Conceptual Understanding for Voltage in a Circuit

[Blissed]

I'm having trouble understanding circuits, particularly voltage. So my question is more of a conceptual dilemma. Please correct my definitions and relations if it is incorrect. Voltage is the energy provided per charge, and this energy is converted to KE once the circuit is closed, or in other words it influences the velocity of the electrons, and in effect influences the current.

For instance, in a simple circuit made up of a battery and a resistor, the voltage drops to 0 once the electrons pass through the resistor, or the energy provided by the battery is all dissipated. Then why is the current still constant? If the voltage drops, doesn't that imply a drop in velocity of the electrons? So shouldn't it lead to a decrease in current as it leaves the resistor?

A follow up question is, why is the voltage drop the same in a resistor as the voltage provided by battery? How does a resistor "know" that it should use up the same amount of voltage provided by the battery?

 

[.Scott]

Voltage is not energy - it is a difference in charge. If you think of water levels, it is a difference in water levels - but the size of the body of water is not relevant.
So, an electrostatic charge may give you a 20Kv voltage compared to you environment, but the energy involved may be slight.

When voltage is applied to a resistor, the current will be proportional to the voltage and inversely proportional to the resistance. So if a battery is connected to a resistor, the current will decay as the voltage provided by the battery decays.

Since the battery and the resistor have both ends tied to each other, they will always have the same voltage across them.

 

[sophiecentaur]

Voltage is not energy - it is a difference in charge.

I think you should do some homework before you try to 'explain' something to someone.
Voltage and charge are different quantities entirely. (You would not say that a head of water in metres is the same as the difference in volume between two tanks of water.)
Voltage is Energy per Unit Charge and it represents the Energy that is available or used up as units of charge move round a circuit. One volt is One Joule per Coulomb.

So, an electrostatic charge may give you a 20Kv voltage compared to you environment, but the energy involved may be slight.

That's more like it. 20kV gives 20kJ to every Coulomb it supplies but, often, a very tiny charge is involved (say 1milliCoulomb) so the actual Energy would only be 20J.

A follow up question is, why is the voltage drop the same in a resistor as the voltage provided by battery? How does a resistor "know" that it should use up the same amount of voltage provided by the battery?

That confuses everyone. The current that flows through a Resistor is determined by its Resistance Value but that is true, only after a certain (short) amount of settling time. Initially, an Electric Pulse flows along the wires and the steady current flowing is given by I=V/r. Things in Science don't "know" what's happening or what part they play in a system - they just behave according to the rules. Those rules have been formulated as a result of many measurements. Resistors behave 'as if they knew' about the rest of the circuit. Does the idea of a settling time satisfy you? It deals with the 'knowing' problem.

 

[Blissed]

So I should just take that it's a given that the voltage will be the same? It's just the current that changes depending on the resistor. I mean I can understand this from the formulas itself, but I'm trying to visually and conceptually understand why this occurs. I imagine an electron moving through a resistor with a certain amount of KE provided by the battery. Once it collides with the molecules in the resistor, the KE provided by battery can drop to 0. However, in a different scenario, what if that electron leaves the resistor with a portion of KE provided by the battery intact, doesn't that imply the voltage drop for that electron isn't exactly equal to the voltage of the battery? So that's where my two questions come from. Why can we comfortably say that the voltage drop in a resistor is equal to the voltage in the battery?

As for my first question, why is the current consistent as it exits and enters the resistor? If the current is affected by the velocity of the electron, and the velocity of the electron correlates with the voltage provided by the battery, doesn't this imply that the current changes since the KE of the electron changes as it traverses through the resistor? So the KE of the electrons just before it enters the resistor is higher than the KE of the electrons as it exits the resistor.

 

[sophiecentaur]

So I should just take that it's a given that the voltage will be the same?

