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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 1 00 . . V 12.0 A (b) Initial current is 1.00 A: V 12 1 00 12 00 12 0 = = . . . A V a fa f V 1 200 1 00 1 200 1 20 = = . . A kV a fb g V L = 1 21 . kV . FIG. P32.26 (c) I I e Rt L = max : dI dt I R L e Rt L = − max and = = L dI dt V I R e L Rt L max . Solving 12 0 1 212 1 212 2 00 . . V V = b g e t so 9 90 10 3 606 . × = e t . Thus, t = 7 62 . ms . P32.27 τ = = = L R 0 140 4 90 28 6 . . . ms I R max . . = = = ε 6 00 1 22 V 4.90 A (a) I I e t = max 1 τ e j so 0 220 1 22 1 . . = e t τ e j e t = τ 0 820 . : t = − = τ ln . . 0 820 5 66 a f ms (b) I I e e = = = max . . . . 1 1 22 1 1 22 10 0 0 028 6 350 e j a f e j A A FIG. P32.27 (c) I I e t = max τ and 0 160 1 22 . . = e t τ so t = − = τ ln . . 0 131 58 1 a f 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
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