Question

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

• A. Zero
• B. $90\mu \text{C}$
• C. $40\mu \text{C}$
• D. $60\mu \text{C}$

Answer 1

The correct option is C $40\mu \text{C}$


Here, $C$ and $2C$ are in series, so same amount of charge will be stored in both the capacitor.

On going from $A$ to $B$, we get,
$V_A+10-\frac{Q}{2C}-\frac{Q}{C}=V_B$

$\Rightarrow 0+10-\frac{3Q}{2C}=0$ [A and B are grounded]

$\Rightarrow Q=\frac{20C}{3}=\frac{20\times 6}{3}=40\mu \text{C}$