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Q46P

Expert-verifiedFound in: Page 1275

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Calculate ${\mathit{d}}{\mathit{\rho}}{\mathbf{/}}{\mathit{d}}{\mathit{T}}$ at room temperature for (a) copper and (b) silicon, using data from Table 41-1.**

- The value of $d\rho /dT$ at room temperature for copper is $8\times {10}^{-11}\Omega .m/K$ .
- The value of $d\rho /dT$ at room temperature for silicon is $-2.1\times {10}^{2}\Omega .m/K$ .

From the value of Table 41-1:

- Resistivity of copper, ${\rho}_{Cu}=2\times {10}^{-8}\Omega .m$
- Temperature coefficient of resistivity of copper, ${\alpha}_{Cu}=4\times {10}^{-3}{K}^{-1}$
- Resistivity of silicon, ${\rho}_{Si}=3\times {10}^{3}\Omega .m$
- Temperature coefficient of resistivity of silicon, ${\alpha}_{Si}=-70\times {10}^{-3}{K}^{-1}$

**The rate of change of resistivity is positive for conductors whereas for semiconductors it is negative. Thus, the resistance of conductors increase with increase in temperature and the resistance of semiconductors decrease with increase in temperature.**

Formula:

The rate of resistivity of a material, $d\rho /dT=\rho \alpha $ (i)

Using the given data in equation (i), we can get the value of for the copper material as follows:

${\left(d\rho /dT\right)}_{Cu}=\left(2\times {10}^{-8}\mathrm{\Omega}.\mathrm{m}\right)\left(4\times {10}^{-3}{\mathrm{K}}^{-1}\right)\phantom{\rule{0ex}{0ex}}=8\times {10}^{-11}\mathrm{\Omega}.\mathrm{m}/\mathrm{K}$

Hence, the value of is $8\times {10}^{-11}\mathrm{\Omega}.\mathrm{m}/\mathrm{K}$ .

Using the given data in equation (i), we can get the value of for the copper material as follows:

${\left(d\rho /dT\right)}_{Si}=\left(3\times {10}^{3}\mathrm{\Omega}.\mathrm{m}\right)\left(-70\times {10}^{-3}{\mathrm{K}}^{-1}\right)\phantom{\rule{0ex}{0ex}}=-2.1\times {10}^{2}\mathrm{\Omega}.\mathrm{m}/\mathrm{K}$

Hence, the value of is $-2.1\times {10}^{2}\mathrm{\Omega}.\mathrm{m}/\mathrm{K}$ .

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