A copper rod of length 20cm and cross sectional area is ${\text{2mm}}^{\text{2}}$ joined with a similar aluminum rod as shown in diagram. Find the resistance of copper wire and aluminum wire if${\rho }_{\text{c}}=1.7\times {10}^{-8}\Omega \text{m},{\rho }_{\text{A}}=2.6\times {10}^{-8}\Omega \text{m}$                                

 

The resistance of the copper rod$=\frac{\rho \text{l}}{\text{A}}$ 

$=\frac{(1.7\times {10}^{-8}\Omega \text{m})\times (20\times {10}^{-2}\text{m})}{2\times {10}^{-6}{\text{m}}^{\text{2}}}=1.7\times {10}^{-3}\Omega$

Similarly, the resistance of the aluminum rod$=2.6\times {10}^{-3}\Omega$

These rods are joined in parallel so equivalent resistor

$\frac{\text{1}}{\text{R}}=\frac{1}{1.7\times {10}^{-3}}+\frac{1}{2.6\times {10}^{-3}\Omega }$

$\text{R}=\frac{1.7\times 2.6}{4.3}\times {10}^{-3}\Omega =1\text{m}\Omega$  i.e., 1 milli-ohm