Voltage in Circuits: Explaining Voltage Drops & Flow

[UMath1]

Charge carrier

 

[jbriggs444]

Charge carrier

Charge carrier density does not affect field strength. The cable as a whole is (to a good approximation) electrically neutral regardless of charge carrier density. The more negative charge carriers you have, the more electro-positive the substrate becomes.

 

[Dale]

So is the charge density in the wire before the resistor constant?

No, the charge density is higher on and near the surface of the wire compared to inside the wire. Also, if the wire has some resistance the surface charge density varies along the length of the wire.

And how is it possible for there to be a density gradient when the current must be constant?

The continuity equation is $\frac{\partial}{\partial t} \rho + \nabla \cdot j =0$. There can be a density gradient, $\nabla \rho \ne 0$, and a constant current, $\frac{\partial}{\partial t} j = 0$ without violating the continuity equation.

 

[Jyothish]

So is the charge density in the wire before the resistor constant? And how is it possible for there to be a density gradient when the current must be constant? Do the electrons in the upper part have a higher velocity? If so why is that?

I have attached the figure provided in the textbook, "Matter and Interactions"

Figure shows the interface between a copper conductor and a carbon resistor during transient.Electron flow is from right hand side to left hand side, given as green arrows.In the initial transient, before steady state is achieved, the electrons tend to pile up in the interface between highly conductive copper and carbon with higher resistance.Hence if you consider the carbon resistor during transient, there is an inward charge flow to the resistor (electron flow from copper to the right hand side interface ) which is greater than outward flow from the resistor (flow from left hand side of the carbon resistor to copper , as shown in figure).Hence the right hand side become more negative compared to left side of the resistor.In the steady state, however the excess charges will be only on surface, but progressively changing from negative to positive as we observe from right side to left side.This is, if there is only one resistor in the circuit across the battery.If more resistors are there ,the surface charge density variation is from more negative to less negative(or less positive to more positive) depending on the circuit arrangement.In any case, this establishes an electric field from left to right inside the resistor(shown as Ecarbon), which helps electrons to move from right to left overcoming the resistance

Now coming to your question, if the copper wires are super conductors, then the surface charge density is constant on the wires.Because the required electric field to maintain current flow is zero if there is no resistance.It is the variation in surface charge that produces electric field.In practical conductors, there will be progressive variation in the surface charge density which establishes the required electric field to overcome wire resistance.This surface charge is created during initial transient when the switch is closed and the steady state will reach in nano seconds.One more screenshot from "matter and Interactions" is attached for your reference.

In steady state, the force due to this electric field, exactly cancels the resistance offered by conductor/resistor(Similar to a force applied against friction when an object is moved at constant velocity).This makes the charges to move at a constant drift velocity, and hence constant current.

 

[Jyothish]

Regarding the first question, I am not talking about the length of the wire, rather the distance,r, from the positive and negative terminals at each point on the wire. This is what I think is used to calculate voltage in kq/r.

For the second question, I researched about permittivities and it turns out the air has among the lowest. So when the circuit is open, why can't the field still act on electrons far away?

Field definitely acts on a conductor which is far away.But if there is no closed circuit the charges in the far away conductor align and reaches equilibrium.This means the surface charges in the far away conductor are aligned and cancels out the electric field inside.However if the conductor is a part of closed circuit, system never reaches equilibrium as the charges can move freely around the circuit and the battery is continuously pushing the charges apart

Figure shows an open circuit and two metallic conductors(shown in grey color) away from the circuit separated by air.The charges inside the grey conductors align and reach equilibrium cancelling out internal electric field.

 

[UMath1]

If I understand correctly then it is the movement of an electron from an area of high surface charge density to an area of low surface charge density that causes a potential drop? If that is true though, how do electrons move across a wire with uniform surface charge density? Doesn't that mean there's no electric field? And why is the surface charge density uniform? Isn't there a greater concentration of electrons in the battery anode than there is in the wire?

 

[Jyothish]

If I understand correctly then it is the movement of an electron from an area of high surface charge density to an area of low surface charge density that causes a potential drop? If that is true though, how do electrons move across a wire with uniform surface charge density? Doesn't that mean there's no electric field? And why is the surface charge density uniform? Isn't there a greater concentration of electrons in the battery anode than there is in the wire?

Yes.It is the movement of electrons from an area of higher surface charge density(more negative) to lower surface charge density(less negative) causes potential drop

If the wire is superconductor, then the surface charge density is uniform even when current is flowing.This is analogous to a body moving in a friction less surface at a constant velocity.There is no force needed to maintain the motion.Force is needed only to start the motion.Similarly, an incrementally small amount of electric field during initial transient is enough to start current in superconducting wires.To maintain the current no electric field is needed.However, practical conductors have resistance and there is a surface charge density variation.The resulting electric field pushes the electrons against wire resistance

I don't think there is greater concentration in battery anode.Why do you think so?

