Q 12-03-163JEE MainJEE Main 2022 (25 Jun, Shift 2)Medium
Two cells of the same emf $E$ but different internal resistances $r_1$ and $r_2$ are connected in series with an external resistance $R$ as shown in the figure. The terminal potential difference across the second cell is found to be zero. The external resistance $R$ must then be:
Answer: (C) $r_2 - r_1$
Current: $i = \dfrac{2E}{R + r_1 + r_2}$. Terminal voltage of the second cell: $E - ir_2 = 0\Rightarrow E = ir_2$.
$$E = \frac{2Er_2}{R + r_1 + r_2}\ \Rightarrow\ R + r_1 + r_2 = 2r_2\ \Rightarrow\ R = r_2 - r_1$$
Solution by Sreeraj P, M.Sc Physics