Question

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

Hint:

To calculate the material constant \( \beta \) for a thermistor, use the thermistor resistance-temperature equation and solve for \( \beta \) by substituting the given resistances and temperatures.

Answer 1

Step 1: Use the thermistor resistance-temperature relation.
The resistance-temperature relation for a thermistor is given by the formula:
\[ R_2 = R_1 \exp \left( \frac{\beta}{T_2} - \frac{\beta}{T_1} \right) \]
where \( R_1 \) and \( R_2 \) are the resistances at temperatures \( T_1 \) and \( T_2 \), respectively, and \( \beta \) is the material constant.
Step 2: Substitute the given values.
We are given:

Substituting these values into the equation:
\[ 1.17 = 2.25 \exp \left( \frac{\beta}{333.15} - \frac{\beta}{303.15} \right) \]
Step 3: Simplify the equation.
We know the temperatures in Kelvin:
\( T_1 = 30 + 273.15 = 303.15 \, {K} \),
\( T_2 = 60 + 273.15 = 333.15 \, {K} \).
Substituting these into the equation:
\[ 1.17 = 2.25 \exp \left( \frac{\beta}{333.15} - \frac{\beta}{303.15} \right) \]
Step 4: Solve for \( \beta \).
Now, simplify the expression inside the exponent:
\[ \frac{1}{333.15} - \frac{1}{303.15} \approx -0.0001006 \]
Thus, the equation becomes:
\[ \frac{1.17}{2.25} = \exp(-0.0001006\beta) \]
Solving for \( \beta \), we take the natural logarithm on both sides:
\[ \ln \left( \frac{1.17}{2.25} \right) = -0.0001006\beta \]
\[ \ln(0.5200) \approx -0.6532 \]
Now solve for \( \beta \):
\[ \beta = \frac{-(-0.6532)}{0.0001006} \approx 2160 \, {K} \]
Step 5: Round off the result.
Rounding to the nearest integer:
\[ \boxed{\beta = 2160 \, {K}} \]