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I am slightly confused on the physics of direct electric circuits. Here is what I have been taught:

  • Batteries (sources of emf) provide a constant potential difference between its terminals;
  • Electrons (charge carriers) gain a potential difference by being forced by the emf to go from the positive terminal to the negative terminal;
  • If the battery is attached to a closed circuit, the battery having a potential difference by definition means an electric field exists in the wires of the circuit; (x )
  • Electrons from the negative terminal of the battery enter the wire, but the electrons already in the wire itself also move due to the presence of this electric field. The circuit is electrically neutral, however, because the same number of electrons entering the circuit are also exiting the circuit into the positive terminal.

If the above is true, something is bothering me. I have seen graphs of voltages through different parts of a circuit, indicating that potential difference is gained from the electron passing through the battery, then in the wire the voltage is constant, then decreases when passing through a load. But this seems odd to me; if there is an electric field in the wire, then there should be equipotential lines through the wire, which would mean that the electric potential of the electron would change, not remain constant, through the wire. Textbooks I've read tend to indicate that the electrons only gain their energy when they pass through the battery, but not inside the actual circuit itself (until it is lost through a load). This would mean there is no field in the wire... but if there were no field in the wire, a current would not exist. Can someone shed some light on this? Thanks.

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  • $\begingroup$ There is a small field in the wires: according to Ohm's law (wires behave pretty much like Ohmic resistors), the voltage loss along a wire is given by $V=RI$ and one can easily measure this with a cheap multimeter. Engineers usually make these losses small compared to the work done on the load, but they are always there and everybody who deals with electric circuits thinks about them all of the time. $\endgroup$
    – CuriousOne
    Jul 2, 2016 at 21:43

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there is an electric field inside the wire, and there is a loss of potential energy, or voltage as they move but this drop in voltage is usually negligible (thought not in some applications) and we only consider that the drop in voltage comes only from the circuits elements o loads. This idealization often fails not with the wires, but within the battery itself. It is very common to add a resistance to the battery when it is in a closed circuit, because the drop of voltage produced by this load inside the battery is sometimes noticeable enough to be included.

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  • $\begingroup$ Thanks, this makes sense, but from an electric energy point of view, if the electron is sent through the wire with a field pointing in the opposite direction to its motion, wouldn't its potential increase? Another way I thought of it was if the electron's potential energy is decreasing, then the voltage should be increasing. $\endgroup$
    – ChrisL
    Jul 3, 2016 at 0:16
  • $\begingroup$ Actually, I just realized my confusion was coming from the fact that an electron would lose potential when moving between the terminals, and later regain the potential when coming back to the positive terminal. Thank you!! $\endgroup$
    – ChrisL
    Jul 3, 2016 at 0:32

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