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So, I tried thinking about currents and circuits from a molecular point of view this morning but I couldn't wrap my head around the following: [See the attached Image]

enter image description here

----When electrons enter the resistor(CD), they will experience
collisions, which would result in a decrease in the KE of the
electrons, which would further result in a decrease in Current.
(Which is not the case, the current at C and D is equal).

Throughout this problem, I have assumed that current, I = nVdAQe and that we are not aware of Ohm's Law.

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2 Answers 2

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  1. There is no difference in potential from A to C, so there can be no acceleration of electrons.

  2. Either $n$ (which I’m assuming is the charge carrier density) or $A$ (which I’m assuming is the cross sectional area of the resister) will change depending on what the resister is. So to keep the current constant the drift velocity will increase. This increase happens due to a difference in potential from C to D.

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  • $\begingroup$ Okay, what about part 2? $\endgroup$
    – SAN
    Commented May 12, 2021 at 15:00
  • $\begingroup$ @SanchitSharma I’ve updated the answer. $\endgroup$
    – jamie1989
    Commented May 12, 2021 at 18:20
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Why does current not increase on passing through a zero-resistance wire?

Because the charge theoretically would not experience any force in a zero-resistance wire due to the voltage between any two points in the wire being zero.

The force $F$ experienced by a charge $q$ moving between two points a distance $d$ apart having a potential difference $V$ is

$$F=\frac{qV}{d}$$

The potential difference $V$ between two points is the work per unit charge required to move the charge between the points. No work is required to move charge between two points if there is no resistance to overcome to move the charge, i.e., $V=0$. Thus the force $F$ on the charge is zero and it does not accelerate.

When electrons enter the resistor(CD), they will experience collisions, which would result in a decrease in the KE of the electrons, which would further result in a decrease in Current.

There is no net decrease in the KE of the electrons moving through the resistor. The net change in KE is zero, and thus the current is constant. To explain:

The electrons flowing through the resistor alternately gain kinetic energy from the loss of the electrical potential energy given them by the battery and lose kinetic energy due to collisions with particles in the resistor. The gain of KE equals the loss of KE for an overall change in KE of zero. The result is constant current.

A mechanical analogy involving friction may be of help in connection with both of the above. Imagine pushing a box (analogous to the electrons) at constant velocity (analogous to constant current) on a floor.

Between points D and C of the floor there is kinetic friction between the box and the floor (analogous to the resistor). For the box to move at constant velocity from D to C, you (analogous to the battery) apply a force that equals the opposing kinetic friction force (analogous to the effect of the collisions in the resistor) for a net force and net work of zero. The work you (the battery) did is dissipated as heat between the box and floor (resistance heating).

Now imagine after pushing the box from D to C, the floor suddenly becomes frictionless. This is analogous to the electrons encountering a wire having no resistance. Since there is nothing opposing the motion of the box, you need not apply any force on the box for the box to continue on at constant velocity on the frictionless surface. This is analogous to no work required (no potential difference, or voltage) to move the electrons at constant speed between any two points in the wire with no resistance.

Of course the reality is all wires (excluding superconductors) will have some resistance. So work per unit charge (voltage) will always be required to move charge through a real wire.

Hope this helps.

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  • $\begingroup$ Thanks this was helpful but how do the electrons gain KE from the battery?. $\endgroup$
    – SAN
    Commented May 13, 2021 at 1:07
  • $\begingroup$ @SanchitSharma They gain KE from the loss of the electrical potential energy given them by the battery. I've revised my answer to clarify that point. $\endgroup$
    – Bob D
    Commented May 13, 2021 at 14:22

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