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Two cells $E_1$ and $E_2$ having equal $EMF$ $E$ and internal resistances $r_1$ and $r_2$ $(r_1 > r_2)$ respectively are connected in series. This combination is connected to an external resistance $R$. It is observed that the potential difference across the cell $E_1$ becomes zero. The value of $R$ will be
- A$r_1 - r_2$
- B$r_1 + r_2$
- C$\frac{r_1 - r_2}{2}$
- D$\frac{r_1 + r_2}{2}$
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View SolutionTwo identical cells each of emf $1.5\,V$ are connected in series across a $10\,\Omega$ resistance. An ideal voltmeter connected across $10\,\Omega$ resistance reads $1.5\,V$. The internal resistance of each cell is $......\Omega$.
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