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905_Physics ProblemsTechnical Physics

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

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Chapter 32 245 P32.22 I I e t = max 1 τ e j : 0 980 1 3 00 10 3 . . = × e τ 0 020 0 3 00 10 0 020 0 7 67 10 3 00 10 3 4 3 . . ln . . . = = − × = × × e τ τ b g s τ = L R , so L R = = × = τ 7 67 10 10 0 7 67 4 . . . e j a f mH FIG. P32.22 P32.23 Name the currents as shown. By Kirchhoff’s laws: I I I 1 2 3 = + (1) + = 10 0 4 00 4 00 0 1 2 . . . V I I (2) + = 10 0 4 00 8 00 1 00 0 1 3 3 . . . . V I I dI dt a f (3) From (1) and (2), + + = 10 0 4 00 4 00 4 00 0 1 1 3 . . . . I I I and I I 1 3 0 500 1 25 = + . . A . FIG. P32.23 Then (3) becomes 10 0 4 00 0 500 1 25 8 00 1 00 0 3 3 3 . . . . . . V A + = I I dI dt b g a f 1 00 10 0 5 00 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: I t e e I I e t t t 3 10 0 1.00 10 1 3 10 5 00 1 0 500 1 1 25 0 500 1 50 0 250 a f a f a f a f = F H G I K J = = + = . . . . . . . V 10.0 A A A H s s P32.24 (a) Using τ = = RC L R , we get R L C = = × = × = 3 00 1 00 10 1 00 6 3 . . . H 3.00 10 F k . (b) τ = = × × = × = RC 1 00 10 3 00 10 3 00 10 3 00 3 6 3 . . . . F s ms e je j P32.25 For t 0 , the current in the inductor is zero . At t = 0 , it starts to
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