Kirchhoff's Laws & Measuring Instruments question from JEE Main 2026 (24 Jan, Shift 2)

Answer: (b)

1. In steady state the capacitor branch (with its $8\ \Omega$) carries no current.
2. The ideal ammeter joins the bottom nodes, so the left $8\ \Omega$ is in parallel with $4 + (8\|8) = 8\ \Omega$.
3. $8\|8 = 4\ \Omega$; total $R = 1 + 4 = 5\ \Omega$, $I = \dfrac{10}{5} = 2\text{ A}$.
4. The current splits equally; the right-hand $1\text{ A}$ returns through the ammeter.
5. So the answer is (b) $1\text{ A}$.