In potentiometer experiment, the balancing length is $8 \mathrm{~m}$ when two cells $E_1$ and $E_2$ are…

In potentiometer experiment, the balancing length is $8 \mathrm{~m}$ when two cells $E_1$ and $E_2$ are joined in series. When two cells are connected in opposition the balancing length is $4 \mathrm{~m}$. The ratio of the e.m.f. of the two cells $\left(\frac{E_1}{E_2}\right)$ is
  1. $1:2$
  2. $2:1$
  3. $1:3$
  4. $3:1$

Solution

$\begin{aligned} & \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{l_1+l_2}{l_1-l_2}=\frac{8+4}{8-4} \\ & \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{12}{4} \\ \therefore \quad \frac{\mathrm{E}_1}{\mathrm{E}_2} & =3\end{aligned}$

Asked in: MHT CET 2023 (14 May Shift 2)

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