In a potentiometer experiment,cells of e.m.f. | vedclass

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(D) Let $k$ be the potential gradient of the potentiometer wire.When the cells are connected in series with the same polarity,the total e.m.f. is $(E_

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$3: 1$

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(D) Let $k$ be the potential gradient of the potentiometer wire.When the cells are connected in series with the same polarity,the total e.m.f. is $(E_{1} + E_{2})$. The balancing condition is $(E_{1} + E_{2}) = k \ell_{1}$,where $\ell_{1} = 64 \ cm$.When the polarity of $E_{2}$ is reversed,the total e.m.f. is $(E_{1} - E_{2})$. The balancing condition is $(E_{1} - E_{2}) = k \ell_{2}$,where $\ell_{2} = 32 \ cm$.Dividing the two equations: $\frac{E_{1} + E_{2}}{E_{1} - E_{2}} = \frac{64}{32} = 2$.$E_{1} + E_{2} = 2(E_{1} - E_{2})$.$E_{1} + E_{2} = 2E_{1} - 2E_{2}$.$3E_{2} = E_{1}$.Therefore,$\frac{E_{1}}{E_{2}} = 3: 1$.

In a potentiometer experiment,cells of e.m.f. $E_{1}$ and $E_{2}$ are connected in series $(E_{1} > E_{2})$,and the balancing length is $64 \ cm$. If the polarity of $E_{2}$ is reversed,the balancing length becomes $32 \ cm$. The ratio $\frac{E_{1}}{E_{2}}$ is:

Similar Questions

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