Suggested languages for you:

Americas

Europe

Q68P

Expert-verified
Found in: Page 769

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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

The temperature of windings now is $57°C$.

See the step by step solution

## Step 1: The given data

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,$\alpha =4.3×{10}^{-3}{K}^{-1}$

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

## Step 2: Understanding the concept of the resistivity

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,$\rho -{\rho }_{0}=\rho \alpha \left(T-{T}_{0}\right)$ (i)

## Step 3: Calculation of the temperature of the windings

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

$T=\frac{\left(\rho -{\rho }_{0}\right)}{\rho \alpha }+{T}_{0}\phantom{\rule{0ex}{0ex}}=\left(\frac{\rho }{{\rho }_{0}}-1\right)\frac{1}{\alpha }+{T}_{0}\phantom{\rule{0ex}{0ex}}=\left(\frac{R}{{R}_{0}}-1\right)\frac{1}{\alpha }+{T}_{0}\left(\because \frac{\mathrm{\rho }}{{\mathrm{\rho }}_{0}}=\frac{\mathrm{R}}{{\mathrm{R}}_{0}},\mathrm{for}\mathrm{area}\mathrm{and}\mathrm{length}\mathrm{being}\mathrm{constant},\mathrm{R}=\frac{\mathrm{\rho L}}{\mathrm{A}}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{58}{50}-1\right)\frac{1}{4.3×{10}^{-3}}+20\left(\because \mathrm{substituting}\mathrm{the}\mathrm{given}\mathrm{values}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{58}{50}-1\right)\frac{1}{4.3×{10}^{-3}}+20\phantom{\rule{0ex}{0ex}}=0.037×{10}^{3}+20\phantom{\rule{0ex}{0ex}}=37+20\phantom{\rule{0ex}{0ex}}=57°C$

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