Question

$\text{ABCD}$ is a square, whose each side is a uniform wire of resistance $1\Omega$. There is point $E$ on $CD$ such that if a uniform wire of resistance $1\Omega$ is connected across $AE$ and a constant potential difference is applied across $A$ and $C$, the points $B$ and $E$ will be equipotential. Value of length ratio $\frac{CE}{ED}$ will be:

• A. $\frac{\sqrt{2}}{1}$
• B. $\frac{1}{1}$
• C. $\frac{\sqrt{3}}{1}$
• D. $\frac{3}{1}$

Answer 1

The correct option is A $\frac{\sqrt{2}}{1}$

The situation is shown in the figure below:


The points $B$ and $E$ will be equipotential, when
$\text{Effective resistance across AE=effective resistance across CE}$

Because resistance across $AB$ and $BC$ are same, so it will be a balanced state of Wheatstone bridge which happens when

$\frac{R_{AB}}{R_{AE}}=\frac{R_{BC}}{R_{CE}}$

$\Rightarrow R_{AE}=R_{CE}.....\left(1\right)[\because R_{AB}=R_{BC}=1\Omega ]$

Where, $R_{AE}$ is the equivalent resistance of the loop $ADEA$. Thus,

$R_{AE}=\frac{(1+x)\times 1}{(1+x)+\left(1\right)}...\left(2\right)$

Using equations $\left(1\right)$ and $\left(2\right)$,

$\frac{(1+x)\left(1\right)}{(1+x)+\left(1\right)}=(1-x)$

$\Rightarrow 1+x=2-2x+x-x^2$

$\Rightarrow x^2+2x-1=0$

Solving we get

$\Rightarrow x=\frac{-2+\sqrt{2^2-4\times 1\times (-1)}}{2\times 1}=\sqrt{2}-1$

or $\Rightarrow x=\frac{-2-\sqrt{2^2-4\times 1\times (-1)}}{2\times 1}=-\sqrt{2}-1$

$\therefore x=(\sqrt{2}-1)$

$\Rightarrow 1-x=1-(\sqrt{2}-1)=2-\sqrt{2}=\sqrt{2}(\sqrt{2}-1)$

$\Rightarrow \frac{CE}{ED}=\frac{1-x}{x}=\frac{\sqrt{2}(\sqrt{2}-1)}{(\sqrt{2}-1)}$

$\therefore \frac{CE}{ED}=\frac{\sqrt{2}}{1}$

Hence, option(a) is the correct answer.