C = Laod Current × Capacitor discharge time ÷ acceptable ripple amplitude

I.e.

C = (I × T÷2) ÷ (ΔV)

Capacitor discharge time is approximated to T/2 for full wave bridge rectifier.

For, I = 2.2 A, ΔV = 2V and,

(T÷2) = 1 ÷ (2 × F) =1 ÷ (2 × 50Hz) = 10ms

Thus substituting these values in above equation, i get,

C= 2.2A × 10ms ÷ 2V = 11000 μF

Do I need such a big Capacitor??

Am i going wrong anywhere..?