Question

In the circuit shown in figure $C_1=C_2=2\mu F$. 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 $40\mu C$

When capacitors are fully charged , no current draws from the cell. So we can remove the capacitors from the circuit and the corresponding circuit is shown in the figure.
The net resistance of the circuit is $R_{eq}=(1+2+3)||(2+1+3)=6||6=3\Omega$
The current drawn from cell is $I=\frac{V}{R_{eq}}=\frac{120}{3}=40A$
As the resistances of top and bottom branches are equal so half of current $I$ will flow through each branch.
here, $V_P-V_A=1\left(I/2\right)=20V$ and $V_P-V_B=2\left(I/2\right)=40V$
Thus, $V_A-V_B=20V$ This is the potential across $C_1$
so, Charge stored in $C_1$ is $Q_1=C_1(V_A-V_B)=40\mu C$
also $V_C-V_S=3\left(I/2\right)=60V$ and $V_D-V_S=3\left(I/2\right)=60V$
Thus, $V_C-V_D=0V$ This is the potential across $C_2$
so, Charge stored in $C_2$ is $Q_2=C_2(V_C-V_D)=0\mu C$