How do circuits work at a subatomic level?

Question

I imagine it’s like this. Let’s say we connect a copper wire to a battery’s terminals. The electrons from the negative side will thus exert a push on the electrons near the wire and it will continue. This push can be seen as voltage, the energy per coulomb. Now let’s say there’s a resistor along the way. Since these resistors resist the flow of charge, therefore, they take in some of that original push. Now it seems as if since the electrons in the resistor can’t move as freely, the current decreases. This causes the current all throughout the wire to decrease since the electrons keep pushing one another and therefore we take current as a constant. Now, the voltage of these electrons i.e the amount of energy they had reduces. Now let’s say they flow through another resistor. My question is, why is it that the sum of voltages across both resistors the original voltage?

Answer 1

Circuits do not work at the subatomic level. Circuits are an approximation to an approximation of what happens at the subatomic level.

Electromagnetism at the subatomic level is governed by quantum electrodynamics. Maxwell's equations in matter come from this level as an approximation which is only valid as the expectation of a large number of particles. At this level there are no individual subatomic particles, but rather there are so many that they are treated as a continuum. The expectations of the various quantum mechanical operators and wavefunctions are summarized in the constitutive relations for Maxwell's equations.

Circuit theory is a further approximation to Maxwell's equations. It assumes that electromagnetic effects happen instantaneously, that there is no net charge on any lumped element, and that any magnetic flux is contained within a single lumped element. With those assumptions you get Kirchoff's voltage and current laws and a lot of complicated behavior becomes linear.

Without Kirchoff's laws you don't have circuit theory, so if you are looking at a scale two levels of abstraction below where Kirchoff's laws happen, I am not convinced that it makes sense to even speak of electrical circuits. You have quantum electrodynamics at that scale, but not circuits.

Answer 2

Since these resistors resist the flow of charge, therefore, they take in some of that original push. Now it seems as if since the electrons in the resistor can’t move as freely, the current decreases..

Collisions between the mobile charge (electrons in metal conductors) and particles of the resistor takes away kinetic energy from the charge. But at the same time the electric field does work to restore that kinetic energy to the charge so that the current remains constant.

The potential difference (voltage) between two points along the conductor is defined as the work required per unit charge to move the charge between the two points. Thus the work that the field does to maintain the kinetic energy between two points of the conductor results in a decrease in voltage along the conductor.

Hope this helps.

Answer 3

The voltage of a circuit is determined by the potential difference of the source. If I take a $4.5 \,\text{V}$ battery and short-circuit it, then theoretically (neglecting the internal resistance of the source and the reaction speed of the chemical processes), all releasable electrons flow from the source to the sink at once and at a voltage of $4.5 \,\text{V}$.

If I now connect the current source to a conductor, I also connect a resistor between the potential difference of the battery. The potential difference of the battery remains the same, as this is caused by its chemical process. What changes is the conductivity for the number of electrons.

If you now measure the potential difference to one pole of the source along your wire at ten equally distant points, you will find that the voltage along the wire changes by $0.45 \,\text{V}$ from measurement to measurement. If you use a real resistor, this, of course, radically changes the game. The ohmic resistance of the lead wire is usually not considered relevant, and the entire potential difference is attributed to the resistor.

My question is, why is it that the sum of voltages across both resistors is the original voltage?

If you measure the potential difference along your resistor at every tenth of a distance, you will also get a voltage difference of $0.45 \,\text{V}$, as in the above example with the wire (which is now neglected). If you divide your resistor into two equal resistors, each will generate a voltage difference of $2.25 \,\text{V}$.