A ring is made of a wire having a resistance `R_(0) = 12 Omega`. Find the point `A` and `B`, as shown in the figure, at which a current carrying conductor should be connected so that the resistance `R` of the subcircuit between these points is equal to `(8)/(3) Omega`

Options

A. `(l_(1))/(l_(2))=(3)/(8)`
B. `(l_(1))/ (l_(2))=(1)/(2)`
C. ` (l_(1))/(l_(2))=(5)/(8)`
D. `(l_(1))/(l_(2))=(1)/(3)`

Answer

2

Explanation

`(R_(1)R_(2))/(R_(1)+R_(2))=(8)/(3)& R_(1)+R_(2)=12rArrR_(1)R_(2)=32`
`rArrR_(2)-R_(1)=sqrt((R_(1)+R_(2))^(2)-4R_(1)R_(2))`
`=sqrt(12^(2)-4xx32)=4Omega`
So `R_(1)=4Omega` and `R_(2)=8Omega`
Hence `(l_(1))/(l_(2))=(4)/(8)=(1)/(2)`