The charge deposited on $4\,\mu F$ capacitor in the circuit is
Medium
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$6 \mu \mathrm{F}$ and $6 \mu \mathrm{F}$ are in series
$\therefore$ Voltage ascross $4 \mu \mathrm{F}=6 \mathrm{V}$
$\therefore Q=6 \times 4 \times 10^{-6}$
$=24 \times 10^{-6} \mathrm{C}$
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