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  4. In a potentiometer experiment two cells of emf E1 and E2 are used in series and in conjunction and the balancing length is found to be 58 cm of the wire. If the polarity of E2 is reversed, then the balancing length becomes 29 cm. The ratio (E1/E2) of the emfs of the two cells is

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Q. In a potentiometer experiment two cells of emf $E_{1}$ and $E_{2}$ are used in series and in conjunction and the balancing length is found to be $58 \,cm$ of the wire. If the polarity of $E_{2}$ is reversed, then the balancing length becomes $29 \,cm$. The ratio $\frac{E_{1}}{E_{2}}$ of the emfs of the two cells is

ManipalManipal 2018

Solution:

$E_{1}+E_{2}=k \times 58 \ldots$ (i)
$E_{1}-E_{2}=k \times 29 \ldots$ (ii)
$\therefore \frac{E_{1}+E_{2}}{E_{1}-E_{2}}=2$
$\Rightarrow \frac{E_{1}}{E_{2}}=\frac{3}{1}$