If the potential difference between B and D is zero, the value of x...

As potential difference between B & D is zero therefore it is a balanced wheatstone bridge. So, we can write

$⇒ \dfrac{{R}_{1}}{{R}_{2}} = \dfrac{{R}_{3}}{{R}_{4}} ⇒ \dfrac{\left(\dfrac{6 \times 3}{6 + 3}\right)}{1 + 2} = \dfrac{\left(\dfrac{x \times 1}{x + 1}\right)}{x}$

$⇒ \dfrac{2}{3} = \dfrac{1}{x + 1}$

$⇒ x = 0 . 5 = \dfrac{1}{2}$

Hence, required $n = 2$.