+ VLT Problem 2: Appearing to the right is again our famil- iar RLC series circuit, this time with a new voltage la- bel

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Vlt Problem 2 Appearing To The Right Is Again Our Famil Iar Rlc Series Circuit This Time With A New Voltage La Bel 1 (35.04 KiB) Viewed 32 times

+ VLT 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 mon L V1 constant. Va + ictv +1 i 1-part (C) = 0 Keat+Kentif a1 +421 in-compt) = Kiem + Kateat, if a1 = az. in(t) = 11-comp (6) + 11-part(t) R W where 2 2 R (1 R 2LV2L C- 1 LC and 02 R 2L V (2)" R 21 1 LC Find an expression for u(t) assuming a *ag. There is no need to solve for Ki and Ky 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, K, K2,a,, and a2 + a, the voltage v10) =