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 extra charge will flow through $\text{AB}$ if the switch $S$ is closed?

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

Answer 1

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


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

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

$C^{\prime }$ and $C_3$ are in series.

$\therefore C_{eq}=\frac{C^{\prime }{C}_3}{C^{\prime }+C_3}=\frac{20\times 10}{30}=\frac{20}{3}\mu \text{F}$

Now using formula, $Q=CV$

$Q=C_{eq}V=\frac{20}{3}\times 1=\frac{20}{3}\mu \text{C}$

When $S$ is closed, capacitors $1$ and $2$ are in parallel, but capacitor $3$ is ineffective due to short circuit with switch being closed.


So, for this arrangement, effective capacitance will be

$C_{eq}=C_1+C_2$

$C_{eq}=(10+10)\mu \text{F}=20\mu \text{F}$

Similarly, charge flowing through the circuit will be

$Q^{\prime }=C_{eq}V=20\mu \text{C}$

Therefore, the charge flow through $\text{AB}$ will be

$\Delta Q=Q^{\prime }-Q$

$\Rightarrow \Delta Q=(20-\frac{20}{3})\mu \text{C}=\frac{40}{3}\mu \text{C}$

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