{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

804_Physics ProblemsTechnical Physics

804_Physics ProblemsTechnical Physics - 144 Direct Current...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
144 Direct Current Circuits P28.34 q t Q e t RC a f = 1 so q t Q e t RC a f = 1 0 600 1 0 900 . . = e RC or e RC = = 0 900 1 0 600 0 400 . . . = 0 900 0 400 . ln . RC a f thus RC = = 0 900 0 400 0 982 . ln . . a f s . *P28.35 We are to calculate e dt RC e dt RC RC e RC e e RC RC t RC t RC t RC −∞ z z = − F H G I K J = − = − = − = + 2 0 2 0 2 0 0 2 2 2 2 2 0 1 2 . P28.36 (a) τ = = × × = RC 1 50 10 10 0 10 1 50 5 6 . . . F s e je j (b) τ = × × = 1 00 10 10 0 10 1 00 5 6 . . . F s e je j (c) The battery carries current 10 0 200 . V 50.0 10 A 3 × = µ . The 100 k carries current of magnitude I I e e t RC t = = × F H G I K J 0 1.00 10 0 . V 100 10 3 s . So the switch carries downward current 200 100 1.00 A A s µ µ + b g 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 = 8 00 . V because of voltage divider:
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}