The given circuit represents a balanced Wheatstone’s bridge. Calculate the value of resistance x.

In the balanced condition,

\(\frac{\text{P}}{\text{Q}}=\frac{\text{R}}{\text{S}}\)

\(\therefore\frac{4+4}{1}=\frac{4\text{+}\text{x}}{2}\)

\(\Rightarrow\text{x}=12\Omega\)