Q126 / 200 · NEET 2006 · Physics

Two cells, having the same e.m.f. are connected in series through an external resistance $R$. Cells have internal resistances $r_1$ and $r_2\left(r_1>r_2\right)$ respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of $R$ is:

  • A
    $r_1+r_2$
  • Correct answer
    B
    $r_1-r_2$
  • C
    $\frac{r_1+r_2}{2}$
  • D
    $\frac{r_1-r_2}{2}$

Solution

According to the question $E-I r_1=0 \text { and } \mathrm{I}=\frac{E+E}{r_1+r_2+R}$ $\begin{array}{l} \therefore \frac{E}{r_1} =\frac{2 E}{r_1+r_2+R} \\ \Rightarrow r_1+r_2+R =2 r_1-r_2 \\ \Rightarrow R =r_1-r_2 \end{array}$

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

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