Conceptual Question about Resistance and Voltage Drop

[Steelerfan]

Hi,

1) Can anyone explain to me from a conceptual standpoint, why increased current leads to an increased voltage drop across a resistor? Mathematically, I know V=IR. But it doesn't make sense to me why increasing a current across a resistor would lead to a greater drop in energy per coulomb of charge? Clearly, if one increased resistance, more energy per coulomb would be dissipated in order to traverse the resistance. But as V=IR tells us, if we increase the number of coulombs flowing through a resistor per second, each coulomb is subject to a greater drop in energy.

It does make sense to me why increasing voltage or decreasing resistance would lead to a greater current flow. But for a given resistor, if you pass more current through it, why would a greater energy be lost per coulomb than if you had passed less current through it?

2) One last question. A greater voltage, or EMF, from say a battery, leads to each coulomb having a greater electric potential energy. A greater voltage also leads to a greater current. Are these interrelated phenomenon? Is there some sort of causation, like a larger energy per charge helps to induce more charge to flow per second? This doesn't really make any sense to me, but I'm just throwing it out there as an illustration of what I'm asking.

Thanks!

 

[berkeman]

The causality is higher voltage --> higher current. It is not causal to say that a higher current causes a higher voltage drop, so that is why it is not making sense to you.

The potential difference (the voltage) across the resistive element is what causes the current to flow (or more accurately, the electrons to flow in the opposite direction).

 

[Steelerfan]

I think I get what you're saying...but it still doesn't answer my question in my mind. Let's be more concrete: If I had a voltage of 10V and a resistor of 2 Ohms, the current flowing would be 5 amperes and going across the resistor each coulomb would experience a drop of 10V. If you increased this voltage to 20 V with the resistor still at 2 Ohms, the current would be 10 amperes, which makes sense, because you have a larger force moving the current forward. But going across the resistor, each coulomb now experiences a drop of 20V. Conceptually, why would an increased current (just more coulombs flowing per second) cause the energy drop of each coulomb across a resistor to be higher? In this second scenario, the only thing that has changed is an increased voltage, causing there to be increased current. With this larger current, why would each coulomb experience a larger drop in voltage than when there was a smaller current?

 

[berkeman]

The increase in current comes from an increase in the average drift velocity of the electrons, which causes them to "hit" atoms harder as they bounce their way through the resistive material. This transfers more heat energy into the resistive material.

 

[ZapperZ]

Because the RATE of current flow increased!

Think as if you have a parallel plate capacitor. The charges moving across it gained in energy as it moves from one plate to another. If you increase the potential difference, the speed (KE) that the charges gained is also higher, meaning it they all went from one plate to another faster. That is the energy gained, and the charges all gained more energy with a higher potential difference.

So the increase in current (which is really the # of charges passing through a X-sectional area per unit time) is due to the increase in "speed" of the charges going through the resistor. However, this "speed" should be taken with some caveat as stated in solid state physics.

Zz.

 

[Steelerfan]

Ahhh, maybe I understand. So Zz, what you are saying is that if you have a 5 amp current going through a 2 ohm resistor, you would need a 10 V drop for each coulomb in order to maintain the 5 amp current, otherwise you'd get a "backup" of charge? And similarly, if you have a 10 amp current going through a 2 ohm resistor, in order to get this increased current through the resistor, you would need to apply twice as much energy per coulomb in order to maintain this faster rate? So connecting this idea to the second question I asked in my original post...the reason for the doubled current in a circuit with doubled voltage is that each coulomb has double the electric potential energy, causing each coulomb to move twice as fast, and vwoila, doubled current?

Thanks for your replies. I just want to make sure that I'm thinking about this correctly.

 

[Steelerfan]

And just to make sure, the reason a current doubles upon a doubling of the voltage is because the RATE at which the coulombs flow doubles. That is, the electric potential energy associated with each coulomb doubles, and this is manifested in terms of increased velocity?

 

[berkeman]

And just to make sure, the reason a current doubles upon a doubling of the voltage is because the RATE at which the coulombs flow doubles. That is, the electric potential energy associated with each coulomb doubles, and this is manifested in terms of increased velocity?

