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Question

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

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$\frac{n(n+1)}{2}$

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(A) The resistance of a wire is given by $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity,$L$ is the length,and $A$ is the cross-sectional area.Since all $n$ wires have the same dimensions,$L$ and $A$ are constant for all wires.When connected in series,the total resistance $R_{eq}$ is the sum of individual resistances: $R_{eq} = R_1 + R_2 + . . . + R_n$.Substituting the expression for resistance: $\rho_{eq} \frac{L}{A} = \rho_1 \frac{L}{A} + \rho_2 \frac{L}{A} + . . . + \rho_n \frac{L}{A}$.Dividing both sides by $\frac{L}{A}$,we get the equivalent resistivity: $\rho_{eq} = \rho_1 + \rho_2 + . . . + \rho_n$.Given $\rho_1 = 1, \rho_2 = 2, . . . , \rho_n = n$,the sum is $\rho_{eq} = 1 + 2 + 3 + . . . + n$.Using the arithmetic series sum formula,$\rho_{eq} = \frac{n(n+1)}{2}$.

$n$ conducting wires of same dimensions but having resistivities $1, 2, 3, . . . , n$ are connected in series. The equivalent resistivity of the combination is

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