P31_103 - 6 s, we have t/ L = 3 / 5, and we apply Eq....

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103. (a) As the switch closes at t = 0, the current being zero in the inductor serves as an initial condition for the building-up of current in the circuit. Thus, at t = 0 the current through the battery is also zero. (b) With no current anywhere in the circuit at t = 0, the loop rule requires the emf of the inductor E L to cancel that of the battery ( E = 40 V). Thus, the absolute value of Eq. 31-37 yields di dt = |E L | L = 40 0 . 050 = 800 A / s . (c) This circuit becomes equivalent to that analyzed in § 31-9 when we replace the parallel set of 20000 Ω resistors with R = 10000 Ω. Now, with τ L = L/R =5 × 10
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Unformatted text preview: 6 s, we have t/ L = 3 / 5, and we apply Eq. 31-43: i = E R 1 e 3 / 5 1 . 8 10 3 A . (d) The rate of change of the current is Fgured from the loop rule (and Eq. 31-37): E i R |E L | = 0 . Using the values from part (c), we obtain |E L | 22 V. Then, di dt = |E L | L = 22 . 050 440 A / s . (e) and (f) As t , the circuit reaches a steady state condition, so that di/dt = 0 and E L = 0. The loop rule then leads to E i R |E L | = 0 = i = 40 10000 = 4 . 10 3 A ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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