Physics domain analysis of Voltage drop in series resistor

[Muhammad Usman]

Hi all,

I apologize if its a very basic question but this is related to my quest to know the things deeply. And it is not homework help at all, it is just a general question.

The main question or objective to ask this question is to understand what is actually happening in a circuit with one resistor and how the voltage drop happened across resistor. I just need some help from expert like you to please read this and let me know if this is exactly what is happening in the circuit.

I generated a test circuit as mentioned below :-

In the above test circuit we have two resistors, one is 800 ohm and the other resistor is 1 ohm, basically one ohm resistor is connected to simulate the behaviour of load in the circuit. I have also labelled the circuit diagram with A1, B1, C1, D1 and E1 etc so it will be easy for me to reference the places with charge accumulations in the figure. The main objective of this analysis is to study how the voltage is dropped against a resistor.

As per my analysis when we turned on the battery the charge will start accumulating at point A1 and B1, since A1 is close to the positive terminal of the battery therefore more positive charges will start accumulating at A1 and more negative charges are accumulated at B1, there are some positive charges are also present at B1 but most of the charge accumulation is negative and I am also concerned about accumulated charge impact therefore only negative charges are drawn, similar case happened with R2 where charges are accumulated across resistor terminal in the same way as happened in R1.

The main analysis

Negative charge accumulation at B1 and C1 exerts coulomb force (which is repulsion for negative charges) to negative charges at D1, E1 and F1. Let's call this force F2. Force generated by the battery electric field is F1. This is a non-coulomb force and this non-coulomb force is actually the main driving force of the circuit.

The force experienced by the electron at E1 is equal to F1 - F2 and this will be the cause of potential drop at R2 and this potential drop is experienced by the load which is in this case R1.

 

[jbriggs444]

Have you learned about the Drude model?

 

[Drakkith]

Is charge accumulation in a circuit actually something that happens in this context? I thought that only occurred for capacitors.

The main objective of this analysis is to study how the voltage is dropped against a resistor.

There's no mystery here. Work is done by the charges as they pass through the resistor. The energy required to do this work comes from the circuit, ultimately going back to the voltage source(s), leading to a voltage drop across the resistor. This voltage drop is the expenditure of energy in the resistor. If 9 out of 10 available volts is dropped across the first resistor, then only 1 volt is available to be dropped across the rest of the circuit. The exact amount dropped across each component in the circuit will depend on each component's resistance and the voltage applied by the voltage source. At least in this simplified circuit where we are only talking about a battery and a couple of resistors in series.

 

[Muhammad Usman]

Is charge accumulation in a circuit actually something that happens in this context? I thought that only occurred for capacitors.There's no mystery here. Work is done by the charges as they pass through the resistor. The energy required to do this work comes from the circuit, ultimately going back to the voltage source(s), leading to a voltage drop across the resistor. This voltage drop is the expenditure of energy in the resistor. If 9 out of 10 available volts is dropped across the first resistor, then only 1 volt is available to be dropped across the rest of the circuit. The exact amount dropped across each component in the circuit will depend on each component's resistance and the voltage applied by the voltage source. At least in this simplified circuit where we are only talking about a battery and a couple of resistors in series.

I watched this video "" and also studied a book that do say charge accumulation happens at the edges of resistor

 

[Drakkith]

I don't remember discussing this in my EE class, but it has been a while, so maybe I just don't remember.

Still, his explanation doesn't really make sense to me. Why is there a separation of charges between the two resistors? Why don't the electrons spread out along the conductor between the resistors since there would be an attractive force between the positive charges at the end of one and the negative charges at the end of the other?

And he talks of the 'speed' of the current, saying electrons go really fast in the wire and much slower in the resistors. This isn't really how current flow works.

The force experienced by the electron at E1 is equal to F1 - F2 and this will be the cause of potential drop at R2 and this potential drop is experienced by the load which is in this case R1.

I don't agree with this. Voltage drop is caused by a loss of energy, specifically as heat if we're talking about a resistor. Simply moving a negative charge closer to other negative charges doesn't lose energy, it just changes it into potential energy. So something else must be causing the voltage drop in the resistor, not electrostatic repulsion.

I'm not going to say that absolutely no charge separation occurs, but I don't think it happens like you and the person in the video have drawn.

 

[Dale]

Is charge accumulation in a circuit actually something that happens in this context?

Yes, it does happen. My favorite reference on this topic is “A Semiquantitative treatment of surface charges in DC circuits”, Mueller, Am. J. Phys. 80 (9), 2012.

https://www.tu-braunschweig.de/inde...oken=2cc8a71e4fdbf159121c6b8ef8348952a2e0c197

Why is there a separation of charges between the two resistors? Why don't the electrons spread out along the conductor between the resistors since there would be an attractive force between the positive charges at the end of one and the negative charges at the end of the other?

In a DC circuit potential is always continuous, but the E field can be discontinuous. Anywhere there is a discontinuity you get a surface charge.

When current goes between two materials with different resistivity the current must be the same so by Ohm’s law the E field must be different. This discontinuity leads to a surface charge between the two materials.

Similarly, the field outside a wire points whatever direction it points. But inside the wire, by Ohm’s law it points along the wire. This is also a discontinuity which leads to surface charges on the outside of the conductor.

please read this and let me know if this is exactly what is happening in the circuit

Not exactly, but you have the general right idea. Read the paper I cited and apply their method. You will be able to get a much closer result. It still won’t be exact, but it will be substantially more accurate.

 

[sophiecentaur]

“A Semiquantitative treatment of surface charges in DC circuits”, Mueller, Am. J. Phys. 80 (9), 2012.

Smashing diagrams.

 

[Drakkith]

Yes, it does happen.

Well, my mistake then. Cheers.