A fullwave rectifier supplies a load of `1 k Omega`. The a.c voltage applied to the diode is 220 volt rms. If diode resistance is neglected, calculate
(i) Average d.c voltage
(ii) Average d.c current
(iii) Ripple voltage (rms).

Explanation

(i) Average d.c voltage `V_(dc)=(2V_(m))/(pi)=0.636V_(m)`
where `V_(m)` = maximum across each half of the secondary winding
If V be the rms voltage across each half of the secondary winding, then `V = (V_(m))/(sqrt(2))`
`:. V_(dc)=0.636xxVsqrt(2)=0.9V=0.9xx220=198` volt
(ii) For fullwave recifier `I_(dc)=(V_(dc))/(R_(L))=(198)/(1000)=198mA`
(iii) `tau=(V_(r("rms")))/(V_(dc))rArr V_(r("rms"))=rxxV_(dc)rArr V_(r("rms"))=0.482xx198=95.436` volt