Question

$ABCD$ is a square where each side is a uniform wire of resistance $1\Omega$. A point $E$ lies on $CD$ such that if a uniform wire of resistance $1\Omega$ is connected across $AE$ and constant potential difference is applied across $A$ and $C$, then $B$ and $E$ are equipotential, if

• A. $\frac{CE}{ED}=1$
• B. $\frac{CE}{ED}=2$
• C. $\frac{CE}{ED}=\frac{1}{\sqrt{2}}$
• D. $\frac{CE}{ED}=\sqrt{2}$

Answer 1

The correct option is D $\frac{CE}{ED}=\sqrt{2}$


As $1\Omega ,(1+x)\Omega$ are in parallel between A,E
Equivalent resistance between A,E is $\frac{1+x}{2+x}$

For B,E to be equipotential it has to be a balanced wheatstone bridge
$\frac{R_{AB}}{R_{AE}}=\frac{R_{BC}}{R_{CE}}$
$\frac{1}{\frac{1+x}{2+x}}=\frac{1}{1-x}$

$\frac{2+x}{1+x}=\frac{1}{1-x}$
$(2+x)(1-x)=1+x$
$2-2x+x-x^2=1+x$
$x^2+2x-1=0$
$(x+1{)}^2=2$
$x=\sqrt{2}-1$

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

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