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 ................ $\Omega$
Medium
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By using $i = \frac{E}{{R + r}}$
$ \Rightarrow 0.5 = \frac{E}{{11 + r}}$ $ \Rightarrow $ $E = 5.5 + 0.5r$ ..… $( i)$
and $0.9 = \frac{E}{{5 + r}}$ $ \Rightarrow $ $E = 4.5 + 0.9r$ ..… $(ii)$
On solving these equation, we have $r = 2.5\,\Omega $
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