What happens to a circuit over a resistor that causes measured voltage to drop? If voltage is related to a charge imbalance creating electric potential between two points, then what does electrons moving through a resistor have to do with a reduction of that charge imbalance? Series circuit: Diagram 1 Negative (10 volts)--> ----------------A1-----1 ohm resistor-----B1-------------Positive terminal Diagram 2 Negative (10 volts)--> ----------------A2-----No resistor---------B2-------------Positive terminal Since I is fixed for the entire circuit, it's the same at points A and B. A1 and B1 would have a lower I value than A2 and B2. The only difference i can think of is that electrons give off energy in the resistor, so their energy levels would be lower at point B than point A. Does that mean that voltage measured is in relation to the energy of the electrons at a given point? How does that work? Is it supposed to represent how much of the electron's potential has been "used up" since the start of the circuit? In diagram 2, there's no resistor, except for the marginal resistance of the wire, which we could consider to be much less than 1 ohm. But if voltage along a circuit is related to electron energy, and all series circuits go from full to 0 voltage over the course of the circuit, then the energy in the electrons in diagram 2 must reach the level of the electrons in diagram 1 by the end. Diagram 2 has a larger current. Does this mean that the electrons lose energy more quickly along the course of the non resistive wire than they would over the same wire in a circuit where there was also a resistor? (i.e. diagram 1) So then increased current means that current will suffer a voltage drop more quickly over a fixed resistance than a slower current?