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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.

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    $\begingroup$ 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. $\endgroup$
    – Javier
    Mar 24 at 23:04
  • $\begingroup$ Please check again, I edited - that wasn't the problem I was trying to get solved $\endgroup$ Mar 24 at 23:06
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    $\begingroup$ Maybe you should add a schematic to clarify the problem a bit. $\endgroup$
    – ZaellixA
    Mar 24 at 23:25
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    $\begingroup$ The battery has a voltage "lift" of 20 volts. $\endgroup$
    – BowlOfRed
    Mar 24 at 23:35

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You messed up the sign of voltage, which is related to the current direction. When a positive charge travels from A to B over the resistor, it experiences a voltage drop of 20 volts, as you defined it. But as soon as the charge travels back from B to A in the battery, it experiences a voltage drop of -20 Volts, which is not a voltage drop, but actually a voltage rise. Hence, the sum of voltages in the loop is zero. Voltage is not just potential difference (which would be arbitrary with respect to sign, and leads to your confusion), but potential difference in relation to the current flow direction (whereby it is irrelevant whether you take flow of positive or negative charges, as long as you stay consistent). Hence, when you consider Kirchhoff, you should never forget to draw the voltage arrows (which should be cyclic of direction).

Kichhoff's voltage law is just energy conservation. If you drive down a hill with your bike, you will have to pedal hard to get back uphill, at least if you want to drive in a loop.

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  • $\begingroup$ I don't understand your third sentence - if we were to take out the resistor, the current would still move from high to low electric potential and its voltage would still decrease, which does not make sense if the battery is supposed to supply voltage. $\endgroup$ Mar 24 at 23:56
  • $\begingroup$ @JeanPierre, if you take out the resistor, no current will flow, because air is a very poor conductor. If you replace the resistor with a piece of wire, then the wire is just another resistor with a very low value and all the math that applies to the case of a resistor connecting the battery terminal still applies. $\endgroup$
    – The Photon
    Mar 25 at 0:12
  • $\begingroup$ So let's say I have only a battery and wire from the positive to negative terminal. If current is flowing from the positive to the negative terminal, there is a voltage drop, but how can this occur if the battery is supplying voltage (increasing electric potential energy)? $\endgroup$ Mar 25 at 0:17
  • $\begingroup$ @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). $\endgroup$
    – oliver
    Mar 25 at 0:27

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