JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Current Electricity

Join the lower-left corner to $A$ (plain wire) and the upper-right corner to $B$ (plain wire). The network then has two inner nodes: $M$ between the top $6\Omega$ and the top $3\Omega$, and $N$ at the foot of the central $3\Omega$. The resistors are $A$-$M=6\Omega$, $M$-$B=3\Omega$, $A$-$N=3\Omega$, $N$-$B=6\Omega$, and the bridge $M$-$N=3\Omega$.

Take $V_A=V$, $V_B=0$. Kirchhoff's current law at $M$ and $N$ gives

which simplify to $5V_M-2V_N=V$ and $-2V_M+5V_N=2V$. Solving, $V_M=\frac{3}{7}V$ and $V_N=\frac{4}{7}V$.

The current drawn from $A$ is


Hence $R_{AB}=\frac{V}{I}=\frac{21}{5}\Omega$, so $x=21$.