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(B) For two wires connected in series,the total resistance $R_{eq}$ is the sum of individual resistances: $R_{eq} = R_1 + R_2$.Since the wires have th

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$\frac{\rho_1 + \rho_2}{2}$

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(B) For two wires connected in series,the total resistance $R_{eq}$ is the sum of individual resistances: $R_{eq} = R_1 + R_2$.Since the wires have the same dimensions (length $l$ and cross-sectional area $A$),the resistance of each wire is given by $R = \rho \frac{l}{A}$.Substituting this into the series combination formula: $\rho_{eq} \frac{2l}{A} = \rho_1 \frac{l}{A} + \rho_2 \frac{l}{A}$.Dividing both sides by $\frac{l}{A}$,we get: $2\rho_{eq} = \rho_1 + \rho_2$.Therefore,the equivalent resistivity is $\rho_{eq} = \frac{\rho_1 + \rho_2}{2}$.

Two wires of the same dimensions but resistivities $\rho_1$ and $\rho_2$ are connected in series. The equivalent resistivity of the combination is

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