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In a circle the voltage drops across a resistor. This means that some electrons lost some of their electrostatic potential energy. Where does that energy go, and how? For potential energy to be lost work must be done opposite to the conservative force causing the potential energy. This implies that there is an acceleration – so my question is: Do electrons speed up or slow down because of a resistor?

Alternatively, E = resistivity x current density, meaning a higher resistivity implies a stronger electric field. The difference in electric field strength resulting from the difference in resisitivities causes a net force, meaning there is acceleration shortly, until the dragging force becomes equal and a new drift velocity is achieved. The drift velocity in a resistor is greater than in the surrounding wires.

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Electrons inside solids have a very random motion. They keep on bumping around here and there. What happens when you put a battery? The motion of these electrons is only slightly disturbed. Free electrons inside the metal move at large speeds, a battery sets up an electric field inside the solid which tries to push the one side but the acceleration caused is very low. This is why electrons do not move like uniform particles, line by line behind each other in a queue but rather they just slightly drift towards one side while still doing that random motion.This means there are a large number of collisions.The electric fields energy is not going completely in pushing but alot is going into making them collide, this generates heat and this is how energy is lost and potential drops, it runs off as HEAT ENERGY. We cannot say if electrons slow down or speed up or since they are moving at random speeds, but we can definitely say that the rate at which the move a little side by side along the solid becomes less. This rate is called DRIFT VELOCITY and it becomes less. If a current of I amperes flows through a resistor of resistance R in voltage V then the amount of heat dissipated or potential energy lost per second is given by: $$ I^2R = V^2/R = VI $$ For more detail you can study about these on wikipedia: $$ $$ ->Drift velocity $$ $$ ->Heat loss in resistors $$ $$ Also here's a helpful diagram to understand the motion of electrons $$ enter image description here

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  • $\begingroup$ So does a resistor reduce the drift velocity along the circuit by making it harder(higher distance to cover, more objects in path, etc?) for the electrons to flow? This reduced drift velocity then reflects in the smaller current $\endgroup$
    – Bandoo
    Mar 8, 2020 at 11:08
  • $\begingroup$ Bandoo - indirectly yes, because it depends on the nature of resistor on how much does it bind the electrons and the collision frequency value, so the reduction in drift speed is caused by the resistor $\endgroup$
    – arnav009
    Mar 8, 2020 at 15:34
  • $\begingroup$ After further discussion, it seemed as if the drift speed inside the resistor should be higher than outside, because I=eVNA and everything is constant except N and V when we compare a cross section of a "wire" against a "resistor". N is smaller for a material with higher resistivity, meaning V is larger as for I to stay the same(conservation of charge) $\endgroup$
    – Bandoo
    Mar 8, 2020 at 15:39
  • $\begingroup$ The equation you are refering to is actual derived from drift speed. The relation between V and I that your equation describes is actually derived from the concept of drift speed. The fundamental equation from which drift speed is derived is V= eEt/m where t is relaxation time, it is the average time between two successive collision of the electron. For resistor this time decreases and so its spees decreases $\endgroup$
    – arnav009
    Mar 8, 2020 at 15:45
  • $\begingroup$ This is a problem that even I used to face. You have to understand, the more fundamental and logical and axiomatic equation rules.it is not the current or the charge carrier density(N) that governs drift speed. But drift speed rather decides current. Charge density and drift speed effect current but they do not affect each other because it is axiomatically wrong. $\endgroup$
    – arnav009
    Mar 8, 2020 at 15:56
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There is a lot of random thermal motion of charges, but on average their speed (drift velocity) through a resistor is constant. Around a given series circuit the current is constant with respect to position, so the average speed depends only on the product of the cross sectional area and the free electron density. Since that is constant across a resistor the drift velocity is constant with respect to position also.

For potential energy to be lost work must be done opposite to the conservative force causing the potential energy. This implies that there is an acceleration

This is faulty logic. Suppose there are two forces, the conservative force from the electric field $F_E$ and non conservative resistive force $F_R$. Since, as you say, on average the work done by $F_R$ is equal and opposite that done by $F_E$ we have: $$F_R d =-F_E d$$ $$F_R =-F_E$$ So on average the forces are equal and opposite which means that the average acceleration is 0.

The drift velocity is given by: $$V_d=\frac{I}{e N A}$$ so it is possible for the drift velocity to change around a circuit even with a fixed current $I$, but it is not due to changes in resistivity. It is due to changes in the cross sectional area $A$ or the density of charge carriers $N$. Both $A$ and $N$ are independent of resistivity. It is clear that $A$ is independent, but an examination of tables of resistivity and charge carrier density shows that $N$ is also independent of resistivity. For example Mercury has a relatively high resistivity and moderate $N$, while Aluminum has a high $N$ but a low resistivity.

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  • $\begingroup$ I had not considered the case that the forces are opposite and equal. This is the same with lifting a weight upwards at a constant velocity. But even in that case, the weight was initially at rest and then started moving implying some tiny acceleration. That acceleration is what I'm talking about; the forces do not instantaneously cancel out. $\endgroup$
    – Bandoo
    Mar 8, 2020 at 12:07
  • $\begingroup$ @Bandoo That type of brief acceleration only happens at locations where the thickness of the wire changes. Elsewhere the velocity is constant with respect to position $\endgroup$
    – Dale
    Mar 8, 2020 at 12:58
  • $\begingroup$ Yes this happens if the electrons travel through a region with different wire thickness, but it also happens in a region with different resistivity, no? I also understand that the drift velocity is the same, but the frequency of collisions with the lattice ions is different (if the factors of resistance change). The drift velocity doesn't change with respect to time: the electrons themselves move faster in different regions but cover the same displacement in the same out of time due to the factors that increase the resistance. $\endgroup$
    – Bandoo
    Mar 8, 2020 at 13:12
  • $\begingroup$ @Bandoo no it does not. If the thickness is constant but the resistivity changes then there is no average acceleration $\endgroup$
    – Dale
    Mar 8, 2020 at 13:25
  • $\begingroup$ If we change the cross sectional area we change the drift velocity such that I remains the same as in I=eNVA. But if we increase the resistivity the drift velocity must also decrease. Does this mean the "N" term increases? $\endgroup$
    – Bandoo
    Mar 8, 2020 at 14:07
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In the stationary state, the current along the whole circuit is the same, due to conservation of charge. To put it simply, take any tiny region of the circuit, and at any given time, in the stationary state, the number of electrons flowing in and out of that region must be the same.