Yes, that is correct. The density of conduction band electrons per cm^3 does not change (assuming a constant temperature), so it is just their velocity increase that gives the increased current flow.

 

[Steelerfan]

Okay...if everything I said thus far is correct, then I have another question. If the battery is imparting some amount of potential energy to each coulomb of charge. And it is this potential energy that is being converted into kinetic energy to cause charges to move at a certain rate. Then if a resistor causes a complete voltage drop eqaul to the voltage gain imparted by the battery, how do charges still have a velocity component, as that would seemingly imply that the charges still have kinetic energy?

For example, if you had a 10 V battery and a 2 ohm resistor, resulting in a 5 Amp current. Would not the 10 V drop across the resistor take away all the energy associated with each coulomb of charge? Where is the velocity of the charge coming from after leaving the resistor if all the energy has been dissapated in the resistor?

 

[berkeman]

Okay...if everything I said thus far is correct, then I have another question. If the battery is imparting some amount of potential energy to each coulomb of charge. And it is this potential energy that is being converted into kinetic energy to cause charges to move at a certain rate. Then if a resistor causes a complete voltage drop eqaul to the voltage gain imparted by the battery, how do charges still have a velocity component, as that would seemingly imply that the charges still have kinetic energy?

For example, if you had a 10 V battery and a 2 ohm resistor, resulting in a 5 Amp current. Would not the 10 V drop across the resistor take away all the energy associated with each coulomb of charge? Where is the velocity of the charge coming from after leaving the resistor if all the energy has been dissapated in the resistor?

I hope that you pursue some EE courses, because you are asking good questions, and you deserve to get good answers to help your mental pictures that you use in whatever the heck kind of work you do. In our EE and physics and solid state physics classes, you learn some mental pictures (based on different levels of physical models -- simplistic/basic, classical, solid state, and then QED, and then other stuff that I don't know yet...) that make this all make a lot more sense.

I'll try to do my best on your present follow-up question. The "complete voltage drop" aspect is just because your scenario has one resistor connected across the power supply terminals. If there are several resistors connected in series and parallel, then the voltage drops will add up to the total supplied voltage, and the currents dividing between the parallel paths will add up to the total current supplied out of the power supply.

As for a kinetic energy (KE) component in the analysis, it is more of a "mean free path" concept, and less of a free-path velocity kind of thing. There is an accelerating force (from the the electric field) on the electrons, and a net "resistance" presented by the mean free path between collisions. The greater the force (the higher the voltage), the faster the electrons go, and the greater the current. Alternately, given the same source voltage, the lower the resistance (the greater the mean free path between collisions), the faster the electrons go between collisions with the lattice, and the more energy that is transferred to the lattice in the form of phonons (heat). Remember that the energy associated with velocity ratios with mv^2.

Ack, but I'm rambling. Did all of that answer your question?

 

[cabraham]

The causality is higher voltage --> higher current. It is not causal to say that a higher current causes a higher voltage drop, so that is why it is not making sense to you.

The potential difference (the voltage) across the resistive element is what causes the current to flow (or more accurately, the electrons to flow in the opposite direction).

Voltage does NOT "cause" current. The relation between current and voltage is not causality. Current and voltage have a relation which is best described as "mutually inclusive", i.e. one does not exist without the other, under time-changing conditions. Under static conditions current can exist without voltage, and voltage can exist without current. Neither one is sufficient nor necessary to "cause" the other one.

This is a mental block that many, including a few degreed EE's, can't seem to get past. In our daily lives, we use independent power sources, i.e. batteries and generators, that exhibit constant-voltage behavior. This is why some believe that voltage is what causes current. The idea that the voltage is "causal" is nothing more than a prejudice, or dogma, adopted as a result of the constant-voltage behavior of batteries and generators employed in electrical work.

Claude

 

[ZapperZ]

Voltage does NOT "cause" current. The relation between current and voltage is not causality. Current and voltage have a relation which is best described as "mutually inclusive", i.e. one does not exist without the other, under time-changing conditions. Under static conditions current can exist without voltage, and voltage can exist without current. Neither one is sufficient nor necessary to "cause" the other one.

