What about the electric field in a circuit with resistors?

[Gabriel Maia]

Hi. I am having some trouble understanding what is the voltage drop in a system with resistors in series.

If there is a difference of electric potential between two points in space, since electric potential is electric potential energy per charge, there is a difference in the electric potential energy between both points, which is to say that there is an electric field there. This electric field will make charges move from one point to the other, right?

Now, if we have a system with a battery and one single resistor, the voltage between both ends of the resistor will be the same provided by the battery. If we add one more resistor, in series with the first one, the voltage between the ends of the series will be the same as for the battery, but the voltage between the endpoints of each individual resistor will be different. Why does the voltage drop little by little from one resistor to the next? This implies that the electric field in the resistors is less strong. What is making it so?

Thank you very much.

 

[Anindya Mondal]

Power loss occurs as we move away from the + ve
to the negative terminals of the battery through a wire.

 

[Gabriel Maia]

Power loss occurs as we move away from the + ve
to the negative terminals of the battery through a wire.

That does not explain the question about the resistor in series vs a single resistor.

 

[Chandra Prayaga]

It is not true that if the voltage drops, the electric field drops. Think of the resistor as a long wire of length L, with uniform cross section area A, and made of the same material throughout, so the resistivity ρ is the same throughout. The voltage drop across the wire is equal to the battery potential, V_B. The resistance of the wire is R, and the current through the wire is I = V_B / R. Now the same current is passing through any cross section of the wire, because of charge conservation. You should also know the relation between resistance and resistivity in this case:

R = ρL / A

Now look at a part of the wire, with say, half the total length. Since the resistivity of the material is the same throughout, the resistance of this portion of the wire is

R' = ρ (L/2) / A = R / 2

The potential drop across this part is

ΔV = I R' = V_B / 2.

If you think of 1/3 of the length of the wire, the potential drop across this will be V_B / 3. That is why the potential reduces in steps as you walk around a circuit. If you have two resistors, think of them as two long wires of the same material joined (welded) together, each with a length proportional to the resistance, and the same argument follows.

 

[Gabriel Maia]

First of all, thank you very much for taking some of your time to help me.

Secondly, I understand the mathematics behind the voltage drop, but the physics underlying it is giving me trouble. I will separate my thought process into blocks and you tell me if anyone of the blocks is wrong, ok?

- Voltage is the difference in electric potential between two points, right? So, if I increase the voltage between two points the work done on the passing particles will increase as well.

- Consequently, the particles will acquire a higher speed at point 2 if the voltage between point 2 and 1 is greater. This implies a greater current.

- When I have a circuit with one resistor the voltage between its endpoints is the same as the voltage between the endpoints of the battery.

- If I add another resistor to the circuit, in series with the first one, the voltage between the endpoints of each resistor will be smaller than the voltage between the endpoints of the battery, but the voltage between the endpoints of the whole association of resistors will be the same as for the battery.

- So, if the voltage between the endpoints of each resistor is smaller than it was before I added the second resistor to the circuit it means that the work done on the electrons as they travel through the resistors is smaller than before.

- Why is that? Thinking of it I can only imagine that the voltage is giving you the effective work that is being done on the electrons. I mean, there is the work of the electric field, but you have to consider that some energy is lost as the electrons bump into the resistor's atoms, so the voltage is giving you information about the net work done on the particles. Does it make sense?

 

[CWatters]

When you add another resistor that increases the total resistance and reduces the current.

 

[Gabriel Maia]

When you add another resistor that increases the total resistance and reduces the current.

Yes, know. But as I said, I'm having trouble with the physics underlying the mathematics and this is just description of the formula. It explains how. Not why. And my question is about why two resistors in series have, individually, a voltage lesser than the one provided by the battery. What causes this drop? Is it the relative distance between charges that changes? What?

 

[CWatters]

Your last paragraph in #5 correct. See first paragraph here..

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

 

[ZapperZ]

Yes, know. But as I said, I'm having trouble with the physics underlying the mathematics and this is just description of the formula. It explains how. Not why. And my question is about why two resistors in series have, individually, a voltage lesser than the one provided by the battery. What causes this drop? Is it the relative distance between charges that changes? What?

If I'm reading you correctly, you are asking for the microscopic explanation for the origin of electrical resistance.

