Q 12-03-081JEE MainJEE Main 2024 (5 Apr, Shift 1)Medium
In the given figure $R_1 = 10\ \Omega$, $R_2 = 8\ \Omega$, $R_3 = 4\ \Omega$ and $R_4 = 8\ \Omega$. The battery is ideal with emf $12\ \text{V}$. The equivalent resistance of the circuit and the current supplied by the battery are respectively
Answer: (B) $12\ \Omega$ and $1\ \text{A}$
$R_2$, $R_4$ and $R_3$ are each connected between the same two nodes (top and bottom wires), so they are in parallel:
$$\frac{1}{R_p} = \frac18 + \frac18 + \frac14 = \frac12 \Rightarrow R_p = 2\ \Omega$$
$$R_{eq} = 10 + 2 = 12\ \Omega,\qquad I = \frac{12}{12} = 1\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics