ECE 201 Spring 2010Homework 40 SolutionsProblem 30(a)Zth= (R1||R2) − j/ωC= (20 − j10) ΩVoc= VsR2R1+ R2= 20 VrmsFor maximum power transfer, ZL= Z∗th= (20 + j10) Ω.(b)Pavg=V2oc4RL= 5 WProblem 37(a)Here we cannot apply the maximum power transfer theorem because RLisfixed. The expression for average power across RLis given byPavg=V2sRL(R + RL)2+ (ωL − 1/ωC)2Clearly, the average power is maximum when the denominator is minimum,which happens when the following two conditions hold,R = 0, ωL =1ωC1C =1ω2L= 2.5 mFPavgmax=V2sRL=5025= 500 W(b)Pavg=V2sRL(R + RL)2+ (ωL − 1/ωC)2For maximum Pavg,dPavgdRL= 0⇒ R2L= R2+ (ωL − 1/ωC)2⇒ RL= 4.011
View Full Document