$R_{1}=\frac{l}{\sigma_{1} A}$         ...$(i)$

and that of the second wire is

$R_{2}=\frac{l}{\sigma_{2} A}$        ....$(ii)$

As they are connected in series, so their effective

resistance is

$R_{s} =R_{1}+R_{2} $

$=\frac{l}{\sigma_{1} A}+\frac{l}{\sigma_{2} A}$       $ \quad(\text { using }(\mathrm{i}) \text { and (ii) })$

$=\frac{l}{A}\left(\frac{1}{\sigma_{1}}+\frac{1}{\sigma_{2}}\right)$       ....$(iii)$

If $\sigma_{\mathrm{eff}}$ is the effective conductivity of the combination, then

$R_{s}=\frac{2 l}{\sigma_{\mathrm{eff}} A}$        ....$(iv)$

Equating eqns. $(iii)$ and $(iv),$ we get

${\frac{2 l}{\sigma_{\mathrm{eff}} A}=\frac{l}{A}\left(\frac{1}{\sigma_{1}}+\frac{1}{\sigma_{2}}\right)} $

${\frac{2}{\sigma_{\mathrm{eff}}}=\frac{\sigma_{2}+\sigma_{1}}{\sigma_{1} \sigma_{2}} \text { or } \sigma_{\mathrm{eff}}=\frac{2 \sigma_{1} \sigma_{2}}{\sigma_{1}+\sigma_{2}}}$