 Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q46P

Expert-verified Found in: Page 1275 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # Calculate ${\mathbit{d}}{\mathbit{\rho }}{\mathbf{/}}{\mathbit{d}}{\mathbit{T}}$ at room temperature for (a) copper and (b) silicon, using data from Table 41-1. 1. The value of $d\rho /dT$ at room temperature for copper is $8×{10}^{-11}\Omega .m/K$ .
2. The value of $d\rho /dT$ at room temperature for silicon is $-2.1×{10}^{2}\Omega .m/K$ .
See the step by step solution

## Step 1: The given data

From the value of Table 41-1:

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

## Step 2: 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)

## Step 3: 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:

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

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

## Step 4: 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:

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

Hence, the value of is $-2.1×{10}^{2}\mathrm{\Omega }.\mathrm{m}/\mathrm{K}$ . ### Want to see more solutions like these? 