I need the answer as soon as possible Find the voltage across the capacitor as a function of time for t>0 for the circuit in Fig.8.72. Assume steady-state conditions exist at t=01a 025 H20 vFigure 8.72For Prob. 8. 18.Chapter 8, Solution 18.When the switch is off, we have a source-free parallel RLC circuit.0.5VLC2,J0.25x12RCa<@,underdamped case o, - Va4-0.25 =1.936L(0) = i(0) = initial inductor current = 20/5 - 4AV0) = v(0) = initial capacitor voltage =0 Vvv(1) =e(4, cosot+ A, sin @,t)=e®*" (A, cos 1.936 + A, sin1.9361)v(0) -0 - A,dv"(-0.5)(4, cos 1.936t + A, sin 1.936) + e0s (-1.9364, sin 1.936t +1.9364, cos1.9361)dtdv(0)diThus,(V_ + RIRC(0+4)--0.54, +1.9364,A--2.066-(v.+IR)%3D41) =-2.066esin 1.936tRR0.50How did he get the valueinside the red circle88ט

Answered: Image /qna-images/answer/9d2a7795-53b3-4bce-a4d2-727b32c81925.jpg

Find the voltage across the capacitor as a function of time for t>0 for the circuit in Fig. 8.72. Assume steady-state conditions exist at 1=0 sa 1a 025 H 20 V Figure 8.72 For Prob. 8. 18. Chapter 8, Solution 18. When the switch is off, we have a source-free parallel RLC circuit. 1 2, 0.5 LC J0.25xl 2RC underdamped case V4 – 0.25 = 1.936 a<@, L(0) = (0) = initial inductor current = 20/5 = 4A VL0) = v(0) - initial capacitor voltage = o v (1) =e*(4, cosot+ A, sin w,t)=e® (4, cos1.936 + A, sin1.9361) v(0) =0 - A, dv -0.5)4, cos1.9361 + A, sin 1.936r)+e(-1.9364, sin 1.936r +1.9364, cos1.936 dt (V + RI RC dv(0) (0+4) --0.54, +1.936A, A --2.066 dt -LR = -(V++IR) Thus, %3D {1) =-2.066e sin 1.936t -0.50

Find the voltage across the capacitor as a function of time for t>0 for the circuit in Fig. 8.72. Assume steady-state conditions exist at 1=0 sa 1a 025 H 20 V Figure 8.72 For Prob. 8. 18. Chapter 8, Solution 18. When the switch is off, we have a source-free parallel RLC circuit. 1 2, 0.5 LC J0.25xl 2RC underdamped case V4 – 0.25 = 1.936 a<@, L(0) = (0) = initial inductor current = 20/5 = 4A VL0) = v(0) - initial capacitor voltage = o v (1) =e*(4, cosot+ A, sin w,t)=e® (4, cos1.936 + A, sin1.9361) v(0) =0 - A, dv -0.5)4, cos1.9361 + A, sin 1.936r)+e(-1.9364, sin 1.936r +1.9364, cos1.936 dt (V + RI RC dv(0) (0+4) --0.54, +1.936A, A --2.066 dt -LR = -(V++IR) Thus, %3D {1) =-2.066e sin 1.936t -0.50

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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