le= 2+ (5 – 2)e-15t= 2+ 3e-15tA,%3Dt >0ExampleAt t = 0, switch 1 is closed, and switch 2 is closed 4 s later, Find i(t) for t > 0. Calculate i for a2 s and t= 5 s.t=042t=440 V2Ω5 H10 VWe need to consider the three time intervals t<0, 0 4 separately. For t<0,switches S1 and S2 are open so that i = 0. Since the inductor current cannot change instantly,

Answered: Image /qna-images/answer/16f4dccb-f0af-46c5-9a8a-fd28697461ff.jpg

lle = 2+ (5 – 2)e-15t = 2+ 3e-15t A, %3D t > 0 %3D Example At t= 0, switch 1 is closed, and switch 2 is closed 4 s later. Find i(t) for t> 0. Calculate i for 1s 2 s and t= 5 s. t= 0 42 6 2 t=4 40 V 2Ω 5 H 10 V We need to consider the three time intervals t<0, 0St54, and t> 4 separately. For t<0, switches S1 and S2 are open so that i = 0. Since the inductor current cannot change instantly,

lle = 2+ (5 – 2)e-15t = 2+ 3e-15t A, %3D t > 0 %3D Example At t= 0, switch 1 is closed, and switch 2 is closed 4 s later. Find i(t) for t> 0. Calculate i for 1s 2 s and t= 5 s. t= 0 42 6 2 t=4 40 V 2Ω 5 H 10 V We need to consider the three time intervals t<0, 0St54, and t> 4 separately. For t<0, switches S1 and S2 are open so that i = 0. Since the inductor current cannot change instantly,

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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