Question

In the circuit shown initially $C_1,C_2$ are uncharged. After closing the switch

• A. The charge on $C_2$ is greater than on $C_1$
• B. The charge on $C_1$ and $C_2$ are the same
• C. The potential drops across $C_1$ and $C_2$ are the same
• D. The potential drops across $C_2$ is greater than that across $C_1$

Answer 1

Answer: (B)

The charge on $C_1$ and $C_2$ are the same

Here $C_1$ and $C_2$ are in series and the total potential between the capacitors is $V_T=12+6=18V$

$C_{eq}=\frac{4\times 8}{4+8}=\frac{8}{3}\mu F$ and

$Q_{eq}=C_{eq}{V}_T=\frac{8}{3}\times 18=48\mu C$

As they are in series so the charge on each capacitors is equal to $Q_{eq}=48\mu C$.

$\therefore Q_1=Q_2=48\mu C$

Potential across $C_1$ is $V_1=\frac{Q_1}{C_1}=\frac{48}{4}=12V$

Potential across $C_2$ is $V_2=\frac{Q_2}{C_2}=\frac{48}{8}=6V$