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(D) For two resistors connected in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$.Since th

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$\frac{{\rho _1}{\rho _2}\left( {{A_1} + {A_2}} \right)}{{{A_1}{\rho _2} + {A_2}{\rho _1}}}$

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(D) For two resistors connected in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$.Since the conductors have the same length $\ell$,their resistances are $R_1 = \frac{\rho_1 \ell}{A_1}$ and $R_2 = \frac{\rho_2 \ell}{A_2}$.The equivalent resistance of the combination is $R_{eq} = \frac{\rho_{eq} \ell}{A_1 + A_2}$.Substituting these into the parallel combination formula:$\frac{A_1 + A_2}{\rho_{eq} \ell} = \frac{A_1}{\rho_1 \ell} + \frac{A_2}{\rho_2 \ell}$.Canceling $\ell$ from both sides:$\frac{A_1 + A_2}{\rho_{eq}} = \frac{A_1}{\rho_1} + \frac{A_2}{\rho_2} = \frac{A_1 \rho_2 + A_2 \rho_1}{\rho_1 \rho_2}$.Therefore,$\rho_{eq} = \frac{\rho_1 \rho_2 (A_1 + A_2)}{A_1 \rho_2 + A_2 \rho_1}$.

Two conductors of same length are connected in parallel as shown in the figure. Their cross-sectional areas are $A_1$ and $A_2$ and their resistivities are $\rho_1$ and $\rho_2$ respectively. The equivalent resistivity of this combination is

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