Extending this to the whole circuit, the intensity is the same everywhere, so for wires of the same cross-section, no matter the resistivity of the material, the drift velocity must stay the same.

This means there is no net drift acceleration inside a resistor. If there was, electrons would bunch up inside it, because more would be flowing in than flowing out, and charge up like a capacitor would, but you can check that doesn't happen.

The difference in potential energy across a resistor is macroscopically turned into heat, and is not spent on accelerating the electrons themselves.

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  • $\begingroup$ I understand that the difference in potential energy ends up as heat, however, my question is how? Because if the potential energy turned into kinetic energy that would essentially mean the particle had an increase in velocity(or mass?) meaning there was some form of acceleration. By any means, the heat you are talking about is reflected through a change in temperature, which is the average kinetic energy. For that to increase velocity must. $\endgroup$
    – Bandoo
    Mar 8, 2020 at 12:10
  • $\begingroup$ One of the first models for electronic movement in metals is the Drude model. It models electronic movement by saying every certain amount of time, electrons collide against lattice ions, and are then accelerated by the electric field, only to crash again some time later. The time between collissions in this model depends on the resistivity of the material at hand. These interactions between the electrons and the lattice ions create heat. Of course, quantum mechanical models are much better to be used, but Drude's model can be good for a first and intuitive understanding of the matter at hand. $\endgroup$ Mar 8, 2020 at 12:14
  • $\begingroup$ is it reasonable to say that a resistor decreases the drift velocity across the whole circuit which is why a circuit with high resistance has less current given a constant potential difference? drift velocity is equal to the small net displacement divided by time between collisions. If this time is larger for a high-resistivity material then it makes sense to conclude that the drift velocity is smaller, and it also means less collisions against the lattice ions which means less energy dissipated per unit time, which is why short circuits are dangerous. $\endgroup$
    – Bandoo
    Mar 8, 2020 at 12:32
  • $\begingroup$ Indeed. You could intuitively think of resistors as electronic bottle necks. They limit the current along the circuit and thus the drift velocity. What is important is that this effect does not only take place inside the resistor, but along the whole circuit, because of conservation of charge. And indeed, the more current, the more collisions against the lattice ions, which generates more heat. $\endgroup$ Mar 8, 2020 at 12:35
  • $\begingroup$ Finally, is it a valid conclusion that electrons come into the resistor with some potential energy. Consequently , they collide so many times inside the resistor and these collisions, more so than inside the wires. These collisions constantly accelerate and decelerate the electrons. These collisions transfer the potential energy of the electrons to kinetic energy of lattice ions, which can be macroscopically seen as heat. Finally, the electrons leave the resistor and continue collisions across the wire, albeit at a much lower rate, leading to minuscule potential drop. $\endgroup$
    – Bandoo
    Mar 8, 2020 at 12:49
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The answer to the above questions is that the drift velocity of electrons must change for the current to remain the same if other factors change. For example, if the wire suddenly becomes thinner, like a bottle neck then the A in $$I=eNV_dA$$ becomes smaller and thus $V_d$ must increase. The drift velocity is also proportional on the relaxation time, so if that changes, the drift velocity must also change, all other things equal. In conclusion, the drift velocity inside a circuit will change depending on the components of the circuit. Because the electrons change their drift velocity, it is plausible to conclude that there is some net acceleration/deceleration through the components of the circuit, all so as to keep the current itself constant throughout the circuit since charge MUST be conserved. Axiomatically, the important statement is that $I$ is constant before and after the resistor.

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  • $\begingroup$ Where is resistance or resistivity in your equation? $\endgroup$
    – Dale
    Mar 9, 2020 at 11:33
  • $\begingroup$ Both the drift velocity and the resistivity depend on the relaxation time. I can make it explicit if you'd like. $\endgroup$
    – Bandoo
    Mar 9, 2020 at 11:37
  • $\begingroup$ $$E=\rho J$$ $$E=\rho Nev_d$$ $$E=\frac{\rho Ne^2E}{m}\tau $$ $$\rho=\frac{m}{Ne^2\tau}$$ In order to obtain resistivity in that equation you would solve for relaxation time in terms of resistivity from above and then use the relationship $V_d = a\tau$ to substitute for drift velocity $\endgroup$
    – Bandoo
    Mar 9, 2020 at 11:40
  • $\begingroup$ $\rho $ and $\tau $ are not independent. This equation isn’t telling you what you think it is. $\endgroup$
    – Dale
    Mar 9, 2020 at 11:47
  • $\begingroup$ This equation is stating that a smaller relaxation time leads to a higher resistivity, which makes sense: the electrons collide with the lattice ions more frequently, and thus they are slower(less drift speed). $$v_d=\frac{E}{Ne\rho}$$ Also, $$ I = \frac{EA}{\rho}$$ $\endgroup$
    – Bandoo
    Mar 9, 2020 at 11:57

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