In the circuit shown in the figure, $C = 6\,\mu F$. The charge stored in the capacitor of capacity $C$ is......$\mu C$
Medium
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Both the capaciotrs are in series. Therefore, charge stored on them will be same.
Net capacity $=\frac{\mathrm{C}(2 \mathrm{C})}{\mathrm{C}+2 \mathrm{C}}=\frac{2}{3} \mathrm{C}=\frac{2}{3} \times 6\, \mu \mathrm{F}=4\, \mu \mathrm{F}$
Potential difference $=10 \mathrm{\,V}$
$\therefore \quad \mathrm{q}=\mathrm{CV}=40\, \mu \mathrm{C}$
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