ECE 201 Spring 2010 Homework 17 Solutions Problem 27 a Let V be the common voltage across C1 and C2 Thus dV dt dV C2 dt C1 C2 is t iC2 t iC2 t C2 is t C1 C2 b Using KVL across the second loop and relation from part a we get rm iC2 t L1 L2 di 0 dt vout t L2 rm di dt L2 C2 is t L1 L2 C1 C2 Problem 38 a Starting from the right end and combining the inductors in series and parallel consecutively we get Leq as Leq 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 combination is in series with 2 4 mH the equivalent of which is in parallel with 7 mH Thus Leq 7 2 4 1 2 0 6 7 2 4 0 4 7 2 8 2 mH Problem 41 a Remember that the equivalent capacitance expressions for series and parallel connections are opposite of that used for resistances The equivalent capacitance can be written as Ceq C1 C2 C3 C4 6 3 4 9 6 F Z t 1 vs t is d Ceq 1 sin 104 t V 6 b Ceq C1 C2 C3 C4 C5 2 1 1 1 C2 C3 C4 C5 18 54 10 8 6 F 1 Ceq 6 60 66 F 1 Zt vs t is d Ceq 1 1 cos 105 t V 660 3
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