University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 31 - Alternating Current - Problems - Exercises - Page 1044: 31.2

Answer

a. 2.97A. b. 1.89A. c. The rms value is larger.

Work Step by Step

a. $I=\sqrt{2} I_{rms}=\sqrt2(2.10A)=2.97A$. b. $I_{rav}=\frac{2}{\pi}I\frac{2}{\pi}(2.97A)=1.89A$ c. Comparing Figures 31.3b, and step 2 in Figure 31.4 , we see that the root-mean-square current is always greater than the rectified average. This is verified here; $2.10A\gt1.89A$. In other words, squaring the sinusoidal current, averaging, and then taking the square root to get the rms value gives a larger value than averaging the current’s absolute value.
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