consider a peak rectifierfed by 60Hz sinusoidal having a peak valueVp=100v. let the load resistance R=10Kohm calculate the fraction of cycle during which the diode is conducting
θ=2VrVp=2×2100=0.2rad0.22π×100%=3.18%\theta=\sqrt{\dfrac{2V_r}{V_p}}=\sqrt{\dfrac{2\times{2}}{100}}=0.2rad\\ \dfrac{0.2}{2\pi}\times100\%=3.18\%θ=Vp2Vr=1002×2=0.2rad2π0.2×100%=3.18%
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