Figure 6.32 shows a meter bridge in the (which is nothing but a particle wheastone bridge), consisting of two resistors `X` and `Y` together in parellel with a meter long constantan wire of uniform cross section.

with the help of a movable contact `d`, one can change the ratio of resistance of the two segments of the wire until a sensitive galvanometer `G` connected across `b` and `D` shows no deflection. The null point is found to be at a distance of `33.7 cm`. The resistor `Y` is shunted by a resistance of `12 Omega`, and the null point is found to shift by a distance of `18.2 cm`. Determine the resistance of `X` and `Y`.

Answer

`x = 6.85 omega` and `y = 13.47 Omega`

Explanation

As the wire is if uniform cross section, the resistances of the two segments `AD` and `DC` of the wire are in the radio of the lengths of `AD` and `Dc`. According to the condition of balance of wheatstone bridge,
`(X)/(y) = (l_(1))/(l_(2))`
Here `l_(1) = 33.7 cm` and `l_(2) = 100 - 33.7 = 66.3 cm`
`(X)/(Y) = (33.7)/(66.3)` (i)
As resistance `Y'` is due to a parallel combination of resistance `Y` and a resistance of `12 Omega`,
`(1)/(Y') = (1)/(Y) + (1)/(12) = (12 + Y)/(12Y)` or `Y' = (12Y)/(12 + Y)`
Since `Y'` is less than `Y`, the ratio `X//Y`' will be greater than `X//Y` and the null point should shift toward end `C`.
`(X)/(Y') = (33.7 + 18.2)/(66.3 - 18.2) = (51.9)/(48.1)`
or `(X(12 + Y))/(12Y)=(51.9)/(48.1)`
or `12+Y=(51.9)/(48.1) xx 12 xx (66.3)/(33.7)`
`= 25.47 Omega`
or `Y= 25.47 - 12 = 13.47 Omega`
putting this value in Eq.(i), we get
`X = (33.7)/(66.3) xx Y = (33.7)/(66.3) xx 13.47 = 6.85 Omega`