In a meter bridge (Fig. 3.27), the null point 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, `(R )/(S)=(33.7)/(66.3)" ….(i)"`
After S is connected in parallel with a resistance of `12Omega`, the resistance across the gap changes from S to `S_(eq)`,
where `S_(eq)=(12S)/(S+12)` and hence the new balance condition gives `(51.9)/(48.1)=(R )/(S_("eq))=(R(S+12))/(12S)`
Substituting the value of R/S from equation (i),
we get `(51.9)/(48.1)=(S+12)/(12).(33.7)/(66.3)`,
which gives `S=13.5Oemga`
Using the value of `R//S` above, we get `R=6.86Omega`.