500 N1000 N160 nF100 mHR = 500 N, R2 = 1000 2,L = 100 mH, C = 160 nF, i,(t) = 4 cos 5000t mAi,(t) = 4 cos 5000t mAconstant current sourceV,(t)Vc(t)ic(t)V,(t)i,(t)VR, (t)VR, (t)voltage on consțant current sourcevoltage across capacitorcurrent through the capacitorinductance voltagecurrent flowing through inductanceR = 500 N Voltage across the resistorR2 = 1000 Q voltage across the resistor+ 10* sIc(s)H(s) =!g(s)s2 + 104 s + 625. 105What is the mathematical expressionof the voltage Ic(t) in the steady(continuous) state?a) Vc(e) = 4,472 cos ( 5000r - ) vb) Vc(t) = 3,5776 cos ( 5000r + ) vc) Ve(t) = 2 cos ( 5000t +d) Ve(e) = 4 cos ( s000r - ) v18,43e) Vc(t) = 6,32455 cos ( 5000tV.

Answered: Image /qna-images/answer/09f24b38-d381-46f0-acf0-9a70515b27c9.jpg

500 N 1000 N 160 nF 100 mH R = 500 N, R2 = 1000 N, L = 100 mH, C = 160 nF, i,(t) = 4 cos 5000t mA i,(t) = 4 cos 5000t mA constant current source V,(t) Vc(t) ic(t) V,(t) i,(t) VR, (t) VR, (t) voltage on consțant current source voltage across capacitor current through the capacitor inductance voltage current flowing through inductance R = 500 NVvoltage across the resistor R2 = 1000 Q voltage across the resistor + 10* s Ic(s) H(s) = Ig(s) s2 + 104 s + 625. 105 What is the mathematical expression of the voltage Ic(t) in the steady (continuous) state? a) Ve(e) = 4,472 cos ( 5000r - 2 ) v b) Ve(t) = 3,5776 cos ( 5000t + ) v c) Ve(t) = 2 cos ( 5000t + d) Ve(e) = 4 cos ( 5000r - ) v 18,43 e) Vc(t) = 6,32455 cos ( 5000t V.

500 N 1000 N 160 nF 100 mH R = 500 N, R2 = 1000 N, L = 100 mH, C = 160 nF, i,(t) = 4 cos 5000t mA i,(t) = 4 cos 5000t mA constant current source V,(t) Vc(t) ic(t) V,(t) i,(t) VR, (t) VR, (t) voltage on consțant current source voltage across capacitor current through the capacitor inductance voltage current flowing through inductance R = 500 NVvoltage across the resistor R2 = 1000 Q voltage across the resistor + 10* s Ic(s) H(s) = Ig(s) s2 + 104 s + 625. 105 What is the mathematical expression of the voltage Ic(t) in the steady (continuous) state? a) Ve(e) = 4,472 cos ( 5000r - 2 ) v b) Ve(t) = 3,5776 cos ( 5000t + ) v c) Ve(t) = 2 cos ( 5000t + d) Ve(e) = 4 cos ( 5000r - ) v 18,43 e) Vc(t) = 6,32455 cos ( 5000t V.

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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