Question

In the circuit shown in figure, $C_1=C_2$. Then charge stored in

• A. Capacitor $C_1$ is zero
• B. Capacitor $C_2$ is zero
• C. Both capacitor is zero
• D. Capacitor $C_1$ is $40\mu C$

Answer 1

Answer: (B), (D)

The correct options are
B Capacitor $C_2$ is zero
D Capacitor $C_1$ is $40\mu C$

$i=\frac{120}{R_{e}q}=\frac{120}{(\frac{1}{2}+\frac{1}{2}{)}^{-1}}=40A$

Now, this $i$ will be divided equally, so $i_1=20A$ and $i_2=20A$

Now, $\Delta V_2=0$

and

$Q_{C}1=20\times 2\mu C=40\mu C$