MCQ
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 
  • A
    $0.5 \,\Omega$
  • $\frac{1}{3}\;\Omega $
  • C
    $\;\frac{1}{4}\;\Omega $
  • D
    $1.0\,\Omega$

Answer

Correct option: B.
$\frac{1}{3}\;\Omega $
b
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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