# How does resistor affect drift velocity?

After reading many question and answers on stack, i got totally confused about how does resistor affect drift velocity.

Consider a simple circuit, a battery and a resistor, connected to each other with wire ( wire with area $A$ and resistivity $\rho_w$ (very small), resistor with area $A$ and resistivity $\rho_r$ ; $\rho_w \ll \rho_r$). From the formula $v=\frac{I}{enA}$, we can conclude that overall drift velocity in circuit decreases by using a larger resistor (as it reduces the current).

Having constant current density $J$, electrical field in wire is $E_w = J \rho_w$ and resistor $E_r = J \rho_r$. As $\rho_w < \rho_r$, it says that electrical inside resistor is stronger than in wire. So the greater the electrical field, the more force on electrons and thus an increase in their speed (drift velocity). ( The increase in drift velocity can be concluded in another way : For both wire and resistor, all parameters in formula $v=\frac{I}{enA}$ are equal except $n$, charge carrier density. So drift velocity in resistor with less charge carrier density must be larger. )

Now some questions :

1. It is correct to say that having a resistor in a circuit, slows down the overall drift velocity comparing to another circuit with a lower resistor, but inside a circuit itself, drift speed in resistor is more than in wire ?
2. It is correct to say that in any element with nonzero resistor, as electrons move in opposite direction of electrical field, they lose electrical potential energy and it's get converted to kinetic energy and heat?
3. Considering the simple circuit, if wires have zero resistance, does electrons flow through ? (Noting that there is no potential difference, so electrons have no electrical potential energy to convert to kinetic energy and start moving toward other terminal of battery)
4. Does voltage drop mean that electrons have lost some electrical potential energy or it means that there are more electrons on one side of the resistor compared to the other side?

I will assume a constant current in the circuit in my answers.

1. a) If you compare two circuits with the same properties, up to a different resistor, which have the SAME VOLTAGE applied, then in the circuit with the lower resistor there will be a higher current. Since all the other properties of the circuit didn't change, the overall drift velocity $v_d$ will also be higher. b) It depends on your resistor wether the drift velocity is bigger in the wire than in the resistor (everything has just to satisfy your formula $I = v_d n A e$. If you find a Resistor with $n_{Resistor}$ being bigger than $n_{Wire}$, and give it the same area $A$ the wire has, then the drift-velocity will be lower in the resistor. This is also pointed out in the answer you've linked, in the last section. Still, usualy resistors have lower charge densities, so they require a higher drift velocity.

2. If by "opposite to the electric field" you mean "in direction of the force that acts uppon them" (eletrons have negative charge), then yes. The situation in the resistor is an equilibrium situation, where the electric field accelerates the electrons (converting its potential energy to kinetic energy) while at the same rate collisions of electrons convert the kinetic energy of the electron into heat.

3. Of course they still move through, they just need a shorter amount of time: Even without resistance of the wire, saying that there is NO potential difference throughout the WHOLE wire is a property of the equilibrium situation, and as long as there is a net flow of charge, this equilibrium situation will not be reached: Let's say the resistor has infinite resistance, and you connect wire and battery. What will happen? In the beginning, there is a potential difference throughout the wire. Electrons become accellerated. They will flow through the wire until they reach the resistor. With the electrons comes the charge, and with the charge comes potential. Since there now are more electrons in the wire, the potential difference has shrinked. As long as there will be a potential difference, electrons will become accelerated, until somewhen the potential is really constant throughout the entire wire.

This won't happen if your resistor has finite resistance, in that case there will be electrons drained out of the wire all the time, which creates a slightly little potential gradient at that part of the wire, and at the same time there are electrons pushed into the wire out of the battery, which also creates a tiny potential gradient.

The resistance being zero will just make the electron movement faster, and therefore they can balance out the potential difference between the start of the resistor and the battery faster. But as long as there is a current flowing, the potential difference between start of the resistor and the battery will not be zero, but a little bit more. You are right, forces can only act on the electron when there is a potential gradient, but since there is no resistance throughout the wire, it is sufficient if the electron is accelerated at the start of the wire and slowed down at the end (for example).

1. Voltage drop means that electrons have lost potential energy throughout the resistor. This doesn't neccesarrily mean that there are more electrons on one side of the resistor compared to the other side, it just can mean that. The question here is the geometry of the resistor: The charges in the wire before and after, and the ones in the resistor have to create a potential (via 1. Maxwell equation) that has the mentioned potential difference throughout the resistor. However, I usualy would also think that there is a potential gradient pushing electrons through a wire when there is more of them on the first side than on the second. So in principle the answer is "yes", but I don't know if it always has to be like that.