Two metal wires of identical dimensions are connected in series. If ${\sigma }_1 and {\sigma }_2$  are the conductivities of the metal wires respectively, the effective conductivity of the combination is

As both metal wires are of identical dimensions, so their length and area of cross-section will be same. Let them be I and A respectively. Then The resistance of the first wire is

$R_1=\frac{l}{{\sigma }_{1}A} ....\left(i\right)$

and that of the second wire is 

$R_2=\frac{l}{{\sigma }_{2}A} ...\left(ii\right)$

As they are connected in series, so their effective resistance is

$R_s=R_1+R_2$

$=\frac{l}{{\sigma }_{1}A}+\frac{l}{{\sigma }_{2}A} (\text{ using (i) and (ii))}$

$=\frac{l}{A}\left(\frac{1}{{\sigma }_1}+\frac{1}{{\sigma }_2}\right) . . .\left(iii\right)$

$If {\sigma }_{eff} is the effective conductivity of the combination, then$

$R_s=\frac{2l}{{\sigma }_{\text{eff }}A} . . . .\left(iv\right)$

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

$\frac{2l}{{\sigma }_{\mathrm{eff}}A}=\frac{l}{A}\left(\frac{1}{{\sigma }_1}+\frac{1}{{\sigma }_2}\right)$

$\frac{2}{{\sigma }_{\text{eff }}}=\frac{{\sigma }_2+{\sigma }_1}{{\sigma }_1{\sigma }_2}\Rightarrow {\sigma }_{\text{eff }}=\frac{2{\sigma }_1{\sigma }_2}{{\sigma }_1+{\sigma }_2}$