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. Then,

Options

A. `(CE)/(ED)=1`
B. `(CE)/(ED)=(1)/(sqrt(2))`
C. `(CE)/(ED)=(1)/(2)`
D. `(CE)/(ED)=sqrt(2)`

Answer

D

Explanation

Equivalent resistance between A and E is
`y=((x+1))/(x+2)`
for B and E to be at equal potential we get
`(R_(AE))/(R_(AB))=(R_(EC))/(R_(BC))implies(x+1)/((x+2)1)=(1-x)/(1)`
solving `x=sqrt(2)-1`
Now `(CE)/(ED)=(1-x)/(x)=sqrt(2)`