Unformatted text preview: Solve: (a) The equivalent capacitance of the and capacitors in parallel is When these two capacitors are replaced by their equivalent we get the network sketched in Figure 18.55. The equivalent capacitance of these three capacitors in series is (b) (c) is the same as Q for each of the capacitors in the series combination shown in Figure 18.55, so Q for each of the capacitors is Figure 18.55 Reflect: The voltages across each capacitor in Figure 18.55 are and The sum of the voltages equals the applied voltage, apart from a small difference due to rounding. 18.56. Set Up: For capacitors in series the voltages add and the charges are the same; For capacitors in parallel the voltages are the same and the charges add; Solve: (a) The equivalent capacitance of the 18.0 nF, 30.0 nF and 10.0 nF capacitors in series is 5.29 nF. When these capacitors are replaced by their equivalent we get the network sketched in Figure 18.56. The equivalent capacitance of these three capacitors in parallel is 19.3 nF, and this is the equivalent capacitance of the original network.of these three capacitors in parallel is 19....
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This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.
 Spring '07
 Shoberg
 Capacitance, Work

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