Good discussion and few good remarks. I must explain myself. At my first post I declared:
I do not want to analyze problem in deep. I just want to show you a direction you can follow and do a precise analyze.
Discussion concentrated on main capacitor value (Ok I can agree wit 10 000 uF and with 1 000 uF, both values has their pro and contras, I personally like low ripple at main capacitor so 4700uF for me is the value I prefer). However there is a significant mistake on the drawing: The voltage at capacitors were drawn twice , once as it were
without capacitor (half of sine) and on second drawing
correct voltage with capacitor.
Finally I will calculate the ripple.
The load current is constant, because the 7805 gives the constant voltage at its output, so the load takes constant current and 4700uF capacitor is discharged with constant 1A current. Hence the voltage drop at capacitor is linear (not exponential)
From definition capacitor voltage is equal U= Q/C (Q= charge)
Q= I * time = 1A *8 msec ( not 10 ms , because for 2 ms capcitor is charged from rectifier ).
U = Q/C = (I*t)/C= 1A * 0,008 s / 0,0047 F = 1,7 Vpp
(1,7V peak to peak . Ripple value is often given as rms value of ac component).
The final remark.
To avoid the power on surge a thermistor can be included at transformer primary.
When cold, thermistors resistance is high and limits current surge , during normal work it worms up so its resistance is lowering and losses in thermistor became insignificant.