Question

Each capacitor shown in the figure has a capacitance of $10\mu \text{F}$. The emf of the battery is $1\text{V}$. How much charge will flow through $AB$ , if the switch $S$ is closed?

• A. $\frac{10}{3}\mu \text{C}$
• B. $\frac{40}{3}\mu \text{C}$
• C. $\frac{20}{3}\mu \text{C}$
  
• D. $20\mu \text{C}$

Answer 1

The correct option is B $\frac{40}{3}\mu \text{C}$

When $S$ is open capacitors $1$ and $2$ are in parallel and they can be replaced by a capacitor whose capacitance is,

$C^{\prime }=(10+10)\mu \text{F}=20\mu \text{F}$


Now $C^{\prime }$ and $C_3$ are connected in series,
$C_{eq}=\frac{C^{\prime }{C}_3}{C^{\prime }+C_3}=\frac{10\times 20}{30}=\frac{20}{3}\mu \text{F}$
Therefore the charge supplied by the battery is,
$Q=C_{\text{eq}}V=\frac{20}{3}\mu \text{C}$

When the switch $S$ is closed, the capacitors $1$ and $2$ are in parallel but capacitor $3$ becomes ineffective due to short circuit.
$C_{\text{eq}}=(10+10)\mu \text{F}=20\mu \text{F}$
The charge on the equivalent capacitor will be,
$Q^{\prime }=20\mu \text{C}$


Therefore the charge flown through $AB$ will be,
$\Delta Q=Q^{\prime }-Q$
$\Delta Q=(20-\frac{20}{3})\mu \text{C}=\frac{40}{3}\mu \text{C}$

Hence, option (b) is the correct answer.