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Two cells, having the same e.m.f, are connected in series through an external resistance R. Cells have internal resistances $r_{1}$ and $r_{2}$ $(r_{1} > r_{2})$ respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of R is

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$\frac{r_{1} + r_{2}}{2}$

$\frac{r_{1} - r_{2}}{2}$

$r_{1} + r_{2}$

$r_{1} - r_{2}$

answer is C.

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Detailed Solution

$i = \frac{2 E}{r_{1} + R_{2} + R}$

P.d across first cell = 0

$R = r_{1} - r_{2}$

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