Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 21 - Electromagnetic Induction and Faraday's Law - General Problems - Page 624: 92

Answer

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Work Step by Step

a) From the given definition of $Q$, $$Q=\dfrac{V_L}{V_R}=\dfrac{IX_L}{IR} $$ $$Q= \dfrac{ X_L}{ R} =\dfrac{2\pi f L}{R}$$ And we know that the resonance frequency is given by $$f=\dfrac{1}{2\pi}\sqrt{\dfrac{1}{LC}}$$ So that; $$Q= \dfrac{2\pi L}{R}\dfrac{1}{2\pi}\sqrt{\dfrac{1}{LC}}$$ $$Q= \dfrac{ L}{R} \sqrt{\dfrac{1}{LC}}= \dfrac{ 1}{R}\sqrt{\dfrac{L^2}{LC}}$$ $$\boxed{Q= \dfrac{ 1}{R}\sqrt{\dfrac{L }{ C}}}$$ ---------------------------------------------------------------------- b) First, we need to find $L$; $$f=\dfrac{1}{2\pi}\sqrt{\dfrac{1}{LC}}$$ $$(2\pi f)^2= \dfrac{1}{LC}$$ $$L = \dfrac{1}{(2\pi f)^2C}$$ Plugging the given; $$L = \dfrac{1}{(2\pi \cdot 1.0\times10^6)^2\cdot 0.01\times10^{-6}}$$ $$L =\color{red}{\bf 2.53\times10^{-6}}\;\rm H$$ Now we need to find $R$ by solving the boxed formula above for it. $$R= \dfrac{ 1}{Q}\sqrt{\dfrac{L }{ C}}$$ Plugging the given; $$R= \dfrac{ 1}{650}\sqrt{\dfrac{ 2.53\times10^{-6} }{ 0.01\times10^{-6}}}$$ $$R= \color{red}{\bf0.0245}\;\rm \Omega$$
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