Question

Three capacitors, each of capacitance $4\mu \text{F}$ are connected to a battery of voltage $10\text{V}$. When key $K$ is closed, the charge which will flow through the battery is

• A. $80\mu \text{C}$
• B. $40\mu \text{C}$
• C. $20\mu \text{C}$
• D. $40/3\mu \text{C}$

Answer 1

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


The capacitance of the parallel combination is $(4+4)\mu \text{F}=8\mu \text{F}$.

When key is open, the total effective capacitance will be

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

So, charge on capacitors,

$Q_1=C_{eq}V=\frac{8}{3}\times 10=\frac{80}{3}\mu C$

When the key closed, capacitors in parallel are short-circuited or ineffective because conducting wire have same potential point and capacitor between same potential points is ineffective.

So, the capacitance of the circuit is now $4\mu \text{F}$.

Therefore, charge $Q_2=40\mu \text{C}$

$\therefore$ charge flown through battery is $(Q_2-Q_1).$

$\Rightarrow Q_2-Q_1=(40-\frac{80}{3})\mu \text{C}=\frac{40}{3}\mu \text{C}$

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

Key concept- Magnitude of charge flown is the difference between initial and final charge on the capacitors.