Q 12-03-104JEE MainJEE Main 2024 (30 Jan, Shift 1)Medium
Two cells are connected in opposition in a single closed loop. Cell $E_1$ has $8\ \text{V}$ emf and $2\ \Omega$ internal resistance; cell $E_2$ has $2\ \text{V}$ emf and $4\ \Omega$ internal resistance. There is no other resistance in the circuit. The terminal potential difference of cell $E_2$ is ______ V.
Numerical value type. Enter your answer.
Answer: 6
Net emf $= 8 - 2 = 6\ \text{V}$, total resistance $= 2 + 4 = 6\ \Omega$, so $I = 1\ \text{A}$.
The current is driven by $E_1$ and enters $E_2$ at its positive terminal (E₂ is being charged), so
$$V_2 = E_2 + Ir_2 = 2 + 1\times4 = 6\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics