If the potential difference between points B and D is zero in the given circuit, the value of x is $\dfrac{1}{n}\ \Omega$. The value n is

The given circuit is a Wheatstone bridge. Since the potential difference between points B and D is zero, the bridge is balanced.

Let the nodes be A (left), B (top), C (right), and D (bottom).

The equivalent resistance of the arm AB is:
$R_{AB} = \dfrac{6 \times 3}{6 + 3} = \dfrac{18}{9} = 2\ \Omega$

The equivalent resistance of the arm AD is:
$R_{AD} = 1 + 2 = 3\ \Omega$

The equivalent resistance of the arm BC is:
$R_{BC} = \dfrac{x \times 1}{x + 1} = \dfrac{x}{x + 1}\ \Omega$

The equivalent resistance of the arm DC is:
$R_{DC} = x\ \Omega$

For a balanced Wheatstone bridge, the condition is:
$\dfrac{R_{AB}}{R_{AD}} = \dfrac{R_{BC}}{R_{DC}}$

Substituting the values:
$\dfrac{2}{3} = \dfrac{\dfrac{x}{x + 1}}{x}$

$\dfrac{2}{3} = \dfrac{1}{x + 1}$

$2(x + 1) = 3$

$2x + 2 = 3$

$2x = 1$

$x = \dfrac{1}{2}\ \Omega$

Given $x = \dfrac{1}{n}\ \Omega$, we get $n = 2$.

Answer: $2$