When a resistance of $2\, ohm$ is connected across the terminals of a cell, the current is $0.5\, A$. When the resistance is increased to $5\, ohm$, the current is $0.25\, A$. The $e.m.f.$ of the cell is ................. $V$
Medium
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Since $i\, = \left( {\frac{E}{{R + r}}} \right),$ we get
$0.5 = \frac{E}{{2 + r}}$ ...... $(i)$
$0.25 = \frac{E}{{5 + r}}$ ..... $(ii)$
Dividing $(i)$ by $(ii)$,
we get $2 = \frac{{5 + r}}{{2 + r}}$ $ \Rightarrow $ $r = 1\,\Omega $
$0.5 = \frac{E}{{2 + 1}}$ $ \Rightarrow $ $E = 1.5\,V$
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