Electrical Engineering: Principles & Applications (6th Edition)

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

Chapter 1 - 1.6 - Problems - Introduction to Circuit Elements - Page 40: P1.61



Work Step by Step

Power delivered to the resistor is: $p(t)=v^2(t)/R=48\cos^2(2\pi t)=24+24\cos(4\pi t)$ Then to find energy delivered: $w=\int_0^2 p(t)dt=\int_0^2 24+24\cos(4\pi t)dt =[24+\frac6\pi\sin(4\pi t)]_0^2=48J$
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