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(B) The current $i$ in a circuit with a cell of $e.m.f.$ $E$ and internal resistance $r$ is given by $i = \frac{E}{R + r}$.For the first case,$R_1 = 2

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$1.5$

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(B) The current $i$ in a circuit with a cell of $e.m.f.$ $E$ and internal resistance $r$ is given by $i = \frac{E}{R + r}$.For the first case,$R_1 = 2\,\Omega$ and $i_1 = 0.5\,A$:$0.5 = \frac{E}{2 + r}$ ...... $(i)$For the second case,$R_2 = 5\,\Omega$ and $i_2 = 0.25\,A$:$0.25 = \frac{E}{5 + r}$ ...... $(ii)$Dividing equation $(i)$ by equation $(ii)$:$\frac{0.5}{0.25} = \frac{5 + r}{2 + r}$$2 = \frac{5 + r}{2 + r}$$2(2 + r) = 5 + r$$4 + 2r = 5 + r$$r = 1\,\Omega$Substituting $r = 1\,\Omega$ into equation $(i)$:$0.5 = \frac{E}{2 + 1}$$0.5 = \frac{E}{3}$$E = 0.5 	imes 3 = 1.5\,V$.

When a resistance of $2\,\Omega$ is connected across the terminals of a cell,the current is $0.5\,A$. When the resistance is increased to $5\,\Omega$,the current is $0.25\,A$. The $e.m.f.$ of the cell is ................. $V$.

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