Two Equivalent RepresentationsS 1210 VC+_vCS closesat t = 0+12C+_vCRRvs(t)+-vstvs(0−) = 0vs(0+) = 10010S 1210 VC+_vCS closesat t = 0+R= 5 K= 2 µFStep 1Find vC(0+) : Replace C by battery with voltage = vC(0−), then calculate vC(0+). 1210 V+_vC(0+) = vC(0−) = 5R= 5 KvC(0−)= 5Step 2Find vC(t∞) : Replace C by open circuit,then calculate vC= vC(t∞). 1210 V+_R= 5 KvC= vC(t∞)= 10 VStep 3Calculate the time constantτ = R CFundamental Behavior of 1st-order CircuitsThe voltage vjk(t) between any pair of nodesand , and the current io(t) at any terminalof any linear resistive circuit is always anexponential waveform, except for the trivialcircuit consisting of a current source (resp., voltage source ) connected across a capacitor(resp., inductor), whose solution is a linear“ramp” function.jkS 1210 VC+_vCS closesat t = 0+5 K= 2 µFStep 1Calculate iR(0+) : Replace C by a 5 V battery :1210 V+_vC(0+) = vC(0−) = 5 V5 KvC(0−)= 5Step 2Calculate iR(t∞) : Replace C by open circuit :1210 V5 KProblem : Calculate iR(t) under the same setting.iR(0+)510(0 ) 15RimAK+−==−iR(t∞)iC(t∞) = 0iR(t∞) = 0iR(t)Calculate : Calculate : iC(t)Step 3Calculate time constantτ = R C = (5 × 103) (2 × 10−6) = 10 mst, msiR0-0.5-11020 30 40 50 60τ = 10 ms0.6310() , 0tmsRit e t−+=−≥t, msiC00.511020 30 40 50 600.6310() , 0tmsCit e t−+=≥Verification of solution310 1001() (0 )tdttCCvt v eC−−−×+=+∫310 103015210tdtte−−×+−=+×∫310 1055 5te−−×=+− +310 1010 5 , 0tet−−+×=− ≥()()CRitit=−Step 3Calculate time constantτ = R C = (5 × 103) (2 × 10−6) = 10 mst, msiR0-0.5-11020 30 40 50 60τ = 10 ms0.6310() 1 , 0tmsRit e t−+=−≥iVerification01() (0 ) ( )tCC Cvt v i dCττ−+=+∫310 103015210tdtte−−×+−=+×∫310 1055 5te−−×=+− +10 5 , 0tetτ−+=− ≥Let us calculate :10() ()tmsCRit it e−=− =Finding Solutions by InspectionS 1210 VC+_vCS closesat t = 0+RProblemAssume vC(0−) = 5 V, where t = 0−denotes the time just beforeswitch S made contact with resistor R.Sketch vC(t) for t ≥ 0+, where t = 0+denotes the instant switch S made contact with resistor R.Solution10() 10 5tCvt e−=−vC(0+) = vC(0−) = 5 V,vC(t∞) = 10 Vτ = RC= 5(103) [2(10-6)] = 10 mst, msvC05101020 30 40 50 60τ = 10 msCalculate : 10() 10 5 , 0tCvt e t−+=−≥1+_vCHow to Find vC(t∞) ?How to Find iL(t∞) ?N21+_N2vC= vC(t∞)iC= 0Step 1 :Open capacitorStep 2 :Calculate : vC|Open capacitorThen() |CCvt v∞=1iLN21+_N2vL= 0iL= iL(t∞)Step 1 :Short capacitorStep 2 :Calculate : iL|Short inductorThen() |LLit i∞=Open capacitorShort inductor1+_voHow to Find ?circuit conditionchanged at2Step 1 :Calculate :Step 2 :+() ()Co Covt vt=()covtHow to Find ?()Loit()Covtott=1+_vocircuit conditionchanged at2Step 1 :Calculate :Step 2 :+() ()Lo Loit it=()Loitott=+ +______How to Find τ = RC?Step 1 :Find Thevenin equivalentcircuit of N.Step 2 :How to Find τ = GL = ?LR1CN2vocReqTheveninequivalent circuitτ = ReqCStep 1 :Find Thevenin (or Norton)equivalent circuit of N.Step 2 :1LN2vocReqTheveninequivalent circuitτ =LReqSubstitution TheoremLet N be a circuit made of a nonlinear resistive one-port NR terminated in an arbitrary one-port NL, as shown in Fig. (a).1. If N has a unique solution for all t, then NLmay be substitutedby a voltage source without affecting the branch voltage and branch current solution inside NR, provided the substituted circuit Nvin Fig. (b) hasa unique solution for all t.ˆ()vvt=ˆ()vt2. If N has a unique solution for all t, then NLmay be substitutedby a current source without affecting the branch voltage and branch current solution inside NR, provided the substituted circuit Niin Fig. (c) hasa unique solution for all t.ˆ()iit=ˆ()itCircuit
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