A current of $2\,A$ flows through a $2\,\Omega$ resistor when connected across a battery. The same battery supplies a current of $0.5\,\, A$ when connected across a $9 \,\,\Omega$ resistor. The internal resistance of the battery is
AIPMT 2011, Medium
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Let $\varepsilon $ be the emf and $r$ be internal resistance of the battery.
In the first case,
$2=\frac{\varepsilon}{2+r}$ ....$(i)$
In the second case,
$0.5=\frac{\varepsilon}{9+r}$ ....$(ii)$
Divide $(i)$ by $(ii),$ we get
${\frac{2}{0.5}=\frac{9+r}{2+r} \Rightarrow 4+2 r=4.5+0.5 r} $
${1.5 r=0.5 \Rightarrow r=\frac{0.5}{1.5}=\frac{1}{3} \,\Omega}$
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