Discussion
A thermistor of 10 k\( \Omega \) at 25 \( ^\circ \)C and \( \beta = 3500 \) K is warmed to 35 \( ^\circ \)C. Its resistance
Answer: C
Correct answer: falls to roughly 6.5 k\( \Omega \)
A negative coefficient thermistor follows \[ R = R_0\exp\left[\beta\left(\frac{1}{T}-\frac{1}{T_0}\right)\right] = 10\,\mathrm{k}\Omega \times \exp\!\left[3500\left(\tfrac{1}{308}-\tfrac{1}{298}\right)\right] = 10 \times e^{-0.381} = 6.8\ \mathrm{k}\Omega. \] So ten degrees removes about a third of the resistance - a far larger change than a platinum sensor would give, which is the thermistor’s advantage, bought at the cost of a strongly non-linear characteristic.
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