Timeline for How can an electron have 0 electric potential after exiting a resistor but have current?
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11 events
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| Jan 27, 2019 at 15:25 | comment | added | Abagon | @ThePhoton - I understand that that the electrons that are exiting the resistor have a small non-zero electric potential because of a small resistance the wire has after the resistor which allows them to flow back to the positive terminal, but how about in a superconductor? The resistance after the resistor is 0, therefore all the electric potential is used by the resistor and there is no more electric potential for the electrons to flow back to the positive terminal. Why is there still current? How can there be current with 0 electric potential and 0 resistance according to Simon B's answer. | |
| Jan 27, 2019 at 9:47 | answer | added | jonk | timeline score: 3 | |
| Jan 27, 2019 at 8:53 | answer | added | wbeaty | timeline score: -1 | |
| Jan 27, 2019 at 8:27 | answer | added | sarthak | timeline score: 3 | |
| Jan 27, 2019 at 6:26 | comment | added | Chris Stratton | @jsotola - your "push not pull" theory is mistaken. It's actually both, something that can be demonstrated with a thought experiment of a positively charged sphere attracting an electron even when there's no negatively charged one to repel it in that direction. | |
| Jan 27, 2019 at 5:19 | comment | added | user57037 | You should try to understand what is happening from the perspective of field theory. The voltage between + and - terminals of a battery sets up electric field gradients. The potential of an electron depends on where it is with respect to the field. After passing through a resistor, the electron is unchanged. But it has less potential because the field is different on the other side of the resistor. | |
| Jan 27, 2019 at 4:55 | comment | added | jsotola | simplistically put: the positive terminal of a battery does not pull the electrons in the wire ..... the positive terminal simply has room for the electrons to go .... the chemical reaction in the battery forces the electrons out of the negative terminal ..... those electrons push against other electrons that are already in the wire .... the push propagates through the wire until it reaches the positive terminal | |
| Jan 27, 2019 at 4:55 | comment | added | The Photon | You should read Simon B's answer to the question you linked to. | |
| Jan 27, 2019 at 4:43 | comment | added | The Photon | If it makes you feel better, remember that the wire from the resistor back to the negative terminal actually has a very small non-zero resistance, maybe a few milliohms. So the electron exiting the resistor actually has a small non-zero potential (by the voltage divider rule), which pushes it through the wire. But in many situations it just makes the math more complicated and doesn't change the results much to worry about a few millivolts across the wire when there were several volts across the resistor. | |
| Jan 27, 2019 at 4:30 | review | First posts | |||
| Jan 27, 2019 at 6:08 | |||||
| Jan 27, 2019 at 4:29 | history | asked | Abagon | CC BY-SA 4.0 |