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. Then, the ratio of lengths

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

Answer 1

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

We can rearrange the circuit as, with CE = 1–x and ED = x

Which is a balanced wheat stance network since B and E are at the same potential. Thus,
$\frac{2+x}{1+x}=\frac{1}{1-x}\Rightarrow x^2-2x-1=0$
$\Rightarrow x=(\sqrt{2}-1)\Omega$
$\frac{CE}{DE}=\frac{1-x}{x}=\frac{2-\sqrt{2}}{\sqrt{2}-1}=\sqrt{2}$