Two cells with the same e.m.f. \(E\) and different internal resistances \(r_1\) and \(r_2\) are connected in series to an external resistance \(R\). The value of \(R\) so that the potential difference across the first cell be zero is

• A

  \(\sqrt{r_1 r_2}\)
• B

  \(r_1+r_2\)
• Correct answer
  C

  \(\mathrm{r}_1-\mathrm{r}_2\)
• D

  \(\frac{r_1+r_2}{2}\)

Solution

Hints:
\(I=\frac{2 E}{R+r_1+r_2}\) Potential difference across first cell \(\mathrm{V}=\mathrm{E}-\mathrm{Ir}_1=0\) \(\begin{aligned} & E-\frac{2 E r_1}{R+r_1+r_2}=0 \\ & {\left[\frac{R+r_1+r_2-2 r_1}{R+r_1+r_2}\right]=0} \\ & \Rightarrow R+r_2-r_1=0 \\ & R=r_1-r_2 \end{aligned}\)

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