18_InstSolManual_PDF_Part21

# 18_InstSolManual_PDF_Part21 - 2 1 c C eq Energy 5 1 2 QV 5...

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18.83. Set Up: Solve: (a) (b) (c) 18.84. Set Up: The stored energy is Solve: (a) (b) so (c) (d) 18.85. Set Up: For capacitors in series, the equivalent resistance is given by For capacitors in parallel, the equivalent capacitance is given by Figure 18.85 Solve: (a) Using the rules for combining capacitors in series and in parallel gives the sequence of equivalent net- works shown in Figure 18.85. The equivalent capacitance of the network is (b) In Figure 18.85d the three capacitors in series have the same capacitance, so the voltage across each is The voltage across in Figure 18.85c is 140 V, so Reflect: We could continue to analyze the networks in Figure 18.85 and find V and Q for each capacitor in the network. Q 2 5 C 2 V 2 5 1 4.6 m F 21 140 V 2 5 644 m C. C 2 Q 1 5 C 1 V 1 5 1 6.9 m F 21 140 V 2 5 966 m C. 1 420 V 2 / 3 5 140 V. 2.3 m F. b b a a C 1 C 1 C 2 C 2 2.3 m F C 1 C 1 ( a ) 2.3 m F C 1 C 1 C 2 a b ( c ) C 1 C 1 6.9 m F ( d ) 2.3 m F a b ( e ) C 1 C 1 C 1 C 2 C 1 6.9 m F ( b ) b a C eq 5 C 1 1 C 2 1 c C eq 1 C eq 5 1 C 1 1 1 C 2 1
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Unformatted text preview: 2 1 c . C eq Energy 5 1 2 QV 5 1 2 1 0.0180 3 10 2 6 C 21 200 V 2 5 1.80 3 10 2 6 5 1.80 m J V 5 Ed 5 1 3.0 3 10 6 V / m 21 1.50 3 10 2 3 m 2 5 4.5 3 10 3 V A 5 Cd P 5 1 9.00 3 10 2 11 F 21 1.50 3 10 2 3 m 2 8.854 3 10 2 12 C 2 / 1 N # m 2 2 5 0.0152 m 2 . C 5 P A d C 5 0.0180 3 10 2 6 C 200 V 5 9.00 3 10 2 11 F 5 90.0 pF 1 2 QV . V ab 5 Ed . C 5 P A d . C 5 Q V ab . Energy 5 1 2 CV 2 5 1 2 1 3.54 3 10 2 11 F 21 300 V 2 2 5 1.59 3 10 2 6 J Q 5 CV ab 5 1 3.54 3 10 2 11 F 21 300 V 2 5 1.06 3 10 2 8 C C 5 1 5.00 2 1 8.854 3 10 2 12 C 2 / 1 N # m 2 221 16.0 3 10 2 4 m 2 2 0.200 3 10 2 2 m 5 3.54 3 10 2 11 F Energy 5 1 2 QV ab 5 1 2 CV ab 2 5 1 2 Q 2 C . C 5 Q V ab . C 5 K P A d . Electric Potential and Capacitance 18-21...
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