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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)
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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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=0 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. 0.5 VLC 2, J0.25x1 2RC a<@, underdamped case o, - Va 4-0.25 =1.936 L(0) = i(0) = initial inductor current = 20/5 - 4A V0) = v(0) = initial capacitor voltage =0 Vv v(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) dt dv(0) di Thus, (V_ + RI RC (0+4) --0.54, +1.9364, A--2.066 -(v.+IR) %3D 41) =-2.066e sin 1.936t R R 0.50 How did he get the value inside the red circle 88 ט
Transcribed Image Text: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=0 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. 0.5 VLC 2, J0.25x1 2RC a<@, underdamped case o, - Va 4-0.25 =1.936 L(0) = i(0) = initial inductor current = 20/5 - 4A V0) = v(0) = initial capacitor voltage =0 Vv v(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) dt dv(0) di Thus, (V_ + RI RC (0+4) --0.54, +1.9364, A--2.066 -(v.+IR) %3D 41) =-2.066e sin 1.936t R R 0.50 How did he get the value inside the red circle 88 ט
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