To know that, one has to start with something called the Drude Model. Here, the charge carriers, in this case, conduction electrons, are considered as a "classical gas". What this means is that they do two things: they respond to an external field, and (ii) they do a lot of bumping around. They bump around with the lattice ions (i.e. the atoms that make up the resistor), they bump into any impurities and defect in the bulk material, and they also bump into one another. These collisions are the origin of electrical resistivity. At higher temperatures, the lattice ions vibrate even more, and causes more collisions with these electrons. This is why the resistivity of a metal increases at higher temperatures.

Because of the existence of non-zero resistivity, by Ohm's Law, if R≠0, then for a set current, there is a potential drop across that resistance, i.e. this isn't an electrical short.

Zz.

 

[gleem]

Consider your resistor to be a long high resistance wire of constant diameter and composition i.e. uniform. Would it not be expected that the decrease in potential at a point from the high potential end be proportional to the length of the segment from the high potential end compared to its total length? Thus if the total length is X then at a point of Y% of X the potential should be y% of the potential difference across the wire. Yes? There is no reason to expect otherwise. Right? We should expect that the electric field to be constant implying the decrease in potential is constant. Thus your two resistor can be thought of as different lengths of a uniform high resistance wire.

 

[Dale]

First of all, thank you very much for taking some of your time to help me.

Secondly, I understand the mathematics behind the voltage drop, but the physics underlying it is giving me trouble. I will separate my thought process into blocks and you tell me if anyone of the blocks is wrong, ok?

- Voltage is the difference in electric potential between two points, right? So, if I increase the voltage between two points the work done on the passing particles will increase as well.

- Consequently, the particles will acquire a higher speed at point 2 if the voltage between point 2 and 1 is greater. This implies a greater current.

- When I have a circuit with one resistor the voltage between its endpoints is the same as the voltage between the endpoints of the battery.

- If I add another resistor to the circuit, in series with the first one, the voltage between the endpoints of each resistor will be smaller than the voltage between the endpoints of the battery, but the voltage between the endpoints of the whole association of resistors will be the same as for the battery.

- So, if the voltage between the endpoints of each resistor is smaller than it was before I added the second resistor to the circuit it means that the work done on the electrons as they travel through the resistors is smaller than before.

- Why is that? Thinking of it I can only imagine that the voltage is giving you the effective work that is being done on the electrons. I mean, there is the work of the electric field, but you have to consider that some energy is lost as the electrons bump into the resistor's atoms, so the voltage is giving you information about the net work done on the particles. Does it make sense?

I think you are way over complicating this. In circuit theory you only care about two things: voltage and current.

Is it really so hard to believe that there just happen to be some circuit elements where the current is proportional to the voltage?

 

[cnh1995]

@Gabriel Maia: As Dale said, circuit analysis is all about the v-i relationships of the components. For a resistor, V=IR and that's all you need to know. See if this insights article is any help.
https://www.physicsforums.com/insights/circuit-analysis-assumptions/.

If you want to dig deeper, google 'surface charges in circuits'. That will answer your questions about the voltage drop across and electric field inside a resistor.

 

[Anindya Mondal]

First of all, thank you very much for taking some of your time to help me.

Secondly, I understand the mathematics behind the voltage drop, but the physics underlying it is giving me trouble. I will separate my thought process into blocks and you tell me if anyone of the blocks is wrong, ok?

- Voltage is the difference in electric potential between two points, right? So, if I increase the voltage between two points the work done on the passing particles will increase as well.

- Consequently, the particles will acquire a higher speed at point 2 if the voltage between point 2 and 1 is greater. This implies a greater current.

- When I have a circuit with one resistor the voltage between its endpoints is the same as the voltage between the endpoints of the battery.

- If I add another resistor to the circuit, in series with the first one, the voltage between the endpoints of each resistor will be smaller than the voltage between the endpoints of the battery, but the voltage between the endpoints of the whole association of resistors will be the same as for the battery.

- So, if the voltage between the endpoints of each resistor is smaller than it was before I added the second resistor to the circuit it means that the work done on the electrons as they travel through the resistors is smaller than before.

- Why is that? Thinking of it I can only imagine that the voltage is giving you the effective work that is being done on the electrons. I mean, there is the work of the electric field, but you have to consider that some energy is lost as the electrons bump into the resistor's atoms, so the voltage is giving you information about the net work done on the particles. Does it make sense?

Greater velocity of electrons doesn't mean greater current flow!