This is a mental block that many, including a few degreed EE's, can't seem to get past. In our daily lives, we use independent power sources, i.e. batteries and generators, that exhibit constant-voltage behavior. This is why some believe that voltage is what causes current. The idea that the voltage is "causal" is nothing more than a prejudice, or dogma, adopted as a result of the constant-voltage behavior of batteries and generators employed in electrical work.

Claude

Maybe you should provide specific examples to prove your point. Please don't include exotic effects such as superconductivity, or even induced current, because that isn't what the OP is asking here.

While I can see a situation where a potential difference results in no real current (as opposed to displacement current) (example: steady state capacitor), can you show me where there is real current when there is no potential difference?

Zz.

 

[cabraham]

Maybe you should provide specific examples to prove your point. Please don't include exotic effects such as superconductivity, or even induced current, because that isn't what the OP is asking here.

While I can see a situation where a potential difference results in no real current (as opposed to displacement current) (example: steady state capacitor), can you show me where there is real current when there is no potential difference?

Zz.

Yes I can, but only under *static* conditions. An inductor or a single turn loop (one turn inductor) with superconducting wire will sustain a *static* current indefinitely without a potential difference or voltage. Superconductivity is not an "exotic" concept at all, anymore than an open circuit. A voltage source or charged capacitor terminated in an "open circuit" can sustain an indefinite "static voltage" without any current. The open circuit can be termed a "super-insulator". Not "exotic" at all, it is just how nature is. It just so happens that superconductivity exists, today anyway, at very low temperatures, whereas superinsulativity exists over a broad range of usable temperatures. In the first case I exists without V, and in the second case, V exists without I. Neither one is necessary nor sufficient to *cause* the other.

All circuits possesses both inductance and capacitance, and store energy magnetically and electrically resp. Once a current is established, the inductance tends to maintain that value of current. Voltage is required to change current from zero to some non-zero value, but to sustain a steady value of current requires zero voltage. A similar analogy holds for capacitance as well. You cannot change the capacitor voltage without current first. But a cap can sustain a steady value of voltage with zero current.

Under time-varying, or dynamic conditions, even a perfect cap with ideal insulation will draw displacement current. You can't ignore it. This current chronologically precedes the voltage, i.e. I leads V. Since a change in current chronologically precedes a change in voltage, in a cap, V cannot possibly be the "cause" of I, because an effect cannot precede its cause. A similar analogy holds for inductors. Since V leads I, I cannot be the cause of V.

You seem to have a desire to completely ignore superconductivity, induction, and displacement current. Unfortunately, e-m field theory makes it clear that one cannot understand circuits without them. Have I explained it well? Peace.

Claude

 

[ZapperZ]

Yes I can, but only under *static* conditions. An inductor or a single turn loop (one turn inductor) with superconducting wire will sustain a *static* current indefinitely without a potential difference or voltage. Superconductivity is not an "exotic" concept at all, anymore than an open circuit. A voltage source or charged capacitor terminated in an "open circuit" can sustain an indefinite "static voltage" without any current. The open circuit can be termed a "super-insulator". Not "exotic" at all, it is just how nature is. It just so happens that superconductivity exists, today anyway, at very low temperatures, whereas superinsulativity exists over a broad range of usable temperatures. In the first case I exists without V, and in the second case, V exists without I. Neither one is necessary nor sufficient to *cause* the other.

All circuits possesses both inductance and capacitance, and store energy magnetically and electrically resp. Once a current is established, the inductance tends to maintain that value of current. Voltage is required to change current from zero to some non-zero value, but to sustain a steady value of current requires zero voltage. A similar analogy holds for capacitance as well. You cannot change the capacitor voltage without current first. But a cap can sustain a steady value of voltage with zero current.