A battery will provide the same voltage whatever load you hang on it (that's the ideal battery we use in problems - a Voltage Source)
A voltage source has no internal resistance but a real battery has a small series resistance which will lose you some of the nominal battery volts as soon as you take current through it (same as a long connecting wire resistance will do).
Ignore the KE of electrons. It is not a significant quantity in the transfer of the energy. Electrons have an average speed of only around 1mm per second and their mass is a tiny proportion of the metal (much less than 1/10,.00th) so KE is not part of things. Also, the electrons are traveling through the resistor at a constant (very low) average speed.

why is the current consistent as it exits and enters the resistor

Think in terms of the links in a bicycle chain. They can be going quite slowly and the same number are being pulled from the back sprocket as are going round the chain wheel every second. Their KE of the links is not very relevant - it's the tension in the chain and the speed it's going over the wheels (Force times speed = Power). In the case of electrons, the KE is an even smaller proportion of the energy transferred and the same number enter at the - end and leave at the + end of the resistor. If the numbers didn't balance, there would be IMMENSE forces of mutual repulsion or attraction.

So the KE of the electrons just before it enters the resistor is higher than the KE of the electrons as it exits the resistor.

Again NO.

 

[anorlunda]

I imagine an electron moving through a resistor with a certain amount of KE provided by the battery

That is a very incorrect way to think. KE plays no significant role, and it is not helpful to think of electrons like billiard balls filled with energy. Accept V=IR as it is and forget completely about electrons.

The current that flows through a Resistor is determined by its Resistance Value but that is true, only after a certain (short) amount of settling time. Initially, an Electric Pulse flows along the wires and the steady current flowing is given by I=V/r.

@sophiecentaur , Remember the recent thread where I ranted about levels 1 QED, 2 Maxwells, 3 Circuits, and how it was wrong to make fractional level metaphors?. I thought you supported that.

I'm sure it is inadvertent, but when you say that current in the resistor starts after "a short time" that is mixing the metaphor, and providing an explanation not 2 and not 3.

• In circuit analysis (3), V=IR in a resistor is instantaneous. There is no first and no next.
• In Maxwells equations (2) there is a wave that propagates in space and time, but there is no such thing as "a resistor" We have 3D space filled with materials of differing conductivity, and where the dimensions of those materials is important (i.e. the size and shape of the resistive stuff).

I am urging all of us PF regulars, to resist the urge to mix the metaphors on PF and to never mention a physical phenomenon (no matter how real) except in the context of 1, 2 or 3. Would you agree?

 

[Blissed]

Thanks for the explanation, it seems like my confusion mainly came from observing the transfer of energy in terms of KE. So my follow up question is, how exactly then is the energy transferred to a resistor?

 

[sophiecentaur]

In circuit analysis (3), V=IR in a resistor is instantaneous. There is no first and no next.

I couldn't agree more but the OP seemed to be needing some reason to accept that in the idealised context. My 'explanation' is a fair enough way to connect the stark, axiomatic approach to circuit analysis with 'real behaviour'. We do tend to take for granted that using the steady state condition as a start in solving problems makes sense to the uninitiated. It's the same with problems about forces on frameworks; how does the bolt in the wall know what its share of the load on the shelf is? With EE it's even worse because we are actually dealing with something moving.

I am urging all of us PF regulars, to resist the urge to mix the metaphors

That's a good principle to aim at. But it would involve cutting short an awful lot of threads.

 

[anorlunda]

So my follow up question is, how exactly then is the energy transferred to a resistor?

Power, rate of delivery of energy, is voltage times current. V*I You need both. That is the point that many students miss.

My 'explanation' is a fair enough way to connect the stark, axiomatic approach to circuit analysis with 'real behaviour'.

Sorry, I disagree. Your explanation is not 'real behavior' as I explained in #6. That's why I object. And yes, cutting short those mixed metaphors in very many PF threads is what I advocate.

We might avoid this debate if students were taught Maxwells Equations first, and circuits second. Then they would appreciate and welcome the simplifying assumptions of circuits analysis. But alas, teaching almost always goes from simplest first to less simple later, leaving many students with some (3) but zero learning about (2).

