# % ripple factor

Discussion in 'Homework Help' started by lynnfaiz, Dec 16, 2012.

1. ### lynnfaiz Thread Starter New Member

Dec 16, 2012
29
1
I have an answer for these questions but do not have the solution steps from my supervisor. However, i manage to get the answer for i & ii. but my answer for iii is different to the one given.
My work are written below. can anyone help to confirm if the answer i have is the correct one.

Question:
The 35 Vrms ac voltage Vs is derived from an ideal 60Hz main
transformer. It is connected to a half-wave rectifier and a
220μF capacitor to form a dc power supply. If the load RL
draws an average current 0.15A, determine:
i. the peak-to-peak capacitor ripple voltage Answer : Vr(pp) = 5.6V
ii. the average or dc voltage across the load Answer : Vdc = 43.82V
iii. the percentage ripple factor, %r. Answer : 12.8%

iii) my solution steps
with formular --> %r = [Vr(rms) /Vdc] x 100%.
I calculated Vr(rms) = Idc/ (4√3 (60)(220u) = 1.64V
and Vdc = Vm - (Idc/2fC) =43.82V
So, the %r = 1.64 / 43.82 (100%) = 3.74%

2. ### lynnfaiz Thread Starter New Member

Dec 16, 2012
29
1
and back to same question -# ii,
Since i have the anwer for Vr(p-p) = 5.68V, why cant i use the formular

Vdc= Vm - 1/2 [Vr(p-p)] to get the Vdc ?

3. ### #12 Expert

Nov 30, 2010
16,345
6,831
I am very old and can not understand modern school speak, but I don't remember any 4 radical3's in the ripple formula. I also did not see accounting for the voltage loss in the rectifier.

I believe your last formula is correct. (Vdc=Vm-1/2[Vr(p-p)]

lynnfaiz likes this.

Dec 16, 2012
29
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