A wire of resistance `10 Omega` is used to form a circular ring of circumference 10 cm. IF two current carrying conductors are connected at any two points, the sub circuit so formed has a resistance of `1 Omega`. Find the positions of the two points.

Explanation

Current carrying conductors are connected at the two points A and B of the circular ring [Fig.1.30] according to the question,
`R_1+R_2=10 Omega`…….(1)
The two parts having resistances `R_1` and `R_2` from a parallel combination at A and B whose equivalent resistance is `1 Omega`.

So, `(R_1R_2)/(R_1+R_2)=1or,(R_1R_2)/10=1or,R_2=10/R_1`
SUbstituting this value of `R_2` in equation (1) we have
`R_1+10/R_1=10or,R_1^2-10R_1+10=0`
or,`R_1=(10pmsqrt(10^2-4.1.10))/2=5pmsqrt15=5pm 3.873`
`So,R_1=8.873Omega(or1.127 Omega)`
and `R_2=(10-R_1)=1.127 Omega(or8.873 Omega)`
Since resistance of the wire of length 10 cm is `10 Omega`, the resistance of the wire of length 1 cm is `1 Omega`
Therefore, the lengths of the two portions of the wire having resistances `R_1` and `R_2` are 8.873 cm and 1.127 cm.