 

[Blissed]

So in the microscopic level, what exactly is happening to the charges as they move through the resistor, what causes them to transfer energy to the resistor? I know power is the rate at which energy is transferred, but I would like a conceptual understanding as to how this occurs, hence I thought it was the collisions due to the KE that caused the energy transfer. Now I know that the KE of the electrons are irrelevant, what causes the energy to be transferred?

 

[cnh1995]

I believe this video will clear it up for you.
https://www.khanacademy.org/science/physics/circuits-topic/circuits-resistance/v/electric-power

 

[cnh1995]

So in the microscopic level, what exactly is happening to the charges as they move through the resistor, what causes them to transfer energy to the resistor? I know power is the rate at which energy is transferred, but I would like a conceptual understanding as to how this occurs, hence I thought it was the collisions due to the KE that caused the energy transfer. Now I know that the KE of the electrons are irrelevant, what causes the energy to be transferred?

The electric and magnetic fields carry the energy, and not the wires. If you want to dig deeper, look up 'surface charge feedback' and 'Poynting Vector'.
These past threads might help you.
https://www.physicsforums.com/posts/5521826/
https://www.physicsforums.com/posts/5667921/
https://www.physicsforums.com/posts/5507882/.

 

[sophiecentaur]

We might avoid this debate if students were taught Maxwells Equations first, and circuits second.

That would exclude the majority of School Students from any work on electric circuits. The Maths involved in Maxwell's equations is not all taught, even at A level.

 

[sophiecentaur]

Sorry, I disagree. Your explanation is not 'real behavior' as I explained in #6. That's why I object. And yes, cutting short those mixed metaphors in very many PF threads is what I advocate.

I just read this again. It is quite justifiable for someone to challenge the notion of a step change of an uninterrupted steady state condition when we all know that we turn on the current at some point and it takes time for the steady state to be reached. Simply saying that it takes time for the situation to settle down is not an arm waving low level analogy. The question from Blissed (?) was about 'how the resistor knows' what to do. If you take time out of the problem then this is not a problem because the question is not allowable. But the approach to Physics that insists on the steady state can be very confusing. What alternative justification would you suggest? Or would you just tell Blissed that he has to accept that that's the way we do it? I think that would be playing right into the hands of those who say that we want Physics taught by rote. I know you don't want that(?),

 

[.Scott]

Voltage and charge are different quantities entirely.

Sorry. I was trying to avoid the term "potential".

 

[lychette]

Sorry. I was trying to avoid the term "potential".

1 volt means 1 joule per coulomb, you cannot avoid energy when referring to voltage.

 

[anorlunda]

I believe this video will clear it up for you.
https://www.khanacademy.org/science/physics/circuits-topic/circuits-resistance/v/electric-power

Thanks for that @cnh1995 . As usual, Khan Academy does an excellent job. But it took them about 1400 words assisted by graphics and animation and ending with an accurate mathematical derivation; much too long for a verbal reply here on PF. I'll use that link the next time the question comes up.

 

[Joseph M. Zias]

I suggest reading "Matter and Interactions" by Chabay and Sherwood, Vol. II. Also, going back a bit in time I always liked "Electrical Science" by Norman Balabanian for a good treatment of the basic circuit laws. As mentioned before there are a LOT of references on surface charges, power flow, and the Poynting vector.

 

[dreens]

So in the microscopic level, what exactly is happening to the charges as they move through the resistor, what causes them to transfer energy to the resistor? I know power is the rate at which energy is transferred, but I would like a conceptual understanding as to how this occurs, hence I thought it was the collisions due to the KE that caused the energy transfer. Now I know that the KE of the electrons are irrelevant, what causes the energy to be transferred?

It's possible to understand this without delving into the subtleties of electromagnetic fields, especially since your question focuses on resistors and not inductors.

