+ VL m L Vi Problem 2: Appearing to the right is again our famil- iar RLC series circuit, this time with a new voltage l

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Vl M L Vi Problem 2 Appearing To The Right Is Again Our Famil Iar Rlc Series Circuit This Time With A New Voltage L 1 (75.82 KiB) Viewed 29 times

+ VL m L Vi Problem 2: Appearing to the right is again our famil- iar RLC series circuit, this time with a new voltage la- bel, 01. Also once again, for your solving convenience, is a solution for this circuit when the voltage source is constant. + Va | + 11 CVC R 11-part (t) = 0 SKjent +K2e4zt, if a1 + a2; 11-comp Kjenit +Kateat, if a1 = 22. 11(t) = 11-comp(t) + 11-part (1) ) W where R 2 2 21 R 2L + V2L ' 1 LC and 02 R R 2LV2L (6) - 1 LC Find an expression for vi(t) assuming a1 + 02. There is no need to solve for Ki and K, in this problem. Hint: Those of you who are still unsure of the difference between a voltage drop and a voltage with respect to ground: learn it now! In terms of R, L,C,va, K1, K2,01, and a2 + aj the voltage vi(t) =