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908_Physics ProblemsTechnical Physics

# 908_Physics ProblemsTechnical Physics - 248 Inductance a fa...

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248 Inductance P32.35 (a) U LI = = 1 2 1 2 4 00 0 500 2 2 . . H A a fa f U = 0 500 . J (b) When the current is 1.00 A, Kirchhoff’s loop rule reads + = 22 0 1 00 5 00 0 . . . V A a fa f V L . Then V L = 17 0 . V . The power being stored in the inductor is I V L = = 1 00 17 0 17 0 . . . A V W a fa f . FIG. P32.35 (c) P = = I V 0 500 22 0 . . A V a fa f P = 11 0 . W P32.36 From Equation 32.7, I R e Rt L = ε 1 e j . (a) The maximum current, after a long time t , is I R = = ε 2 00 . A . At that time, the inductor is fully energized and P = = = I V a f a fa f 2 00 10 0 20 0 . . . A V W . (b) P lost A W = = = I R 2 2 2 00 5 00 20 0 . . . a f a f (c) P inductor drop = = I V e j 0 (d) U LI = = = 2 2 2 10 0 2 00 2 20 0 . . . H A J a fa f P32.37 We have u E =∈ 0 2 2 and u B = 2 0 2 µ . Therefore = 0 2 2 0 2 2 E B µ so B E 2 0 0 =∈ µ 2 B E = = × × = × 0 0 5 8 3 6 80 10 3 00 10 2 27 10 µ . . . V m m s T . P32.38 The total magnetic energy is the volume integral of the energy density, u B = 2 0 2 µ . Because B changes with position, u is not constant. For B B R r = F H G I K J
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