JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Current Electricity

In steady state, no current can flow through a branch containing a capacitor once it is fully charged, since a capacitor blocks direct current. So the branch carrying the $1\mu F$ capacitor — together with the resistors that carry current only into that dead-end branch — draws zero current and can be dropped from the circuit for the purpose of finding the current supplied by the cell.

Reducing what remains by standard series–parallel combination gives an equivalent resistance across the 2 V source of
$R_{eq}=2\Omega$

so the net current drawn from the cell is
$I=\frac{V}{R_{eq}}=\frac{2}{2}=1\text{ A}$