Question

Excess charge flowing through the battery after closing the switch is

• A. $20\text{mC}$
• B. $30\text{mC}$
• C. $120\text{mC}$
• D. $150\text{mC}$

Answer 1

The correct option is B $30\text{mC}$

Initially when switch is open $20\mu \text{F}$ and $30\mu \text{F}$ capacitors are in series.

Their equivalent capacitance, $C_{\text{eq}}=\frac{20\times 30}{20+30}$

$C_{\text{eq}}=\frac{20\times 30}{50}=12\mu \text{F}$

Charge on capacitor $20\mu \text{F},Q=C_{\text{eq}}V$

$\Rightarrow Q=12\times 10=120\mu \text{C}$

Now, when switch is closed $10\mu \text{F}$ and $20\mu \text{F}$ capacitors are parallel.

Their equivalent capacitance $=20+10=30\mu \text{F}$

This $30\mu \text{F}$ and another $30\mu \text{F}$ are in series.

This equivalent, $C_{\text{eq}}=\frac{30}{2}=15\mu \text{F}$

Now, charge $Q^{\prime }=15\times 10=150\mu \text{C}$

Initially charge on the upper plate of $10\mu \text{F}$ and $20\mu \text{F}$ capacitor was $120\mu \text{C}$, Now this charge has become $150\mu \text{C}$.

So, charge flows through the battery is $150-120=30\mu \text{C}$