Q 12-03-068JEE MainJEE Main 2025 (22 Jan, Shift 2)Medium
The net current flowing in the given circuit is ______ A.
Numerical value type. Enter your answer.
Answer: 1
In steady state no current flows through the $1\ \mu\text{F}$ capacitor, so the branch $2.5\ \Omega$–$1\ \mu\text{F}$ and everything to its right ($1\ \Omega$, $5\ \Omega$, $8\ \Omega$, $4\ \Omega$) carry no current.
The $3\ \Omega$ and $6\ \Omega$ resistors both connect the second top junction to the bottom wire, so they are in parallel:
$$\frac{3\times6}{3+6} = 2\ \Omega$$
This is in series with the $2\ \Omega$: $2 + 2 = 4\ \Omega$. That branch is in parallel with the diagonal $4\ \Omega$ across the cell:
$$R = \frac{4\times4}{4+4} = 2\ \Omega$$
$$I = \frac{2\ \text{V}}{2\ \Omega} = 1\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics