Two cells of emfs ε₁ and ε₂ (ε₁ > ε₂) 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 $\frac{ε_1}{ε_2}$is :

3 : 2

When potentiometer is connected between A and B, then it measures only ε₁ and when connected between A and C, then it measures ε₁ - ε₂.

∴ $\frac{\varepsilon_1}{\varepsilon_1-\varepsilon_2}=\frac{l_1}{l_2}$,          $\frac{\varepsilon_1-\varepsilon_2}{\varepsilon_1}=\frac{l_2}{l_1}$

or  $1-\frac{\varepsilon_2}{\varepsilon_1}=\frac{100}{300}$          or          $\frac{\varepsilon_2}{\varepsilon_1}=1-\frac{1}{3}$          or          $\frac{\varepsilon_2}{\varepsilon_1}=\frac{2}{3}$          or          $\frac{\varepsilon_1}{\varepsilon_2}=\frac{3}{2}$