Q 12-03-091JEE MainJEE Main 2024 (29 Jan, Shift 2)Easy
In the given circuit, the current in resistance $R_3$ is:
Answer: (A) $1\ \text{A}$
$R_2 \parallel R_3 = 2\ \Omega$, so the total resistance is $2 + 2 + 1 = 5\ \Omega$.
$$I = \frac{10}{5} = 2\ \text{A}$$
This divides equally between the equal resistors $R_2$ and $R_3$, so the current in $R_3$ is $1\ \text{A}$.
Solution by Sreeraj P, M.Sc Physics