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: P1.26

Answer

a) Power is given by: $P(t) = 20sin(100 \pi t)$ b) $.127 J$ c) 0 J

Work Step by Step

a) Power is equal to the voltage times the current. Thus, we find: $P(t) = 2 \times 10sin(100 \pi t) = 20sin(100 \pi t)dt$ b) The energy is given by the integral of the power over time. Thus, we find: $w = \int_{t_0}^{t} P(t)dt = \int_{0}^{.01}20sin(100 \pi t)dt $ Using a calculator to evaluate the integral, we find: $\fbox{w = .127 J}$ c) The energy is given by the integral of the power over time. Thus, we find: $w = \int_{t_0}^{t} P(t)dt = \int_{0}^{.02}20sin(100 \pi t)dt $ Using a calculator to evaluate the integral, we find: $\fbox{w = 0J}$
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