In the given figure, the charge stored in \(6 \ \mu \mathrm{F}\) capacitor, when points \(A\) and \(B\) are joined by a connecting wire is _______ \(\mu \mathrm{C}\).

At steady state, capacitor behaves as an open circuit and current flows in circuit as shown in the diagram.
\( \mathrm{R}_{\text {eq }}=9 \Omega \)
\( \mathrm{i}=\frac{9 \mathrm{~V}}{9 \Omega}=1 \mathrm{~A} \)
\( \Delta \mathrm{V}_{6 \Omega}=1 \times 6=6 \mathrm{~V} \)
\( \mathrm{~V}_{\mathrm{A}}=3 \mathrm{~V}\)
So, potential difference across \(6 \mu \mathrm{F}\) is \(6 \mathrm{~V}\).
Hence
\(Q =C \Delta V \)
\( =6 \times 6 \times 10^{-6} \mathrm{C} \)
\( =36 \mu \mathrm{C}\)