In a simple electric circuit consisting of a power supply and a resistor;
My textbook tells me that the potential difference across the lead wires is always 0, and the reasoning given is that since the resistance of these wires is negligible, negligible amounts of electric potential energy is dissipated as heat.
But i understand potential difference to be the change in electric potential energy between the two points (per coulomb); if this is so, even though the electric potential energy isn't converted to heat across the wires, it is at least converted to kinetic energy as the charge carriers move from one point to the other and so there must be a change in electric potential between the two points and therefore a non-zero potential difference?
Correct me if I'm wrong.
Answers & Comments
Verified answer
Yeah you're wrong in theory, though not wrong in practice.
The theory being that electrons in a conductor are completely free to move in the conduction band so they require zero field to get them moving, and so we say that a conductor supports zero field drop. The conduction band is viewed as a sea of electrons, the energy to get them from valence to conduction band being supplied by the temperature to the latice of the conductor. Its the thermal energy that's supplying the energy that you are (astutely) looking for, or so the theory goes.
In practice this in not wholey possible and even superconductors require some energy input to get the electrons moving.
You would also expect to have to overcome a potential caused by the work function difference if you have different metals connected together but the simple model is ignoring this.
The resistance of the wires is so low that, for all practical purposes, the voltage drop across them can be safely ignored. (Yes, there will be some small non-zero potential difference.)
first of all, you choose to degree for voltage not amperage and if your voltmeter is desperate to AC volts then it somewhat is in fact what you would be measuring. 2nd, because of the fact that is AC that is not proper which lead is going on the black twine and which on the white twine. The voltage could examine 120v regardless. A voltmeter could be extremely overkill for this sort of application (whether that is going to do the job). an hardship-free circuit tester (see hyperlink for one occasion) could be handier.
V=IR and you are overthinking this