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The halfwave rectifier supplies power to...

The halfwave rectifier supplies power to be `1 k Omega`. The input supply voltage is 220 V neglecting forward resistance of the diode, calculate
(i) `V_(dc)`
(ii) `I_(ac)` and
(iii) Ripple voltage (rms value)

Text Solution

Verified by Experts

(i) `V_(dc)=(V_(m))/(pi)=(sqrt(2)V_("rms"))/(pi)=(sqrt(2)xx220)/(3.14)=99` volt
(ii) `I_(dc)=(V_(dc))/(R_(L))=(99)/(1000)=99mA`
(ii) `r=((V_(r))_("rms"))/(V_(dc))or(V)_("rms")=rxxV_(dc)=1.21 xx99 = 119.79` volt
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