In a meter bridge, the null point is found at a distance of *l*_{1} cm from A. If now a resistance of X is connected in parallel with S, the null point occurs at *l*_{2} cm. Obtain a formula for X in terms of *l*_{1}, *l*_{2} and S.

#### Solution

Initially, when *X* is not connected

`R/S = l_1/(100 -l_1)` [Condition for balance ] ....... (1)

The equivalent resistance (*R*_{eq}) of the combination of *X* and *S* is

`1/R_(eq) = 1/X +1/S`

`R_(eq) = (SX)/(X+S)`

`R/R_(eq) = (l_2)/(100 - l_2)`

`(R(X+S))/(SX) = (l_2)/((100 - l_2))` ........ (2)

On dividing (i) by (ii), we obtain

`R/(R(X +S)) xx (SX)/X = (l_1(100 - l_2))/(l_2(100 - l_1)) `

`X /(X +S) = (l_1(100 - l_2))/(l_2(100-l_1)`

`Xl_2 (100 - l_1) = (X +S)l_1(100 - l_2)`

`XI_2 (100 - l_1) = Xl_1(100 - l_2) + Sl_1 (100 - l_2)`

`X = (Sl_1 (100 - l_2))/([l_2(100 -l_1) - l_1 (100 - l_1)])`

This is the expression for *X* in terms of *S*, *l*_{1} and *l*_{2}.