Two wires of equal diameters,of resistivities | vedclass

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(A) The resistance of a wire is given by $R = \rho \frac{l}{A}$,where $\rho$ is resistivity,$l$ is length,and $A$ is the cross-sectional area.Since th

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$\frac{\rho_1 l_1 + \rho_2 l_2}{l_1 + l_2}$

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(A) The resistance of a wire is given by $R = \rho \frac{l}{A}$,where $\rho$ is resistivity,$l$ is length,and $A$ is the cross-sectional area.Since the diameters are equal,the cross-sectional areas $A$ are equal for both wires.The resistances are $R_1 = \frac{\rho_1 l_1}{A}$ and $R_2 = \frac{\rho_2 l_2}{A}$.When joined in series,the equivalent resistance is $R_{eq} = R_1 + R_2$.The equivalent wire has total length $(l_1 + l_2)$ and the same cross-sectional area $A$,so $R_{eq} = \frac{\rho_{eq}(l_1 + l_2)}{A}$.Equating the expressions: $\frac{\rho_{eq}(l_1 + l_2)}{A} = \frac{\rho_1 l_1}{A} + \frac{\rho_2 l_2}{A}$.Canceling $A$ from both sides,we get $\rho_{eq}(l_1 + l_2) = \rho_1 l_1 + \rho_2 l_2$.Therefore,the equivalent resistivity is $\rho_{eq} = \frac{\rho_1 l_1 + \rho_2 l_2}{l_1 + l_2}$.

Two wires of equal diameters,of resistivities $\rho_1$ and $\rho_2$ and lengths $l_1$ and $l_2$,respectively,are joined in series. The equivalent resistivity of the combination is

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