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20 V 8 ΚΩ t = 0 2 ΚΩ 10 μF Boundary conditions: v(0) = 4 V, v(∞) = 0 V. i(0+) = 2 mA, i(∞) = 0 A. ODES: dv di +50v = 0. +50i = 0. dt dt Responses: v(t) = 4e-50t V₁t > 0. -50t i(t) = 2e mA, t> 0. t = 0 VR 6 ΚΩ 3 ΚΩ 12 V 1/3 mF VC Boundary conditions: vc(0+) = 0 V, vc(∞) = 12 V. = VR(0) 12 V, VR (∞) = 0 V. ODES: dvc dvR +vc = 12 V. +VR = 0. dt dt Responses: vc(t) = 12(1e−t) V, t> 0. VR(t) 12et V,t> 0.

20 V 8 ΚΩ t = 0 2 ΚΩ 10 μF Boundary conditions: v(0) = 4 V, v(∞) = 0 V. i(0+) = 2 mA, i(∞) = 0 A. ODES: dv di +50v = 0. +50i = 0. dt dt Responses: v(t) = 4e-50t V₁t > 0. -50t i(t) = 2e mA, t> 0. t = 0 VR 6 ΚΩ 3 ΚΩ 12 V 1/3 mF VC Boundary conditions: vc(0+) = 0 V, vc(∞) = 12 V. = VR(0) 12 V, VR (∞) = 0 V. ODES: dvc dvR +vc = 12 V. +VR = 0. dt dt Responses: vc(t) = 12(1e−t) V, t> 0. VR(t) 12et V,t> 0.

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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please show best and easy way/working to solve for boundary conditions, odes, and general solutions for both parts

20 V 8 ΚΩ t = 0 2 ΚΩ 10 μF Boundary conditions: v(0) = 4 V, v(∞) = 0 V. i(0+) = 2 mA, i(∞) = 0 A. ODES: dv di +50v = 0. +50i = 0. dt dt Responses: v(t) = 4e-50t V₁t > 0. -50t i(t) = 2e mA, t> 0. t = 0 VR 6 ΚΩ 3 ΚΩ 12 V 1/3 mF VC Boundary conditions: vc(0+) = 0 V, vc(∞) = 12 V. = VR(0) 12 V, VR (∞) = 0 V. ODES: dvc dvR +vc = 12 V. +VR = 0. dt dt Responses: vc(t) = 12(1e−t) V, t> 0. VR(t) 12et V,t> 0.
Transcribed Image Text:20 V 8 ΚΩ t = 0 2 ΚΩ 10 μF Boundary conditions: v(0) = 4 V, v(∞) = 0 V. i(0+) = 2 mA, i(∞) = 0 A. ODES: dv di +50v = 0. +50i = 0. dt dt Responses: v(t) = 4e-50t V₁t > 0. -50t i(t) = 2e mA, t> 0. t = 0 VR 6 ΚΩ 3 ΚΩ 12 V 1/3 mF VC Boundary conditions: vc(0+) = 0 V, vc(∞) = 12 V. = VR(0) 12 V, VR (∞) = 0 V. ODES: dvc dvR +vc = 12 V. +VR = 0. dt dt Responses: vc(t) = 12(1e−t) V, t> 0. VR(t) 12et V,t> 0.
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