Two cells of emf $E_{1} and E_{2}$ are to be compared in a potentiometer $(E_{1} > E_{2})$ . When be the cells are used in series correctly, the balancing length obtained is 400 cm. When they are used in series but $E_{2}$ is connected with reverse polarities, then find the balancing length obtained is 200 of emf of cells ....

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Solution

According to the question
$\frac{e_{1} + e_{2}}{e_{1} - e_{2}} = \frac{400}{200}$
$\frac{e_{1} + e_{2}}{e_{1} - e_{2}} = \frac{2}{1}$
Applying compand & dividend
$\frac{e_{1} + e_{2} + e_{1} - e_{2}}{e_{1} + e_{2} - e_{1} + e_{2}} = \frac{2 + 1}{2 - 1}$
$\frac{e_{1}}{e_{2}} = \frac{3}{1}$ .

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