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This answer sums up the reason why a resistor does not alter the current flowing through it in a circuit.

Let's say that we have a $10$ $V$ battery connected with two resistors of resistance $3$ and $2$ $ohm$ respectively in series.

Then we can use ohm's law to find out that a current of $2 A$ flows through the circuit and that we have a potential drop of $4$ $V$ and and 6 $V$ across the $2$ $\Omega$ and $3$ $\Omega$ resistors respectively.

The above process can be and is conveniently described by Ohms law.

However I am curious that how on a microscopic level do these resistors individually manage to lower the Voltage across them , is it due to some internal electric field created inside the resistors or something else ?

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  • $\begingroup$ Related if not duplicate question $\endgroup$ Commented Oct 29, 2019 at 18:58
  • $\begingroup$ try hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmmic.html $\endgroup$
    – anna v
    Commented Oct 29, 2019 at 19:00
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    $\begingroup$ I wrote a answer to an earlier question where I examined the things that happen (very quickly in most cases) when a circuit is switched on or otherwise changed in configuration. The nice steady-state flows that we study in intro circuits are established by the interactions of very basic rules, but if you are unwilling to simply accept the results you have to study those interactions. $\endgroup$ Commented Oct 29, 2019 at 19:04
  • $\begingroup$ Also note that any potential drop by definition comes from electric fields. $\endgroup$ Commented Oct 29, 2019 at 19:18
  • $\begingroup$ I disapprove of the wording "individually manage". Resistors are not individuals. Resistors do not manage. Resistors resist. Managers manage. $\endgroup$
    – my2cts
    Commented Oct 29, 2019 at 20:28

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