I need the answer as soon as possible -5A, + 19.365A,A21.033dt19.365i(f) = e (4cos19.365t +1.033 sin19.365t)a) Find the roots of the characteristic equation thatgoverns the transient behavior of the voltageshown in Fig. 8.5 if R = 200 0, L = 50 mH, andC = 0.2 µF.icb) Will the response be overdamped, underdamped,How did he get the valueb) The voltage responsew3 < a?.inside the red circlec) For R = 312.5 0,Solution106a) For the given values of R, L, and C,a =(625)(0.2)10664 x 1061= (%3Dα= 1.25 x 104 rad/s,2 RC(400)(0.2)As w remains at 10 1(10°)(10º)S1 = -80001= 10° rad?/s?.%3DLC(50)(0.2)S, = -8000From Eqs. 8.14 and 8.15,d) For critical damping1E05 x 10%108-2RCor= -12,500 + 7500 = -5000 rad/s,2RandS2 = -1.25 x 104 - V1.5625 × 108 – 10*10|R =(2 × 10Given the circuit in Fig 8 95 fnd ilt) and vft) for t

Answered: Image /qna-images/answer/62b13848-87b6-43ff-8393-4cee7068464b.jpg

a) Find the roots of the characteristic equation that governs the transient behavior of the voltage shown in Fig. 8.5 if R = 200 Q, L = 50 mH, and C = 0.2 µF. + Vo b) Will the response be overdamped, underdamped, How did he get the value b) The voltage response wi < a?. inside the red circle c) For R = 312.5 Q, %3D Solution 106 ) For the given values of R, L, and C, a (625)(0.2) 106 a = 64 x 106 = ( 1 a = %3D %3D = 1.25 x 104 rad/s, 2RC (400)(0.2) As wó remains at 10 1 (10°)(10) S1 = = -8000 1 = 10° rad?/s². LC (50)(0.2) S2 = -8000 %3D From Eqs. 8.14 and 8.15, d) For critical damping 05 x 10 108 2RC or 1 = -12,500 + 7500 = -5000 rad/s, 2R and S2 = -1.25 x 10 - V1.5625 x 10 – 10 10 R = (2 × 10

a) Find the roots of the characteristic equation that governs the transient behavior of the voltage shown in Fig. 8.5 if R = 200 Q, L = 50 mH, and C = 0.2 µF. + Vo b) Will the response be overdamped, underdamped, How did he get the value b) The voltage response wi < a?. inside the red circle c) For R = 312.5 Q, %3D Solution 106 ) For the given values of R, L, and C, a (625)(0.2) 106 a = 64 x 106 = ( 1 a = %3D %3D = 1.25 x 104 rad/s, 2RC (400)(0.2) As wó remains at 10 1 (10°)(10) S1 = = -8000 1 = 10° rad?/s². LC (50)(0.2) S2 = -8000 %3D From Eqs. 8.14 and 8.15, d) For critical damping 05 x 10 108 2RC or 1 = -12,500 + 7500 = -5000 rad/s, 2R and S2 = -1.25 x 10 - V1.5625 x 10 – 10 10 R = (2 × 10

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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