Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 24 - Alternating-Current Circuits - Problems and Conceptual Exercises - Page 867: 15

Answer

(a) $1.16K\Omega$ (b) $0.137\mu F$ (c) $0.86mA$ (d) $4.3mA$

Work Step by Step

(a) We know that $I_{rms}=\omega CV_{rms}$ This can be rearranged as: $\frac{1}{\omega C}=\frac{V_{rms}}{I_{rms}}$ $\implies X_C=\frac{0.5}{0.43\times 10^{-3}}$ $\implies X_C=1.16\times 10^3\Omega=1.16K\Omega$ (b) We know that $C=\frac{1}{\omega X_C}$ We plug in the known values to obtain: $C=\frac{1}{2000\times \pi \times 1.162\times 10^3}$ $\implies C=0.137\mu F$ (c) The rms current can be determined as $\omega=2\pi f$ $\omega=2\pi \times 2000=4000\pi s^{-1}$ Now $I_{rms}=\omega C\times V_{rms}$ $I_{rms}=4000\pi \times 0.137\times 10^{-6}\times 0.5$ $\implies I_{rms}=0.860mA$ (d) We know that $\omega=2\pi f$ $\omega=2\pi (10000)=20000\pi s^{-1}$ Now $I_{rms}=20000\pi\times 0.137\times 10^{-6}\times 5$ $\implies I_{rms}=4.3mA$
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