Temperature of Motor Copper Winding from Resistance Change

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Electric Circuits Intermediate Electric Current

Source: High school physics (Chinese)

Problem

When a motor is not running, the resistance of its copper winding at $t_0 = 20\,^{\circ}\!\text{C}$ is $R_0 = 50\ \Omega$. After running for several hours, the resistance rises to $R = 58\ \Omega$. Take the temperature coefficient of resistance for copper to be $\alpha = 4.3 \times 10^{-3}\ {}^{\circ}\!\text{C}^{-1}$.

What is the temperature of the copper winding at this time?
$t \approx 57\,^{\circ}\!\text{C}$

The linear temperature dependence of resistance gives

$$R = R_0 \left[ 1 + \alpha (t - t_0) \right],$$

so

$$t = t_0 + \frac{R - R_0}{\alpha R_0}.$$

Substituting:

$$t = 20 + \frac{58 - 50}{4.3 \times 10^{-3} \times 50} = 20 + \frac{8}{0.215} \approx 20 + 37.2 \approx 57\,^{\circ}\!\text{C}.$$