JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Current Electricity

In the steady state the capacitor is fully charged and passes no current, so the branch holding the $10\mu \text{F}$ capacitor in series with its $8\Omega$ resistor is effectively open. The ideal ammeter has no resistance, so the node where it sits lies at the battery's negative terminal potential. Measured from the node just after the $1\Omega$ resistor, two routes reach that terminal: the lone $8\Omega$ resistor, and the $4\Omega$ resistor followed by the two $8\Omega$ resistors in parallel ($8\Vert 8=4\Omega$), i.e. $4+4=8\Omega$. These two $8\Omega$ routes in parallel give $4\Omega$, and with the series $1\Omega$ the battery sees $5\Omega$ and drives $\frac{10}{5}=2$ A, putting that node at $10-2(1)=8$ V. The ammeter carries only the current of the $4\Omega +4\Omega$ route, $\frac{8}{8}=1$ A. The ammeter reads $1$ A.