Question

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.

Short answer

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

Step-by-step solution

The given data

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, ${\alpha }_{Si}=-70\times {10}^{-3}{K}^{-1}$

Understanding the concept of rate of resistivity

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)

a) Calculation for the rate of resistivity of copper

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

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

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

b) Calculation for the rate of resistivity of silicon

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

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

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