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Alexander-Sadiku Fundamentals of Electric Circuits Chapter 8 Second-Order Circuits Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 Second-Order Circuits Chapter 8 8.1 8.2 8.3 8.4 8.5 Examples of 2nd order RCL circuit The source-free series RLC circuit The source-free parallel RLC circuit Step response of a series RLC circuit Step response of a parallel RLC 2 1 8.1 Examples of Second Order RLC circuits (1) What is a 2nd order circuit? A second-order circuit is characterized by a secondorder differential equation. It consists of resistors and the equivalent of two energy storage elements. RLC Series RL T-config RLC Parallel RC Pi-config 3 8.2 Source-Free Series RLC Circuits (1) The solution of the source-free series RLC circuit is called as the natural response of the circuit. The circuit is excited by the energy initially stored in the capacitor and inductor. The 2nd order of expression d 2i dt 2 R di L dt i 0 LC How to derive and how to solve? 4 2 8.2 Source-Free Series RLC Circuits (2) Method will be illustrated during the lecture 5 8.2 Source-Free Series RLC Circuits (3) There are three possible solutions for the following 2nd order differential equation: d 2i dt 2 => d 2i di 2 2 dt dt 2 0 R di L dt i 0 i 0 LC where R 2L and 0 1 LC General 2nd order Form The types of solutions for i(t) depend on the relative values of and 6 3 8.2 Source-Free Series RLC Circuits (4) There are three possible solutions for the following 2nd order differential equation: d 2i di 2 2 dt dt 1. If > i (t ) A1e 2. If = i (t ) ( A2 3. If i (t ) e < t o, s1t o, 2 0 over-damped case A2 e s2t ( B1 cos 2 2 where s1, 2 0 critical damped case A1t )e o, i 0 t where s1, 2 under-damped case d t B2 sin d t ) where d 2 0 2 7 8.2 Source-Free Series RLC Circuits (5) Example 1 If R = 10 , L = 5 H, and C = 2 mF in 8.8, find , 0, s1 and s2. What type of natural response will the circuit have? Please refer to lecture or textbook for more detail elaboration. Answer: underdamped 8 4 8.2 Source-Free Series RLC Circuits (6) Example 2 The circuit shown below has reached steady state at t = 0-. If the make-before-break switch moves to position b at t = 0, calculate i(t) for t > 0. Please refer to lecture or textbook for more detail elaboration. Answer: i(t) = e 2.5t[5cos1.6583t 7.538sin1.6583t] A 9 8.3 Source-Free Parallel RLC Circuits (1) 0 Let i ( 0) I 0 1 v(t )dt L v(0) = V0 Apply KCL to the top node: t v 1 dv vdt C 0 R L dt Taking the derivative with respect to t and dividing by C The 2nd order of expression d 2v dt 2 1 dv RC dt 1 v 0 LC 10 5 8.3 Source-Free Parallel RLC Circuits (2) There are three possible solutions for the following 2nd order differential equation: d 2v dt 2 1. If dv dt 2 2 0 v > o, A1 e s1t 2. If = o, v(t ) ( A2 3. If < o, t ( B 1 cos v(t ) v (t ) e 0 1 2 RC where and 1 LC 0 over-damped case A2 e s2t where 2 2 s1, 2 0 critical damped case A1t ) e where t s1, 2 under-damped case d t B 2 sin d t ) where d 2 0 2 11 8.3 Source-Free Parallel RLC Circuits (3) Example 3 Refer to the circuit shown below. Find v(t) for t > 0. Please refer to lecture or textbook for more detail elaboration. Answer: v(t) = 66.67(e 10t e 2.5t) V 12 6 8.4 Step-Response Series RLC Circuits (1) The step response is obtained by the sudden application of a dc source. d 2v dt 2 The 2nd order of expression R dv L dt v LC vs LC The above equation has the same form as the equation for source-free series RLC circuit. The same coefficients (important in determining the frequency parameters). Different circuit variable in the equation. 13 8.4 Step-Response Series RLC Circuits (2) The solution of the equation should have two components: the transient response vt(t) & the steady-state response vss(t): v (t ) v t (t ) v ss ( t ) The transient response vt is the same as that for source-free case vt (t ) A1e s1t A2 e s2t vt (t ) ( A1 A2t )e vt ( t ) e t (over-damped) t ( A1 cos (critically damped) d t A2 sin d t ) (under-damped) The steady-state response is the final value of v(t). vss(t) = v( ) The values of A1 and A2 are obtained from the initial conditions: v(0) and dv(0)/dt. 14 7 8.4 Step-Response Series RLC Circuits (3) Example 4 Having been in position for a long time, the switch in the circuit below is moved to position b at t = 0. Find v(t) and vR(t) for t > 0. Please refer to lecture or textbook for more detail elaboration. Answer: v(t) = {10 + [( 2cos3.464t vR(t)= [2.31sin3.464t]e 1.1547sin3.464t)e 2t]} V 2t V 15 8.5 Step-Response Parallel RLC Circuits (1) The step response is obtained by the sudden application of a dc source. The 2nd order of expression d 2i dt 2 1 di RC dt i LC Is LC It has the same form as the equation for source-free parallel RLC circuit. The same coefficients (important in determining the frequency parameters). Different circuit variable in the equation. 16 8 8.5 Step-Response Parallel RLC Circuits (2) The solution of the equation should have two components: the transient response vt(t) & the steady-state response vss(t): i (t ) it (t ) iss (t ) The transient response it is the same as that for source-free case it (t ) A1e s1t A2 e s2t it (t ) ( A1 A2t )e it (t ) e t ( A1 cos (over-damped) t (critical damped) d t A2 sin d t) (under-damped) The steady-state response is the final value of i(t). iss(t) = i( ) = Is The values of A1 and A2 are obtained from the initial conditions: i(0) and di(0)/dt. 17 8.5 Step-Response Parallel RLC Circuits (3) Example 5 Find i(t) and v(t) for t > 0 in the circuit shown in circuit shown below: Please refer to lecture or textbook for more detail elaboration. Answer: v(t) = Ldi/dt = 5x20sint = 100sint V 18 9 This document was created with Win2PDF available at http://www.daneprairie.com. 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