# What happens to the charges when a resistor is short-circuited?

let's consider a parallel between a short circuit and a resistor $$R$$, supplied by a voltage source $$V_S$$ with a series resistor $$R_S$$.

Circuit Theory description: the resistor $$R$$ is completely bypassed by the short circuit, since the short circuit has no voltage between its terminals (A and B). So, no voltage across $$R$$, and no current flows on $$R$$. In an equivalent way, the resistance of such a parallel is 0 ($$R\times0 /(R+0) = 0$$).

But, What physically happens to the charges when the nodes $$A$$ and $$B$$ are connected to the voltage source?

My analysis:

1. The voltage source $$V_S$$ has the role of separating positive charges (which will be accumulated on the upper terminal) and negative ones (accumulated on the lower terminal.

2. When we connect the nodes A and B (with R and short circuit in parallel) to the voltage source terminals, the charges in each terminal have the opportunity to distribute along the media connected to the voltage source. Let's focus on the positive charges accumulated on the upper terminal.

3. Positive charges will move (thanks to Coulomb force) towards node A. What do they do at this point? I'd say, there is no reason not to flow both in the short circuit and in the resistor. If I were a charge, whose purpose is only that of separating from charges with its same sign, why shouldn't I use also the path with the resistor?

4. Next, charges flowing on the short circuit won't find any obstacle to their movement, while charges flowing on the resistor will be slowdowned because of their collision with the medium particles. This will decrease the current, since it's proportional to drift velocity which will be slowdowned (am I right?)

5. Well, this will decrease current on the resistor during time. But why will it be exactly zero at steady state (in which circuit theory is a perfect description of the phenomenon)? Why will charge decide not to flow more on the resistor?

• "Circuit Theory" doesn't talk about "charges." It doesn't talk about their speed, it doesn't talk about their "collision with the medium particles." In describing a real, physical circuit, with real metal wires you can talk about those things, but then in that case, you don't have a "voltage source;": you have a battery with all of its chemical messiness, or you have some other physical artifact that moves "charges" in its own special way. Commented Mar 9, 2021 at 23:53

What physically happens to the charges when the nodes $$A$$ and $$B$$ are connected to the voltage source?
I think this isn't a short circuit but rather a circuit with resistors in parallel. That is, you look at Kirchhoff's circuit laws and create a loop that goes through $$R_{s}$$ and $$R$$ as one and another that goes through $$R_{s}$$ and then bypasses $$R$$. That is, this will act like a circuit with resistors in parallel where one equivalent resistor has $$R_{1} = R_{s} + R$$ and the other has $$R_{2} = R$$ then you add them up in the usual inverse manner, i.e., $$R_{tot}^{-1} = R_{1}^{-1} + R_{2}^{-1}$$
So the short answer is that the charges will flow in the normal sense and nothing spectacular will happen. There will be different amounts of current in each loop, i.e., $$I_{1} = V_{s}/R_{1}$$ and $$I_{2} = V_{s}/R_{2}$$, where $$I_{1} \neq I_{2}$$.