Voltage in Circuits: Explaining Voltage Drops & Flow

[UMath1]

Sorry but I can't change it to basic level at this point and I am still confused. So if I understand correctly, to calculate the total electric potential you must sum the effects of each individual charge in the wire and this would be affected by the charge density. So the charge density in the resistor is different from that in the wire, and so the change in potential is different?

I don't see the clear connection between charge density and voltage though.

 

[Dale]

Sorry but I can't change it to basic level at this point and I am still confused.

I don't think that you will be able to get un-confused yet. I think that you are asking a question that you are not yet ready to learn the answer. That is causing your confusion. In my opinion you need to take a step back and fill in some gaps first. We can answer your questions, and have already done so, but understanding requires asking the right questions too.

So if I understand correctly, to calculate the total electric potential you must sum the effects of each individual charge in the wire and this would be affected by the charge density.

Yes, to calculate the potential from first principles requires knowing the exact charge density throughout the circuit. Alternatively you can just use Kirchoff's voltage and current laws. You get the same answer with a lot less effort.

So the charge density in the resistor is different from that in the wire, and so the change in potential is different?

Yes. The charge density in the entire circuit is different with vs without the resistor.

I don't see the clear connection between charge density and voltage though.

That connection is given by the first equation in the section "origin from retarded potentials" that I pointed to earlier. The one that shows exactly how to calculate $\varphi$ given $\rho$

 

[Dale]

Can I ask what your background and goals are regarding physics and math? Looking at your questions I don't see a clear pattern.

 

[UMath1]

I am currently in AP Calc BC and AP Physics 2. I have completed AP Physics 1 and a self-study of AP Physics C: Mech.

I just want to gain a better understanding of the concepts we are covering in class. Right now I feel as if I can apply a formulas and get the right answers but I don't really feel strong on the concepts. I want to have a unified understanding of voltage in electrostatics and circuits because right now the two seem to be two different ideas.

 

[Dale]

I am currently in AP Calc BC and AP Physics 2. I have completed AP Physics 1 and a self-study of AP Physics C: Mech.

Are the AP physics courses calculus-based or algebra based?

 

[UMath1]

AP Physics 1 & 2 are algebra based. But last year I self studied AP Physics C: Mechanics which was calc-based. This year I am also self-studying AP Physics C: Electricity and Magnetism which is calc-based.

 

[Dale]

OK, so one general point. You are looking to understand unified concepts. That generally requires a mathematical framework which unifies the different concepts. So, when you are studying some physics using algebra you are going to find that a lot of things which you hope would be unified cannot be unified strictly algebraically. Have patience, as you develop the mathematical framework necessary you will revisit those concepts and see how they are unified. Until you do so there is not much that can be done.

For classical EM, the unifying framework is Maxwell's equations:
https://en.wikipedia.org/wiki/Maxwell's_equations#Conventional_formulation_in_SI_units
which can be expressed in terms of potentials as
https://en.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field#Lorenz_gauge
which is heavily based on vector calculus.

The concepts are unified, but you are probably several semesters of calculus away from being able to follow it. Until then all I can do is to urge patience. You are quite far ahead of where you need to be already, so don't feel upset that you cannot do everything today. Give yourself time to learn in the proper order and time.

 

[AlejandroDes]

I also feel stuck on how voltage is POTENTIAL energy per unit charge. How can friction play a role on potential energy which is based soley only location? If it was losing kinetic energy across the resistor I would understand because friction and resistance slow things down.

I think you're getting confused because you are trying to compare the voltage (electric potencial) with the energy of a mass at certain height (potencial energy). They are not the same thing (the units are not the same). Imagine you have a mass at a certain height the Potencial energy would be given by PE = m*g*h, now divide this PE by the height and you'll get PE/h = m*g [J/m] (Weigth). Now you get a usefull quantity that will offer you information on how the mass will behave under any heigth variation.

