A thermistor is connected across a supply that maintains a constant potential difference of 6.00 V across it. At a temperature of 20 °C, the thermistor has resistance R1 = 1200. The thermistor is then warmed to a higher temperature, at which its resistance falls to R2 = 300. (a)…

Answer: ((a)) I1 = 5.00×10^-3 A ((b)) I2 = 2.00×10^-2 A ((c)) In the semiconductor, warming the thermistor releases many more free charge carriers, so the number density n increases sharply; since I = Anvq, a much larger n (with A and q fixed, and v not falling enough to compensate) means a larger current flows for the same V, so R = V/I f…. Part (a): Current at 20 degrees Celsius The supply maintains a constant potential difference V = 6.00 V across the thermistor. Using Ohm's law, I = V/R, with R1 = 1200: I1 = 6.00/1200 = 5.00×10^-3 A Part (b): Current at the higher temperat…

AS Level Physics 9702: Thermistor Resistance and Charge Carrier Density

Part (a): Current at 20 degrees Celsius

The supply maintains a constant potential difference $V = 6.00\text{ V}$ across the thermistor. Using Ohm’s law, $I = V/R$ , with $R_1 = 1200\,\Omega$ : $I_1 = \frac{6.00}{1200} = 5.00\times10^{-3}\text{ A}$

Part (b): Current at the higher temperature

The p.d. is still $V = 6.00\text{ V}$ (unchanged, since it is held constant by the supply), but the resistance has fallen to $R_2 = 300\,\Omega$ : $I_2 = \frac{6.00}{300} = 2.00\times10^{-2}\text{ A}$

Check: $R_2 = R_1/4$ , so $I_2$ should be $4\times I_1$ : $4 \times 5.00\times10^{-3} = 2.00\times10^{-2}\text{ A}$ , which matches. $\checkmark$

Part (c): Explaining the falling resistance using I = Anvq

The current in the thermistor is related to its charge carriers by: $I = Anvq$

A thermistor is made from a semiconductor, in which (unlike a metal) the number density $n$ of free charge carriers is not fixed. At low temperature relatively few charge carriers are free to move, but as the temperature rises, thermal energy releases significantly more of them from the semiconductor’s atoms, so $n$ increases sharply.

With the cross-sectional area $A$ and the carrier charge $q$ unchanged, this large increase in $n$ means that, for the same applied potential difference $V$ (and hence roughly similar drift velocity $v$ ), a much greater current $I$ can flow. Since resistance is $R = V/I$ , a larger current at the same $V$ corresponds to a smaller resistance. This is why the thermistor’s resistance falls as it warms up. The effect of the rapidly increasing number of charge carriers outweighs any increase in carrier collisions (which would otherwise tend to reduce $v$ and raise resistance, as happens in a metal).

Final answers

(a) Current at $20\,^{\circ}\text{C}$ $=$ $5.00\times10^{-3}\text{ A}$
(b) Current at the higher temperature $=$ $2.00\times10^{-2}\text{ A}$
(c) Resistance falls because warming the semiconductor releases many more free charge carriers, so the number density $n$ in $I = Anvq$ rises sharply, increasing $I$ for the same $V$ and hence decreasing $R = V/I$ .