# Voltage drop across a circuit misunderstanding

Consider a circuit with a 20 volt battery and a resistor (no other components).

Next, let's say I choose a point A on the positive terminal of the battery. The current flows from point A, around the circuit, back to point A, which results in an electric potential difference of 0 (since we're moving back to A).

However, if we were to add a resistor, wouldn't this violate Kirchhoff's Voltage Law? Let's say the resistor has a voltage drop of 20 volts. From earlier, the battery has a voltage drop of 20 volts.

Over the course of the circuit, the electric change in potential appears to be -40, not 0 (talking about current here, not electrons). Can someone tell me where I messed up? Thanks.

• You don't get to choose the voltage drop of the resistor. It's going to be 20 volts, because that's the battery voltage. Mar 24 at 23:04
• Please check again, I edited - that wasn't the problem I was trying to get solved Mar 24 at 23:06
• Maybe you should add a schematic to clarify the problem a bit. Mar 24 at 23:25
• The battery has a voltage "lift" of 20 volts. Mar 24 at 23:35

• @JeanPierre: a (real) wire is just a resistor, so nothing special in there: just a voltage drop at the wire (and possibly a very high current due to the low resistance of the wire), and a reverse voltage drop/voltage rise at the battery. Where it gets more interesting: if you replace the wire by a superconductor. The SC's zero resistance naively means also zero voltage drop, but that would violate Kirchhoff. But actually you cannot neglect (self-)induction anymore in that case, which would equilibrate the constant battery voltage by a linearly rising current ($U_{ind}=-L\dot I$, L=inductance). Mar 25 at 0:27