Under time-varying, or dynamic conditions, even a perfect cap with ideal insulation will draw displacement current. You can't ignore it. This current chronologically precedes the voltage, i.e. I leads V. Since a change in current chronologically precedes a change in voltage, in a cap, V cannot possibly be the "cause" of I, because an effect cannot precede its cause. A similar analogy holds for inductors. Since V leads I, I cannot be the cause of V.

You seem to have a desire to completely ignore superconductivity, induction, and displacement current. Unfortunately, e-m field theory makes it clear that one cannot understand circuits without them. Have I explained it well? Peace.

Claude

Considering that I did my PhD in superconductivity, no, I am not ignoring it. It is just that within the scope of the question and the explanation given, it isn't relevant!

Furthermore, if you want to get technical about it, superconductivity cannot be considered as a "current flow", because there is no such thing as a classical current in a supercurrent. The long-range coherence of the supercurrent means that when there is no potential being applied, there is equal superposition of the supercurrent going in one direction and in another (refer to the Stony Brook/Delft SQUID experiments). Thus, it is also not correct, if we want to be exact about this, to say that there is a currrent "flow" at all.

When we confine it to the classical current that is relevant in the OP, then what has been described by berkeman is perfectly valid.

Zz.

 

[berkeman]

Voltage does NOT "cause" current. The relation between current and voltage is not causality. Current and voltage have a relation which is best described as "mutually inclusive", i.e. one does not exist without the other, under time-changing conditions.

Of course it is causal. The voltage causes the force on the electrons, which makes them move and that's the current. Just as in F=ma, where the force causes acceleration. You wouldn't claim that acceleration causes force, would you?

 

[cabraham]

Of course it is causal. The voltage causes the force on the electrons, which makes them move and that's the current. Just as in F=ma, where the force causes acceleration. You wouldn't claim that acceleration causes force, would you?

Of course I would not claim that a causes F. When you say that voltage causes force on electrons, you are making a declaration with nothing to back it up. The force acting on charges is an inherent property associated with all charged particles. This force is postulated. Centuries of empirical obseration have consistently confirmed the presence of said force. What causes this force is not yet known.

The Coulomb force equation is well known as:

F = k * q1 * q2 / (r^2).

This equation is a postulate, and is not derived from anything more fundamental.

If we move a charge, q2, from a to b, in the presence of another charge q1, of the same polarity, where the path from a to b moves q2 closer to q1, we must exert a force, since the natural tendency for two like charges is repulsion. This force when integrated over the path from a to b results in the energy put into the system. This represents the increase in potential energy. Dividing by the charge gives potential, or energy per unit charge, aka "voltage". Voltage is a derived quantity, defined in terms of force, charge, and distance. Voltage does not cause the force. Voltage is a measure of the force, integrated along the path length, divided by the charge. To say that voltage is what causes the force is a circular definition. Force is more fundamental, and voltage is derivative. Any e-m fields reference text will affirm this.

To say that voltage causes force implies that voltage can exist without force, which isn't the case. The "force" which moves charges is - FORCE.

Another example. Say that two billiard balls on a pool table are positively charged. The 8 ball is sitting near a pocket. The cue ball is far enough away such that the repusive force is insufficient to overcome the static friction between the 8 ball and the table's cloth surface. If I grab the cue ball and move it closer to the 8 ball, it will eventually be repulsed and roll into the pocket. The instant I moved the cue ball I began to increase the force and the voltage. But at the same time, I also gave rise to a current as well, since moving a charge constitutes current. Did this current "cause" the 8 ball to roll into the pocket, or was it the voltage change, or was it the force? The only logical position is that the work/force I exerted in moving the cue ball is what caused the 8 ball to move. My expenditure of energy is what moved the 8 ball, and the change in current and voltage were simultaneous and mutual. Neither I nor V was the "cause". Energy was transferred. Enough for now. Peace.

Claude

 

[Steelerfan]

Thanks for your responses. I'm actually a medical student, but for one reason or another, I find myself getting obsessed with certain things that don't make sense in my head when I read about it, whether or not it's in my coursework. And the majority of people in med school, students and teachers alike, aren't too interested in this kind of stuff, so I'm really glad that I found this forum.

