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Here's what I think I know:

Voltage is the result of an electric potential difference of some charge that perhaps came about by a battery cell or generator or whatever. In this situation, the charge - let's say an electron - tends to travel to the low potential rigorously (with the ability to do work) due to the nature of electric fields. In a circuit, as it does so (relatively slowly), all the other weakly bonded electrons on the conductive wire travel as well. This simultaneous electron migration results in a current (flow of charge through a certain point within a certain time).

Now, here are the premises that I think underlie my confusion:

Apparently all of this electric energy must be converted to some other form by the time it reaches the low potential point. So if you have a simple light-bulb circuit at 12 volts, there is a voltage drop of 12 volts for electrons that travel through the bulb. If you have a 12 volt series circuit with two identical light-bulbs, there will be two 6 volt drops.

My confusion:

What is the reason for this volt drop distribution? Intuitively, I would think that the voltage drop due to a resistor would be a fixed amount depending on the light's material, regardless of how many other resistors become involved.

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    $\begingroup$ I'm almost certain this has already been asked and answered here. Have you searched the site or the "Related" questions listed on the right? $\endgroup$ – Alfred Centauri Sep 25 '16 at 23:14
  • $\begingroup$ @AlfredCentauri There doesn't seem to be information answering my question in those posts. Perhaps the answers are there, but hidden in language that I do not understand. If it's not a bother, could you refer me to any posts you think would be helpful? $\endgroup$ – Arman Som Sep 26 '16 at 6:14
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    $\begingroup$ Possible duplicate of How do electrons "know" to share their voltage between two resistors? $\endgroup$ – ACuriousMind Sep 26 '16 at 11:54