Two sources of equal emf are connected in series. This combination is 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

$R = r_2-r_1$

The correct answer is Option (1) → $R = r_2-r_1$

Let the EMFs of the sources be $E$ each, internal resistances $r_1$ and $r_2$, and external resistance $R$.

Current in series: $I = \frac{E + E}{R + r_1 + r_2} = \frac{2E}{R + r_1 + r_2}$

Potential difference across source with $r_2$:

$V_2 = E - I r_2$

Given $V_2 = 0$:

$0 = E - I r_2 \Rightarrow I r_2 = E \Rightarrow I = \frac{E}{r_2}$

Also, $I = \frac{2E}{R + r_1 + r_2}$

Equate currents:

$\frac{E}{r_2} = \frac{2E}{R + r_1 + r_2} \Rightarrow R + r_1 + r_2 = 2 r_2$

$R = 2 r_2 - r_1 - r_2 = r_2 - r_1$