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

Correct Answer: 2160

Step 1: Use the thermistor resistance–temperature relation.

The resistance–temperature relation for a thermistor is given by:
\[ R_2 = R_1 \exp\left( \frac{\beta}{T_2} - \frac{\beta}{T_1} \right) \]
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.

Given:
\( R_1 = 2.25~\text{k}\Omega \) at \( T_1 = 30^\circ\text{C} \)
\( R_2 = 1.17~\text{k}\Omega \) at \( T_2 = 60^\circ\text{C} \)

Convert temperatures to Kelvin:
\( T_1 = 30 + 273.15 = 303.15~\text{K} \)
\( T_2 = 60 + 273.15 = 333.15~\text{K} \)

Substitute into the equation:
\[ 1.17 = 2.25 \exp\left( \frac{\beta}{333.15} - \frac{\beta}{303.15} \right) \]

Step 3: Simplify the equation.

Compute the difference of reciprocals:
\[ \frac{1}{333.15} - \frac{1}{303.15} \approx -0.0001006 \]

Thus, \[ \frac{1.17}{2.25} = \exp(-0.0001006\,\beta) \]

Step 4: Solve for \( \beta \).

Take natural logarithm on both sides:
\[ \ln\left(\frac{1.17}{2.25}\right) = -0.0001006\,\beta \]

\[ \ln(0.5200) \approx -0.6532 \]

\[ \beta = \frac{0.6532}{0.0001006} \approx 2160~\text{K} \]

Step 5: Final Answer.

\(\beta = 2160~\text{K}\)