In the figure, a long uniform potentiometer wire AB is having a constant potential gradient along its length. The null points for the two primary cells of emfs ε₁ and ε₂ connected in the manner shown are obtained at a distance of 120 cm and 300 cm from the end A. Find (i) ε₁/ε₂ and (ii) position of null point for the cell ε₁. How is the sensitivity of a potentiometer increased?

Let k be the potential gradient of the potentiometer ,for null point=120cm,as the cells are opposite to each other and are in series ,ε1−ε2=120k..eq1 ,for null point=300cm,as the cells are supporting to each other and are in series ,ε1+ε2=300k..eq2 ,solving eq1 and eq2 ,ε1=210kandε2=90k,(i) thereforeε1/ε2=210k/90k=7/3,(ii) position of null point forε1=kl1=210×k=210cm.The potentiometer is said to be sensitive when a small displacement of jockey from null point , produces a large deflection in galvanometer . For that potential gradient (potential drop per unit length of wire of poentiometer) should be small , as potential gradient is given by ,k=V/l, where l is the length of wire of potentiometer and V is the potential drop across it ,so larger the l, smaller the k . hence we should use a wire of large length to increase the sensitivity.