In a meter bridge, the null point is found at a distance of `40 cm` from A. If a resistance of `12 Omega` in connected in parallel with `S` the null point occurs at `50.0 cm` from A Determine the value of R and S

Answer

`6 OMega and R = 4 Omega `

Explanation

When bridge is balance , `(R )/(s) = (1)/(100 - 1)`
For first condition
`(R )/(S) = (40)/(100 - 40) = (40)/(60) = (2)/(3) or R = (2)/(3) S`
For second condition
`(R )/(2 xx 12(S + 12)) = (50)/(100 - 50) = 1`
or `R (1 + 12) = 12 S or (2)/(3) S (S + 12) = 125`
On solving we get `S = 6 Omega and R = 4 Omega `