The copper windings of a motor have a resistance of$\mathbf{50}\mathit{\Omega }$at$\mathbf{20}\mathbf{°}\mathbf{C}$when the motor is idle. After the motor has run for several hours, the resistance rises to$\mathbf{58}\mathit{\Omega }$.What is the temperature of the windings now? Ignore changes in the dimensions of the windings. (Use Table 26-1.)

a) Resistance of the windings,$R_0=50\Omega$

b) Initial temperature of the motor,$T_0=20°C$

c) New resistance,$R=58\Omega$

d) Coefficient of expansion,$\alpha =4.3\times {10}^{-3}{K}^{-1}$

e) Resistivity of the copper material,$\rho =1.69\times {10}^{-8}\Omega .m$

By using the equation 26-17, the parameter, initial temperature is selected as a reference temperature and the resistivity at that temperature. Using the relationship between the temperature, resistance, and resistivity, we can find the temperature.

Formulae:

The resistivity relation due to change in resistance due to temperature for resistivity being directly proportional to resistance,$\rho -{\rho }_0=\rho \alpha (T-T_0)$ (i)

Rearranging the resistivity relation of equation (i), we can get the value of the temperature of the windings of the motor now as follows:

$$\begin{gathered} T=\frac{\left(\rho -{\rho }_0\right)}{\rho \alpha }+T_0 \\ =\left(\frac{\rho }{{\rho }_0}-1\right)\frac{1}{\alpha }+T_0 \\ =\left(\frac{R}{R_0}-1\right)\frac{1}{\alpha }+T_0\left(\because \frac{\mathrm{\rho }}{{\mathrm{\rho }}_0}=\frac{\mathrm{R}}{{\mathrm{R}}_0},\mathrm{for}\mathrm{area}\mathrm{and}\mathrm{length}\mathrm{being}\mathrm{constant},\mathrm{R}=\frac{\mathrm{\rho L}}{\mathrm{A}}\right) \\ =\left(\frac{58}{50}-1\right)\frac{1}{4.3\times {10}^{-3}}+20\left(\because \mathrm{substituting}\mathrm{the}\mathrm{given}\mathrm{values}\right) \\ =\left(\frac{58}{50}-1\right)\frac{1}{4.3\times {10}^{-3}}+20 \\ =0.037\times {10}^3+20 \\ =37+20 \\ =57°C \end{gathered}$$

Hence, the value of the now temperature is $57°C$.