Question

In the circuit shown in figure, $C=6\mu F$. The charge stored in the capacitor of capacity $C$ is

• A. zero
• B. $90\mu c$
• C. $40\mu c$
• D. $60\mu c$

Answer 1

Answer: (C)

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,
$q=C_s\times V$
$=\frac{2C}{3}V$
$=\frac{2\times 6}{3}\times 10$
$=40\mu C$