Experiment set up of a meter bridge .When the two unknown resistance `X and Y ` are intserted, the null point `D` is obtained `40 cm` from the end A .When a resistance of `30 Omega` is connected in series with `X` the mill point shift by `10 cm` .Find the position of the point when the `30 Omega` resistance is connected in series with X, the null point shifts by 10 cm. Find the position of the null point when the `30 Omega` resistance is connected in series with resistance `Y` instead of `X`. Determine the values of resistances `X and Y`

Answer

` 90 Omega , 60 Omega ,33.3 cm`

Explanation

When mull point is at `D` the bridge is balance
so `(X)/(Y) = (40)/(100 - 50) = (50)/(50) = 1 or m Y= X + 30 `…..(ii)
from (i) and (ii) `Y = (2)/(3) Y + 30`
or `1//3 = 30 or Y = 90 Omega `
`:. X = (2)/(3) xx 90 = 60 Omega `
When `30 Omega` resistance is connected in series with `X ` then `(X)/(Y + 30) = (1)/((100 - 1)) or (1)/(2) = (1)/((100 - 1))`
On solving we get `1 = 33.3 cm`