In a meter bridge , the null point is found at a distance of 36.7 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 value 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) = (12 S)/(S + 12)`
and hence the new balance condition now gives
`(51.9)/(48.1) = (R )/(S_(sq)) = (R(S + 12))/(12 s)`
Substituting the have of R/S from Equation (1), we get
`(51.9)/(48.1) = (S + 12)/(12) . (33.7)/(66.3)`
Which gives `S = 11.31Omega`. Using the value of R/S above, We get `R = 6.36Omega`