A cell of emf E and internal resistance r is | vedclass

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(C) The current $I$ flowing through the circuit is given by $I = \frac{E}{R + r}$,where $R = nr$ is the external resistance.Substituting the values,we

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$n/(n+1)$

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(C) The current $I$ flowing through the circuit is given by $I = \frac{E}{R + r}$,where $R = nr$ is the external resistance.Substituting the values,we get $I = \frac{E}{nr + r} = \frac{E}{r(n+1)}$.The terminal potential difference $V$ across the cell is given by $V = E - Ir$.Substituting the value of $I$,we get $V = E - \left( \frac{E}{r(n+1)} \right) r = E - \frac{E}{n+1}$.Simplifying this,$V = E \left( 1 - \frac{1}{n+1} \right) = E \left( \frac{n+1-1}{n+1} \right) = E \left( \frac{n}{n+1} \right)$.Therefore,the ratio of the terminal potential difference to $emf$ is $\frac{V}{E} = \frac{n}{n+1}$.

$A$ cell of $emf$ $E$ and internal resistance $r$ is connected in series with an external resistance $nr$. Then,the ratio of the terminal potential difference to $emf$ is

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