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Why is it that potential difference decreases in thermistor when temperature of circuit is increased? From my understanding when temperature increases, there are more free electrons ($n$) so according to $I=navq$, current increases as $n$ is greater. But why is it that potential difference decreases; if $V=IR$ shouldn't pd also increase?

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You are confusing cause and effect. The current is a function of the applied voltage, or the other way around. It is incorrect to say that a thermistor changes in voltage or current as a function of temperature. That's dependent on the circuit it is connected to.

The external property that changes in thermistors as a function of temperature is the resistance. R goes up or down with increased temperature, depending on the kind of thermistor you have. Most things just called a "thermistor" exhibit decreased resistance with rising temperature. There are also such things as PTC (positive temperature coefficient) thermistors that exhibit the opposite effect.

As you say, V = IR. If R goes down, the V will go down at the same I. Conversely, if V is held constant, the I goes up. Both can be legitimate ways to run a thermistor. Probably the most common way is to put the thermistor in series with a fixed voltage and a fixed resistance. In that case, both I and V go down as R goes down.

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if the temperature decreases the resistance of the thermistor will increase so the potential different will increase because V=IR so if R is increased then the total answer for V will increase

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  • $\begingroup$ The resistance of a thermistor goes down with temperature, so this is incorrect. $\endgroup$ – Chris Apr 14 at 16:30

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