Question

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 ${\epsilon }_1$ and ${\epsilon }_2$ connected in the manner shown are obtained at a distance of $120cm$ and $300cm$ from the end $A$. Find (i) ${\epsilon }_1/{\epsilon }_2$ and (ii) position of null point for the cell ${\epsilon }_1$.
How is the sensitivity of a potentiometer increased?

Answer 1

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 ,

${\epsilon }_1-{\epsilon }_2=120k$ ..........eq1 ,

for null point $=300cm$ ,

as the cells are supporting to each other and are in series ,

${\epsilon }_1+{\epsilon }_2=300k$ ..........eq2 ,

solving eq1 and eq2 ,

${\epsilon }_1=210k$ and ${\epsilon }_2=90k$ ,

(i) therefore ${\epsilon }_1/{\epsilon }_2=210k/90k=7/3$ ,

(ii) position of null point for ${\epsilon }_1=k{l}_1=210\times k=210cm$.

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.