I'm just trying to get a conceptual grounding for the circuit problems in our book. So I understand that a circuit maintains a potential difference via redox reactions. The electrons have a desire to flow, so they do so. The potential difference and the conducting wire allow the electrons to follow a path and reach the higher potential - conventionally opposite, but it doesn't matter. The electric field inside the circuit is related to the current density by j = σE. From this, we can derive ohm's law. This really is my question. The resistors are going to cause voltage drops due to the loss of kinetic energy by the electrons, correct? Afterward, what happens to electric field? There's no difference in potential between the bottom of the resistor and the lower potential of the battery. If so, the electric field must be 0 then? And if that, what pushes the charges to the other electrode? The repelling force of the other charges? And then, the electric field isn't constant in the circuit. I'm pretty confused, and would be very appreciative if you could enlighten me. Thanks for taking time to read.