Electrical Engineering: Principles & Applications (6th Edition)

Published by Prentice Hall
ISBN 10: 0133116646
ISBN 13: 978-0-13311-664-9

Chapter 1 - 1.3 - Problems - Power and Enery - Page 36: P1.24


22.5 J is absorbed

Work Step by Step

Here, passive reference configuration is not followed, so power = -vi = -(-15)3$e^{-2t}$ = 45$e^{-2t}$ W, is always positive. This implies energy is delivered to the element. E = $\int P dt $ Energy transferred from t = 0 to t = $\infty$ = $$\int_{0}^{\infty} P dt$$ = $$\int_{0}^{\infty} 45e^{-2t} dt$$ = 22.5 J
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