Two sources of equal emf are connected in series. This combination is, in turn connected to an external resistance $R$ . The internal resistance of two sources are $r_{1}$ and $r_{2}$ ( $r_{2} > r_{1}$ ). If the potential difference across the source of internal resistance $r_{2}$ is zero, then $R$ equals to

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

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

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r_{2} - r_{1}

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

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Solution

The correct option is C $r_{2} - r_{1}$
According to the situation given

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

$V = E - i r$

$\Rightarrow 0 = E - i r_{2}$

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

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

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