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

Explanation

When resistances R and S are connected then balance point is found at a distance of 40 cm form the zero.
`(R)/(S)=(40)/(100-40) [because (P)/(Q) =(I)/(100-I)]`
When a resistance of `12 omega ` is connected is parallel with S then total resistance in the right hadn gap is ` S_1 =(12S)/(S+12) ...........(ii) `
Since, balance point is obtained at a distance of 50 cm from the zero. `therefore (R)/(S_1) -(50)/(100-50)rArr (R)/(S_1)=(50)/(50)`
`therefore (R)/(S_1)=1`.
Dividing (i) by (iii) we get.
`( (R)/(S))/((R)/(S_1))=(2)/(3) =(R)/(S).(S_1)/(R)=(2)/(3)rArr (S_1)/(5)=(2)/(3)rArr S_1=(2)/(3)4`.
Putting the value of `S_1` in (ii), we get,
` (2)/(3)S= (12S)/(S+12)rArr (2)/(3) =(12)/(S+12)`
` rArr 2S+24=36rArr 2S= 12 therefore S=6`.
Putting the value of S in (i), we get
` (R)/(6)=(2)/(3) rArr R =(2)/(3)xx 6=5`
` R=4Omega and S=6 Omega`.