Question

In a meter bridge circuit, resistance in the left-hand gap is 2 Ω and an unknown resistance  X is in the right-hand gap as shown in the figure below. The null point is found to be 40 cm from the left end of the Wire. What resistance should be connected to X so that the new null point is 50 cm from the left end of the wire?

Answer 1

According to Fig.
                       `2/X = 40/(100-40) = 4/6`                           
                           X = 3 Ω
Let R Ω be connected with X in parallel to obtain the null point at 50 cm from left.
Now, if S Ω be in equivalent resistance of X and R then,
                      `2/S = 50/(100-50) = 1`
∴                      S = 2 Ω
Now, to obtain 2 Ω as equivalent resistance between X and R, we have

                       `1/2 = 1/3 +1/R`

∴                        R = 6 Ω
∴    6  Ω resistance is to be connected in parallel to X.