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(C) The current $i$ in a circuit with an electric cell of $EMF$ $E$ and internal resistance $r$ connected to an external resistor $R$ is given by $i =

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$2.5$

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(C) The current $i$ in a circuit with an electric cell of $EMF$ $E$ and internal resistance $r$ connected to an external resistor $R$ is given by $i = \frac{E}{R + r}$.Case $1$: When $R_1 = 11 \, \Omega$,$i_1 = 0.5 \, A$.$0.5 = \frac{E}{11 + r} \Rightarrow E = 0.5(11 + r) = 5.5 + 0.5r$ ... $(i)$Case $2$: When $R_2 = 5 \, \Omega$,the current increases by $0.4 \, A$,so $i_2 = 0.5 + 0.4 = 0.9 \, A$.$0.9 = \frac{E}{5 + r} \Rightarrow E = 0.9(5 + r) = 4.5 + 0.9r$ ... $(ii)$Equating $(i)$ and $(ii)$:$5.5 + 0.5r = 4.5 + 0.9r$$5.5 - 4.5 = 0.9r - 0.5r$$1.0 = 0.4r$$r = \frac{1.0}{0.4} = 2.5 \, \Omega$.Thus,the internal resistance of the cell is $2.5 \, \Omega$.

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