As there are 10 V, for Vp1, that is the peak-voltage of the source:
[tex]Vp1=10*\sqrt{2}=14.14 V[/tex]
Then, transformer's theory says that the relation of transformations is:
V1/V2=a
Where V1 is the voltage in the primary and V2 in the secondary.
V1=14.14 V
V2=8.55 V
a=1.65
Then, with the 8.5 V, we find the real peak-voltage, taking in account that in the diodes we have a drop of 0.7 V each, so:
8.5 -1.4=7.1 V
And this will be called VpL
Now we proceed to calculate the mean voltage:
[tex]V_{mean}=VpL-(\frac{Vr}{2} )[/tex]
Where Vr is the ripple voltage, we asume that is 1 V
So, Vmean = 6.6 V
Then we have
Vmean/R= I mean
We have that R=1000 Ohm
Imedia=6.6 V/1000 Ohm
Imedia=6.6 mAmps
Finally, we can calculate the capacitor:
C=Q/Vr
C=Imean/(Vr*2f)
Where f is 60Hz
C=6.6mA/(1V*120)
C=5.5 uFarads
Therefore:
C=5.5 uFarads that works at 12 V
