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 r₁ and r₂ (r₁ > r₂). If the potential difference across the source of internal resistance r₁ is zero then the value of R will be

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$r_{1} - r_{2}$

$r_{2} - r_{1}$

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

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

answer is A.

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Detailed Solution

$\begin{array}{l} I = \frac{2 E}{r_{1} + r_{2} + R} \\ IR = E - {Ir}_{2} \\ I (R + r_{2}) = E \\ I = \frac{E}{R + r_{2}} \\ \frac{2 E}{r_{1} + r_{2} + R} = \frac{E}{R + r_{2}} \\ 2 R + 2 r_{2} = r_{1} + r_{2} + R \\ R = r_{1} - r_{2} \end{array}$

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