ECE 201 Spring 2010Homework 17 SolutionsProblem 27(a)Let V be the common voltage across C1and C2. Thus(C1+ C2)dVdt= is(t)C2dVdt= iC2(t)⇒ iC2(t) =C2C1+ C2is(t)(b)Using KVL across the second loop and relation from part (a), we getrmiC2(t) − (L1+ L2)didt= 0vout(t) = L2didt= rmL2C2(L1+ L2)(C1+ C2)is(t)Problem 38(a)Starting from the right end and combining the inductors in series and parallelconsecutively, we get LeqasLeq= 4 + ((36)||(10 + ((1 + 5)||(3))))= 4 + ((36)||(10 + ((6)||(3))))1= 4 + ((36)||(10 + (2)))= 4 + ((36)||(12))= 4 + 9= 13 mH(b)Clearly, the given circuit has 1.2 mH and 0.6 mH in parallel, and this com-bination is in series with 2.4 mH, the equivalent of which is in parallel with7 mH. ThusLeq= 7||(2.4 + (1.2||0.6))= 7||(2.4 + 0.4)= 7||2.8= 2 mHProblem 41(a)Remember that the equivalent capacitance expressions for series and par-allel connections are opposite of that used for resistances. The equivalentcapacitance can be written asCeq= C1||(C2+ (C3||C4))= 4 +6 × 39= 6 µFvs(t) =1CeqZt−∞is(τ) dτ=16sin(104t) V(b)Ceq= C1||(C2+ (C3||C4) + C5)2(C2+ (C3||C4) + C5) =118+154+110.8−1= 6 µF⇒ Ceq= 6 + 60= 66 µFvs(t) =1CeqZt−∞is(τ) dτ=1660(1 − cos(105t))
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