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24_Ch 18 College Physics ProblemCH18 Direct-Current Circuits

24_Ch 18 College Physics ProblemCH18 Direct-Current...

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122 CHAPTER 18 18.35 From ( ) max 1 t e Q Q τ = , we have at 0.900 s t = , 0.900 s max Q e Q = 1 0 τ = .600 Thus, 0.900 s 0.400 e τ = , or ( ) ln 0.400 ( ) 0.900 τ s = giving the time constant as 0. 982 s ln 900 s 0. 0.400 = τ = − 18.36 (a) max R I ε = , so the resistance is 4 -3 max 48.0 V 9.60 10 0.500 10 A R I ε = = = × × RC The time constant is τ = , so the capacitance is found to be 5 4 0.960 s 10 F 10.0 F 9.60 10 C R 1.00 τ µ = = × = = × (b) ( )( ) max 10.0 F 48.0 V 480 C Q C µ µ ε = = = 1.92 s t = , so the charge stored in the capacitor at is ( ) ( ) ( ) ( ) 1.92 s 2 0.960 s max 1 480 C 1 480 C 1 415 C t e e e τ Q Q µ µ µ = = =
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