the level of photon production per given number of elemental atoms, thus, the greater reduction in the available current, resulting in a greater voltage drop.
I am having trouble wrapping my head around the above statement. Silver (1), copper (2), mercury (2) are good conductors. Carbon (4), Sulfur (6) are poor conductors.

A phonon is just another concept based on the photon which is, of course, still not all that well defined (as in, "wave particle" as described in quantum physics).
Agreed. It is more of a mathematical artifice that a real particle.

Ratch
 

Ratchit

Well-Known Member
That still doesn't explain why the voltage drops in exact ratios. Let's have a 6V battery powering a circuit with 2 series resistors, R and 2R, the voltage drop at 2R will be 4V, and the voltage drop at R will be 2V. However, if you use a single resistor R, the voltage drop will be 6V at that single resistor. My question is, if the energy lost at the single circuit resistor is 6 Joules per Coulomb, why doesn't the same energy drops across the first resistor in the series arrangement?
Single, double, or triple resistors make no difference. It matters not how many resistors are in the circuit, it depends on their value. In the above examples you gave, both have a total resistance value of six ohms. That determines the current of one amp for both circuits. Since the current is the same through all resistors in a series circuit, naturally the energy lost will be proportional to the resistance value through which the charge flow passes.

Why does voltage know ahead that it has to give only 4 Joules per Coulomb at the first resistor and only 2 J/C at the second?
It doesn't. It doesn't have to know anything. If the same series current exists through two resistors, the larger valued resistor will lose more energy and register a higher energy density change (voltage drop). I thought I explained that in my previous posting to you.

This seems just strange. If you use 2 resistors, then the energy lost at each one is less than if you use a single one. It seems voltage is planning ahead.
That is because the resistance value of each of the two resistors is less that the single resistor. Get it now?

Can anyone explain this? I know it has to be this way because this is the logical way. Of course it would make no sense if the voltage dropped all across the first resistor and none at the second. But it's still strange that it can divide perfectly among them both.
Logic does not determine how something works. Logic indicates the best explanation choice. It divides perfectly because the energy density loss/charge amount (volts} is proportional to the resistance value.

I also understand that 2 resistors are equivalent to one large resistor. So in this case, if we use one single resistor then the voltage drop at each point in the resistor will be different. Why is this?
See above

Let's say we have an electric field inside the resistor, which comes from the negative end of the battery (Assuming positive current). Then at each point in the resistor will the electric field be variable or constant?
Constant, if the material of the resistor has a constant dielectric constant.

If voltage is the line integral of the electric field with respect to distance, then we multiply the electric field at each point by an infinitesimal distance dr, and add the small pieces of voltage together to find the total voltage drop.
Yes.

This can of course be divided into a sum of integrals or voltage drops across the resistor.
No, a sum of differentials. Integration is the process of adding the differentials

If the electric field is constant, then the integral will simply be E*r where r is a single dimensional length (assuming 1 dimension).
Yes.

This is proportional to the distance travelled. Is the electric field constant ?
Question already asked, already answered.

Who can explain this in terms of the electric fields created by the voltage source and through the resistor?
Explain what?

Ratch
 
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