p33_036 - (c) Consider Eq. 32-28 with E = 1 2 E m . In...

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36. (a) The circuit consists of one generator across one capacitor; therefore, E m = V C . Consequently, the current amplitude is I = E m X C = ωC E m = (377 rad / s)(4 . 15 × 10 6 F)(25 . 0V)=3 . 91 × 10 2 A . (b) When the current is at a maximum, the charge on the capacitor is changing at its largest rate. This happens not when it is fully charged ( ± q max ), but rather as it passes through the (momentary) states of being uncharged ( q = 0). Since q = CV , then the voltage across the capacitor (and at the generator, by the loop rule) is zero when the current is at a maximum. Stated more precisely, the time-dependent emf E ( t ) and current i ( t )havea φ = 90 phase relation, implying E ( t )=0when i ( t )= I .Th efa c ttha t φ = 90
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Unformatted text preview: (c) Consider Eq. 32-28 with E = 1 2 E m . In order to satisfy this equation, we require sin( d t ) = 1 / 2. Now we note that the problem states that E is increasing in magnitude , which (since it is already negative) means that it is becoming more negative. Thus, dierentiating 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 = ( 3 . 91 10 3 A ) 3 2 ! = 3 . 38 10 2 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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