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(C) For two wires of length $l$ and cross-sectional area $A$, the resistance is given by $R = \frac{l}{\sigma A}$.When connected in series, the equiva

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$\frac{2 \sigma_{1} \sigma_{2}}{\sigma_{1}+\sigma_{2}}$

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(C) For two wires of length $l$ and cross-sectional area $A$, the resistance is given by $R = \frac{l}{\sigma A}$.When connected in series, the equivalent resistance is $R_{eq} = R_{1} + R_{2}$.Substituting the values, we get $\frac{2l}{\sigma_{eq} A} = \frac{l}{\sigma_{1} A} + \frac{l}{\sigma_{2} A}$.Dividing both sides by $\frac{l}{A}$, we get $\frac{2}{\sigma_{eq}} = \frac{1}{\sigma_{1}} + \frac{1}{\sigma_{2}}$.$\frac{2}{\sigma_{eq}} = \frac{\sigma_{1} + \sigma_{2}}{\sigma_{1} \sigma_{2}}$.Therefore, the effective conductivity is $\sigma_{eq} = \frac{2 \sigma_{1} \sigma_{2}}{\sigma_{1} + \sigma_{2}}$.

Two metal wires of identical dimensions are connected in series. If $\sigma_{1}$ and $\sigma_{2}$ are the electrical conductivities of the metal wires respectively, the effective conductivity of the combination is

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