Two identical capacitors have same capacitance C . One of them is...

The common potential after connection would be, ${V}_{C} = \dfrac{{q}_{net}}{{C}_{net}} = \dfrac{CV + C\left(2V\right)}{2C}$

$⇒ {V}_{C} = \dfrac{3V}{2}$

Therefore, loss of energy

$= \dfrac{1}{2}C{V}^{2} + \dfrac{1}{2}C{\left(2V\right)}^{2} - \dfrac{1}{2}2C{\left(\dfrac{3V}{2}\right)}^{2}$

$= \left(\dfrac{C{V}^{2}}{4}\right)$