We can do the same for the electric problem. Imagine we have a bunch of charged particles (negative and positive) and we take the time and work to separate all the positive particles from the negative particles (you'll get escentially a battery). Now, that we invested some work (energy) into the separation from it's natural state (like putting work into lifting something) we have stored potencial energy in this "battery". Let's say that this Energy is proportional to the amount of particles we had: PE = k*Q. If we divide this PE by the amount of charge we'll get PE/Q = k [J/Co] (Voltage), so k = V.

Now for the resistor part, resistors do not cause a voltage drop by themselves. What causes the voltage drop is given by a current flowing through a resistor. Also don't take into account kinetic energy. Electrons in a wire do not move at the speed of ligth that would be impossible. In electricity we are transfering energy from one place to another but it doesn't mean that the electron(charge) has to cover the distance. Is way easier to just pass "energy" to the next electron and so on. Now say that in a wire you have a cross section with a finithe amount of electrons (I would use 100). So you have 100 electrons passing energy to the next 100 which will then pass it to the next 100 and so on. Suddenly, you get to a part where the system isn't closed anymore (electrons can "lose" some of the energy while passing it) so each time an electron passes the energy to the next one, some of it gets out and you end up with a lesser amount of energy. Less energy is equal to less voltage. So the voltage drop is given by the amount of energy that the electrons lost.

I hope this helped.

 

[Jesra]

I feel that I don't understand how voltage works in a circuit. I understand voltage to be electric potential energy per unit charge (kq/r). In the case of a circuit, electrons flow from low potential to high potential. But I don't understand how resistors cause a voltage drop. Isn't voltage based on position? How can the resistors cause a drop in potential energy? I can understand the resistor causing a drop in kinetic energy slowing the electrons down, but how does it lower the voltage? And if the total voltage is equal to the sum of the voltages of the resistors, then the voltage difference in wire after the last resistor and the positive terminal would be zero, right? Then how would the electrons be able to flow back to the positive terminal? Wouldn't they just stop?

The best analogy I can think of is a river flowing downhill that turns a turbine. But in this case the turbine doesn't cause a drop in the water's potential energy. It only takes some of the water's kinetic energy.

My understanding is voltage does not change with resistance, but amperes do. Voltage is an assist to predict what distance an amplitude will travel through x ohm resistance.

''
Ohm's law states that the current through a conductor between two points is directly proportional to the potential difference across the two points. Introducing the constant of proportionality, the resistance,[1] one arrives at the usual mathematical equation that describes this relationship:[2]

where I is the current through the conductor in units of amperes, V is the potential difference measured across the conductor in units of volts, and R is the resistance of the conductor in units of ohms. More specifically, Ohm's law states that the R in this relation is constant, independent of the current.[3]''

https://en.wikipedia.org/wiki/Ohm's_law

So, doing variable relation conversions:

amperes=voltage/resistance, resistance (amperes)=volts, resistance=volts/amperes

Electrons flow from negative potential to positive potential, but we need to keep in mind that the interpretation of electron flow is temporally dependent (e.g. how can an alternating current have direction when the 'current' is both positive and negative in an oscillating relation (but that relation is with the return line to its generator). The direction of an electron flow is as indicative to voltage polarity as is the ampere difference at two points in the flow is to resistance. Different voltage generating systems may have different rules, such as battery voltage variations and power line voltage variations. e.g. I have measured 'dead' batteries with voltages only a volt less than live batteries and I am reasonably certain two dead batteries in series add up to one dead battery at 24 volts. :-)

 

[Chandra Prayaga]

See but then shouodn't voltage drop be uniform because it is only due to change in position. I dom't see why it is more across resistors and almost zero in the wire. Can you show me how this would work with kq/r?

If you can explain the potential energy graph on this page :
https://community.dur.ac.uk/p.m.johnson/electric_circuits/11_bulb_circuit.htm that would be very helpful.

I also don't understand why the circuit must be closed for the field to be set up. My textbook says that the electrons do not rush to the lightbulb as soon as the circuit is closed, but instead the field acts on the electrons near the lightbulb. But why can't the field act even if the circuit is open?

