WHomework set 5_solns_2011

# WHomework set 5_solns_2011 - Homework set # 5 Problems:...

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Homework set # 5 Problems: 16.27, 32, 34-5, 37, 42; 17.1, 13 Solutions: 16.27 (a) For series connection, 12 11 1 eq eq CC C C C C   = 0.050 F 0.100 F 400 V 13.3 C on each 0.050 F 0.100 F eq QC V V          (b)      = 0.050 F 400 V 20.0 C QC V      22 = 0.100 F 400 V 40.0 C 16.32 (a) The combination reduces to an equivalent capacitance of 12.0 F in stages as shown below. (b) From Figure 2,    4 4.00 F 36.0 V 144 C Q    2 2.00 F 36.0 V 72.0 C Q and    6 6.00 F 36.0 V 216 C Q Then, from Figure 1, 2 486 216 C QQ Q  4.0 m F 2.0 m F 24.0 m F 8.0 m F 36.0±V Figure 1 4.0 m F 2.0 m F 6.0 m F 36.0±V Figure ± 12.0 m F 36.0±V Figure 3

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16.34 From   QCV  , the initial charge of each capacitor is    10 10.0 F 12.0 V 120 C Q  and   00 xx QC After the capacitors are connected in parallel, the potential difference across each is 3.00 V V  , and the total charge of 10 120 C x QQ Q  is divided between the two capacitors as
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## This note was uploaded on 05/12/2011 for the course PHYS 1112 taught by Professor Seaton during the Spring '08 term at University of Georgia Athens.

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WHomework set 5_solns_2011 - Homework set # 5 Problems:...

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