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(B) The resistance of a wire is given by $R = \frac{L}{\sigma A}$,where $L$ is the length,$A$ is the cross-sectional area,and $\sigma$ is the conducti

Vedclass · Accepted Answer

$\frac{2 \sigma_1 \sigma_2}{\sigma_1+\sigma_2}$

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(B) The resistance of a wire is given by $R = \frac{L}{\sigma A}$,where $L$ is the length,$A$ is the cross-sectional area,and $\sigma$ is the conductivity.For two wires connected in series,the total resistance is $R_{eq} = R_1 + R_2$.The total length of the combined wire is $2L$,and the area remains $A$.Substituting the expressions for resistance: $\frac{2L}{\sigma_{eq} A} = \frac{L}{\sigma_1 A} + \frac{L}{\sigma_2 A}$.Canceling $L$ and $A$ from both sides,we get: $\frac{2}{\sigma_{eq}} = \frac{1}{\sigma_1} + \frac{1}{\sigma_2}$.Simplifying the right side: $\frac{2}{\sigma_{eq}} = \frac{\sigma_1 + \sigma_2}{\sigma_1 \sigma_2}$.Therefore,the effective conductivity is $\sigma_{eq} = \frac{2 \sigma_1 \sigma_2}{\sigma_1 + \sigma_2}$.

Two metal wires of the same length and same area of cross-section have conductivities of their material $\sigma_1$ and $\sigma_2$. If they are connected in series,the effective conductivity is:

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