- M R L EXTRA BONUS 20PTS Given the RLC circuit in the figure with R = 20;L = 0.14; C = 0.1F 1. Find the initial conditi

[answerhappygod]

M R L Extra Bonus 20pts Given The Rlc Circuit In The Figure With R 20 L 0 14 C 0 1f 1 Find The Initial Conditi 1 (35.17 KiB) Viewed 61 times

- M R L EXTRA BONUS 20PTS Given the RLC circuit in the figure with R = 20;L = 0.14; C = 0.1F 1. Find the initial conditions (0*), pv (0*) when :(0) = 2A, vc(0") = 5V, and V.CC) = u(t) HINTS Att = 0+: because of the inductor and the capacitor (0*) = 2 vc(0*) = 5 Then wy(0) + At any timet Fr(t) = Ri(t) = 2(t) • Then pyy(t) = 2pt(t) = {v_(t) = 2002 (1) • pv (0*) = 200,00) By KVL: V.(t) = 2(t) + (1) + c(t) = 1-(0*) = V(0*) - 1x(0) - We(0*) = 1 - 4-5=-8 Finally pur (0*) = 20v_(0*) = -160; 2. Find the DFQ governing val). > 0 Clearly, V.() = RICE) 87 KV: v.ce) = 2px(e) + Ri(e) + ) ( Then, LCp?i(t) + RCp\(e) + 2(t) = CV, or (p2+p+)) = px (e) Finally, multiply by R: (p? +20p + 100).(t) - 20px;(1) 3. Use any approach to solve the DFQ in part 2 for the ICs in part 1 Solve G* + 20p + 100),(t) = 20px() = 2000) = 0, > 0 • Characteristic roots: 52 + 20s + 100 = (s + 10), thus s = -10 -10 • The particular component is • The total solution is vy(t) = Ke-+ K tejust the homogeneous component) • Use the IC: (0) - Ky; and ope() = -10K e-"* + KC-100e-+39) so that pv (0) = -160 = -106, +Kz Then Kg = -160 + 10(4)=-120 Finally (t) = (4 **** - 1204e).(o) () +