906_Physics ProblemsTechnical Physics

906_Physics ProblemsTechnical Physics - 246 Inductance I=...

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246 Inductance P32.26 (a) I R == = ε 12 0 100 . . V 12.0 A (b) Initial current is 1.00 A: ∆Ω V 12 1200 120 .. . A V af a f V 1 00 1 200 1 20 k V bg V L = 121 . k V . FIG. P32.26 (c) II e Rt L = max : dI dt I R L e Rt L =− max and −= = L dI dt VIR e L Rt L max . Solving 12 0 1 212 1212 200 . . V V = e t so 990 10 36 0 6 . ×= −− e t . Thus, t = 762 m s . P32.27 τ = L R 0140 490 28 6 . . . m s I R max . . = 600 122 V 4.90 A (a) II e t max 1 ej so 0220 1221 e t e t = 0820 . : t = ln . . 566 ms (b) e e = = max 1 1 22 1 1 22 10 0 0 028 6 350 A FIG. P32.27 (c) t = max and 0160 122 = e t so t = ln . . 0131 581 ms . P32.28 (a) For a series connection, both inductors carry equal currents at every instant, so dI dt is the same for both. The voltage across the pair is L dI dt L dI dt L dI dt eq =+ 12 so LL L eq . (b) L dI dt L dI dt L dI dt V L eq === 1 1 2 2
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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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