Question

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

• A. $0$
• B. $90\mu C$
• C. $40\mu C$
• D. $60\mu C$

Answer 1

Answer: (C)

$40\mu C$

Using KVL ,

$V_1-\frac{Q}{C}+10-\frac{Q}{2C}=V_2$

$0-\frac{Q}{C}+10-\frac{Q}{2C}=0$

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