Question

Two parallel plate capacitors $\text{A}$ and $\text{B}$ having capacitance $2\mu \text{F}$ and $3\mu \text{F}$ are charged separately up to $20\text{V}$ and $10\text{V}$ respectively. Now, the positive plate of $\text{A}$ is connected to the positive plate of $\text{B}$ and negative plate of $\text{A}$ is connected with negative plate of $\text{B}$. Find how much charge flows in the circuit?

• A. $40\mu \text{C}$
• B. $42\mu \text{C}$
• C. $12\mu \text{C}$
• D. $28\mu \text{C}$

Answer 1

The correct option is C $12\mu \text{C}$

$\text{Before connection}:$

Let, $Q_1$ and $Q_2$ be initial charge on capacitors $\text{A}$ and $\text{B}$ respectively.

So,
$Q_1=2\mu \text{F}\times 20\text{V}=40\mu \text{C}$
$Q_2=3\mu \text{F}\times 10\text{V}=30\mu \text{C}$

$\text{After Connection:}$


Let, due to the charge distribution both the capacitors attain a common potential $V$ and their charges become $Q_1^{\prime }$ and $Q_2^{\prime }$ respectively. So, using $Q=CV$, we get

$Q_1^{\prime }=2V$ and $Q_2^{\prime }=3V$

Using principle of conservation of charge,

$Q_1+Q_2=Q_1^{\prime }+Q_2^{\prime }$

$\Rightarrow 40+30=2V+3V$

$\therefore V=14\text{V}$

Therefore,

$Q_1^{\prime }=2V=2\times 14=28\mu \text{C}$

$Q_2^{\prime }=3V=3\times 14=42\mu \text{C}$

On capacitor $\text{A}$, initially $40\mu \text{C}$ charge is present and after connection it can hold upto $28\mu \text{C}$ only. Therefore $12\mu \text{C}$ from $\text{A}$ flows through the circuit.


Hence, option (c) is the correct answer.