Voltage in Circuits: Explaining Voltage Drops & Flow

[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