Physics: Principles with Applications (7th Edition)

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

Chapter 19 - DC Circuits - General Problems - Page 558: 91

Answer

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Work Step by Step

a. Use the formula for the equivalent capacitance of capacitors in parallel. $$C_{eq}= 0.20\mu F +0.45\mu F =0.65\mu F $$ Use the equivalent capacitance to calculate the stored energy. $$PE=\frac{1}{2}C_{eq}V^2=\frac{1}{2}(0.65\times10^{-6} F)(24V)^2=1.9\times10^{-4}J$$ b. Use the formula for the equivalent capacitance of capacitors in series. $$C_{eq}= (\frac{1}{0.20\mu F }+\frac{1}{0.45\mu F })^{-1} =0.1385\mu F $$ Use the equivalent capacitance to calculate the stored energy. $$PE=\frac{1}{2}C_{eq}V^2=\frac{1}{2}(0.1385\times10^{-6} F)(24V)^2=4.0\times10^{-5}J$$ c. Use Q=CV to find the charge. $$Q_{parallel}=CV=(0.65\times10^{-6}F)(24V)=1.6\times10^{-5}C$$ $$Q_{series}=CV=(0.1385\times10^{-6}F)(24V)=3.3\times10^{-6}C$$
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