Problem 3: Appearing to the right yet again is our familiar RLC series circuit. The voltage source, va, is a constant voltage and because of the capacitor in the loop the particular solution for i is zero. elle + Va C = i1-part = 0 (Kiedit + K2e2t, Kjeait + K2teait, if a1 = a2. if a1 + a2; R i1-c -comp i1(t) = i1-comp(t) + i1-part (t) where R ai = - 2L R 1 R R 1 and a2 2L LC 2L 2L LC The initial conditions for this circuit are vc(0¯) = 2va and iL(0~) = R: Provide a sketch for your expectations of i1 given these initial conditions and assuming that the circuit is underdamped. Solve for K1 and K2 given these initial conditions and for the case a1 # a2. K1 and K2 in terms of a1, a2, R, L, C, Va. Be sure to show
Problem 3: Appearing to the right yet again is our familiar RLC series circuit. The voltage source, va, is a constant voltage and because of the capacitor in the loop the particular solution for i is zero. elle + Va C = i1-part = 0 (Kiedit + K2e2t, Kjeait + K2teait, if a1 = a2. if a1 + a2; R i1-c -comp i1(t) = i1-comp(t) + i1-part (t) where R ai = - 2L R 1 R R 1 and a2 2L LC 2L 2L LC The initial conditions for this circuit are vc(0¯) = 2va and iL(0~) = R: Provide a sketch for your expectations of i1 given these initial conditions and assuming that the circuit is underdamped. Solve for K1 and K2 given these initial conditions and for the case a1 # a2. K1 and K2 in terms of a1, a2, R, L, C, Va. Be sure to show
Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter6: Power Flows
Section: Chapter Questions
Problem 6.61P
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