In a meterbridge, the null point is found at a distance of 33.7cm from A. If a resistance of 12W is connected in parallel with S, the null points occurs at 51.9cm Determine the values of R and S.

Explanation

Before connecting 12 `Omega` resistance, null point is obtained at 33.7 cm and so according to condition of balanced Wheatstone bridge
`(R)/(S) = (l_(1))/(100 - l_(1)) = (33.7)/(100 - 33.7)`
`therefore (R)/(S) = (33.7)/(66.3) " " ` .... (1)
After connecting `12 Omega` resistance in parallel with S, equivalent resistance of this parallel combination would be,
`S_(eq) = (12 xx S)/(12 + S)`
In this case, null point is obtained at 51.9 cm and so for balancing of Wheatstone bridge,
`(R)/(S_(eq)) = (l_(1))/(100 - l_(1)) = (51.9)/(100 - 51.9)`
`therefore (R(12 + S))/(12 xx S) = (51.9)/(48.1) `
`therefore ((12 + S)/(12) ) ((R)/(S)) = (51.9)/(48.1)`
`therefore ((12 + S)/(12) ) ((33.7)/(66.3)) = (51.9)/(48.1) ` [ From equation (1) ]
`therefore 12 + S = (51.9)/(48.1) xx (66.3)/(33.7) xx 12`
`therefore 12 + S = 25.47`
`therefore S = 13.47 Omega approx 13.5 Omega " "` .... (2)
From equations (1) and (2),
`(R)/(13.5) = (33.7)/(66.3) `
`therefore R = 13.5 xx (33.7)/(66.3)`
`therefore R = 6.86 Omega`