It helps to think of a fluid analogy. Consider water flowing in a river. Even though at the microscopic level the fluid flow is composed of water molecules with kinetic energy jostling each other, the way to understand the system is in terms of its macroscopic properties, it's pressure and it's velocity. You wouldn't solve for the speed of water flowing into the atlantic ocean from the Mississippi river by taking the average initial gravitational potential energy of the rain water when it landed on the continent and converting to kinetic energy, would you? No, because fluid flow is an inherently frictional process. The speed of the water as it flows is related to the resistive friction the fluid experiences with it's channel and the height decrease along it's flow path. The energy transfer is all frictional at the microscopic level, but it doesn't mean the fluid doesn't have a well-defined mass-flow rate everywhere along it's course.

The situation is nearly identical in a circuit. In fact, you could just think of it as a fluid flow of electrons. The electrons constantly jostle each other and their micromotion is random and frictional, but they can be well described by the bulk properties voltage and current. The electron flow is most impeded during it's travel through the resistor- often because the flow is forced through a narrow and long channel (a wire-wound resistor), maximizing collisions between electrons and the material. Because of this, the voltage drop is largest there. But since it is a fluid flow, this doesn't mean the fluid somehow runs out of kinetic energy and stops flowing at the other side of the resistor. It just means that the fluid pressure (voltage) has dropped back to its initial value.

 

[rumborak]

Ha, I almost suspected my earlier thread on the topic was going to get mentioned :D

I too am still left unsatisfied. On the one hand I know that it is the electrons that *facilitate* all that is happening in a circuit, and at the same we also know that it is the Poynting vector that shows the flow of energy, and for the most part that vector is flowing *outside* the wire, and only enters into the resistor.

Take for example the fact that the Poynting has the magnetic field as a key factor in its definition. But, how is the magnetic field even relevant to the electrons?The Khan video was *very* unsatisfying for the topic at hand, because he already started with the fact there's a voltage drop over a resistor. After that it was algebraic shuffling, but the question remains, why is there a drop? What happens inside the resistor that causes this?

 

[cnh1995]

But, how is the magnetic field even relevant to the electrons?

Isn't it current that creates the magnetic field?

 

[rumborak]

Of course the current creates a magnetic field, that is without doubt. But, how is this magnetic field a key ingredient to the power dissipation inside the resistor? That's what the Poynting vector says after all.

 

[anorlunda]

The Khan video was *very* unsatisfying for the topic at hand, because he already started with the fact there's a voltage drop over a resistor. After that it was algebraic shuffling, but the question remains, why is there a drop? What happens inside the resistor that causes this?

This is an I level thread, so some math is appropriate. Try this article.

https://en.m.wikipedia.org/wiki/Free_electron_model

 

[cnh1995]

But, how is this magnetic field a key ingredient to the power dissipation inside the resistor?

It is not the magnetic field alone. Power is supplied by the combination (cross product) of electric and magnetic fields. E-field is proportional to voltage and B-field is proportional to current. Hence, power is voltage*current.

Surface charges create electric field. The created electric field (which depends on resistance) and resistance decide the current. The current creates a magnetic field "such that" the cross product of E and B fields at various locations will ensure continuous power supply to the load through the Poynting vector.

 

[rumborak]

Yes, but if the magnetic field is zero, the Poynting vector suggests that no energy would flow. What does the *static* magnetic field do to the electrons in the resistor that causes them to transform the energy into heat, that they otherwise wouldn't if the magnetic field was absent?

 

[cnh1995]

I don't claim to be an expert in this but here's what I think.

What does the *static* magnetic field do to the electrons in the resistor that causes them to transform the energy into heat, that they otherwise wouldn't if the magnetic field was absent?

It's the electric field that moves the electrons. Electric and magnetic fields can be thought of as a "feedback" to the power source. The current creates a magnetic field. The cross product of E and B fields around a resistor indicates that the energy is being absorbed. Thus, magnetic field along with electric fields at various locations ensures continuous power supply to the load via the Poynting vector.

In a simple series circuit, the current is same throughout the circuit. In this circuit, it's the direction of E-field that decides the direction of Poynting vector and power flow.

