Question

Two unknown resistances `X` and `Y` are placed on the left and right gaps of a meter bridge. The null point in the galvanometer is obtained at a distance of `80 cm` from left. A resistance of `100 Omega` is now connected in parallel across `X`. The null point is then found by shifting the sliding contact toward left by `20 cm`. Calculate `X` and `Y`.

Answer 1

From the first null point : `(X)/(Y) = (80)/(20)` (i)
From the second null point : `(((100 X)/(100 + X)))/(Y) =(60)/(40)` (ii)

From Eqs. (i) and (ii), we get
`X = (500)/(3) Omega` and `Y = (125)/(3) Omega`