p33_034 - in magnitude , which (since it is already...

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34. (a) The circuit consists of one generator across one inductor; therefore, E m = V L . The current ampli- tude is I = E m X L = E m ω d L = 25 . 0V (377 rad / s)(12 . 7 H) =5 . 22 × 10 3 A . (b) When the current is at a maximum, its derivative is zero. Thus, Eq. 31-37 gives E L = 0 at that instant. Stated another way, since E ( t )and i ( t )havea90 phase diFerence, then E ( t ) must be zero when i ( t )= I .Thefac ttha t φ =90 = π/ 2 rad is used in part (c). (c) Consider Eq. 32-28 with E = 1 2 E m . In order to satisfy this equation, we require sin( ω d t )= 1 / 2. Now we note that the problem states that
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Unformatted text preview: in magnitude , which (since it is already negative) means that it is becoming more negative. Thus, diFerentiating Eq. 32-28 with respect to time (and demanding the result be negative) we must also require cos( d t ) < 0. These conditions imply that t must equal (2 n 5 / 6) [ n = integer] . Consequently, Eq. 33-29 yields (for all values of n ) i = I sin 2 n 5 6 2 = ( 5 . 22 10 3 A ) 3 2 ! = 4 . 51 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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