L Va C V1 W Problem 4: Appearing to the right is not our famil- iar RLC series circuit. Two nodes are labeled with volta

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L Va C V1 W Problem 4: Appearing to the right is not our famil- iar RLC series circuit. Two nodes are labeled with volta

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L Va C V1 W Problem 4 Appearing To The Right Is Not Our Famil Iar Rlc Series Circuit Two Nodes Are Labeled With Volta 1
L Va C V1 W Problem 4 Appearing To The Right Is Not Our Famil Iar Rlc Series Circuit Two Nodes Are Labeled With Volta 1 (24.62 KiB) Viewed 28 times
L Va C V1 W Problem 4 Appearing To The Right Is Not Our Famil Iar Rlc Series Circuit Two Nodes Are Labeled With Volta 2
L Va C V1 W Problem 4 Appearing To The Right Is Not Our Famil Iar Rlc Series Circuit Two Nodes Are Labeled With Volta 2 (49.8 KiB) Viewed 28 times
L Va C V1 W Problem 4: Appearing to the right is not our famil- iar RLC series circuit. Two nodes are labeled with voltages, vi and U2. (a) Write KCL node equations for those nodes and put them in normal form. Use the p operator notation R1 for « (Remember p"#(t) dx() for n > 0 and a(t) = S «(t)dt.) Each voltage should be multiplied by a polynomial in p such as (ap + B + r) KCL node equations in normal form using p operator notation: C2 R₂ dt dem V2
(6) Let G denote a matrix of the coefficients of the equations above when written in normal form. (See Homework 4, https://www.ece. Isu.edu/ee2120/2022/hw04.pdf.) The determinant of G provides the characteristic equation used to find the complementary (transient) solution for each voltage. (For those that are curious: Given GV = S, and using the fact that G-1 = adj G/ det G, we get a set of non-simultaneous equations using det(G)V = adj(G)S, where adj(G) is the adjugate of G. Note that det G multiplies each element of V whereas adj(G)S is a matrix/vector product.) For this problem with R1 R2 + R and C + C + C, the determinant is: 4C 2 1 det G =Cp2+ + L - RP+ RLP Multiplying by p/C2 and equating with zero yields the characteristic equation: 4 2 LCP+ RCP²+ = 0 pit LCP* RLCP For R=100K2, L = 20 mH, and C = 30/4F the factored characteristic equation is (p +0.6667)(p+0.3333 - j1291)p+.3333 +j1291) = 0 Based on this characteristic equation show the complementary solution for U, and provide a sketch of vi over time. (The answer is exactly the same for 12.) (Do not attempt to solve for the constants K1, K2, K3.) Could an RLC circuit have such a complementary (transient) response? Complementary solution for : Sketch appearance and indicate if appearance is possible for an RLC circuit in which their is one inductor and one capacitor.
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