A cell of $emf$ $E$ and internal resistance $r$ is connected in series with an external resistacne $nr.$ Then, the ratio of the terminal potential difference to $emf$ is
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$I=\frac{E}{r+n r}=\frac{E}{r(n+1)}$
$V=E-I r=E-\frac{E}{r(n+1)} r=\frac{n E}{n+1} \Rightarrow \frac{V}{E}=\frac{n}{n+1}$
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