In this case, I guess I'm trying to extrapolate difficult concepts from what I learned in my intro physics class from 4 years ago. But I still believe that I can make sense of all this, even with my limited knowledge base, because it's conceptual. Do those of you with higher-level course work disagree with me and think that I should just put off thinking about this, because it would take up too much of my time? I thought there would be simple answers to my questions, but it seems that a lot of the contradictions that are in my "mental map" may be from my lack of education in these areas, more than my not thinking about it correctly. At this point am I just limited in what I know (intro physics) from really understanding what's going on in a circut?

What would be really helpful, is if someone could construct a cause and effect narrative of what happens when you have a battery, connected to one resistor for simplicity. Let's say you have a battery of 10 V and a resistor of 2 ohms. What's the first thing that happens when you connect the wires to complete the circuit? Why does the current flow at 5 A? How does the current know to flow at this rate? What type of energy is stored in each coulomb as it is forced through the battery and travels through the wires: does the energy change from potential to kinetic throughout its journey, does it remain as one type...what's going on here? Upon encountering the resistor, what happens to the current? When faced with resistance, does the kinetic energy of each coulomb always drop to 0, and then regain its velocity due to the stored electric potential energy, thereby conserving the current between resistor entry and exit? This in and of itself, doesn't make complete sense...let's say you had again, a 10 V battery and a 2 ohm resistor. If the energy for each coulomb is stored as EPE (10 J/C), then where is the kinetic energy that is causing the movement. Furthermore, when going across the resistor, if there is a 10 V drop, then all the energy that is associated with that Coulomb should be dissapated as heat. Therefore, where is the energy that is keeping the current at 5 A coming from? Those 10 J for each coulomb was lost by the resistor, yet the current remains. How?

If someone could piece together a narrative from step 1 of circuit construction, then maybe it'd be easier for me to understand what's precisely occurring.

Thanks! Sorry to bother all of you, but I really appreciate your help, and hope that this is at least somewhat interesting to you. I plan to keep asking questions, until my mental picture is clear. Hopefully, you'll keep responding.

 

[berkeman]

Of course I would not claim that a causes F.

Good. Thank you for that.

When you say that voltage causes force on electrons, you are making a declaration with nothing to back it up. The force acting on charges is an inherent property associated with all charged particles. This force is postulated. Centuries of empirical obseration have consistently confirmed the presence of said force. What causes this force is not yet known.

The Coulomb force equation is well known as:

F = k * q1 * q2 / (r^2).

This equation is a postulate, and is not derived from anything more fundamental.

You want me to come up with a fundamental "proof" of the Columb force equation, in order to confirm my assertion of F --> ma causality?

I don't think that us debating reality is going to help the OP very much in this case. Hopefully their question has been answered by now.

 

[berkeman]

What would be really helpful, is if someone could construct a cause and effect narrative of what happens when you have a battery, connected to one resistor for simplicity. Let's say you have a battery of 10 V and a resistor of 2 ohms. What's the first thing that happens when you connect the wires to complete the circuit? Why does the current flow at 5 A? How does the current know to flow at this rate? What type of energy is stored in each coulomb as it is forced through the battery and travels through the wires: does the energy change from potential to kinetic throughout its journey, does it remain as one type...what's going on here? Upon encountering the resistor, what happens to the current? When faced with resistance, does the kinetic energy of each coulomb always drop to 0, and then regain its velocity due to the stored electric potential energy, thereby conserving the current between resistor entry and exit? This in and of itself, doesn't make complete sense...let's say you had again, a 10 V battery and a 2 ohm resistor. If the energy for each coulomb is stored as EPE (10 J/C), then where is the kinetic energy that is causing the movement. Furthermore, when going across the resistor, if there is a 10 V drop, then all the energy that is associated with that Coulomb should be dissapated as heat. Therefore, where is the energy that is keeping the current at 5 A coming from? Those 10 J for each coulomb was lost by the resistor, yet the current remains. How?

