Question

A capacitor $C_1=1.0\mu F$ is charged up to a voltage $V=60V$ by connecting it to a battery $B$ through the switch (1). Now $C_1$ is disconnected from battery and connected to a circuit consisting of two uncharged capacitors $C_2=3.0\mu F$ and $C_3=6.0\mu F$ through a switch (2) as shown in the figure. The sum of final charges on $C_2$ and $C_3$ is:

• A. $40\mu C$
• B. $36\mu C$
• C. $20\mu C$
• D. $54\mu C$

Answer 1

The correct option is A $40\mu C$

$\begin{array}{c}Q=1\mu F\times 60\\ =60\mu C\\ 1\left(x-0\right)+2\left(x-0\right)\\ \Rightarrow 3x=60\\ \Rightarrow x=20\\ \therefore Q2\mu F=20\times 2=40\mu C\end{array}$

Hence,

option $\left(A\right)$ is correct answer.