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

Options

A. `(CE)/(ED) =1`
B. `(CE)/(ED)= (1)/(sqrt2)`
C. `(CE)/(ED) = (1)/(sqrt2)`
D. `(CE)/(ED) = sqrt2`

Answer

D

Explanation

Equivalent resistance between A and E is
`y = (x+1)/(x+2)`

For B and E to be equipotent, we get
`(R_(AE))/(R_(AB))=(R_(EC))/(R_(BC))or(x+1)/((x+2)xx1)=(1-x)/(1)`
Solve to get `x = sqrt2 - 1 Omega`
Now `(CE)/(ED) = (1-x)/(x) = sqrt2 Omega`