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The solution are attached. Can you please explain if the circuit is a natural or step responce. I;m having difficultu understanding. I think it should be natural responce because once the switch is closed, all the electrical components will discharge through the resistor? Thank you

The solution are attached. Can you please explain if the circuit is a natural or step responce. I;m having difficultu understanding. I think it should be natural responce because once the switch is closed, all the electrical components will discharge through the resistor? Thank you

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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The solution are attached. Can you please explain if the circuit is a natural or step responce. I;m having difficultu understanding. I think it should be natural responce because once the switch is closed, all the electrical components will discharge through the resistor?

Thank you

8.44 In the circuit in Fig. P8.44, the resistor is adjusted for critical damping. The initial capacitor voltage is 15 V, and the initial inductor current is 6 mA. a) Find the numerical value of R. (Ans: 1250 (2) b) Find the numerical values of i and di/dt immedi- ately after the switch is closed. (Ans: 60 A/s) c) Find vc(t) for t≥ 0. (Ans: 56,250 t e-5,000t + 15×e-5,000t) Figure P8.44 + VC t = 0) 320 nF R 125 mH
Transcribed Image Text:8.44 In the circuit in Fig. P8.44, the resistor is adjusted for critical damping. The initial capacitor voltage is 15 V, and the initial inductor current is 6 mA. a) Find the numerical value of R. (Ans: 1250 (2) b) Find the numerical values of i and di/dt immedi- ately after the switch is closed. (Ans: 60 A/s) c) Find vc(t) for t≥ 0. (Ans: 56,250 t e-5,000t + 15×e-5,000t) Figure P8.44 + VC t = 0) 320 nF R 125 mH
P 8.44 [a] w² = α = R 2L ... R = (5000) (2) L = 1250 N di 1 LC [c] vc dt (0) [b] i(0) — iz(0) = 6 mA = 10⁹ (125)(0.32) wwwww Wo VL vL (0) — 15 — (0.006) (1250) = 7.5 V 60 A/s + D₂e-5000t 7.5 0.125 Dite-5000t vc (0) = D₂ = 15 V - - W = A 5000 rad/s duc (0) dt .. D₁ = 56,250 V/s VC 56,250te ²(0) = D₁ – 5000D₂ = -5000t 25 x 106 = ic(0) C V₁ + 15e-5000t V. -iL (0) с = -18,750 t≥0.
Transcribed Image Text:P 8.44 [a] w² = α = R 2L ... R = (5000) (2) L = 1250 N di 1 LC [c] vc dt (0) [b] i(0) — iz(0) = 6 mA = 10⁹ (125)(0.32) wwwww Wo VL vL (0) — 15 — (0.006) (1250) = 7.5 V 60 A/s + D₂e-5000t 7.5 0.125 Dite-5000t vc (0) = D₂ = 15 V - - W = A 5000 rad/s duc (0) dt .. D₁ = 56,250 V/s VC 56,250te ²(0) = D₁ – 5000D₂ = -5000t 25 x 106 = ic(0) C V₁ + 15e-5000t V. -iL (0) с = -18,750 t≥0.
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