Two conductors of same length are connected in parallel as shown in figure. Their cross-sectional areas $A_1$ and $A_2$ and their resistivities are ${\rho _1}$ and  ${\rho _2}$ respectively. The equivalent resistivity of this combination is
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$\frac{1}{\mathrm{R}_{\mathrm{p}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$

$\frac{\mathrm{A}_{1}+\mathrm{A}_{2}}{\rho_{\mathrm{eq}} \ell}=\frac{\mathrm{A}_{1}}{\rho_{1} \ell}+\frac{\mathrm{A}_{2}}{\rho_{2} \ell}$

$\rho_{\mathrm{eq}}=\frac{\rho_{1} \rho_{2}\left(\mathrm{A}_{1}+\mathrm{A}_{2}\right)}{\rho_{2} \mathrm{A}_{1}+\rho_{1} \mathrm{A}_{2}}$

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