Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16EE201 Lecture 17 P. 1Capacitors and Inductors (cont.)Inductors in Series-voltage division (see lect. 15, P.9)vin(t)Leqvn(t) =LnExample: Find Leq and v1(t) 16e-10t V12mH+_v1(t)+ - 4mH 16mHLeqStep 1: Find Leq Leq = 4 mH + 16 mH + 12 mH = 32 mHStep 2: Find v1(t) v1(t) = (4 mH/32 mH) 16e-10t V = 2e-10t VEE201 Lecture 17 P. 2Inductors in Parallel-current division (see lect. 15, P.13)iLj = 1Lj 1Lnniin(t)Example: Find Leq and i3(t). 100e-t mALeqi3(t)16mH32mH32mHEE201 Lecture 17 P. 3Step 1: Leq computation. 1 1 1 1 4 Leq 16mH 32mH 32mH 32mH Leq =8mHStep 2: Current division i3= is = 100e-t mA = + + =1/32 mH1/8 mH14 i3 = 25e-t mAEE201 Lecture 17 P. 4Example: Find Leq and iout(t) 6H Leq = 6H + 3H // 6H Leq = 6H + 2H = 8H iout = 9sin(2 t) = 3sin(2t) Av1(t)+ -6H9sin(2 t)A3Hiout 1/61/6 +1/3EE201 Lecture 17 P. 5Capacitors in Series-voltage division (see lect. 16, P.9)vCn= 1/Cn1/CeqvCExample: Find Ceq and v2(t) in the circuit. +_16 cos(t) V16F16F+_v216F10F30FStep 1: 1 1/16F + 1/16F +1/56F Ceq = 7 F Ceq =Step 2: Find v2(t) using voltage division. 1/16 7 1/7 16v2(t) =vs = 16cos(t) v2(t) = 7 cos(t) VEE201 Lecture 17 P. 6Example: Find Ceq and vout. 8(24) 60(12)32 72 Ceq = 6 mF + 10mF = 16 mF+_18 cos(t) V60mF12mF+_vout24mF8mF Ceq = mF + mFStep 1:vout = [18cos(t)] = [18cos(t)]1/6 vout= 3 cos(t) VStep 2: 1/601/60+1/12EE201 Lecture 17 P. 7Capacitors in Parallel-current division (see lect. 16, P.11)iCn= CnCeqiCExample: Find i2. 16 cos(t) A60mF12mFi224mF8mF12Branch:EE201 Lecture 17 P. 8C1 = 6 mF C2 = 10mF Ceq = 16 mFStep 1: From example on P.6, equivalent capacitances for branch 1 and 2 are knownStep 2: Calculate current through branch 2i2= C2CeqiCi2= 10mF16mF16 cos(t) Ai2 = 10 cos(t) AEE201 Lecture 17 P. 9Introduction to First Order CircuitsRL and RC Circuits:Inductor-resistor (RL) CircuitCombine an inductor with a resistor to examine how initial inductor current decays with time.+-vLiLLRv = LdiLdt= -RiLL diLdt+ RiL = 0First-order differential equation describing how inductor current behaves with time:diLdt+ RLiL = 0EE201 Lecture 17 P. 10Capacitor-resistor (RC) CircuitCombine a capacitor with a resistor to examine how initial capacitor voltage decays with time.ic+-vCRic = CdvCdt= -vCRdvCdt+ 1RCvC = 0Solutions for Differential EquationsGeneral Differential Eqn:dxdt+1x = 0x = iL or vC = time constant = L/R (RL) = RC (RC)CEE201 Lecture 17 P. 11Solution form for general differential equation:x(t) = [e- (t-to) /x(t0)x(t) = [e-K (t-to)x(t0)where K-frequency K = 1Note:1) In RL circuit driven by an independent source for a long time, the inductor behaves as a short.2) In RC circuit driven by an independent source for a long time, the capacitor behaves as an open.General strategy: reduce complicated circuit (attached to inductor or capacitor) to Thevenin equivalent resistor, attach inductor or capacitor to resistor, and use Eqn (17.1).(17.1)EE201 Lecture 17 P. 12Example: Find vL(t) for t 0+_t = 01iL(t)_+vL(t)0.5 H911 V0.1Step 1: Find initial condition iL(0+).For all times t < 0, circuit is configured as below.+_t < 01iL(t)_+vL(t)0.5 H911 V0.1Assume long time operation because voltage source has been on for long time. Therefore, inductor behaves as a short.iL(0-) = 11 V1.1= 10A = iL(0+)EE201 Lecture 17 P. 13 iL(0-) = iL(0+) because of the continuity property of inductor current.Was assumption of long time operation valid?Calculate time constant for circuit when t < 0 s. = L/RTH = 0.5 Vs/A1.10 0.45 sA long time is about 3-5 time constants. Therefore, 2-3 sec after the voltage source turned on (in this case at t = -), the inductor behaves as a short.EE201 Lecture 17 P. 14Step 2: Find iL(t) t 0 s.Configuration of circuit for t 0 s:iL(t) = exp[-RTHLt]iL(0+)iL(t) = 10 e-20t AStep 3: Calculate vL(t), t 0vL(t) = L diLdt= - 100e-20t VRTH = 10 +_t 01iL(t)_+vL(t)0.5 H911 V0.1EE201 Lecture 17 P. 15Example: Find vC(t), t 0-++_vC(t)0.1F94.572t=1t=0110VStep 1: Find vC(0+) = vC(0-) For all t < 0 capacitor is open vC(0-) = 9V = vC(0+) Step 2: Find vC(t) 0 t 1 RTH =8+vC(t)_vC(t) = vC(0+) e-(t-to) / RCRTHC = 0.8 s1/RC = 1.25 s-1EE201 Lecture 17 P. 16Therefore,vC(t) = 9 exp[ -1.25t] VStep 3: Find vC(t) 1 tRTH = 3+vC(t)_0.1FvC(1-) = vC(1+) = 9e-1.25 vC(1-) = vC(1+) = 2.58 V9122.58 = 0.3 s = 0.8 stvC(t) = vC(1+) e-(t-to) / RCvC(t) = 2.58 e-(t-1) / 0.3 VStep 4: Plot vC(t) vs.
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