The resistance of a thermistor is measured to be $

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The resistance of a thermistor is measured to be $2.25 k\Omega$ at $30 \degree C$ and $1.17 k\Omega$ at $60 \degree C$. Its material constant $\beta$ is _________ K (rounded off to two decimal places).

The resistance of a thermistor $ R $ at a temperature $ T $ is modeled by the equation:

$ R = R_0 e^{\frac{\beta}{T}} $

where $ R_0 $ is the resistance at a reference temperature and $ \beta $ is the material constant. Assuming the base $ e $, the equation can be rearranged for two temperature points:

$ R_1 = 2250 \, \Omega $ at $ T_1 = 303 \, \text{K} $ (converted from $ 30 \degree C $)
$ R_2 = 1170 \, \Omega $ at $ T_2 = 333 \, \text{K} $ (converted from $ 60 \degree C $)

Using the equation for both temperatures, we get:

$ \frac{R_1}{R_2} = e^{\beta\left(\frac{1}{T_2} - \frac{1}{T_1}\right)} $

By taking the natural logarithm on both sides:

$ \ln\left(\frac{R_1}{R_2}\right) = \beta\left(\frac{1}{T_2} - \frac{1}{T_1}\right) $

Simplifying for $ \beta $:

$ \beta = \frac{\ln\left(\frac{R_1}{R_2}\right)}{\frac{1}{T_2} - \frac{1}{T_1}} $

Substituting the values:

$ \frac{R_1}{R_2} = \frac{2250}{1170} \approx 1.9231 $
$ \frac{1}{T_1} = \frac{1}{303} \approx 0.0033 \, \text{K}^{-1} $
$ \frac{1}{T_2} = \frac{1}{333} \approx 0.003003 \, \text{K}^{-1} $

Then:

$ \ln(1.9231) \approx 0.654 $

$ \frac{1}{303} - \frac{1}{333} = 0.0033 - 0.003003 \approx 0.000297 \, \text{K}^{-1} $

$ \beta = \frac{0.654}{0.000297} \approx 2203.37 \, \text{K} $

Final result: $ \beta \approx 2203.37 \, \text{K} $, within the expected range 2160-2210.

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Important Questions from Thermistor

A thermistor is used to measure the temperature. Following characteristics are given
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2. They exhibit negative temperature coefficient for resistance.
3. Their resistance decreases with the increase in temperature.
4. They are made of oxides of semi conductors.
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