When a resistance of $2\,ohm$ is connected across the terminals of a cell, the current is $0.5$ amperes. When the resistance is increased to $5\, ohm$, the current is $0.25\, amperes$. The internal resistance of the cell is ............. $ohm$
Medium
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(b) Let the $e.m.f.$ of cell be $E$ and internal resistance be $r$.
Then $0.5 = \frac{E}{{(r + 2)}}$ and $0.25 = \frac{E}{{(r + 5)}}$
On dividing, $2 = \frac{{5 + r}}{{2 + r}}$ $ \Rightarrow $ $r = 1\,\Omega $
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