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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. 2, JLC V0.25xl 0.5 2RC underdamped case V4 –0.25 = 1.936 a

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. 2, JLC V0.25xl 0.5 2RC underdamped case V4 –0.25 = 1.936 a

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= o 0.25 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 -JIc J0.25xl 2, 0.5 2RC a<@. underdamped case 4–0.25 = 1.936 L(0) = i(0) = initial inductor current 20/5 = 4A V0) = v(0) = initial capacitor voltage = 0 V v(1) =e(4, cosot+ A, sin ot)=es (A, cos 1.936 + A, sin1.9361) v(0) -0= A, dv s"(-0.5X4, cos1.936t + A, sin L92 +1.9364, cos 1.9361) dr dv(0) dt (V + RI A =-2.066 --0.54, +1.936A, RC = =V-LR -(vo+IR) R Thus, (1) --2.066e sin 1.936 How did he get the value inside the red circle 88 i)
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= o 0.25 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 -JIc J0.25xl 2, 0.5 2RC a<@. underdamped case 4–0.25 = 1.936 L(0) = i(0) = initial inductor current 20/5 = 4A V0) = v(0) = initial capacitor voltage = 0 V v(1) =e(4, cosot+ A, sin ot)=es (A, cos 1.936 + A, sin1.9361) v(0) -0= A, dv s"(-0.5X4, cos1.936t + A, sin L92 +1.9364, cos 1.9361) dr dv(0) dt (V + RI A =-2.066 --0.54, +1.936A, RC = =V-LR -(vo+IR) R Thus, (1) --2.066e sin 1.936 How did he get the value inside the red circle 88 i)
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