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final_solution - Prob 5 :n h8 fu frlon'ing rpetitivepattern...

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Prob. 5. Tha periodicwaveform dpw:n h8 fu frlon'ing r€petitive pattern: 5V for l0;rsec, ard -5V for 2psec. Set up the integral and find the m1s value of the waveform. V&, = Avrea4e value o{ Vttt), )o,^ CRnt 9*e &o^ *he' wp,vsf,okp^ I hq* -Yz is a (w atrt 71 6o VRr, = $ /\od, d.* i,"+ "{9* L r \ . ' V*, = E*,\ \(-t>".t+ r _ L O lZtt , l + ) (+s)'J+l = Zysec r J* llra ZyttL a6tl+ '7"7 ^ s b l t VTnc ) 0 v.z =(+..-)(as) (/*-6t +tQ,ry.dd = 7s vz l4rv,a* r.lflrYe*t\- '\[.* |Zysc u: Page 5 of9
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Prob. 6. The tl€guency ot tn€ luvnns volage soulpe ln me clwtul [rerow rs aqJusraurE. wlrnu fiequency f = 0, no current flows because the capacitor is an open circuit at DC. As f is slowly increased from zero, crurent begins to flow. Assuming that the voltage source magnituile remains l0Vrms, use phasor analysis to find a frequency f (in Hertz) where the current magnittrde equals 0.707Arms. t o l 0 l0O l00mH -. 3 - l-rA^-l.rn-t +L- rovrmsO *to* t l r( +3t"1* -l- j*e tilz ' -tqp , l o o + X t [o; fuL-h) g , S X : 9c lDo 0,sf= AJorut u)L-*= I0-a- w"L_ * = I7W wz-Pf -b =c 2 t i i - l 0 0 * * * l o b = 0 L,U ; w ; 1 0 0 k'+(wu- *)- (..#-*\^J defi"e ns X = p,s(roo+ x") ) IDD;50{ 1CI0* 5o/ xz * | oo, x='i,,0 - (0,'lo'1\z= Q, { l05l ) -qg I stol l*, ,1 0"0 J il(r L
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prob.8.
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  • Spring '08
  • Zhang
  • Trigraph, Ih wp,vsf,okp^ q*, ./*-6t +t Q,ry.dd, volage soulpe ln, tl€guency ot tn€, increasedrom zero,crurentbeginsto flow

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