Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 35 - AC Circuits - Exercises and Problems - Page 1054: 59

Answer

a) ${\bf 11.6}\;\rm pF $ b) ${\bf 1.49}\;\rm m\Omega$

Work Step by Step

$$\color{blue}{\bf [a]}$$ We know, for $RLC$-circuits, that the resonance frequency occurs when $$X_L=X_C$$ $$2\pi f L=\dfrac{1}{2\pi f C}$$ Hence, $$C=\dfrac{1}{4\pi f^2 L}$$ Plug the known; $$C=\dfrac{1}{4\pi (104.3\times 10^6)^2 (0.20\times 10^{-6})}$$ $$C=\color{red}{\bf 1.16\times 10^{-11}}\;\rm F$$ $$\color{blue}{\bf [b]}$$ The peak current of the original station is given by $$I=\dfrac{\varepsilon_0}{Z} =\dfrac{\varepsilon_0}{\sqrt{(X_L-X_C)^2+R^2}}$$ where at the resonance frequency $X_L=X_C$, so $Z=R$ $$I =\dfrac{\varepsilon_0}{R}$$ The current of the out-of-tune station is $$I'=\dfrac{0.1}{100}I=\dfrac{0.1}{100}\dfrac{\varepsilon_0}{R}$$ where $I'=\varepsilon_0/Z=\varepsilon_0/\sqrt{(X_L-X_C)^2+R^2}$ $$ \dfrac{ \color{red}{\bf\not} \varepsilon_0}{\sqrt{(X_L-X_C)^2+R^2}} =\dfrac{0.1}{100}\dfrac{ \color{red}{\bf\not} \varepsilon_0}{R}$$ Squaring both sides; $$ \dfrac{ 1}{ (X_L-X_C)^2+R^2} = \dfrac{ 10^{-6}}{R^2}$$ Hence, $$R^2=10^{-6}\left[(X_L-X_C)^2+R^2\right]$$ $$10^{ 6}R^2-R^2= (X_L-X_C)^2 $$ $$R =\sqrt{ \dfrac{(X_L-X_C)^2 }{(10^{ 6} -1)}}$$ $$R =\dfrac{|X_L-X_C|}{\sqrt{ (10^{ 6} -1)}}$$ where $X_L=\omega L=2\pi fL$, and $X_C=1/2\pi fC$ $$R =\dfrac{\left|2\pi f'L-\dfrac{1}{2\pi f' C}\right|}{\sqrt{ (10^{ 6} -1)}}$$ Plug the known; $$R =\dfrac{\left|2\pi (103.9\times 10^{6})(0.2\times 10^{-6})-\dfrac{1}{2\pi (103.9\times 10^{6}) (1.16\times 10^{-11})}\right|}{\sqrt{ (10^{ 6} -1)}}$$ $$R=\color{red}{\bf 1.49}\;\rm m\Omega$$
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