Question

Two cells of emf ${\epsilon }_1$ and ${\epsilon }_2$ $({\epsilon }_1>{\epsilon }_2)$ are connected as shown in figure.
When a potentiometer is connected between A and B, the balancing length of the potentiometer wire is 300 cm. on connecting the same potentiometer between A and C, the balancing length is 100 cm. The ratio ${\epsilon }_1:{\epsilon }_2$ is

• A. 3 : 1
• B. 1 : 3
• C. 2 : 3
• D. 3 : 2

Answer 1

The correct option is D 3 : 2

When potentiometer is connected between A and B, then it measures only ${\epsilon }_1$ and when connected between A and C, then it measures ${\epsilon }_1-{\epsilon }_2$.
$\therefore \frac{{\epsilon }_1}{{\epsilon }_1-{\epsilon }_2}=\frac{l_1}{l_2},\frac{{\epsilon }_1-{\epsilon }_2}{{\epsilon }_1}=\frac{l_2}{l_1}or1-\frac{{\epsilon }_2}{{\epsilon }_1}=\frac{100}{300}or\frac{{\epsilon }_2}{{\epsilon }_1}=1-\frac{1}{3}or\frac{{\epsilon }_2}{{\epsilon }_1}=\frac{2}{3}or\frac{{\epsilon }_1}{{\epsilon }_2}=\frac{3}{2}$