NEED AHNDWRITTEN SOLUTION PLEASE vs (t)+ I402wwR₁iL (t)17 cos(500t+45°) V R₁₂102L8mHC0.4mFAfter transforming the series (vs (t) + R₁) to a parallel current source and resistor(is (t) || R1),Total admittance YT-Ex: 0.11 -jSInductor current IL=iL(t):=Ex: 11.1 A0cos (Ex: 1t-°) A

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Answered: vs (t) + I 402 ww R₁ iL (t) 17…

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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Questions 16 to 20 (Write True or False) A16 A larger inductor produces a smaller rate of change of current. A17 KVL states that E IR + E EMF = 0 for any open circuits. A18 For a sine waveform, the value of form factor is 1.414. A19 In a purely inductive circuit, the current lags the voltage by II/4 rads. In an ideal transformer, the rate of change of flux is the same for both primary A20 and secondary.
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Find the theoretical values and answer the questions. please answear about all this questions.
4. A capacitor now is connected in parallel with R1 to reduce the ripple. Find the value of the capacitance needed to make the following ripple index less than 5% through PSpice simulation study. V -×100% -V o,min 0,max rpp RFP = -x100% = V dc dc D1 Dbreak R1 V1 1k VAMPL = 169.71 FREQ = 50 Vo
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Calibri (Body) -11 by AaBbCcDx AaBbCcD AaBbC AaBbCc AaB AaBbCc. mat Painter BIU 1Normal 1No Spaci. Heading 1 Heading 2 Title Subtitle Chang Styles d Font Paragraph Styles IMPEDANCE QUIZ # 2 1. What capacitance when connected in series with a 500Q resistor will limit the current drawn from a 48-mV 465-kHz source to 20µA? * a. 144pF b. 145pF c. 146pF d. 147pF MY ANSWER: c.146pF 2. What is the total impedance at 20kHz of a series circuit consisting of a 1.5mH inductance, a 100Q resistance, and a 0.08uF capacitance? * a. 1340 41.7 b. 1930 L-37.2" C. 920 290 d. 530 2-12.6°
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vs (t) + I 402 ww R₁ iL (t) 17 cos(500t+45°) V R₁₂102 L8mH C 0.4mF After transforming the series (vs (t) + R₁) to a parallel current source and resistor (is (t) || R1), Total admittance YT-Ex: 0.11 -j S Inductor current IL = iL(t): = Ex: 11.1 A 0 cos (Ex: 1 t- °) A