A cell of internal resistance $r$ is connected across an external resistance $n r$. Then the ratio of the terminal voltage to the emf of the cell is
A$\frac{1}{n}$
B$\frac{1}{n+1}$
C$\frac{n}{n+1}$
D$\frac{n-1}{n}$
AIIMS 2019, Medium
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C$\frac{n}{n+1}$
c The ratio of terminal voltage and emf is calculated as,
$V=E-I r$
$=E-\frac{E r}{(n+1) r}$
$=E\left(1-\frac{1}{n+1}\right)$
$\frac{V}{E}=\frac{n}{n+1}$
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