Question

Two capacitors $\text{A}$ and $\text{B}$ with capacitances $10\mu \text{F}$ and $20\mu \text{F}$ are charged to a potential difference of $80\text{V}$ and $40\text{V}$ respectively. Consider a wire connected to each capacitor as shown (with free ends). The upper plate of $\text{A}$ is positive and that of $\text{B}$ is negative, an uncharged $10\mu \text{F}$ capacitor $\text{C}$ with lead wires falls on the free ends to complete the circuit. Find the final charge on each capacitor

• A. $480\mu \text{C}$, $480\mu \text{C}$, $320\mu \text{C}$
• B. $320\mu \text{C}$, $320\mu \text{C}$, $480\mu \text{C}$
• C. $480\mu \text{C}$, $320\mu \text{C}$, $320\mu \text{C}$
• D. $480\mu \text{C}$, $480\mu \text{C}$, $320\mu \text{C}$

Answer 1

The correct option is B $320\mu \text{C}$, $320\mu \text{C}$, $480\mu \text{C}$

Given,

Initially capacitor $\text{A}$ and $\text{B}$ are charged and $\text{C}$ is uncharged. So, the initial charges on capacitor $\text{A}$ and $\text{B}$ are,

$Q_A=10\times 80=800\mu C$ and $Q_B=20\times 40=800\mu C$

Let us consider after connecting capacitor $\text{C}$, $q$ amount of charge flow through the circuit.


So, the final charges on the capacitors are,

$Q_A^{\prime }=800-q$
$Q_B^{\prime }=800-q$ and
$Q_C^{\prime }=q$

Using $\text{KVL}$, we get

$\frac{800-q}{10}-\frac{q}{10}+\frac{800-q}{20}=0$

$\Rightarrow 1600-2q-2q+800-q=0$

$\Rightarrow 5q=2400$

$\Rightarrow q=480\mu \text{C}$

Therefore final charges on $\text{A}$, $\text{B}$ and $\text{C}$ are as follows,

$Q_B^{\prime }=800-480=320\mu \text{C}$ ; $Q_A^{\prime }=800-480=320\mu \text{C}$ ; and
$Q_C^{\prime }=480\mu C$

Hence, option (d) is correct answer.