a
Let \(X\) be the equivalent resistance between \(S\) and \(6 \,\Omega\).

\(\therefore \quad \frac{1}{X}=\frac{1}{S}+\frac{1}{6}\)      .......\((i)\)

Therefore, the equivalent circuit diagram drawn below.

For a balanced Wheatstone bridge, we get

\(\frac{P}{Q}=\frac{R}{X} \quad \text { or } \quad \frac{2}{2}=\frac{2}{X} \quad \Rightarrow \quad X=2\, \Omega\)

From eqn. \((i)\), we get 

\(\frac{1}{2}=\frac{1}{S}+\frac{1}{6} \quad\) or, \(\quad \frac{1}{S}=\frac{2}{6} \quad\) or, \(\quad S=3 \,\Omega\)