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___________ $\mu \mathrm{C}$.
JEE MAIN 2024, Diffcult
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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}$
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