905_Physics ProblemsTechnical Physics

905_Physics ProblemsTechnical Physics - Chapter 32 P32.22 e...

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Chapter 32 245 P32.22 II e t =− max 1 τ ej : 0980 1 300 10 3 . . −× e 00200 767 10 3 4 3 . . ln . . . = × e bg s = L R , so LR == × = 100 767 4 .. . af mH FIG. P32.22 P32.23 Name the currents as shown. By Kirchhoff’s laws: III 123 =+ (1) +−−= 10 0 4 00 4 00 0 12 . V (2) +−−− = 10 0 4 00 8 00 1 00 0 13 3 . . V dI dt a f (3) From (1) and (2), +− + = 10 0 4 00 4 00 4 00 0 113 . . and 0500 125 A . FIG. P32.23 Then (3) becomes 10 0 4 00 0 500 1 25 8 00 1 00 0 33 3 . . . . V A −+ = dI dt 500 3 3 . H V a f a f dI dt I F H G I K J += . We solve the differential equation using Equations 32.6 and 32.7: It e e e t t t 3 10 0 1.00 10 10 10 5 0 0 1 1 25 0 500 1 50 0 250 a f = F H G I K J −= = . . . . . V 10.0 A A A H s s P32.24 (a) Using RC L R , we get R L C × = 300 100 10 6 3 . H 3.00 10 F k ΩΩ . (b) × × = × = −− RC 1 00 10 3 00 10 3 00 10 3 00 36 3 . . F s m s e j P32.25 For t 0 , the current in the inductor is zero . At t = 0 , it starts to grow from zero toward 10.0 A with time constant = × L R 10 0 100
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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