I'll try to offer a brief explanation. It really does come down to F=ma. The force comes from the EMF (electro-motive force):

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

That is the force that a voltage source like a battery or a transformer exerts on electrons. Before a load is connected to the voltage source, the electrons are only pumped a little bit one way (the opposite direction that the positive "current" flows), and that pumping action stops when the output capacitance of the voltage source is charged up to the open-circuit output voltage. When you connect the load resistance, the voltage potential generates a force on the electrons, pushing them through the external circuit. The voltage across the load resistance causes a voltage gradient throughout the resistor, which causes the force that pushes the electrons along. The electrons "bounce" off of the solid lattice atoms (I'm simplifying here), and the bounces are what transfers energy/heat to the resistor material. The higher the resistance of the material, the shorter the "mean free path" between bounces.

The voltage source can be viewed as the "pump" in the system, giving the electrons energy that they then transfer to the resistive material through collisions.

 

[cabraham]

Good. Thank you for that.



You want me to come up with a fundamental "proof" of the Columb force equation, in order to confirm my assertion of F --> ma causality?

I don't think that us debating reality is going to help the OP very much in this case. Hopefully their question has been answered by now.

No, I don't "want" you to prove Coulomb's force law because you can't. Nor can I. I never ask the impossible of anyone.

Regarding the battery, you simply proclaim that the voltage causes the force which then moves the electrons. So far, all you've done is to declare that V causes F, and have given no support. You accept causality as a given, and proceed from there. Of course, everything cannot be proven, as circular arguments would inevitably result. But in order to establish "A" as the cause, and "B" as the effect, what are the criteria for doing so? Please, just answer that one for me. What about my 2-billiard-ball pool table scenario? Do you have a response for that? I'm just interested. Thanks.

Claude

 

[berkeman]

I'll just try to briefly answer your two present questions. F=qE is the force that I'm referring to. The F comes from the E field.

As to the billiard ball question, you did mechanical work with your hand that caused an increase in force on the 2nd ball via the electric field. I suppose that you can call the motion of the first ball a current -- it fits the definition. Kind of the same way that the current in a circuit into a capacitor increases the electric field between the plates.

Actually, there's an interesting example that might just be useful here in terms of causality... Think of a low-loss LC tank circuit. At peak capacitor voltage, no current is flowing, and the voltage across the inductor causes the series current to start ramping up. At zero capacitor voltage, the series current is flowing through the inductor, and that charges up the cap the other way. So at least in this example, there is a case where the energy stored in the inductor's B field supplies the force to push the electrons along... Interesting.

 

[cabraham]

I'll just try to briefly answer your two present questions. F=qE is the force that I'm referring to. The F comes from the E field.

As to the billiard ball question, you did mechanical work with your hand that caused an increase in force on the 2nd ball via the electric field. I suppose that you can call the motion of the first ball a current -- it fits the definition. Kind of the same way that the current in a circuit into a capacitor increases the electric field between the plates.

Actually, there's an interesting example that might just be useful here in terms of causality... Think of a low-loss LC tank circuit. At peak capacitor voltage, no current is flowing, and the voltage across the inductor causes the series current to start ramping up. At zero capacitor voltage, the series current is flowing through the inductor, and that charges up the cap the other way. So at least in this example, there is a case where the energy stored in the inductor's B field supplies the force to push the electrons along... Interesting.

Very good reply! Honestly, I'm not being patronizing.

You've concurred with what I've been driving at. The mechanical work done with my hand is absolutely indeed what gives rise to the 2nd ball's motion, as you've correctly stated. During the course of my exerting force on the cue ball, the electric field between the two balls, the force, the voltage, and the current, all changed. However, neither the voltage change, nor the current change, is what "caused" the 2nd ball to move. These two changes are naturally incurred because I inputted force/energy into the system. You've explained it quite correctly.

Yes, the LC tank is also a good way of looking at it. Just as energy stored in the form of an E field (capacitor) can be transferred to an inductor, which stores energy in the form of an H field, the converse is true. In other words, the inductor's current can give rise to the capacitor's voltage, just as vice-versa. The fact that either can come first, and give rise to the other clearly demonstrates that what we're observing is merely the transfer of energy in various forms. There is no "pecking order" between I and V. The existence of either one can lead to the rise of the other in no specific order.

