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 - Problems - Page 554: 46

Answer

a) $V_1=6.63V$ $V_2=2.37V$ b) $Q_1=Q_2=3.32\mu C$ c) $V_1=V_2=9.0V$ $Q_1=4.5\mu C$ $Q_2=12.6\mu C$

Work Step by Step

a, b) $C_1=0.50\mu F$ $C_2=1.4\mu F$ $V_1+V_2=9.0V$ $Q_1=Q_2=Q_{eq}$ $C_{eq}=\frac{1}{\frac{1}{C_1}+\frac{1}{C_2}}=0.368\mu F$ $Q_{eq}=C_{eq}V=(0.368\mu F)(9.0V)=3.32\mu C$ $Q_1=Q_2=Q_{eq}$ $V_1=\frac{Q_1}{C_1}=\frac{3.32\mu C}{0.50\mu F}=6.63V$ $V_2=V-V_1=2.37V$ c) $C_1=0.50\mu F$ $C_2=1.4\mu F$ $V_1=V_2=9.0V$ $Q=Q_1+Q_2$ $Q_1=C_1V=(0.50\mu F)(9.0V)=4.5\mu C$ $Q_2=C_2V=(1.40\mu F)(9.0V)=12.6\mu C$
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