 

[jbriggs444]

The charge density will naturally adjust itself so that the resulting electric field will result in a charge flow rate that is consistent through the length of the wire.

If this were not so -- if the electric field resulted in a higher flow rate upstream and a lower flow rate downstream, for instance, then you would get a charge build-up in between. That charge build-up would reduce the potential difference upstream and increase the potential difference downstream. For reasonable conductors, this reduces the current flow rate upstream and increases the current rate downstream so that the charges do not build up.

If an equilibrium is possible at all, it will result in a consistent current flow rate throughout the conductor.

 

[UMath1]

Because the anode is the source of all the charges, so there should be a higher concentration in it? And what about the wire on the cathode side? Does it have a uniform surface cation density? How can that be though? I thought only electrons move, not cations.

Another question I have is how do the electrons have energy to flow from the last resistor to the cathode. If I understand correctly, at the anode an electron has n elecron volts of energy. It loses all n electron volts as it goes through the resistors. Then how does it have enough energy to make it back to the anode?

 

[jbriggs444]

Because the anode is the source of all the charges, so there should be a higher concentration in it? And what about the wire on the cathode side? Does it have a uniform surface cation density? How can that be though? I thought only electrons move, not cations.

Another question I have is how do the electrons have energy to flow from the last resistor to the cathode. If I understand correctly, at the anode an electron has n elecron volts of energy. It loses all n electron volts as it goes through the resistors. Then how does it have enough energy to make it back to the anode?

I thought this one was asked and answered. What is the resistance of the wire between last resister and anode? What is the current between the last resistor and anode? What field strength is needed to drive that much current through that resistance?

 

[Jyothish]

Because the anode is the source of all the charges, so there should be a higher concentration in it? And what about the wire on the cathode side? Does it have a uniform surface cation density? How can that be though? I thought only electrons move, not cations.

Another question I have is how do the electrons have energy to flow from the last resistor to the cathode. If I understand correctly, at the anode an electron has n elecron volts of energy. It loses all n electron volts as it goes through the resistors. Then how does it have enough energy to make it back to the anode?

The surface charge variation depends on the shape of conductor.It also gets affected by the bends present in the conducting wire.In anycase, during steady state they align in such a way to produce the required electric field.If the conductor is cylindrical with uniform cross section and zero resistance(this is the assumption I made) , surface charge density will be uniform.The exact surface density of anode/cathode depends on the shape of anode/cathode and the alignment with respect to connecting wires etc.

After the last resistor, there is no voltage drop and there is no loss of energy when the electron comes back to anode

 

[UMath1]

But isn't energy/work required for the electron to travel the distance from the last resistor to the cathode? Work is F dx, so there has to be some energy needed for the electron to travel a distance dx.

 

[sophiecentaur]

But isn't energy/work required for the electron to travel the distance from the last resistor to the cathode? Work is F dx, so there has to be some energy needed for the electron to travel a distance dx.

You really don't wan to let this one go, do you?
The "F", in this case, is Zero. If you can't get on with the idea of ideal components then you have to assign the wire some very low value of resistance and the power dissipated will be correspondingly small. The "distance" involved in an ideal wire is of no consequence; it's a free ride, wherever the wire is routed and however long it is.

If you were presented with a problem about a bucket of water on a rope, would you instantly start to introduce the fact that some water would be evaporating from (or condensing onto) to bucket? I suspect you would quite happily go along with the assumption that there is no change in the amount of water. So why not accept that there is no change of Energy when the charges flow through an ideal wire? Whenever Maths is used in a problem, you have to make assumptions and approximations - if your sheet of paper is of finite area.

 

[UMath1]

There might be no change in energy, but then the electron had to have some energy after leaving the last resistor. But according to kirchhoffs law, the electron loses all of it energy after the last resistor, so it doesn't have any energy to move.

 

[jbriggs444]

There might be no change in energy, but then the electron had to have some energy after leaving the last resistor. But according to kirchhoffs law, the electron loses all of it energy after the last resistor, so it doesn't have any energy to move.

Kirchoff's law makes no mention of electrons.

 

[sophiecentaur]

so it doesn't have any energy to move.

Why does it need energy to move? That idea went out by Newton's time.
Perhaps your problem is that you want a completely classical model to satisfy your questions. If the 'electron' emerges from the other end with exactly the same KE as when it entered that wire, what energy would it lose? (I am briefly dipping into the Drude model here - which has been superceded)

 

[Jyothish]

There might be no change in energy, but then the electron had to have some energy after leaving the last resistor. But according to kirchhoffs law, the electron loses all of it energy after the last resistor, so it doesn't have any energy to move.

