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Question

Two sources of equal emfs are connected in series. This combination is connected to an external resistance R. The internal resistances of the two sources are $r_{1}$ and $r_{2}$ $\left(r_{1}>r_{2}\right)$. If the potential difference across the source of internal resistance $r_{1}$ is zero, then the value of R will be :

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

Solution

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