Derivation of the potential difference across a resistor

Question

I am trying to understand how the voltage drop across a resistor comes about in a DC circuit. I have no concerns over the relation described by Ohm's law: the current through a resistor will be proportional to the potential difference applied to it. However, I am confused about how the voltage drop across a resistor takes effect in the context of it being placed in a circuit, as I have not been able to find a derivation of this result that both conceptually and mathematically makes sense.

I have read about a derivation that uses Ohm's law to equate the work done on a charge carrier by the battery and the thermal energy that is lost by that charge carrier in going through a resistance. However, after this derivation I am still confused about the potential difference that is set up on the resistance. The reason for my confusion is because I don't understand how thermal energy and potential energy can be equated to each other.

I have read about surface charges, and it makes sense to me for the potential difference to be due to surface charges, yet I haven't a found a direct derivation of the potential difference from the surface charges.

Edit: Here is my derivation. I have doubts about it. Namely, I don't see how this derivation delineates how this potential difference forms.

Derivation based on the idealized DC circuit described by Ohm's law:

1. The electric field is constant across the wires, which are assumed to have a resistivity = 0.

2. The electric field is also constant across the resistor since if it weren't there would be charge buildup within the resistor. The electric field within the resistor also needs to be greater than that within the wires for there to be no charge buildup.

For the wires we have: E1 = ρ1A1/I . The current isn't zero, so the electric field must be zero. (Again, this is a highly idealized circuit).

For the resistor: E2= p2A2/I.

But, since the electric field is constant across both of these wires, there is also the relation E = V/L.

For the wires we find that the potential difference is 0 from the terminal of the battery up to the resistor. So the terminals of the resistor need to be at the potential of the battery.

Answer 1

I think you're making this far too difficult. Ohm's law is a relation between quantities, not a definition of cause and effect. If you apply a voltage to a resistor, Ohm's law tells you the current. But, if you apply a current to a resistor, Ohm's law tells you the voltage.

For networks of resistors, Kirchoff's laws give you additional relationships, allowing you to construct a system of equations whose solutions give the voltages and currents.

The earliest transistor amplifiers amplified voltage by driving a current into the low resistance emitter connection, transferring that current through the device, and applying that current to a high resistance collector load. Voltage->current through low resistance, followed by current->voltage through high resistance amplifies the voltage. The transistor got its name for its special capability of transferring the current from low to high resistance.

Answer 2

Let's start by looking at a common equation which is used to describe resistance. If you have a wire of length $L$, cross-sectional area $A$, and some material specific property $\rho$ which describes how inherently resistive a material is then we can write the total resistance of the wire as $$R = \frac{\rho L}{A}$$

Now let's consider how this resistance value $R$ is connected to the voltage drop across said resistor. If you look at just the $L$ term you can imagine breaking it into shorter and shorter wires until we have n total "sub wires", such that you have something like $$R = \frac{\rho}{A}(l_1 + l_2 + \, ... \, + \, l_i + \, ... + \, l_n)$$ where all the $l$'s add back up to the original length of the wire $l_1 + l_2 + \, ... \, + l_i + \, ... + \, l_n = L$. I'm using wires here but you could also think about breaking up a ceramic resistor into little slices if you're more used to working with hands on circuits.

Now that we've broken it up into pieces we can ask ourselves how the voltage changes across each short wires. Because current is constant throughout a simple closed circuit, the voltage just follows Ohm's law, with each short wire having a voltage drop across it which is proportional its length, $V_i = I \frac{\rho}{A}(l_i)$. When we connect all the wires back together, it is equivalent to summing up all the voltages $\sum_i V_i = I \frac{\rho}{A} \sum_i l_i = IR$.

So the overall answer is that in the simplest approximation, the voltage drops linearly through a resistor when you pop it into a circuit.

One final comment I want to make is how thermal energy and potential energy are relatedto each other. If you think about things in a conservation of energy picture, it makes sense that the potential energy has to go somewhere. When a ball rolls down a hill, potential energy is converted into kinetic energy. As the electrons physically travel through a resistor you can think about it qualitatively like them moving through a high friction environment, with them constantly bumping into the material and running into walls. All of this friction drags on them, causing them to loose energy which is gets converted into heat.

A fun cartoon which I like to go back to is this one