I will try to explain it in another way..In the transient state ,electrons acquire the required kinetic energy they need in the steady state.What battery does during steady state is to supply exactly the 'extra' potential energy needed for the electrons to overcome the resistance.This 'extra' potential energy is being used up in the resistance as heat.Again the electrons come out of resistor with the initial kinetic energy and goes back to battery without any loss.Kirchoffs law deals with steady state and it speaks about this 'extra' energy given by the battery per unit charge each time the charge goes around the circuit and the energy used up in various components like resistors

 

[UMath1]

I think I understand it better. But I am not sure exactly what you mean by steady and transient state. Is transient state when the you start the circuit and charges start moving and steady state when flow out of anode= flow in cathode?

 

[sophiecentaur]

I think I understand it better.

Have you accepted that it actually takes no energy (i.e. no loss of energy) for a charge to enter and exit a zero resistance? Remember, they don't need to 'rush' through the wire, how ever long it is. Charge in one end = charge out the other end. I think you are still thinking in terms of individual particles, rather than Charge.

 

[UMath1]

I know that energy does not need to be lost for a charge to enter and exit a zero resistance. However, I do think the charge must possesses energy to do so. To move, energy is required. The energy does not have to be lost.

For example, for a ball to move through a frictionless surface with no resistance of any kind, it must possesses energy although it does not lose any energy

 

[sophiecentaur]

I do think the charge must possesses energy

That would be Kinetic Energy? What Kinetic Energy does a Coulomb of Charge have?

 

[Jyothish]

I think I understand it better. But I am not sure exactly what you mean by steady and transient state. Is transient state when the you start the circuit and charges start moving and steady state when flow out of anode= flow in cathode?

Transient state is, as you said, when the battery is connected to the circuit through switch.Then the following events occur
1) Charges(electrons) start moving throughout the circuit
2) Due to charge separation caused by battery, some parts of the circuit start accumulating negative surface charge and other parts positive surface charge

Within nano seconds, steady state is reached where
1) There is no change in surface charge densities at various points.This means there is no further charge accumulation
2) Charges will reach a constant drift velocity at a given point in the circuit.But steady state velocities at different points in the circuit (means at different resistors in the same circuit) can vary depending on carrier density, cross section of resistor wire etc
3) The surface charge variation(spatial) in resistances establishes the exact electric field needed to overcome the resistance and to maintain the steady state drift velocity

Assuming the electrons to be particles with mass, the kinetic energy needed for the electrons is very small owing to their small mass.This KE is acquired during transient state.Now in the steady state, as they move around the circuit, you can see that , inside the battery, electrons have to move against the opposing force due to electric field.Because they have to move from positive plate to negative plate inside the battery.This work is done by battery and the resulting extra potential energy is added to electrons.This energy is utilized in resistances.Kirchoff's law implies that the energy supplied is equal to what is used up, in steady state

 

[Dale]

However, I do think the charge must possesses energy to do so. To move, energy is required.

Not potential energy.

 

[Mister T]

There might be no change in energy, but then the electron had to have some energy after leaving the last resistor. But according to kirchhoffs law, the electron loses all of it energy after the last resistor, so it doesn't have any energy to move.

In that model the assumption is that the wire has zero resistance. In fact, what is happening is that the wire has a resistance that is much much smaller than the resistance of that last resistor. Or any other resistor in the circuit. It is therefore safe to call it zero for purposes of Kirchhoff Law calculations, but it is not safe to consider it zero for purposes of understanding how the electrons move through the wire.

The only way an electron can move through a wire is if one end is at a higher potential than the other.

It's like a mechanics problem where a taught string connects two object that are accelerating. The usual approximation is to assume the string is massless, allowing one to calculate things like the acceleration of the objects or the tension in the connecting string. But using this model you can't analyze the acceleration of the string. The string's mass is very small compared to the mass of the objects, so it's safe to ignore for purposes of calculating their acceleration. But it's not safe to ignore for purposes of understanding what happens to the string. The only way it can accelerate is if the force applied to one end is larger than the force applied to the other end. Thus we have to realize that the tension along the string decreases from one end to the other. And when we subtract the force at one end from the force at the other, we get a net force. Divide that by the string's mass and you have the string's acceleration.

In a wire the conduction electrons are transferring kinetic energy to the wire's atoms, causing an increase in the wire's temperature. The electrons are speeding up due the electric field applied by the battery, and slowing down due their interaction with the atoms. Average it out and you get a drift velocity that's responsible for the reading on the ammeter.

 

[sophiecentaur]

The situation with the wire is comparable in that the conduction electrons are transferring kinetic energy to the wire's atoms, causing an increase in the wire's temperature.

This is referring to the Drude model (?), which has been superceded by a more universal Quantum model. Drude cannot cope with superconductivity, afaiaa and we are treating the connecting wires as having zero conductivity.

