Q 12-03-179JEE MainJEE Main 2022 (27 Jul, Shift 1)Medium
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
Answer: (A) $r_1 - r_2$
$$I = \frac{2E}{R + r_1 + r_2}$$
Terminal voltage of the first source: $E - Ir_1 = 0$, so $I = \dfrac{E}{r_1}$.
$$\frac{2E}{R + r_1 + r_2} = \frac{E}{r_1} \Rightarrow R + r_1 + r_2 = 2r_1 \Rightarrow R = r_1 - r_2$$
Solution by Sreeraj P, M.Sc Physics