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I had a question regarding the current along a resistor with variable resistivity. If you have a normal resistor with constant resistivity throughout its length, and you apply a constant DC voltage across it, the current at any point on the restitor is the same, the average quantity of charges travelling per second remains the same. However, if I have a resitor with some length $L$, and with variable resistivity across its length, and I apply a constant DC voltage across the ends of the resistor, will the current be constant, or will it change as a function of its position on the resistor? i.e lower towards the one end and higher towards the other? Or is the current constant throughout the resitor? If so why?

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  • $\begingroup$ A better word than varying might be differing (or similar; maybe an English native can help out here). The term varying is often used for values that change over time but you are refering to changes over distance. So, you resistance is actually constant but just with different constant values at different points. That makes a crucial difference to the answer. Not sure if this is a big deal or if it is just me, but I misread initially. $\endgroup$
    – Steeven
    Dec 28, 2021 at 17:02
  • $\begingroup$ Probably it would be useful to close down the question as there are good asnwers added. $\endgroup$
    – zltn.guba
    Dec 30, 2021 at 8:49

4 Answers 4

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The current will still be constant, otherwise there would be a violation of conservation of charge.

What will be different is the electric field in the resistor. For a uniform resistor the electric field is uniform in the resistor with a magnitude of

$$E=\frac{V}{d}$$

Where $V$ is the voltage drop across the entire resistor and $d$ is the length of the resistor.

If the resistance is not uniform along its length, then the voltage drop per unit length of the resistor will not be constant, even though the total voltage drop is the same. Therefore the electric field will not be uniform along the length of the resistor.

Hope this helps.

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  • $\begingroup$ I accidently downvoted this answer. Could you please modify it a bit so that I can delete my downvote? Thanks in advance :) $\endgroup$
    – zltn.guba
    Dec 30, 2021 at 8:51
  • $\begingroup$ @zltn.guba I just did. Was wondering why it was downvoted. Thanks for letting me know $\endgroup$
    – Bob D
    Dec 30, 2021 at 11:08
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Whatever goes into a branch of a DC resisitive circuit MUST come out. The current cannot 'stop' or 'slow' at different points.

You could break your resistor down to a series of infinitesssimal resistors of different values and see that the current must flow through them all.

The water flow analogy works here - there is no 'accumulator' in the circuit so the current is constant.

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  • $\begingroup$ This is only true for time-independent case, as there the continuity equation suggests $\nabla \mathbf{j}=0$ , which is just the first law of Kirchhoff. But in general charge can accumulate in a non-uniform resistance which results in a non-uniform charge distribution. $\endgroup$
    – zltn.guba
    Dec 28, 2021 at 17:05
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If the incoming current is not equal to the outgoing current for an infinitesimal volume then charge gets accumulated at that infinitesimal volume. The relation between the charge density and the current density is given by the continuity equation \begin{equation} \partial_t \rho + \nabla \mathbf{j} = 0 \end{equation} Now if the absolute value of the incoming current density is not the same as the absolute value of the outgoing current density then $\partial_t \rho$ is nonzero. This suggests that we are talking about a time-dependent process. In contrast to the previous answers, it might be possible to accumulate charge in a circuit, but in a static process the charge density cannot change, so \begin{equation} \nabla \mathbf{j} = 0 \end{equation} for a static process, which is just a statement that the absolute value of the incoming current is the same as the absolute value of the outgoing current.

The reason why we are not dealing with processes where charge accumulation happens is that the time-scale for them is very short. Let's say that you have a setup in which you can have a nontrivial charge distribution. Shortly after you turn on the current generator, the charge distribution in the system is already static, which means you can use $\nabla \mathbf{j} =0 $. This statement is also known as the first law of Kirchhoff.

So in a nutshell, the continuity equation gives the connection between the current density and the charge density and in a static process you can use the first law of Kirchhoff, but in general time-dependent process this law does not hold.

To explicitly answer your questions: in the static limit the current will be the same along the resistor, while the voltage drop will vary along it. Also the electric field will depend on the position along the resistor, and a non-trivial charge distribution will be created.

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I interpret the question as asking to a resistor of gradually different resistances across its geometry. In other words, the resistance is constant (over time) but differes from point to point along its length.

Will the current be the same at all points throughout your resistor?

Yes.

Here are two arguments for that conclusion:

The resistance perspective

A resistor in a circuit will slow down a charge that tries to move through. Intuitively we can imagine charges as "queueing up" behind one another, when the first charge is slowed down. Thus they all slow down behind it. When leaving the resistor, they will all keep moving with this new slower drift speed (and if slowed down further, then the ones behind will keep pushing to keep this drift speed constant). So, now this new, lower current is the new circuit-wide current.

  • Firstly, note that the current in the wire segments is the same as through the resistor. Those wire segments do corresponds to resistors with a very low (zero) resistance. So, we are dealing with a circuit where the current becomes constant even across different components.

  • Secondly, note that adding a another resistor will slow down the charges even more. They will thus "queue up" behind one another at this resistor as well, effectively slowing down the total circuit-wide current even more and this new slower current is again the same throughout and in both resistors.

So, the currect along any series connection of components in a circuit at steady state will be constant at all points. Your scenario with a resistor of gradually differing resistance along its length corresponds to a circuit with many resistors in series. Just imagine those resistances squeezed closely together and each being very, very thin, constituting just a slice each. The current must be the same throughout such component of the same logic.

This was the resistance-view on the question. There is also the, maybe simpler, current-view:

The current perspective

If, say, 10 charges enter a point each second, then also 10 charges must leave each second. Otherwise, where would they go or where would they come from?

  • If fewer leave than enter each second, then some must accumulate within that point. Such a point will thus gain a larger and larger repulsion against new incoming charge, slowing it down until just as many enter as leave each second.

  • If more leave than enter each second then... well, where would they come from? Charge cannot just appear out of thin air, so this is impossible.

In conclusion, for steady state we know that all current in must equal all current out at any point. This includes points within resistors. So, the current must be constant throughout your resistor.

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