A capacitor $\mathrm{C}_1$ is charged up to a voltage $\mathrm{V}=$ $60 \mathrm{~V}$ by connecting it to…
A capacitor $\mathrm{C}_1$ is charged up to a voltage $\mathrm{V}=$ $60 \mathrm{~V}$ by connecting it to battery $\mathrm{B}$ through switch (1), Now $\mathrm{C}_1$ is disconnected from battery and connected to a circuit consisting of two uncharged capacitors $\mathrm{C}_2=3.0 \mu \mathrm{F}$ and $\mathrm{C}_3=6.0 \mu \mathrm{F}$ through a switch $(2)$ as shown in the figure. The sum of final charges on $\mathrm{C}_2$ and $\mathrm{C}_3$ is:
$36 \mu \mathrm{C}$
$20 \mu \mathrm{C}$
$54 \mu \mathrm{C}$
$40 \mu \mathrm{C}$
Solution
The sum of final charges on $\mathrm{C}_2$ and $\mathrm{C}_3$ is 36 $\mu \mathrm{C}$.
