In a meter bridge,

the null points is found at a distance of 33.7 cm from A. If now a resistance of `12 Omega` is connected in parallel with S, the null point occurs at 51.9 cm. Determine the values of R and S.

Explanation

From the first balance point, we get
`(R)/(S)=(33.7)/(66.3)`
After S is connected in parallel with a resistance of `12 Omega`, the resistance across the gap changes from S to `S_(eq)`, where
`S_(eq)=(12S)/(S+12)`
and hence the new balance condition now gives
`(51.9)/(48.1)=(R)/(S_(eq))=(R(S+12))/(12S)`
Subsituting the value of `R//S` from Eq. (2.87), we get
`(51.9)/(48.1)=(S+12)/(1).(33.7)/(66.3)`
which gives `S= 13.5 Omega`. Using the value `R//S` above, we get `R = 6.86 Omega`