A triangle is constructed using wires AB, BC and CA of same material and of resistance $α, 2 α a n d 3 α,$ respectively. Another wire of resistance $α / 3$ from A can make a sliding contact with wire BC. Find the maximum resistance of the network between points A and the point of sliding wire with BC.

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$\frac{5}{11} α$

$\frac{3}{8} α$

$\frac{3}{11} α$

$\frac{5}{8} α$

answer is C.

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Detailed Solution

$\begin{array}{l} ∴ \frac{1}{R_{eq}} = \frac{1}{(α + x)} + \frac{3}{α} + \frac{1}{(5 α - x)} \\ = \frac{α (5 α - x) + 3 (α + x) (5 α - x) - α (α + x)}{α (α + x) (5 α - x)} \\ \frac{1}{R_{eq}} = \frac{5 α^{2} - αx + 15 α^{2} - 3 x^{2} + 12 αx + a^{2} + αx}{α (α + x) (5 α - x)} \\ R_{eq} = \frac{α (α + x) (5 α - x)}{21 α^{2} + 12 αx - 3 x^{2}} \Rightarrow \frac{{dR}_{eq}}{dx} = 0 \\ R_{eq} (\max) = \frac{3}{11} α \end{array}$