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I'm given to understand from my textbook that resistance is inversely proportional to the current (seems straightforward from Ohm's law). I also understand that resistance is a property of the given object. However, the textbook also mentions the law of conservation of charge, stating that the amount of charge in the wire before going through the resistor is equal to the charge after passing through the resistor. I'm having trouble wrapping my head around this.

For example, let us consider a simple circuit with a single resistor. If the current in the wire before entering the resistor is I, then the current after exiting the resistor must also be I. Given that resistance is a property of the material, the current inside the resistor is some Io such that Io < I (assuming the resistor of the wire is smaller). From the definition of current, there would be N number of charged particles before and after exiting the resistor, and No number of charged particles in the resistor (such that No < N). How exactly does this decrease and increase work?

(Please feel free to correct me if I'm interpreting the whole thing wrong)

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2 Answers 2

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You are equating current with number of charges. But you are forgetting the time aspect.

Current is not just the no. of/amount of charge passing through a Cross section but the amount of charge passing through a Cross section per time unit: $$I=\frac{\mathrm dq}{\mathrm dt}.$$

You can reduce the current both by reducing the amount of charge and by slowing them down.

When this is said then we can think of the scenario: Say that a lot of charge arrives at the resistor. It is then "slowed down" inside the resistor and thus comes out in the other side as "slower charge", meaning the current has reduced. But the amount is the same since no charge stayed inside the resistor - if it did so, then the resistor would quickly get a high negative net charge that would repel any further incoming charge and stop all current flow. So charge conservation is fulfilled.

But the incoming "fast charge" will now have to "queue up" in front of the resistor, which will necessarily also slow it down before the resistor. In fact it will be slowed down to the exact same drift speed. Thus the current before the resistor reduces until it is equal to the correct after the resistor. So we have current conservation as well. This takes a brief moment, and when this has been reached then we have what we call steady state.

So, in summary:

  • Charge conservation is always true.
  • Current conservation is true during steady state (which in usual electronics is always reached).
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  • $\begingroup$ Thank you for your response! Super helpful. $\endgroup$
    – Mr.A
    Dec 6, 2022 at 6:54
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I guess you are confused that current somehow lessens inside the resistor specifically. No.

The current is immediately determined by the potential source and resistance of the circuit. If we ignore the resistance of connecting wires, we can deduce $I=V/R$ or $I=(n_{resistor})Aev_d$

whatever elements the circuit has, the current flowing in the circuit must remain the same considering we have no branchings in between. Feel free to ask doubts

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  • $\begingroup$ I see! Thank you for your reply. To make sure I'm getting this correctly, given some circuit with a voltage source, we calculate the resistance for that circuit, and the current is then uniform across all the components of the circuit (assuming no branches). Would this be correct? $\endgroup$
    – Mr.A
    Nov 25, 2022 at 7:31
  • $\begingroup$ @AdityaIyer, Yes make sure the circuit is D.C supplied and there are no capacitors or inductors, which is entirely different ball game. $\endgroup$
    – Kshitij Kumar
    Nov 25, 2022 at 7:35
  • $\begingroup$ Amazing, this clears it up. Thank you! $\endgroup$
    – Mr.A
    Nov 25, 2022 at 7:36

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