Question

Resistance \( R \) of a thermistor varies as a function of temperature \( T \) such that \[ R(T) = R_0 \exp \left[ \beta \left( \frac{1}{T} - \frac{1}{T_0} \right) \right] \] where \( \beta = 3100 \, {K} \), \( R_0 \) and \( T_0 \) are positive constants. If the relative error in measuring \( R \) is 10%, what is the relative error (in percentage) in measuring 310 K temperature? (rounded off to the nearest integer)

Hint:

When finding the relative error in a dependent variable, use logarithmic differentiation to link the relative errors of both variables.

Answer 1

We are given the equation for the resistance of the thermistor: \[ R(T) = R_0 \exp \left[ \beta \left( \frac{1}{T} - \frac{1}{T_0} \right) \right] \] The goal is to find the relative error in temperature, \( T \), when the relative error in resistance \( R \) is given as 10%. 
Step 1: Use logarithmic differentiation. To find the relative error in temperature, we first take the natural logarithm of both sides of the equation: \[ \ln R(T) = \ln R_0 + \beta \left( \frac{1}{T} - \frac{1}{T_0} \right) \] Next, we differentiate with respect to \( T \): \[ \frac{d}{dT} \ln R(T) = \frac{d}{dT} \left[ \beta \left( \frac{1}{T} - \frac{1}{T_0} \right) \right] \] This gives: \[ \frac{1}{R(T)} \frac{dR(T)}{dT} = -\frac{\beta}{T^2} \] Step 2: Relating relative errors. The relative error in resistance is given by: \[ \frac{\Delta R}{R} = 10% = 0.1 \] Using the relationship between relative errors, we can express the relative error in temperature: \[ \frac{\Delta T}{T} = \left| \frac{-\beta}{T^2} \cdot \frac{1}{R(T)} \cdot \Delta R \right| \] Substitute the values of \( \beta = 3100 \, {K} \), \( T = 310 \, {K} \), and the given relative error in resistance (0.1): \[ \frac{\Delta T}{T} = \left| \frac{-3100}{(310)^2} \cdot \Delta R \right| \] \[ \frac{\Delta T}{T} = 0.01 \] Thus, the relative error in temperature is approximately **1%**.