Chapter 18, Problem 14P

To determine

The ratio of the resistance of $10m$ of aluminium wire $2.0mm$ in diameter, to $20.0m$ of copper wire $2.5mm$ in diameter.

Answer to Problem 14P

  $R_{Al}=0.0844\Omega$ and $R_{Cu}=0.0342\Omega$

Explanation of Solution

Given info:

Length of aluminium wire, $L_{Al}=10.0m$

Diameter of aluminium wire, $D_{Al}=2.0mm=2.0\times {10}^{-3}m$

Resistivity of Aluminium wire, $\rho =2.65\times {10}^{-8}\Omega \cdot m$

Length of copper wire, $L_{Cu}=20.0m$

Diameter of copper wire, $D_{Cu}=2.5mm=2.5\times {10}^{-3}m$

Resistivity of Copper wire, $\rho =1.68\times {10}^{-8}\Omega \cdot m$

Formula used:

It is known that resistance:

  $R=\frac{\rho L}{A}$ or $R=\frac{\rho L}{\pi {\left(\frac{D}{2}\right)}^2}$

Where,

  $\rho$ is the resistivity.

  $L$ is the length of wire

  $A$ is the cross-sectional area of the wire.

  $\pi$ is constant and $r$ is a radius

  $d$ is a diameter.

Calculation:

Calculation for resistance of aluminium wire by using the $R_{Al}=\frac{\rho L_{Al}}{\pi {\left(\frac{D_{Al}}{2}\right)}^2}$ :

  $R_{Al}=\left(2.65\times {10}^{-8}\Omega \cdot m\right)\frac{\left(10.0m\right)}{\left[\pi {\left(\frac{2.0\times {10}^{-3}}{2}\right)}^2\right]}=0.0844\Omega$

Calculation for resistance of copper wire by using the $R_{Cu}=\frac{\rho L_{Cu}}{\pi {\left(\frac{D_{Cu}}{2}\right)}^2}$ :

  $R_{Cu}=\left(1.68\times {10}^{-8}\Omega \cdot m\right)\frac{\left(20.0m\right)}{\left[\pi {\left(\frac{2.5\times {10}^{-3}}{2}\right)}^2\right]}=0.0342\Omega$

Conclusion:

Resistance of aluminium wire $R_{Al}=0.0844\Omega$

Resistance of copper wire $R_{Cu}=0.0342\Omega$