Question

Two capacitors $\text{A}$ and $\text{B}$ with capacitances $20\mu \text{F}$ and $50\mu \text{F}$ are charged to a potential difference $10\text{V}$ and $20\text{V}$ respectively. Consider a wire connected to each capacitor as shown (with free ends). The upper plate of both $\text{A}$ and $\text{B}$ are negative. An uncharged $100\mu \text{F}$ capacitor $C$ with lead wires falls on the free ends to complete the circuit. Find the final charge on each capacitor (in $\mu \text{C}$)

• A. $2500,1500,1000$
• B. $1000,1500,500$
• C. $\frac{1100}{3},\frac{2500}{3},\frac{500}{3}$
  
• D. $\frac{2000}{3},\frac{1000}{3},\frac{500}{3}$
  

Answer 1

The correct option is C $\frac{1100}{3},\frac{2500}{3},\frac{500}{3}$


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

$Q_A=20\times 10=200\mu \text{C}$, $Q_B=50\times 20=1000\mu \text{C}$

Let us, consider after connecting the capacitor $\text{C}$ with $\text{A}$ and $\text{B}$, $q$ amount of charge flows through the circuit.


Applying $\text{KVL}$, to the loop $\text{PQRS}$.

$-\frac{200+q}{20}+\frac{q}{100}+\frac{1000-q}{50}=0$

$\Rightarrow -1000-5q+q+2000-2q=0$

$\Rightarrow 1000=6q$

$\therefore q=\frac{500}{3}$

Therefore final charge on capacitors $\text{A, B, C}$ as follows.

$Q_A^{\prime }=200+q=200+\frac{500}{3}=\frac{1100}{3}\mu \text{C}$

$Q_B^{\prime }=1000-q=1000-\frac{500}{3}=\frac{2500}{3}\mu \text{C}$

$Q_C^{\prime }=q=\frac{500}{3}\mu \text{C}$

Hence, option (c) is correct answer.