Charging of the capacitor:

Considering the below diagram, with slight approximation the ripple voltage can be supposed as triangular. To the cut-out point from the cut-in point, whatever charge the capacitor achieves is equivalent to the charge the capacitor has lost during the period of non-conduction, that is, from cut-out point to the next cut-in point.

The charge it has acquired

The charge it has lost

V_{r p-p} * C = I_dc *T₂

It can be supposed that the time T₂ is equal to half the periodic time of the waveform if the capacitor's value is fairly large, or the value of the load resistance is extremely large.

T₂=T/2 = 1/2f

Then

V_{r p-p} =I_dc / 2fC

From the above statements, the ripple waveform will be triangular and its rms value is described by

V_{r rms} =V_{r p-p}/ 2 √3

V_{r rms} = I_d.c. = 4√3 f C

= V _d.c. /4√3 f C R_L   

I_d.c. = V_d.c. / R_L

Ripple, ϒ = V_{r rms} / V_d.c. = 1/4√3 f C R_L  

I_d.c. = V_d.c. / R_L

The ripple might be decreased via increasing C or R_L (both) with a resultant increase in the dc Output voltage.