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Q.

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

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a

$\frac{CE}{ED} = 1$

b

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

c

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

d

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

answer is D.

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Detailed Solution

Question Image

Equivalent resistance between A and E is
$y = \frac{(x + 1)}{x + 2}$
For B and E to be at equal potential , we get
$\frac{R_{AE}}{R_{AB}} = \frac{R_{EC}}{R_{BC}} \Rightarrow \frac{x + 1}{(x + 2) 1} = \frac{1 - x}{1}$
Solving $x = \sqrt{2} - 1$
Now $\frac{CE}{ED} = \frac{1 - x}{x} = \sqrt{2}$

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