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 1053: 40

Answer

${\bf 3.15}\rm\;\mu C$, ${\bf 5.53}\rm\;\mu C$

Work Step by Step

When the two capacitors are in parallel, the equivalent capacitance is given by $$C_{eq,1}=C_1+C_2\tag 1$$ And when the two capacitors are in series, the equivalent capacitance is given by $$C_{eq,2}=\left[\dfrac{1}{C_1}+\dfrac{1}{C_2}\right]^{-1}=\dfrac{C_1C_2}{C_1+C_2}\tag 2$$ So the rms current, in the first case, is given by $$I_{\rm rms,1}=\dfrac{\varepsilon_{\rm rms}}{X_{C}}=\dfrac{\varepsilon_{\rm rms}}{1/\omega C_{eq,1}}$$ Plug from (1) $$I_{\rm rms,1}=2\pi f \varepsilon_{\rm rms} (C_1+C_2)$$ So, $$C_1+C_2=\dfrac{I_{\rm rms,1}}{2\pi f \varepsilon_{\rm rms}}$$ Plug the known; $$C_1+C_2=\dfrac{(545\times 10^{-3})}{2\pi (1\times 10^3)(10) }$$ $$C_1+C_2=\dfrac{109}{4\times 10^6\pi}\tag 3$$ And the rms current, in the second case, is given by $$I_{\rm rms,2}=\dfrac{\varepsilon_{\rm rms}}{X_{C}}=\dfrac{\varepsilon_{\rm rms}}{1/\omega C_{eq,2}}$$ Plug from (2); $$I_{\rm rms,2}=2\pi f \varepsilon_{\rm rms} \dfrac{C_1C_2}{C_1+C_2}$$ So, $$\dfrac{C_1C_2}{C_1+C_2} =\dfrac{I_{\rm rms,2}}{2\pi f \varepsilon_{\rm rms} }$$ $$ C_1C_2 =(C_1+C_2)\dfrac{I_{\rm rms,2}}{2\pi f \varepsilon_{\rm rms}}$$ Plug the known and plug from (3); $$ C_1C_2 =\dfrac{109}{4\times 10^6\pi}\cdot \dfrac{ (126\times 10^{-3})}{2\pi (1\times 10^3) (10) }$$ $$ C_1C_2 =\dfrac{6867}{4\times 10^{13}\pi^2} $$ Hence, $$ C_2 =\dfrac{6867}{4\times 10^{13}\pi^2C_1} $$ Plug into (3); $$C_1+\dfrac{6867}{4\times 10^{13}\pi^2C_1} =\dfrac{109}{4\times 10^6\pi} $$ Thus, the two capacitors are: $$C_1=3.15\times 10^{-6}\;\rm C=\color{red}{\bf 3.15}\;\mu C$$ Or, $$C_2=5.53\times 10^{-6}\;\rm C=\color{red}{\bf 5.53}\;\mu C$$
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