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)

$40\,\mu c$

Answer 2

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