JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Current Electricity

Zero potential difference between $B$ and $D$ means the arrangement is a balanced Wheatstone bridge. The arm from the left corner up to $B$ is $6\Omega$ parallel with $3\Omega$, i.e. $2\Omega$, while the arm from $B$ down to the right corner is $x$ parallel with $1\Omega$, i.e. $\frac{x}{x+1}$. On the lower side the arm from the left corner through $D$ is $1\Omega +2\Omega =3\Omega$, and the arm from $D$ to the right corner is $x$. The balance condition equates the ratios of the upper and lower arms, $\frac{2}{x/(x+1)}=\frac{3}{x}$, giving $2(x+1)=3$, so $x=\frac{1}{2}\Omega$; comparing with $\frac{1}{n}$, the value of $n$ is $2$.