When cell of E.M.F. ' $E_1$ ' is connected to potentiometer wire, the balancing length is $l_1$. Another…
When cell of E.M.F. ' $E_1$ ' is connected to potentiometer wire, the balancing length is $l_1$. Another cell of E.M.F. ' $E_2$ ' $\left(E_1 \gt E_2\right)$ is connected so that two cells oppose each other, then the balancing length is $l_2$. The ratio $\mathrm{E}_1: \mathrm{E}_2$ is
$\frac{l_1}{l_1+l_2}$
$\frac{l_1}{l_1-l_2}$
$\frac{l_1+l_2}{l_1}$
$\frac{l_1+l_2}{l_1-l_2}$
Solution
Using sum and difference method,
$\frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{l_1}{l_1-l_2}$
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