To my understanding, if a resistor reduces the potential energy of the current across it, then the current that leaves the resistor will have less potential energy and thus, less pressure or voltage.
If one resistor only drops one volt from the original voltage (only drops one volt because of its resistance value) and we add another resistor in series with the same value, then that second resistor would also drop one volt. Then, the current leaving the second resistor would be 2 volts less. In reality, this is not the case and somehow, in the middle of the two resistors the voltage would be dropped to half the original voltage. How is this possible if each resistor is made to only drop 1 volt?
My image below shows how each resistor would lower the potential energy of the current in order to end up with 4 volts, since each only drops one.
I know this is incorrect in reality and I also know how to calculate the voltage drops, but physically it doesn't make sense to me.
Best Answer
The starting point of this question makes it hard to give a short, satisfying answer; maybe someone will come along and do a standout one with diagrams, but I don't have time for it. So I'll kick off with this:
With two resistors in series, there's more total resistance on the circuit, which means that less current flows. With less current flow, a given resistor drops less voltage because V=IR. Voltage drop isn't an intrinsic property of the resistor; resistance is.