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Q.
In the circuit shown in figure, $C = 6 \,\mu F$. The charge stored in the capacitor of capacity $C$ is
UP CPMTUP CPMT 2011Electrostatic Potential and Capacitance
Solution:
In figure, the two capacitors are in series. Therefore, their equivalent capacitance is
$\frac{1}{C_{s}}= \frac{1}{C_{1}}+\frac{1}{C_{2}} $
$=\frac{1}{C}+\frac{1}{2C} = \frac{3}{2C}$ or $C_{s} = \frac{2C}{3}$
As the capacitors are connected in series, therefore, charge on each capacitor is same. Hence,
$q=C_{s} \times V$
$ = \frac{2C}{3}V $
$=\frac{2\times 6}{3} \times 10$
$= 40 \,\mu C$
