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

Q: $ABCD$ 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.

d Let resistance $D E=x \Omega$ and that between $C E=(1-x) \Omega .$ The points $\mathrm{B}$ and $\mathrm{E}$ will be equipotential, when effective resistance acros $\mathrm{AE}=$ effective, resistance across $CE$ i.e., $\frac{(1+x) \times 1}{(1+x)+1}=(1-x)$ or $(1+x)=(1-x) \times(2+x)$ or $1+x=2+x-2 x 2 x-x^{2}$ or $x^{2}+2 x-1=0$ On solving it, we get, $x=\sqrt{2}-1$ Now, $1-x=2-(\sqrt{2}-1)=2-\sqrt{2}=(\sqrt{2}-1)$ $\therefore \frac{C E}{E D}=\frac{1-x}{x}=\frac{\sqrt{2}(\sqrt{2}-1)}{(\sqrt{2}-1)}=\sqrt{2}$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

$ABCD$ 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.

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

Advanced

STD 11 Science
JEE
STD 12 - 3. current electricity
Physics

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