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: 31.9


See explanation.

Work Step by Step

a. $X_L=2 \pi fL=2 \pi (80Hz)(3.00H)=1510\Omega $. b. Solve $X_L=2 \pi fL$ for the inductance. $$L=\frac{X_L}{2 \pi f}=\frac{120\Omega }{2 \pi (80Hz)}=0.239H$$ c. $X_C=\frac{1}{2 \pi fC}=\frac{1}{2 \pi (80Hz)(4.00\times10^{-6}F)}=4.97\times10^{2}\Omega$. d. Solve $X_C=\frac{1}{2 \pi fC}$ for the capacitance. $C=\frac{1}{2 \pi fX_C}=\frac{1}{2 \pi (80Hz)( 120\Omega)}=1.66\times10^{-5}F$
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