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} *T_{2}
It can be supposed that the time T_{2} 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.
T_{2}=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.