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% Thanking you guys in advance!
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 ? my answer will be 46.65V
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)]
Thanks for your reply #12. I am not from the States. Ans English is my second language, so it's maybe myself that are not having a good English here The Vdc= Vm - 1/2 [Vr(p-p)] doesn't get me the correct answer. I think it has something to do with the half-wave rectifier. Really appreciate if someone can help