Initially when switch ‘s’ is closed
Total Initial Energy $=\Big(\frac{1}{2}\Big)\text{CV}^2+\Big(\frac{1}{2}\Big)\text{CV}^2=\text{CV}^2\ \dots(1)$
When switch is open the capacitance in each of capacitors varies, hence the energy also varies.
i.e. in case of ‘B’, the charge remains.
Same i.e. cv
$\text{C}_{\text{eff}}=3\text{C}$
$\text{E}=\frac{1}{2}\times\frac{\text{q}^2}{\text{c}}$
$=\frac{1}{2}\times\frac{\text{c}^2\text{v}^2}{3\text{c}}=\frac{\text{cv}^2}{6}$
In case of 'A'
$\text{C}_{\text{eff}}=3\text{c}$
$\text{E}=\frac{1}{2}\times\text{C}_{\text{eff}}\text{v}^2$
$=\frac{1}{2}\times\text{3c}\times\text{v}^2=\frac{3}{2}\text{cv}^2$
Total final energy $=\frac{\text{cv}^2}{6}+\frac{3\text{cv}^2}{2}=\frac{10\text{cv}^2}{6}$
Now, $\frac{\text{Initial Energy}}{\text{Final Energy}}=\frac{\text{cv}^2}{\frac{10\text{cv}^2}{6}}=3$
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