Timeline for Resistance and drift velocity
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 3, 2020 at 23:01 | vote | accept | DJTS | ||
| Dec 3, 2020 at 14:20 | comment | added | DJTS | Thank you. So I am assuming that this relationship $R\propto v$ is just a manifestation of the present example and not a fundamental relationship between resistance and drift velocity. This makes sense. So what is happening is a larger potential difference is being applied across the larger resistance, in order to keep the current constant. This, in turn, increases the drift velocity, which lead me to the fallacious conclusion that resistance is dependent on drift velocity. In that case, the correct answer should be $R_X = R_Y / 2$, from the resistivity formula. Is that correct? | |
| Dec 3, 2020 at 14:15 | comment | added | Roger V. | The situations with a current bias (as we have here) may seem counter-intuitive at first, since most introductory discussions of circuits and electricity usually deal with the voltage bias (i.e., when the voltage is kept constant). | |
| Dec 3, 2020 at 14:12 | comment | added | Roger V. | Resistance is independent on the drift velocity - it is a characteristic of a conductor. If you have two conductors with different resistance and you want to the same current flowing in both, you need to apply higher potential difference to the one with higher resistance. | |
| Dec 3, 2020 at 14:09 | comment | added | DJTS | Yes, I can now see the relationship between Drude's model and resistance, thanks. However, what I am left confused by is the intuition of the relationship $R\propto v$. How do these formulae explain the rise in drift velocity with an increase in resistance? Do they imply a larger electric field acting on the electrons? | |
| Dec 3, 2020 at 14:04 | comment | added | Roger V. | @DJTS That current density is proportional to the electric field is known as Ohm's law: $j=\sigma E$, can be converted to the current and voltage bias as: $I=jA=\sigma A/l V=V/R$. Perhaps this resolves the confusion? | |
| Dec 3, 2020 at 13:58 | comment | added | DJTS | Thank you for the link, but I still don't understand why a higher resistance would lead to higher drift velocity. The main point of the model seems to be that the current density is proportional to the electric field. How does this relate to the resistance? Also, I'm not quite sure which ratio you are saying is correct. Are you saying $R_X = 2R_Y$ after all? | |
| Dec 3, 2020 at 11:34 | history | answered | Roger V. | CC BY-SA 4.0 |