Circuit with three capacitors 2

The relation between charge $Q$ , voltage $V$ and capacitance $C$ is $Q=CV$ .

The energy charged on a capacitor is $E=QV=\cfrac{Q^2}{C}$ .

When capacitors are connected in series, $Q$ on each capacitor is same. When the switch $S$ is closed, the equivalent capacitance:

$C_1=\cfrac{1}{\frac{1}{C_0}+\frac{1}{2C_0}+\frac{1}{2C_0}}= \frac{1}{2}C_0$

$\Rightarrow Q_1= C_1V = \cfrac{1}{2}C_0V \quad \Rightarrow E_1= \cfrac{Q_1^2}{C_0}= \cfrac{1}{4}C_0V^2$

Similarly,

$C_2=\cfrac{1}{\frac{1}{C_0}+\frac{1}{2C_0}}= \frac{2}{3}C_0$

$\Rightarrow Q_2= C_2V = \cfrac{2}{3}C_0V \quad \Rightarrow E_2= \cfrac{Q_2^2}{C_0}= \cfrac{4}{9}C_0V^2$

Therefore, $E_1:E_2= \cfrac{1}{4} : \cfrac{4}{9} = \boxed{9:16}$

When the switch is open the the net capacitance is $\frac {C}{2}$ . So the charge in A in first case is $\frac{C}{2}×E$ . So the energy stored in the capacitor is $\frac {CE^{2}}{8}$ .

In the second case when the switch is closed the net capacitance is $\frac {2C}{3}$ . So the energy stored in the capacitor is $\frac {2CE^{2}}{9}$ .

So the ratio of stored energy in capacitor in two cases is $\frac {9}{16}$ .

when the capacitor are in series they all accumulate the same amount of charge Q=C.V so the capacitor with less capacity will have a higher voltage drop the switch when closed cuts capacitor B out of the circuit when the switch is open the capacitor A will have 1/2 V of the power supply
when the switch is closed the capacitor A will have 2/3 V of the power supply

the energy stored E=1/2 C.V^2 E1=1/4 E E2=4/9 E so E1:E2 = 9:16