There are two questions here. I will address the first one first. Each resistor in your circuit is the equivalent of several miles of a conducting wire. You can easily calculate how long a wire you will need to replace a 1 ohm resistor. If you did that, you will see that the potential drop is indeed uniform across that very long wire. You are effectively taking a very very long wire, packaging it into a tiny wound bundle, and calling it a resistor. This is a model. Real resistors are made differently, but the model is pretty good.

Now the second one:

If you don't close the circuit, the electron no longer has a low resistance path to go through. The air gap has a very high resistance. If you increase the potential difference to very high values, you can even cause a breakdown through the air gap.

 

[davenn]

My understanding is voltage does not change with resistance, but amperes do. Voltage is an assist to predict what distance an amplitude will travel through x ohm resistance.

that's incorrect, voltage and current change, consider this series resistor circuit ...

The total resistance sets the current through the circuit and individual resistances drop the voltage
as determined by the resistance value and the current through it

you ALWAYS get a voltage drop across a resistor
as an exercise, you can work out the resistor values Dave

 

[sophiecentaur]

Ohm's law states that the current through a conductor between two points is directly proportional to the potential difference across the two points. Introducing the constant of proportionality, the resistance,[1] one arrives at the usual mathematical equation that describes this relationship:[2]

Ohm's Law does not say that. Go and read a formal statement of the Law (you can find a reference as easily as I) and you will notice that it refers only to metals at constant temperature. Just because the manufacturers go to the trouble of providing us with resistors 'out of the drawer' that exhibit a constant resistance over a large range of temperatures, doesn't change the true meaning of the 'Law'.
Resistance is just a ratio which applies sometimes. It is no more fundamental than velocity or how many buns I can eat per hour and it proves nothing, one way or another.
Resistance is a very useful derived quantity, of course, and we use it every day but it's only because we make circuits out of mainly metals, that the terms is used in such a cavalier way. R (the ratio) can very easily vary with time or with current so you just can't rely on it to be a fixed value - if you are trying to get a fundamental grasp of EE. Isn't that what this thread is trying to do?

PS What is the resistance of a diode?

 

[cjl]

Ohm's Law does not say that. Go and read a formal statement of the Law (you can find a reference as easily as I) and you will notice that it refers only to metals at constant temperature. Just because the manufacturers go to the trouble of providing us with resistors 'out of the drawer' that exhibit a constant resistance over a large range of temperatures, doesn't change the true meaning of the 'Law'.
Resistance is just a ratio which applies sometimes. It is no more fundamental than velocity or how many buns I can eat per hour and it proves nothing, one way or another.
Resistance is a very useful derived quantity, of course, and we use it every day but it's only because we make circuits out of mainly metals, that the terms is used in such a cavalier way. R (the ratio) can very easily vary with time or with current so you just can't rely on it to be a fixed value - if you are trying to get a fundamental grasp of EE. Isn't that what this thread is trying to do?

PS What is the resistance of a diode?

A diode is not a conductor. It is a device based on either a semiconductor or a vacuum tube. A formal statement of Ohm's Law should not say anything about metals either - there are plenty of non-metallic conductors.

I see nothing wrong with that definition at all - resistance is indeed the proportionality constant between voltage and current, and current through a conductor is indeed proportional to the potential drop across it.

