Q114 / 120 · VITEEE 2012 · Physics

A current of 2 A flows through a $2 \Omega$ resistor when connected across a battery. The same battery supplies a current of $0.5 \mathrm{~A}$ when connected across a $9 \Omega$ resistor. The internal resistance of the battery is

  • A
    $1 / 3 \Omega$
  • B
    $1 / 4 \Omega$
  • C
    $1 \Omega$
  • Correct answer
    D
    $0.5 \Omega$

Solution

$$ \begin{array}{l} \text { Current } i=\frac{E}{R+r} \\ 2=\frac{E}{2+r}...(i) \\ 0.5=\frac{E}{9+r}...(ii) \end{array} $$ From Eqs. (i) and (ii), we have $$ \begin{array}{l} \frac{2}{0.5}=\frac{9+r}{2+r} \Rightarrow 4=\frac{9+r}{2+r} \\ 3 r=1 \therefore r=\frac{1}{3 \Omega} \end{array} $$

Hence, the correct answer is option: 'D'.

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