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problem24_20

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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24.20: a) and b) The equivalent resistance of the combination is 6.0 , F μ therefore the total charge on the network is: C. 10 16 . 2 ) V 36 )( F 0 . 6 ( 4 eq eq - × = = μ V C Q This is also the charge on the F 0 . 9 μ capacitor because it is connected in series with the point b. So: . V 24 F 10 0 . 9 C 10 16 . 2 6 4 9 9 9 = × × = = - - C Q V Then . V 12 V 24 V 36 9 6 12 11 3 = - = - = + = = V V V V V V C. 10 6 . 3 ) V 12 )( F 0 . 3 ( 5 3 3 3 - × = = = μ V C Q . C 10 32 . 1 ) V 12 )( F 11 ( 4 11 11 11 - × = = = μ V C Q 11 3 12 6 Q Q Q Q Q - - = = C. 10 32 . 1 C 10 6 . 3 C 10 16 . 2 4 5 4 - - - × - × - × = C. 10 8 . 4 5 - × = So now the final voltages can be calculated: V. 4 F 10 12 C 10 8 . 4 V. 8 F 10 0 . 6 C 10 8 . 4 6 5 12 12 12 6 5 6 6 6 = × × = = = × × = = - - - - C Q V C Q V c) Since the 3 F 6 and F 11 , F μ μ μ capacitors are connected in parallel and are in
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