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