Q 12-03-195JEE MainJEE Main 2021 (24 Feb, Shift 1)Medium
A cell $E_1$ of emf 6 V and internal resistance $2\ \Omega$ is connected with another cell $E_2$ of emf 4 V and internal resistance $8\ \Omega$ (as shown in the figure). The potential difference across points $X$ and $Y$ is:
Answer: (B) 5.6 V
The two cells are connected in opposition in a single loop:
$$I = \frac{6 - 4}{2 + 8} = 0.2\ \text{A}$$
The larger emf $E_1$ drives the current, so the current passes through $E_2$ against its emf (charging it). The p.d. across $E_2$, i.e. between $X$ and $Y$:
$$V_{XY} = E_2 + Ir_2 = 4 + 0.2\times 8 = 5.6\ \text{V}$$
(Check through $E_1$: $6 - 0.2\times2 = 5.6$ V.)
Solution by Sreeraj P, M.Sc Physics