Question

An uncharged capacitor with capacitance $2\mu \text{F}$ is connected to two charged capacitors as shown in figure.


After connecting them, final charge appearing on the initially uncharged capacitor will be:

• A. $\frac{135}{6}\mu \text{C}$
• B. $\frac{40}{3}\mu \text{C}$
• C. $\frac{25}{2}\mu \text{C}$
• D. $\frac{125}{3}\mu \text{C}$

Answer 1

The correct option is C $\frac{25}{2}\mu \text{C}$

Let the notation of capacitance of uncharged $2\mu \text{F}$ be $C_1$.

For $5\mu \text{F}$ capacitor, $C_2$, initial charge on it is

$Q_2=C_{2}V=5\times 20=100\mu \text{C}$

Similarly, the initial charge on changed $2\mu \text{F}\left(C_3\right)$ capacitor;

$Q_3=C_3{V}_3=2\times 5=10\mu \text{C}$

After the uncharged capacitor is connected, all the three capacitor form a closed loop.

Let a charge $q$ from capacitor $C_2$ moves to capacitor $C_1$;
Now applying $KVL$ in above closed loop along clockwise directions

$(\Delta V{)}_1+(\Delta V{)}_3+(\Delta V{)}_2=0$

Here, $(\Delta V{)}_1,(\Delta V{)}_3,(\Delta V{)}_2$ represents potential drop across capacitors $C_1,C_3,C_2$ respectively.

$\Rightarrow -\left(\frac{q}{C_1}\right)-\left(\frac{10+q}{C_3}\right)+\left(\frac{100-q}{C_2}\right)=0$

[Potential raise has been taken +ve]

$\Rightarrow -\frac{q}{2}-\left(\frac{10+q}{2}\right)+\left(\frac{100-q}{5}\right)=0$

$\therefore q=\frac{200-50}{12}=\frac{75}{6}=\frac{25}{2}\mu \text{C}$

This is the required charge present on the capacitor.

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

$$\begin{array}{c}\text{Why this question?}\\ \text{Tip:Where capacitor are connected in closed}\\ \text{loop, choose any of the charged capacitor}\\ \text{and let that releases charge q. Now, this}\\ \text{charge will circulate through the entire loop.}\\ \text{It is in accordance with KCL(charge conservation)}\end{array}$$