Question

$ABCD$ is a square where each side is uniform wire of resistance $1\Omega$. Find a point $E$ on $CD$ such that a uniform wire of resistance $1\Omega$ is connected across $AE$ and a potential difference is applied across $A$ and $C$, the points $B$ and $E$ will be equipotential.

Answer 1

Let the length of sides of square be $1cm$.

let $E$ be $xcm$ distance from $D$.

$\therefore CE=1-x$

$\therefore$ Resistance of $DE=\frac{x}{1+x}\Omega$

$\therefore$ Resistance of $EC=\frac{1}{1+x}\Omega$

Resistance along $BCE=[1+\frac{1}{1+x}]\Omega$

$=\frac{x+2}{x-1}\Omega$

Equivalent resistance between $AE=\left(\frac{2x+1}{3x+2}\right)\Omega$

Equivalent resistance between $BE=\frac{5x+1}{3x+2}\Omega$

Equating the $2$ equivalent resistance along path $BCE$ and $BAE$ makes $\left\{\frac{(x+2)}{(x+1)}\right\}=\frac{5x+1}{3x+2}$

$⟹x=\frac{1}{\sqrt{2}}$

$\therefore E$ is $\frac{1}{\sqrt{2}}$ units from $D$.