A half-wave rectifier has a resistive load RL=1000 X, with an ac voltage 325 V (peak value) and the diode has a forward resistance of 100 X. Calculate (a) peak average and RMS value of current, (b) dc output power, (c) ac input power and (d) efficiency of the rectifier

Question

A half-wave rectifier has a resistive load [latex]R_L=1000 \Omega[/latex], with an ac voltage 325 V (peak value) and the diode has a forward resistance of [latex]100 \Omega[/latex]. Calculate (a) peak average and RMS value of current, (b) dc output power, (c) ac input power and (d) efficiency of the rectifier

Accepted Answer

(a) Peak value of current,
[latex]I_{m} =\frac{V_m}{r_f+ R_L} =\frac{325}{100+1000} =295.45mA[/latex]
Average current [latex]I_{ds}=\frac{I_m}{\pi }=\frac{295.45}{3.14} =94.046mA[/latex]
RMS current [latex]I_{rms}=\frac{I_m}{2}=147.725mA[/latex]
(b) dc power output=[latex]I_{dc}^2 imes R_L[/latex]
[latex]=(94.046 imes 10^{-3})^2 imes 1000[/latex]
[latex]=8.845W[/latex]
(c) ac input power=[latex]I_{rms}^2(r_f+R_L)[/latex]
[latex]=(147.725 imes 10^{-3})^2 (100+1000)[/latex]
[latex]=24W[/latex]
(d) Efficiency[latex](\eta )=\frac{P_{dc}}{P_{ac}}= \frac{8.845}{24} imes 100[/latex]%
[latex]=26.85[/latex]%

Question 2.10: A half-wave rectifier has a resistive load RL=1000 X, with a...

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