804_Physics ProblemsTechnical Physics

804_Physics ProblemsTechnical Physics - 144 Direct Current...

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144 Direct Current Circuits P28.34 qt Q e tRC a f =− 1s o Q e af 1 0600 1 0900 . . e RC or e RC = 1 0 600 0 400 . .. = 0400 . ln . RC thus RC = = 0982 . ln . . s. *P28.35 We are to calculate ed t RC e dt RC RC e RC ee RC RC −∞ zz F H G I K J − = + 2 0 2 0 2 0 0 2 2 22 2 01 2 . P28.36 (a) τ == × × = RC 150 10 100 10 150 56 . F s ej e j (b) × = 100 . s e j (c) The battery carries current 10 0 200 . V 50.0 10 A 3 × = µ . The 100 k carries current of magnitude II e e t × F H G I K J −− 0 1.00 10 0 100 10 3 s . So the switch carries downward current 200 100 1.00 A A s µµ + bg e t . P28.37 (a) Call the potential at the left junction V L and at the right V R . After a “long” time, the capacitor is fully charged. V L = 800 . V because of voltage divider: I V L L = 10 0 200 10 0 2 00 1 00 8 00 . . . . V 5.00 A V A V Likewise, V R = +
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