A capacitor of capacitance $$900$$ $$\mu$$F is charged by a $$100$$ V battery. The capacitor is disconnected from the battery and connected to another uncharged identical capacitor such that one plate of uncharged capacitor connected to positive plate and another plate of uncharged capacitor connected to negative plate of the charged capacitor. The loss of energy in this process is measured as $$x \times 10^{-2}$$ J. The value of $$x$$ is ______.

A 900 μF capacitor initially charged to 100 V is connected in parallel to an identical uncharged capacitor. The initial energy stored in the first capacitor is $$U_i = \frac{1}{2}CV^2 = \frac{1}{2} \times 900 \times 10^{-6} \times (100)^2 = 4.5$$ J.

After connection, the total charge is $$Q = CV = 900 \times 10^{-6} \times 100 = 0.09$$ C and the combined capacitance is $$C_{total} = 2C = 1800$$ μF. The final voltage across each capacitor is $$V_f = \frac{Q}{C_{total}} = \frac{0.09}{1800 \times 10^{-6}} = 50$$ V, resulting in a final energy of $$U_f = \frac{1}{2}C_{total}V_f^2 = \frac{1}{2} \times 1800 \times 10^{-6} \times 2500 = 2.25$$ J.

The energy loss during the redistribution of charge is $$\Delta U = U_i - U_f = 4.5 - 2.25 = 2.25$$ J $$= 225 \times 10^{-2}$$ J, hence $$x = \boxed{225}$$.