Question

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

• A. $\frac{3}{7}\alpha$
• B. $\frac{2}{7}\alpha$
• C. $\frac{2}{11}\alpha$
• D. $\frac{3}{11}\alpha$

Answer 1

The correct option is D $\frac{3}{11}\alpha$

From the question, we can draw the circuit as shown, and consider $D$ as the contact point

Let $R_{eq}$ be the effective resistance between points $A$ and $D$
$\frac{1}{R_{eq}}=\frac{1}{\alpha +x}+\frac{3}{\alpha }+\frac{1}{5\alpha -x}--\left(i\right)$

On differentiating and equating $i$ to zero we get maximum or minimum.
$\frac{d}{dx}\left(\frac{1}{R_{eq}}\right)=0$ for ${\left(R_{eq}\right)}_{max}$

$\Rightarrow \frac{1}{(5\alpha -x{)}^2}-\frac{1}{(\alpha +x{)}^2}=0$

$\Rightarrow x=2\alpha$

On substituting $x$ in $\left(i\right)$, we get $R_{eq}=\left(\frac{3}{11}\right)\alpha$