23_Ch 18 College Physics ProblemCH18 Direct-Current Circuits

23_Ch 18 College Physics ProblemCH18 Direct-Current...

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Direct-Current Circuits 121 18.32 (a) () ( ) 63 100 20.0 10 F 2.00 10 s 2.00 ms RC τ −− == Ω × = × = (b) ( ) 64 max 20.0 10 F 9.00 V 1.80 10 C 180 C QC µ ε == × = (c) max max max 1 11 1 1 1 t e Q e Q e τττ 4 C QQ  =− =   18.33 ( ) max 5.0 10 F 30 V 1.5 10 C == × ( , and )( ) 66 1.0 10 5.0 10 F 5 ==×Ω × = 10 s 2 t .0 s Thus, at ( ) ( ) 42 max 1 1.5 10 C 1 t e e = × 4 1.3 10 C = × 18.34 The charge on the capacitor at time t is ( ) max 1 t e , where ( ) QC V =∆ and Q max C = . Thus, ( ) 1 t Ve ∆= or ( ) 1 t eV =−∆ We are given that, 12 V = , and at t 1.0 s = , 10 V V = Therefore, 1.0 e s 10 12 1 12 12 10
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