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 $$r_2$$ ($$r_1 > r_2$$). If the potential difference across the source of internal resistance $$r_1$$ is zero then the value of $$R$$ will be

Two sources of equal EMF $$\varepsilon$$ are connected in series with an external resistance $$R$$ and internal resistances $$r_1$$ and $$r_2$$ (with $$r_1>r_2$$). Given that the potential difference across the source with internal resistance $$r_1$$ is zero, we first determine the current in the circuit.

Since the total EMF is $$2\varepsilon$$ and the total resistance is $$R + r_1 + r_2$$, the circuit current is $$I = \frac{2\varepsilon}{R + r_1 + r_2}$$.

Next, the condition of zero potential difference across the first source (with EMF $$\varepsilon$$ and internal resistance $$r_1$$) implies $$V_1 = \varepsilon - I r_1 = 0$$, which yields $$\varepsilon = I r_1$$.

Substituting $$I = \frac{2\varepsilon}{R + r_1 + r_2}$$ into this relation gives $$\varepsilon = \frac{2\varepsilon\,r_1}{R + r_1 + r_2}$$. From here, multiplying both sides by $$R + r_1 + r_2$$ leads to $$R + r_1 + r_2 = 2r_1$$, and hence $$R = 2r_1 - r_1 - r_2 = r_1 - r_2$$.

Therefore, the correct answer is Option A: $$r_1 - r_2$$.