Question

3. (40 pts total) Eigenvalues of Systems of Equations Application: Series RLC Circuit, Natural, or Transient Response (Remember EE280, maybe not) M SR v(t) Consider a series RLC circuit, with a resistor R, inductor L, and capacitor C in series. The same current i(t) flows through R, L, and C. The switch S1 is initially closed and S2 is initially open allowing the circuit to fully charge. At t=0 the switch S1 opens and S2 closes as shown above.

Answer 1

d'i«) +R dilt) + 4°C) = 0 Dť 'I dt + Tai Lt) = 0 1 the Series RLL Eruction. dt The first de totes ]+d'! + t)=0 dict) {CE+() =0 eruction lineer 1st order ODE is. diet Ce-Zzy t) di; (4) -R dilt) - pct) ----6) SI dt dt The second lineey ist_order ODE is. di(t) = () ----(2) linear ist order ODEs have only one verioble. these two governing these two linear The state ist-order. variable ODES Eruction will be

di - 2016. ) These Fooruchons two Eructions cen be into modeled motrices. theme sons, Form E / 2 607 ] di Mas FLE L dt Therefore, Matrix a will be. (4) eigenvalue will be Eigenvolues of metrin a 1 [A-21]] =0 16 32% 3)-

( ( 2 - 2 + 1/20 은이 This ruadratic equation should be algenvelees of metrin A. Thем оче, solled to get two a 1 + 19 = 1/2 2 - - 16 h As Eigenvolues oxe the roots of the cheretexestic Eruction and we know it the roots of the Cheracterstic Eruction lie in the left helf of s-plene then system is stable. As given in Eruction that R, L, C ene ell positive and non- z yo 1. Hence, system is stable. at the eigenvelus of a to be neel then.

eigenvalues of Ait R=40 2 2 = 300 ml, C=400 MB 1, - - JEMYE sem og Score you to 400 = -400 +43 74116 6 =-660667+$62.36 ^,- 13- you to them to the Colby Scanned with

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