 

[Jyothish]

Let me share my understanding about how electric field works in circuits ,what a voltage source does and what happens inside resistor as well as conducting wires.
Consider the 9V battery as shown in the circuit diagram.Assume that the circuit is open.The function of battery/dynamo or any voltage source is separation of positive and negative charges.In a dynamo this task is accomplished by lorenz force in a moving conductor in magnetic field.In a battery it is through 'chemical forces'.Whatever be the charge separation mechanism, the voltage source separates negative and positive charges and pushes them to opposite terminals.So in the example shown, the 9V battery pushes negatively charged electrons from top to the bottom terminal through inside of the battery

Let Fb, be the force with which battery pushes electrons to the bottom terminal.As the negative charge starts accumulating in the bottom terminal and positive charge on top terminal, it will become increasingly difficult for battery to further push electrons from top to bottom terminal, as top terminal attracts electrons back and bottom terminal repels electrons away.Let, Fc be this opposing force due to accumulated charges that makes it difficult for the battery to push the charges.As battery continues to push charges the opposing force Fc gets increasing and will eventually balance Fb.Thus the charge accumulation stops when Fc becomes equal to Fb.For a 9V battery the force Fb is in such magnitude that , equilibrium is reached when the potential difference between terminals become 9V due to accumulated charges. This explains why there is no current conduction in open circuit, although there exists electric field or potential difference between terminals.Even very small charge accumulation can balance the force exerted by voltage source/battery.But in circuit analysis,for small frequencies, this momentary charge accumulation is neglected as its value is so small.Or in other words, the capacitance of the terminals is negligible(charge required to create unit potential difference) .However in high frequency circuits, the charges have to adjust and readjust so quickly that the charging current is no more negligible(I=dq/dt).Same is the case of long distance power transmission lines where we have to consider the charging current/capacitance of the line even during open circuitNow let us see what happens in a closed circuit.

Let us assume the case when the conductors are ideal with zero resistance.

1) When the circuit is closed by a switch , the electrons in the external circuit are pushed away by the negative terminal and attracted by positive terminal. The electrons near the negative terminal pushes the section of electrons adjascent to it in the external circuit.This section of electrons will in turn pushes the next section of electrons adjacent to it. This way, charge carriers start moving in the entire circuit. Here the action is similar to water flowing in a pipe, though the mechanism is due to electromagnetic force. Hence the positive charge tend to get ‘diluted’ in upper positive plate of the battery and negative charge tends to get ‘diluted’ in the negative plate of the battery.This will tend to decrease the force Fc exerted by accumulated charges and the equilibrium between Fb and Fc is affected.Hence, Fb will again manage to push the electrons towards negative terminal and this process continues.If the resistance R =0,or short circuit there is no opposing force for the acceleration of charges created by battery(Fb).Or Fc=0 as there is no accumulation of charges. Then the charge carriers has to reach infinite speed, theoretically(if this is permitted).Hence there is infinite current in case of short circuit!

2) If the conductors are ideal, the inside part of the conductor will be still neutral even though charges are moving.This is because there is no opposition to the motion of charges and hence no charge accumulation anywhere inside the conductor.There is no charge density gradient inside the conductor. However there will be surface charges as shown in the figure which will be uniform.This is positive in upper conductor and negative in the lower conductor.Actually this surface charges has the role of facilitating energy transfer, as they create the required electric field in the external circuit

3) There is no electric field inside the ideal conductor even while it is conducting.You may ask, then how charges are moving.The ‘gentle push’ from neighbouring electrons is enough for these charge carriers to move.This is because there is no resistance to their motion.Here the situation is similar to water flowing at a constant velocity in an ideal frictionless pipe.There is no pressure difference between two points and there is no accumulation of water.The net force in any given section of water is zero
Now let us look into what happens inside finite resistance(say 10 ohms) as given in the diagram.1) Since there is an opposition to the flow of charges in a resistor there will be an higher density of electrons in the lower part of the resistor and the upper part is deprived of electrons.This is due to charge accumulation resulting from the opposition to smooth flow of electrons.Or in other words there is a charge density gradient both on surface and inside the resistor. This makes lower end of the resistor negatively charged(more electrons) and upper end positively charged.Charge density progressively changes from negative to zero and then increases from zero to positive value as we observe from lower end of the resistance to upper end

There will be an electric field inside the resistor due to this charge density difference, as shown as E in the figure.This electric field helps the charges to overcome the resistance and move across the resistor.The work done against resistance will be generated as heat.The potential difference due to this electric field is known as voltage dropAs you can see, the battery does a work against the electric field created due to separation of charges.While in a resistor, electric field does the work against the resistance.The potential energy given to the electrons , by the battery is creating the charge density variation/electric field in the resistance.And the energy is ‘dropped’ by working against resistance.Voltage created by battery is the work done/unit charge against the electric field and voltage drop is the work done per unit charge by electric field against resistance

 

[sophiecentaur]

A diode is not a conductor. It is a device based on either a semiconductor or a vacuum tube. A formal statement of Ohm's Law should not say anything about metals either - there are plenty of non-metallic conductors.

