P31_058 - 1 e t/ L 2 . At t = 0 . 10 s, this yields P...

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58. (a) From Eq. 31-51 and Eq. 31-43, the rate at which the energy is being stored in the inductor is dU B dt = d ( 1 2 Li 2 ) dt = Li di dt = L µ E R ³ 1 e t/τ L ´ ¶µ E R 1 τ L e t/τ L = E 2 R ³ 1 e t/τ L ´ e t/τ L . Now, τ L = L/R =2 . 0H / 10 Ω = 0 . 20 s and E = 100 V, so the above expression yields dU B /dt = 2 . 4 × 10 2 Wwhen t =0 . 10 s. (b) From Eq. 27-22 and Eq. 31-43, the rate at which the resistor is generating thermal energy is P thermal = i 2 R = E 2 R 2 ³ 1 e t/τ L ´ 2 R = E 2 R ³
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Unformatted text preview: 1 e t/ L 2 . At t = 0 . 10 s, this yields P thermal = 1 . 5 10 2 W. (c) By energy conservation, the rate of energy being supplied to the circuit by the battery is P battery = P thermal + dU B dt = 3 . 9 10 2 W . We note that this could result could alternatively have been found from Eq. 28-14 (with Eq. 31-43)....
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