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(C) The current $i$ drawn from a cell is given by $i = \frac{E}{R+r}$,where $E$ is the electromotive force,$R$ is the external resistance,and $r$ is t

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$2.5$

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(C) The current $i$ drawn from a cell is given by $i = \frac{E}{R+r}$,where $E$ is the electromotive force,$R$ is the external resistance,and $r$ is the internal resistance.Case $I$: When $R_1 = 11 \ \Omega$,$i_1 = 0.5 \ A$.$0.5 = \frac{E}{11+r} \implies E = 0.5(11+r) \quad (i)$Case $II$: When $R_2 = 5 \ \Omega$,the current increases by $0.4 \ A$,so $i_2 = 0.5 + 0.4 = 0.9 \ A$.$0.9 = \frac{E}{5+r} \implies E = 0.9(5+r) \quad (ii)$Equating $(i)$ and $(ii)$:$0.5(11+r) = 0.9(5+r)$$5.5 + 0.5r = 4.5 + 0.9r$$5.5 - 4.5 = 0.9r - 0.5r$$1.0 = 0.4r$$r = \frac{1.0}{0.4} = 2.5 \ \Omega$.

When a resistor of $11 \Omega$ is connected in series with an electric cell,the current flowing in it is $0.5 \ A$. Instead,when a resistor of $5 \Omega$ is connected to the same electric cell in series,the current increases by $0.4 \ A$. The internal resistance of the cell is (in $\Omega$)

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