Question

In the circuit shown in the figure, initially the switch is open. When the switch is closed, the charge passing through the switch is

• A. $600\mu \text{C}$; From $\text{B}$ to $\text{A}$
• B. $60\mu \text{C}$; From $\text{A}$ to $\text{B}$
• C. $120\mu \text{C}$; From $\text{B}$ to $\text{A}$
• D. $120\mu \text{C}$; From $\text{A}$ to $\text{B}$

Answer 1

The correct option is B $60\mu \text{C}$; From $\text{A}$ to $\text{B}$

When switch is open:

Since, both capacitors are in series, so effective capacitance will be

$C_{eq}=\frac{C_1{C}_2}{C_1+C_2}=\frac{2\times 3}{2+3}=\frac{6}{5}\mu \text{F}$

Charge on both capacitor will be same and that will be

$Q=C_{eq}V=\frac{6}{5}\times 120=144\mu \text{C}$


So, total charge on plates connected to point $\text{B}=144-144=0$

When switch is closed:


When switch is connected potential difference across both capacitor will be $60\text{V}.$

Therefore, charge on lower plate of $2\mu \text{F}$ capacitor became,

$Q_{2\mu F}=-2\times 60=-120\mu \text{C}$

And, charge on upper plate of $3\mu \text{F}$ capacitor is

$Q_{3\mu F}=3\times 60=180\mu \text{C}$

So, total charge on plates connected to $\text{B}=-120+180=60\mu \text{C}$.

Hence, $60\mu \text{C}$ charge flows from $\text{A}$ to $\text{B}$.

Hence, option $\left(d\right)$ is correct.