A single-phase single-pulse diode rectifier (Fig. 3.1), with a resistive load (R=100Q), is fed from an AC voltage source v,(1) = Vs - /2 -sin(@t), where Vs=240 V, f=50Hz.VdiodeiFig. 3.1. A single-phase single-pulse diode rectifier with a resistive loada) Calculate the average and RMS values of the current and powerdissipation in the load resistor;b) To evaluate the performance of the rectifier, calculate the efficiency of therectification and the power factor;

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A single-phase single-pulse diode rectifier (Fig. 3.1), with a resistive load (R=100 N), is fed from an AC voltage source v,(t) =Vs ·/2•sin(@), where Vs=240 V, f=50 Hz. Fig. 3.1. A single-phase single-pulse diode rectifier with a resistive load

A single-phase single-pulse diode rectifier (Fig. 3.1), with a resistive load (R=100 N), is fed from an AC voltage source v,(t) =Vs ·/2•sin(@), where Vs=240 V, f=50 Hz. Fig. 3.1. A single-phase single-pulse diode rectifier with a resistive load

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Question

A single-phase single-pulse diode rectifier (Fig. 3.1), with a resistive load (R=100
Q), is fed from an AC voltage source v,(1) = Vs - /2 -sin(@t), where Vs=240 V, f=50
Hz.
Vdiode
i
Fig. 3.1. A single-phase single-pulse diode rectifier with a resistive load
a) Calculate the average and RMS values of the current and power
dissipation in the load resistor;
b) To evaluate the performance of the rectifier, calculate the efficiency of the
rectification and the power factor;

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