6. R = 10, 2, C = F and L = 1, H Compute the current i(t) at the steady-state for Ao = 6, A, e(t) = 10 cos(3t), V,
6. R = 10, 2, C = F and L = 1, H Compute the current i(t) at the steady-state for Ao = 6, A, e(t) = 10 cos(3t), V,
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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please solve asap and in detail
![6.
Compute the current i(t) at the steady-state for Ao = 6, A, e(t) = 10 cos(3t), V,
R = 10, N2, C = F and L = 1, H
i(t)
L
m
e(t)
T
R
(↑) Ao](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F25b98205-ee11-4340-b2c6-069b84298498%2F42d40f01-ff0b-464b-81c0-3d1caf99d211%2Fojll6x_processed.png&w=3840&q=75)
Transcribed Image Text:6.
Compute the current i(t) at the steady-state for Ao = 6, A, e(t) = 10 cos(3t), V,
R = 10, N2, C = F and L = 1, H
i(t)
L
m
e(t)
T
R
(↑) Ao
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i dont understand where the formula for i2 came from please explain that also the values for E W L AND C also how did i2 become -5j? and how did -5j turn into 5cos(3t-90)? the answer should be 5sin(3t) are they identical? please solve in a more detailed response. also why was L short circuited and C open circuited? please answer all my inquiries and ofcourse i will give a thumbs up, thank you in advance!
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