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Conductors are ohmic if they have a constant resistance given that the physical conditions, such as temperature, are constant.

A filament bulb and thermistor are considered to be non-ohmic because they have a varying resistance. However, they only vary their resistance when they heat up, so aren't they technically ohmic?

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  • $\begingroup$ What do you mean by "technically ohmic"? $\endgroup$ Commented Jul 2, 2017 at 12:21
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    $\begingroup$ I mean aren't they actually ohmic. At a constant temperature a thermistor obeys ohm's law. Why then is it considered a non-ohmic conductor? $\endgroup$ Commented Jul 2, 2017 at 12:29
  • $\begingroup$ Right, but a thermistor is designed to have relatively large change in resistance with temperature which is then exploited in normal operation. Contrast with as 'good' resistor which is designed to have relatively small change in resistance with temperature such that, in normal operation, the resistance is effectively constant as would be the case if the resistor were ideal. $\endgroup$ Commented Jul 2, 2017 at 12:37
  • $\begingroup$ Would Electrical Engineering be a better home for this question? $\endgroup$
    – Qmechanic
    Commented Jul 2, 2017 at 13:43
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    $\begingroup$ I suppose it would fit in either stack exchange. It's both a physics and electronics question. $\endgroup$ Commented Jul 2, 2017 at 14:09

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In the case of the filament bulb, the test that is commonly used to produce the curvature in it's IV graph does not regulate the temperature of the device. This is technically an improper test for ohmic behaviour as Ohm's law states that the current is proportional to the voltage with all other physical conditions controlled (the main one being temperature). If you regulated the temperature, as a metal, the filament would probably produce ohmic behaviour and a constant resistance.

Similar logic would apply for a thermistor.

A truly non-Ohmic device would be a diode which does not let current flow when a negative voltage is applied.

Another important thing to note is that a real IV characteristics curve for an ohmic conductor where it's temperature is not controlled would not be perfectly linear. Heating effects of the current would cause changes in its resistivity. It is only when the temperature is controlled that you would get a linear graph.

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A thermistor is designed to have a relatively large change in resistance with temperature which is then exploited in normal operation.

Contrast this with a 'good' resistor which is designed to have relatively small changes in resistance with temperature such that, in normal operation, the resistance is effectively constant (as would be the case if the resistor were ideal).

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  • $\begingroup$ This actually answers the question why some components (e.g. a resistor) are considered omic and others (eg. a filament bulb) are not. This should be the accepted answer. $\endgroup$
    – Somanna
    Commented Jun 15, 2021 at 12:06
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Actually the most correct statement of Ohm's Law is that a conductor is Ohmic if the current through it is directly proportional to the potential difference across it,$V=RI$, provided other physical conditions remain constant , in a thermistor the change in current is non-linear with voltage thus it's not an Ohmic conductor( the I-V graph of a thermistor is not a straight line). Same reason for a filament lamp, because as the current in a filament lamp increases it heats up so the resistance changes so again the I-V graph is not a straight line. Thus, these components are not Ohmic conductors.

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  • $\begingroup$ But that's true for all conductors, barring superconductors of course. All conductors heat up when a current passes through them, which will mean their graph is actually not a straight line, but rather if you increase the current enough the graph will begin to curve ever so slightly. $\endgroup$ Commented Jul 2, 2017 at 17:21
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    $\begingroup$ Yes precisely, so ideally all conductors are non- Ohmic just like ideal gases( how all gases are non ideal ) but some are very close thus they are considered roughly Ohmic. $\endgroup$ Commented Jul 3, 2017 at 5:47

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