I see nothing wrong with that definition at all - resistance is indeed the proportionality constant between voltage and current, and current through a conductor is indeed proportional to the potential drop across it.

If a diode is not a conductor then it is of no use in an electric circuit. When forward biased, it conducts very well.
There is no point in saying how Ohm's Law should be stated. Ohm got there long before you did and the law (description of behaviour) applies very well to metals (which is what he was discussing in the first place). If you want to be re-writing things like Ohm's Law then you would be setting yourself a massive task of adjusting all the rest of EE, just to keep everything consistent. The fact is that, with the exception of the sort of circuits that elementary EE students are set as problems, there are so many examples of non-Ohmic conductors in circuits that you can never get away with assuming that you apply Ohms Law to any black box component you might find. It would be complete nonsense.
I really don't know what you are arguing about, apart from trying to maintain that you are not wrong about this detail. (Actually, more than just a detail). If you can't understand why all components can't be treated according to Ohm's Law then you just need to look at a simple diode rectifier circuit and work out how it responds to a fixed AC voltage supply and a varying load or even what happens to a filament light bulb when the supply volts drop a bit.

 

[UMath1]

There are two questions here. I will address the first one first. Each resistor in your circuit is the equivalent of several miles of a conducting wire. You can easily calculate how long a wire you will need to replace a 1 ohm resistor. If you did that, you will see that the potential drop is indeed uniform across that very long wire. You are effectively taking a very very long wire, packaging it into a tiny wound bundle, and calling it a resistor. This is a model. Real resistors are made differently, but the model is pretty good.

Now the second one:

If you don't close the circuit, the electron no longer has a low resistance path to go through. The air gap has a very high resistance. If you increase the potential difference to very high values, you can even cause a breakdown through the air gap.

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?

 

[Chandra Prayaga]

kq/r is the potential of a point charge. The potential in a circuit does not follow that formula.

I am not sure what you have in mind about the second part. Certainly the electric field acts on every charge that is there, including the electrons and protons in all the air molecules around. But they are all bound together inside the atoms with much stronger forces than you are exerting with a battery.

 

[Dale]

This is what I think is used to calculate voltage in kq/r.

Again, this is not the general formula. It only works for the special case of a single static point charge. The general equation is the first of Jefimenko's equations which I already pointed you to earlier. I thought that you had already gotten past that.

 

[UMath1]

Now let us look into what happens inside finite resistance(say 10 ohms) as given in the diagram.1) Since there is an opposition to the flow of charges in a resistor there will be an higher density of electrons in the lower part of the resistor and the upper part is deprived of electrons.This is due to charge accumulation resulting from the opposition to smooth flow of electrons.Or in other words there is a charge density gradient both on surface and inside the resistor. This makes lower end of the resistor negatively charged(more electrons) and upper end positively charged.Charge density progressively changes from negative to zero and then increases from zero to positive value as we observe from lower end of the resistance to upper endView attachment 90153

There will be an electric field inside the resistor due to this charge density difference, as shown as E in the figure.This electric field helps the charges to overcome the resistance and move across the resistor.The work done against resistance will be generated as heat.The potential difference due to this electric field is known as voltage drop

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?

 

[jbriggs444]

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?

Charge density or charge carrier density?

 

[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.

 

[Herman Trivilino]

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.

 

[Herman Trivilino]

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.

 

[Herman Trivilino]

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.