A cell sends a current of 0.9, A through a re | vedclass

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(B) Let $E$ be the electromotive force $(EMF)$ and $r$ be the internal resistance of the cell.According to Ohm's law for a circuit with an internal re

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$0.5$

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(B) Let $E$ be the electromotive force $(EMF)$ and $r$ be the internal resistance of the cell.According to Ohm's law for a circuit with an internal resistance,the current $I$ is given by $I = \frac{E}{R + r}$.For the first case: $0.9 = \frac{E}{2 + r}$  --- $(i)$For the second case: $0.3 = \frac{E}{7 + r}$ --- $(ii)$Dividing equation $(i)$ by equation $(ii)$:$\frac{0.9}{0.3} = \frac{E / (2 + r)}{E / (7 + r)}$$3 = \frac{7 + r}{2 + r}$$3(2 + r) = 7 + r$$6 + 3r = 7 + r$$2r = 1$$r = 0.5\, \Omega$.

$A$ cell sends a current of $0.9\, A$ through a resistance of $2\, \Omega$ and a current of $0.3\, A$ through a resistance of $7\, \Omega$. The internal resistance of the cell is ............. $\Omega$.

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