When a cell of e.m.f. E_1 is connected to a p | vedclass

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(B) In a potentiometer,the e.m.f. $E$ is proportional to the balancing length $\ell$,i.e.,$E = k\ell$,where $k$ is the potential gradient.For the firs

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$\frac{\ell_1}{\ell_1-\ell_2}$

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(B) In a potentiometer,the e.m.f. $E$ is proportional to the balancing length $\ell$,i.e.,$E = k\ell$,where $k$ is the potential gradient.For the first cell,$E_1 = k\ell_1$.When the two cells are connected in opposition,the effective e.m.f. is $(E_1 - E_2)$. The new balancing length is $\ell_2$,so $(E_1 - E_2) = k\ell_2$.Dividing the two equations: $\frac{E_1}{E_1 - E_2} = \frac{k\ell_1}{k\ell_2} = \frac{\ell_1}{\ell_2}$.Cross-multiplying gives $E_1\ell_2 = \ell_1(E_1 - E_2) = \ell_1E_1 - \ell_1E_2$.Rearranging terms: $\ell_1E_2 = E_1(\ell_1 - \ell_2)$.Therefore,the ratio $\frac{E_1}{E_2} = \frac{\ell_1}{\ell_1 - \ell_2}$.

When a cell of e.m.f. $E_1$ is connected to a potentiometer wire,the balancing length is $\ell_1$. Another cell of e.m.f. $E_2$ $(E_1 > E_2)$ is connected such that the two cells oppose each other,and the balancing length is $\ell_2$. The ratio $E_1 : E_2$ is:

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