Timeline for Does V=IR define resistance for any conductor?
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| Mar 16, 2019 at 21:16 | comment | added | Beyza Yıldırım | Thank you! I wish they've phrased the question as "does R=V/I define the resistance for any conductor" so that students like me wouldn't fall into the bait of nonohmic resistors | |
| Mar 7, 2019 at 20:46 | comment | added | untreated_paramediensis_karnik | Let us continue this discussion in chat. | |
| Mar 7, 2019 at 18:14 | comment | added | BioPhysicist | @thermomagneticcondensedboson No. I am saying this would be more of an operational definition. Where you apply a voltage, measure the current that results, and this ratio would be the resistance. | |
| Mar 7, 2019 at 18:00 | comment | added | untreated_paramediensis_karnik | Does your answer mean that the resistivity depends on the current? Because the resistance is equal to the integral of the resistivity with respect to the spatial dimensions of the sample. It does seem a bit strange to me. | |
| Mar 7, 2019 at 15:57 | comment | added | untreated_paramediensis_karnik | I agree, it is exactly similar, for most (or all) metals, semiconductors, etc. The resistance depends on the current direction, so R is never uniquely defined. You're right that it's fine and there's no problem. It just looked like R=V/I yield a single well defined value while it isn't the case. | |
| Mar 7, 2019 at 12:51 | comment | added | BioPhysicist | @thermomagneticcondensedboson Ok? So then what's wrong with having a piecewise resistance if we are talking about steady state? If you look at my footnote a give a (much simpler) example of resistance for when the potential has different polarity. | |
| Mar 7, 2019 at 12:43 | comment | added | untreated_paramediensis_karnik | By the way, this also holds for a homogeneous metal such as nickel or platinum (if not all metals). While the Joule heat doesn't change with current reversal, the Thomson heat does. Since the resistivity is temperature dependent, when the current is passed in a direction, a part of the sample is cooled, while the other part is heated up (let's say the temperature of the ends of the sample are kept at a fixed temperature, it may not even need to be the same). In that case the voltage measured depends on the direction of the current. The temperature distribution is not symmetric with respect to | |
| Mar 7, 2019 at 12:37 | comment | added | untreated_paramediensis_karnik | I should have written "Since the resistivity is usually temperature dependent..." in my last sentence. | |
| Mar 7, 2019 at 12:28 | comment | added | untreated_paramediensis_karnik | No, I am talking about a steady-state, but I do not ignore thermoelectric effects, that's the only difference. I use a non-homogeneous semiconductor to ensure there will be a non homogeneous heating across the (resistive) material, regardless of whether the Seebeck coefficient depends on temperature. Since the resistance is usually temperature dependent, there will be a different resistance measured (as V/I) if the current is reversed. | |
| Mar 7, 2019 at 12:15 | comment | added | BioPhysicist | @thermomagneticcondensedboson Do you mean to say we can't use the definition if there is time dependence for a constant $V$? | |
| Mar 7, 2019 at 12:03 | comment | added | untreated_paramediensis_karnik | I am not sure it is always possible to define R as V/I. Take the example of a non-homogeneous doped semiconductor and consider the generalized Ohm's law $\vec J = -\sigma \nabla V - \sigma S \nabla T$. There will be (in addition to the Seebeck effect), a non-homogeneous Thomson heating across the resistor, as well as (possibly) a Peltier effect at the ends of the material. I think this can lead to two different values for R, if R is defined as V/I, according to the direction of the current. Am I missing something? | |
| Mar 7, 2019 at 11:54 | comment | added | Steeven | +1 for "poor math talk" ... and for a good answer. | |
| Mar 7, 2019 at 11:41 | history | edited | BioPhysicist | CC BY-SA 4.0 |
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| Mar 7, 2019 at 9:12 | comment | added | my2cts | With this definition $ P=VI = I^2R$ still holds so I agree. | |
| Mar 7, 2019 at 7:29 | comment | added | user137289 | I agree. For a diode in the forward direction, there is also the differential resistance, which is about $r \approx 25/I \ \Omega$ where the current is in mA. physics.stackexchange.com/questions/387030/… | |
| Mar 7, 2019 at 4:36 | history | answered | BioPhysicist | CC BY-SA 4.0 |