 

[rumborak]

No offense, but that sounds very hand-wavy, especially the "feedback" part. Also, as pointed out earlier, the kinetic energy transferred to the electrons by the electric field is only a tiny fraction of the energy absorbed. Clearly, something else is happening, crucially involving the magnetic field, and I would like to know what.

 

[cnh1995]

No offense, but that sounds very hand-wavy, especially the "feedback" part. Also, as pointed out earlier, the kinetic energy transferred to the electrons by the electric field is only a tiny fraction of the energy absorbed. Clearly, something else is happening, and I would like to know what.

It is only the E-field inside the wire that can move the electrons. There is a potential difference across the resistor (because of the E field, or the other way around). The current is determined by V (i.e.electric field) and R.

Due to the electric field inside the resistor (thanks to surface charges), electrons acquire electrostatic potential energy. This energy gets converted into heat because of the resistance. Magnetic field has no role in this.
It's the cross product of magnetic field with the E-field around the resistor that "supplies" the power to the resistor.
(I think I should stop here before experts chime in and warn me for hand-waving).

 

[rumborak]

You know what I'm talking about though, right? There's two disparate aspects of this, and how they connect is totally nebulous. "The cross-product supplies the energy" is totally hand-wavy. And I'm not accusing you, as I have no better explanation. But frankly, I have not gotten an answer from anybody here on this, which leads me to believe that we, as a whole, do not have an actual clue how circuits work at the lowest level. We know how to mathematically describe the two parts, sure, but how they connect down to actual constituent parts, I don't think so.

 

[brotherStefan]

Wow. Someone wants to know what time it is, and we proceed to provide instructions on building an atomic clock.

Getting back to the original question about understanding voltage. Voltage is the "prime mover" in an electrical system, just as a "push" is the prime mover in a mechanical system, or "pressure difference" is the prime mover in a fluid system. The voltage is a result of the potential energy that is stored in the battery. Packing the battery with electrons (by charging the battery) is similar in concept to pre-loading a spring by compressing (and holding) it, filling a tank up with water, or filling a party balloon with air.

You can think of voltage as an electrical pressure that can push existing electrons around in a closed circuit -- like when placing a resistor across the terminals of a battery. Normally such a circuit would benefit from the inclusion of a switch to make it easy to energize or de-energize the circuit at will. When the switch is closed, electrons will flow in the circuit from the negative terminal of the battery, through the resistor, and back into the positive terminal of the battery. When the switch is opened, the electron flow stops because the circuit path has been interrupted. The electrons don't go away -- they are still present in the molecular structure of the circuit -- they have just been isolated from the electrical pressure, or electro-motive force (EMF) provided by the battery.

I thought the billiard ball analogy was off to a good start, but then it never got developed in the context of our electrical circuit. When the switch is closed to complete the circuit, the battery will force electrons into the circuit (as fast as the circuit will allow) just like billiard balls being pushed through a tube (the circuit) that is already filled with billiard balls -- when a new billiard ball (electron) enters it bumps the next ball, which bumps the next, and the next, and so on until the one at the very end of the line is crowded out of the tube. In the case of electrons flowing in a circuit, that last electron goes back into the positive terminal of the battery. The continuous current flow (caused by the voltage delivered by the battery) is the result of electrons being bumped along from one atom to the the next in the circuit. The lower the value of circuit resistance, the higher the current flow.

As each electron flows out of the battery's negative terminal, the battery's capacity (and its voltage) is incrementally diminished just as a tank full of water would have its capacity (and pressure) incrementally diminished as water was allowed to flow out (drop by drop) of the bottom of the tank.

The reason that the voltage drop across the resistor is the same as the battery voltage is because the connecting circuit wires have, for all practical purposes (unless they are very, very long) no resistance. In the circuit described, the voltage drop across the resistor is a product of the resistance (in ohms) times the current flow (in amps), and has to be equal to the battery terminal voltage.

The amount of current flow the resistor allows will determine how long the battery will be able to push electrons through the circuit (just like the rate of flow out of the water tank will determine how long the every decreasing water pressure will cause water to flow out of the tank.