To make the potential difference between points $B$ and $D$ zero, the bridge must be balanced.

1. Condition for a balanced Wheatstone bridge

If no current flows through the branch joining $B$ and $D$, then $V_B = V_D$ which implies $\frac{R_{AB}}{R_{BC}} = \frac{R_{AD}}{R_{DC}}$

2. Use the resistances from the figure

From the given bridge, the four arms are in the ratio such that $\frac{R_{AB}}{R_{BC}} = \frac{R_{AD}}{R_{DC}}$ Substituting the resistances shown in the diagram and solving for $x$, we get $x = \frac{1}{2}\,\Omega$

3. Compare with the form given

Given, $x = \frac{1}{n}\,\Omega$ So, $\frac{1}{n} = \frac{1}{2}$ which gives $n = 2$

4. Final answer

$\boxed{2}$