EE201 Lecture 24 P 1 Undriven RLC Circuits continued General form of second order differential equation representing behavior of RLC circuits dx t d2x t b c x t 0 dt2 dt For nontrivial solutions characteristic equation must be satisfied s2 bs c 0 Solutions to this equation are separated into 3 categories Case 1 s1 s2 but s1 s2 are real waveform is overdamped and general solution is of the form x t K1es1 t to K2es2 t to t t0 EE201 Lecture 24 P 2 Case 2 s1 s2 s1 and s2 are complex conjugates Waveform is underdamped and general solution is of form x t e t to Acos d t t0 Bsin d t t0 t t0 Case 3 s1 s2 Waveform is critically damped and general solution is of form x t K1 K2 t t0 es1 t to t t0 EE201 Lecture 24 Example Find vc t for t 0 iL t 1u t A 2H vC t P 3 1 18 F 8u t V vL t 3 Step1 Find initial conditions For t 0 inductor is shortened capacitor is open Continuity iL 0 1A iL 0 principles for L vC 0 8V vC 0 and C EE201 Lecture 24 iC 0 0A P 4 vL 0 0V At t 0 independent sources are deactivated Because of source deactivation circuit appears as iR 0 3 iL 0 1A 2H 1 18F Since elements are in parallel vR 0 vL 0 vC 0 8V iC 0 vC 0 EE201 Lecture 24 P 5 Calculate iR 0 from vR 0 using Ohm s Law vR 0 8V 8 iR 0 R 3 3 A Use KCL to find iC 0 iC 0 iL 0 iR 0 1 8 3 iC 0 5 3 A Initial conditions iC 0 5 3 A vC 0 8V iL 0 1A vL 0 8V Step2 Analyze RLC parallel circuit For parallel RLC circuits general solution is Kest s2 bs c 0 EE201 Lecture 24 1 Where b RC s 2 and c P 6 1 LC 1 1 s 0 RC LC s2 6s 9 0 s 3 2 s 3 s 1 Therefore s1 s2 3 s 1 case 3 vC t K1 K2t e 3t t 0 Step3 Solve for unknown constants and obtain expression for vC t vC 0 K1 K2 t e 3t K1 8 V EE201 dvC 0 dt dvC 0 dt Lecture 24 1 C P 7 iC 0 18 x 5 3 30 3 K K t e 1 2 3t K2 e 3t dvC 0 3 K1 K2 30 dt K2 6 V s Final solution vc t 8 6t e 3t V t 0 EE201 Lecture 24 P 8 Example Find iL t for t 0 25 vL t vR t vc t 6 25 H t 0 15 8mF Step1 Find initial conditions For t 0 inductor is shorted capacitor is open iL 0 2A iL 0 25 80V 50V v 0 C vC 0 40 80V EE201 Lecture 24 For t 0 iL 0 2A P 9 vR 0 25 8mF 6 25 H vC t 50V iC t iL t 2A vR 0 50V vL 0 2 25 50 0 Step 2 Set up solution for series circuit diL t d2iL t 1 R iL t 0 L LC dt dt2 1 R s bs c s s 0 L LC 2 2 EE201 Lecture 24 P 10 s2 4s 20 0 s b b2 4c 2 s 2 j4 s1 2 j4 4 16 80 2 j 1 s2 2 j4 Solution is of form x t e t A cos dt B sin dt e 2t A cos 4t B sin 4t EE201 Lecture 24 P 11 Step3 Find values for unknown constants iL t e 2t A cos 4t B sin 4t t 0 A A iL 0 2A iL 0 diL 0 dt 1 L vL 0 0 diL 0 e 2t d A sin dt d B cos dt dt e t A cos dt B sin dt EE201 diL 0 dt Lecture 24 A dB 2A 4B 0 B 1 Step 4 Write solution iL t e 2t 2 cos 4t sin 4t A P 12
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