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(B) The resistance of a wire is given by $R = \frac{\ell}{\sigma A}$,where $\ell$ is the length,$\sigma$ is the conductivity,and $A$ is the cross-sect

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 $R = \frac{\ell}{\sigma A}$,where $\ell$ is the length,$\sigma$ is the conductivity,and $A$ is the cross-sectional area.For two wires connected in series,the equivalent resistance is $R_{eq} = R_1 + R_2$.Since the wires have identical dimensions,let the length of each wire be $\ell$ and the cross-sectional area be $A$.The equivalent resistance of the combination is $R_{eq} = \frac{2\ell}{\sigma_{eq} A}$.Substituting the expressions for $R_1$,$R_2$,and $R_{eq}$ into the series formula:$\frac{2\ell}{\sigma_{eq} A} = \frac{\ell}{\sigma_1 A} + \frac{\ell}{\sigma_2 A}$Dividing both sides by $\frac{\ell}{A}$:$\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 conductivities of the metal wires respectively,the effective conductivity of the combination is

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