JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Current Electricity

The network is a balanced Wheatstone bridge: along the top the ratio is $\frac{1}{2}$ and along the bottom $\frac{2}{4}$, equal, so no current passes through the central $1\Omega$ and it can be removed. What remains is $(1+2)\Omega$ in parallel with $(2+4)\Omega$, that is $3\Vert 6=2\Omega$ between $A$ and $B$. Including the internal resistance, the battery current is $i=\frac{9}{2+1}=3\mathrm{A}$, and the heat produced in the network in one minute is $i^2{R}_{AB}t=3^2\times 2\times 60=1080\mathrm{J}$.