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(A) The resistance of a wire is given by $R = \rho \frac{L}{A}$.Since the wires are joined in series,the total resistance $R_{eq}$ is the sum of indiv

Vedclass · Accepted Answer

$\frac{{\rho _1}{x_1} + {\rho _2}{x_2}}{{{x_1} + {x_2}}}$

Vedclass · Answer

(A) The resistance of a wire is given by $R = \rho \frac{L}{A}$.Since the wires are joined in series,the total resistance $R_{eq}$ is the sum of individual resistances: $R_{eq} = R_1 + R_2$.Let the equivalent resistivity be $\rho$ and the cross-sectional area be $A$ (which is the same for both wires).Thus,$\rho \frac{(x_1 + x_2)}{A} = \rho_1 \frac{x_1}{A} + \rho_2 \frac{x_2}{A}$.Canceling $A$ from both sides,we get $\rho (x_1 + x_2) = \rho_1 x_1 + \rho_2 x_2$.Therefore,the equivalent resistivity is $\rho = \frac{\rho_1 x_1 + \rho_2 x_2}{x_1 + x_2}$.

Two wires of equal diameters with resistivities ${\rho _1}$ and ${\rho _2}$ and lengths $x_1$ and $x_2$ are joined in series. The equivalent resistivity of the combination is

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