Find the charge on the capacitor $C$ in the given circuit .................. $\mu C$
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The capacitor branch will not allow charge to pass through it.

$\therefore $ Current in the circuit $\mathrm{I}=\frac{12}{6+2}=\frac{3}{2} \mathrm{\,amp}$

Potential difference across $2 \,\mu \mathrm{F}$ is same as across $6\, \Omega$ resistance.

So, potential difference across capacitor of $2\, \mu \mathrm{F}$

$=6 \times \frac{3}{2}=9$ $\mathrm{volt}$

$\therefore $ Charge on capacitor $=2 \times 10^{-6} \times 9\, \mu C=18\, \mu C$

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