ECE 201 – Lecture 23

201 3 prof bermel 025 h solution 20 1030ut 01

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Unformatted text preview: /19/2012 ECE 201-3, Prof. Bermel Solution 20 Ω 0.1 mF + VC - [-1+3u(t)]/2 0.25 H • From earlier definitions: 1 Γ= = = 1 ∙ 0 ∙ 10 1 0. 5 ∙ 10 • Circuit is overdamped 10/19/2012 ECE 201-3, Prof. Bermel = 50 = 00 Solution 20 Ω 0.1 mF + VC - [-1+3u(t)]/2 0.25 H • For t<0, inductor looks like short and VC(t<0)=0 • For t>0, non-zero voltage temporarily possible; start by solving for IL in overdamped regime: = + + Here, Γ = 250, and Γ= 10/19/2012 Γ− = 250 − 200 = 150 ECE 201-3, Prof. Bermel Solution 20 Ω 0.1 mF + VC - [-1+3u(t)]/2 0.25 H • Given that: ≥0 = = + • Since IL(0)=-1/2, and VC(0)=0: + +1 +1 −= + +1 =− − 0.25 −100 − 400 =0 = −4 • Thus, = 1/2, and = −2 10/19/2012 ECE 201-3, Prof. Bermel Solution 20 Ω 0.1 mF + VC - [-1+3u(t)]/2 0.25 H 1 ≥ 0 = −2 + 2 • Since circuit elements are in parallel: +1 ≥0 = = 0.25 −2 −100 = 50 10/19/2012 − ECE 201-3, Prof. Bermel 1 + −400 2 Solution Voltage (VC) 25 20 15 10 5 0 -5 0 5 10 15 20 Time t (ms) 10/19/2012 ECE 201-3, Prof. Bermel 25 30 35 40 Homework • HW #22 due today by 4:30 pm in EE 326B • HW #23 due Mon.: DeCarlo & Lin, Chapter 9: – Problem 11(a) [Correction: the second iL (0-) should be iL(0+)] – Problem 18 – Problem 27 10/19/2012 ECE 201-3, Prof. Bermel...
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