In potentiometer experiment, cells of e.m.f. E and E 2 are connected in series (E $_{1}>\mathrm{E}_{2}$ ),…
In potentiometer experiment, cells of e.m.f. E and E 2 are connected in series (E $_{1}>\mathrm{E}_{2}$ ), the balancing length is $64 \mathrm{~cm}$ of the wire. If the polarity of $\mathrm{E}_{2}$ is reversed, the balancing length becomes $32 \mathrm{~cm}$. The ratio $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}$ is
$1: 2$
$2: 1$
$1: 3$
$3: 1$
Solution
In the potentiometer experiment, cells of e.m.f. $E_1$ and $E_2$ are connected in series $\left(E_1>E_2\right)$. the balancing length is 64 cm of the wire. If the polarity of $E_2$ is reversed, the balancing length becomes 32 cm . The ratio $\frac{E_1}{E_2}$ is $\underline{\mathbf{3}: \mathbf{1}}$.
$\frac{E_1}{E_2}=\frac{\ell_1+\ell_2}{\ell_1-\ell_2}=\frac{64+32}{64-32}=\frac{96}{32}=3$