*Assuming vin(t)=2sin(1000ft) volts and using simplified (2nd) diode approximation, the waveforms ofVin(t) and the load voltage VRL(t) are shown asVin(t)2V-2V+VRL(t)1.5V-OV 1W1t21314 t, ms4t, msNOTES:The diode only conducts from t1 to t2 for every input cycle when vin(t)2vf. When the diode isconducting, VRL(t)=vin(t)-vf=2sin(1000nt)-0.5 volts, When the diode is not conducting, VRL(t)=0V.DETERMINE AVERAGE VALUE AND RMS OF VRL(t): SHOW COMPLETE SOLUTION (SUGGESTED ANSWERS 0.4066 AND 0.6941)(c) The average value and RMS value of VRL(t). Hint: The formulas for half-wave rectified sine cannot be used. Solve by evaluating the general solutions that involve integration.

Answered: Image /qna-images/answer/8e6fa293-7e73-48b0-89b4-9d54304d683d.jpg

* Assuming vin(t)=2sin(1000ft) volts and using simplified (2nd) diode approximation, the waveforms of Vin(t) and the load voltage VRL(t) are shown as Vin(t) 2V W m VRL(t) 1 t2 -2V+ 1.5V- OV 1 13 14 t, ms 4t, ms NOTES: The diode only conducts from t1 to t2 for every input cycle when vin(t)2vf. When the diode is conducting, VRL(t)=vin(t)-vf=2sin(1000xt)-0.5 volts, When the diode is not conducting, VRL(t)=0V. DETERMINE AVERAGE VALUE AND RMS OF VRL(t): SHOW COMPLETE SOLUTION (SUGGESTED ANSWERS 0.4066 AND 0.6941) (c) The average value and RMS value of VRL(t). Hint: The formulas for half-wave rectified sine can not be used. Solve by evaluating the general solutions that involve integration.

* Assuming vin(t)=2sin(1000ft) volts and using simplified (2nd) diode approximation, the waveforms of Vin(t) and the load voltage VRL(t) are shown as Vin(t) 2V W m VRL(t) 1 t2 -2V+ 1.5V- OV 1 13 14 t, ms 4t, ms NOTES: The diode only conducts from t1 to t2 for every input cycle when vin(t)2vf. When the diode is conducting, VRL(t)=vin(t)-vf=2sin(1000xt)-0.5 volts, When the diode is not conducting, VRL(t)=0V. DETERMINE AVERAGE VALUE AND RMS OF VRL(t): SHOW COMPLETE SOLUTION (SUGGESTED ANSWERS 0.4066 AND 0.6941) (c) The average value and RMS value of VRL(t). Hint: The formulas for half-wave rectified sine can not be used. Solve by evaluating the general solutions that involve integration.

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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