Energy, or work is what gets things to happen. Peace.

Claude

 

[Jayzar]

Thanks for your responses. I'm actually a medical student, but for one reason or another, I find myself getting obsessed with certain things that don't make sense in my head when I read about it, whether or not it's in my coursework. And the majority of people in med school, students and teachers alike, aren't too interested in this kind of stuff, so I'm really glad that I found this forum.

In this case, I guess I'm trying to extrapolate difficult concepts from what I learned in my intro physics class from 4 years ago. But I still believe that I can make sense of all this, even with my limited knowledge base, because it's conceptual. Do those of you with higher-level course work disagree with me and think that I should just put off thinking about this, because it would take up too much of my time? I thought there would be simple answers to my questions, but it seems that a lot of the contradictions that are in my "mental map" may be from my lack of education in these areas, more than my not thinking about it correctly. At this point am I just limited in what I know (intro physics) from really understanding what's going on in a circut?

What would be really helpful, is if someone could construct a cause and effect narrative of what happens when you have a battery, connected to one resistor for simplicity. Let's say you have a battery of 10 V and a resistor of 2 ohms. What's the first thing that happens when you connect the wires to complete the circuit? Why does the current flow at 5 A? How does the current know to flow at this rate? What type of energy is stored in each coulomb as it is forced through the battery and travels through the wires: does the energy change from potential to kinetic throughout its journey, does it remain as one type...what's going on here? Upon encountering the resistor, what happens to the current? When faced with resistance, does the kinetic energy of each coulomb always drop to 0, and then regain its velocity due to the stored electric potential energy, thereby conserving the current between resistor entry and exit? This in and of itself, doesn't make complete sense...let's say you had again, a 10 V battery and a 2 ohm resistor. If the energy for each coulomb is stored as EPE (10 J/C), then where is the kinetic energy that is causing the movement. Furthermore, when going across the resistor, if there is a 10 V drop, then all the energy that is associated with that Coulomb should be dissapated as heat. Therefore, where is the energy that is keeping the current at 5 A coming from? Those 10 J for each coulomb was lost by the resistor, yet the current remains. How?

If someone could piece together a narrative from step 1 of circuit construction, then maybe it'd be easier for me to understand what's precisely occurring.

Thanks! Sorry to bother all of you, but I really appreciate your help, and hope that this is at least somewhat interesting to you. I plan to keep asking questions, until my mental picture is clear. Hopefully, you'll keep responding.

Dude; so far you've been reading my mind! I really like your line of questioning.

Where has the energy gone after the resister has consumed it all? Well I can only assume that there must be some energy left, sufficient to restock the other terminal of the battery with electrons, if only very slowly. I shall also assume that just as the cell pushes out electrons from one end it must "suck" them back in at the other.

This would mean that the free-electron density following the resister would be greater than the higher speed higher energy electron density before it...

Kind of like a river where it's high energy and narrow, slower moving and wide??

We need a reply from the professionals!

 

[ZapperZ]

Dude; so far you've been reading my mind! I really like your line of questioning.

Where has the energy gone after the resister has consumed it all? Well I can only assume that there must be some energy left, sufficient to restock the other terminal of the battery with electrons, if only very slowly. I shall also assume that just as the cell pushes out electrons from one end it must "suck" them back in at the other.

This would mean that the free-electron density following the resister would be greater than the higher speed higher energy electron density before it...

Kind of like a river where it's high energy and narrow, slower moving and wide??

We need a reply from the professionals!

But this thread (which had its last activity 3 years ago!) HAS been dealt with by "professionals"!

The origin of Ohm's Law, and many other standard circuit formulas, are adequately described in the Drude model. It gives you a microscopic description of the statistical nature of the free electron gas in a metal under ordinary situation. It is a standard treatment in any solid state physics text, so none of these are really new or even "surprising" (see Chapter 1 of Ashcroft and Mermin's classic text).

The only question here is how much of this do you want explained. Or if you have some knowledge of it, what part do you want further explanation on?

Zz.