Q 12-03-132JEE MainJEE Main 2023 (25 Jan, Shift 2)Medium
Two cells are connected between points $A$ and $B$ as shown. Cell 1 has emf of $12\ \text{V}$ and internal resistance of $3\ \Omega$. Cell 2 has emf of $6\ \text{V}$ and internal resistance of $6\ \Omega$. An external resistor $R$ of $4\ \Omega$ is connected across $A$ and $B$. The current flowing through $R$ will be ______ A.
Numerical value type. Enter your answer.
Answer: 1
The cells are connected with opposite polarities. Treat each as a source between $A$ and $B$ and use the parallel-cell formula with one emf negative:
$$\varepsilon_{eq}=\frac{\frac{12}{3}-\frac{6}{6}}{\frac13+\frac16}=\frac{3}{1/2}=6\ \text{V},\qquad r_{eq}=\frac{3\times6}{3+6}=2\ \Omega$$
$$I=\frac{6}{2+4}=1\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics