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CHAPTER 9 PROBLEM SOLUTIONSSOLUTION TO PROBLEM 9.1.If we can compute expressions for K and q that are real,then these quantities exist by construction. Consider that A, B, K and q must satisfy thefollowing relationship:Kcos(ϖt+ θ)=Kcos(θ)(29cos(ϖt)+-Ksin(θ)(29sin(ϖt)≡Acos(ϖt)+Bsin(ϖt)Therefore Kcos(θ)=Aand -Ksin(θ)=B. Consequently,Kcos(θ)(292+-Ksin(θ)(292=K2=A2+B2in which case K=A2+B2. Further,Ksin(θ)Kcos(θ)=tan(θ)=-BAin which caseθ =tan-1-BAwith due regard to quadrant.SOLUTION TO PROBLEM 9.2.For the inductor,WL(t)=12L⋅iL2(t)=12L VoCLsin1LCt2=CV022sin21LCtand for the capacitor,WC(t)=12C⋅vC2(t)=12C Vocos1LCt2=CV022cos21LCt.Hence,WC+WL=12C⋅vC2(t)+12L⋅iL2(t)=CV022sin21LCt +cos21LCt=CV022SOLUTION TO PROBLEM 9.3.Since x(t)=(K1+K2t)e-αt,x' (t)= - α ⋅K1e-αt+K2e-αt- α ⋅t⋅K2e-αtandx' ' (t)= α2⋅K1e-αt- α ⋅K2e-αt- α ⋅K2e-αt+ α2⋅t⋅K2e-αtSubstituting into the differential equation, we haveα2⋅K1e-αt-2α ⋅K2e-αt+ α2⋅t⋅K2e-αt+2α - α ⋅K1e-αt+K2e-αt- α ⋅t⋅K2e-αt+ α2K1e-αt+K2te-αt=0
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