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Unformatted text preview: .---2 -J'( v *i4O jzo Ac-iL(-14-'t r fFl 1 ) Class Homework Problems #19 Monday, Apr. 6,2009-.j /u 'c ft 7= eL ru ('i/-,c) fz 144Y Name 4lll}009 I l:05:00 AM-l/ot'C R (,t4"L) tZ+JaoL 3. In the circuit shown below find the voltage v6(f) across the resistor. Note that in this circuit the two voltage sources have different frequencies. This means that you rza^rt use superposition to solve this problem. We can't use mesh analysis, for example, because the impedances of the inductor and capacitor will be different for the two different voltege sources. L=2H C=tpF vr(t) =l0cos377t AC vr(t) = 5sin754t l-:zu (= 17. l= C' /rl= .ib,L + V, 'J/b( + ,2 KQ AC u ,r 777 lb/0 H 5p' 4 r= 7s'4 rL- -L t<i, o I r j,u,lLc Vo= /ou t/- - r-/-2,'+ = V=* /r". vcz ,_ Z._1 4"k) = Voz* /6t (L" t t d"-l \, k) = 4,. cu; (a,/ * A-.) u,lfr) j-L ++ *:, /.=ZH = 4. /c) t q"kJ...
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This note was uploaded on 03/05/2010 for the course EGR 240 taught by Professor Michael,p during the Winter '08 term at Oakland University.
- Winter '08