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(C) Let the emf of the cell be $E$ and its internal resistance be $r$. The external resistance is $R = n r$.The total resistance of the circuit is $R_

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$n/(n+1)$

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(C) Let the emf of the cell be $E$ and its internal resistance be $r$. The external resistance is $R = n r$.The total resistance of the circuit is $R_{total} = R + r = n r + r = r(n+1)$.The current $I$ flowing through the circuit is given by Ohm's law: $I = E / R_{total} = E / [r(n+1)]$.The terminal voltage $V$ across the external resistance is $V = I R = [E / (r(n+1))] 	imes (n r)$.Simplifying the expression,we get $V = E 	imes [n / (n+1)]$.Therefore,the ratio of the terminal voltage to the emf is $V/E = n / (n+1)$.

$A$ 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

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