 

[Mister T]

I'm talking about the wires in the circuits encountered in an introductory physics textbook, classroom, and laboratory. I took that to be the context in which the OP is conversing.

I am not talking about superconductors or semiconductors. You don't need a superconducting wire to verify Kirchhoff's Laws in the laboratory. You merely arrange things so that the wires have negligible resistance.

 

[Dale]

@Mister T and @sophiecentaur it doesn't really matter whether you are using the Drude model or QM. UMath1 is mistaken about energy regardless of the underlying model of conductivity and even regardless of the nature of the charge carriers. The voltage is a measure of the potential energy, and it simply does NOT require any potential energy to move. His idea is wrong even in mechanics where you could easily envision large mechanical systems where potential energy does not change even as large massive objects move from place to place.

 

[sophiecentaur]

I am not talking about superconductors

Is there a difference between zero conductivity, 'ideal' connecting wires and superconductors? The drude model would have to treat the connecting wires as having no collisions and hence would have no voltage drop.
But, if you want to ignore super and semi conductors then you would presumably have to limit any description to circuit behaviour to exclusively resistive components. That could be a bit limiting. You would need a separate description for what happens to electrons, depending which component they happen to be flowing through.
I agree that it is nice to discuss 'School Physics' and it is useful as a step towards better understanding. I just think that it is not necessarily helpful to try to 'explain' what is not really explicable within the realm of School Science. For instance, I have read frequent 'explanations' of the effect of temperature on resistance in terms of atoms jiggling about and providing larger targets for electrons to collide with. That is clearly nonsense and far too simplistic.

 

[sophiecentaur]

you could easily envision large mechanical systems where potential energy does not change even as large massive objects move from place to place.

Absolutely. I already made that point but he seems to have a problem about where and when the Energy is relevant.

 

[Mister T]

Is there a difference between zero conductivity, 'ideal' connecting wires and superconductors?

Yes. Although it's zero resistance, not zero conductivity. Ideal connecting wires have negligible resistance, not zero resistance. There's a difference. Even if the phenomenon of superconductivity had never been observed or modeled, the "zero-resistance" wires used in the first-approximation circuit models appearing in textbooks and in the literature would be alive and well. Moreover, those models can and are used by scientists and engineers to study, design, and build stuff; where applicable.

Just as there is no such thing as a massless string. They are still used as viable models in both education and industry, where applicable. It's better to use "negligible" rather than "zero" to avoid this very type of confusion.

 

[Mister T]

UMath1's point is simply wrong regardless of the underlying model of conductivity and even regardless of the nature of the charge carriers. The voltage is a measure of the potential energy, and it simply does NOT require any potential energy to move. His idea is wrong even in mechanics where you could easily envision large mechanical systems where potential energy does not change even as large massive objects move from place to place.

I understand the validity of your point. Perhaps you could help me understand its relevance.

When moving macroscopically large objects that are interacting with other objects, a force is required. If a potential energy function doesn't exist for that force, then the application of that force causes no change in potential energy. But of course if a potential energy function does exist for that force, there will be a change in potential energy.

In the case of moving (microscopically small) charge-carriers through a wire, a force is required because those charge-carriers are interacting with the wire. In the OP's case; as is typical of the case where the connection of batteries, bulbs, and wires are studied; that force is the conservative electrostatic force. A corresponding potential energy function does exist. So the only relevant way to move the charge-carriers is through a difference in potential energy. Or in other words, a difference in electric potential.

 

[Dale]

I understand the validity of your point. Perhaps you could help me understand its relevance.

The point is that UMath1 has a conceptual error about potential and kinetic energy here:

But according to kirchhoffs law, the electron loses all of it energy after the last resistor, so it doesn't have any energy to move.

In other words, the problem isn't that UMath1 is confused about whether a wire has negligible or zero resistance or superconductivity, but rather that he has a mistaken belief that an object which has zero potential energy also has zero kinetic energy. His stated belief is that Kirchoff's law says that the electron will get to the end of the last resistor and then stop because it runs out of energy. He seems to be confounding having zero potential energy ("loses all of its energy") with having zero kinetic energy ("doesn't have any energy to move").

This mistaken concept of energy is followed up in his subsequent posts.

 

[sophiecentaur]

Sorry about the zero conductivity gaff. What a plonker.

 

[Mister T]

Sorry about the zero conductivity gaff. What a plonker.

Let's speak of it no mho.

 

[UMath1]

The only reason I have this confusion is because initially in the anode the electrons have only potential energy, if all this potential energy is used up, how can there possibly be any kinetic energy? In other words, any kinetic energy must originate from the potential energy and if all the potential is used up none can be converted. Total Energy in the beginning was eV. In the end it was 0.