A capacitor of capacitance C is initially charged to a potential difference of V volt. Now it is connected to a battery of 2V volt with opposite polarity. The ratio of heat generated to the final energy stored in the capacitor will be

Initial charge Q = CV (plate A is positive)
 

Final charge Q’ = 2CV = 2Q (plate B is positive)

Charge flow through the 2V emf cell is

 $\Delta Q=3Q=3CV$

$W_{cell}=\left(\Delta Q\right)\left(2V\right)=6C{V}^2$

$U_f=\frac{1}{2}C{\left(2V\right)}^2=2C{V}^2$

Heat  $=W_{cell}+U_i-U_f$

   $=6C{V}^2+\frac{1}{2}C{V}^2-2C{V}^2=4.5C{V}^2$

$\frac{Heat}{U_f}=\frac{4.5C{V}^2}{2C{V}^2}=2.25$