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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

Engineering 43. MTE2 Review 2 nd Order Transient Ckt. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu. 33 μ F. 300 mH. t = 0. 125 mA. 110 Ω. 8 V. 180 Ω. The Problem. Find for the Circuit Below v C (t) i o (t). + v C (t) −. i o (t).

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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

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  1. Engineering 43 MTE2Review2nd OrderTransient Ckt Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

  2. 33 μF 300 mH t = 0 125mA 110 Ω 8 V 180 Ω The Problem • Find for the Circuit Below • vC(t) • io(t) + vC(t) − io(t)

  3. MATLAB Code - 1 % Bruce Mayer, PE % ENGR43 * MTE2 Review Problem % File = E43_MTE2_2ndOrder_Vc_100402.m % % Component Values R1 = 110 R2 = 180 L = 300e-3 C = 33e-6 % % % Calulate the Roots of the Characteristic Equation by Quadratic Formula % * Yields the Time Constants for Exponential Decay of OverDamped Ckt Splus = (-(R1+R2)*C + sqrt(((R1+R2)*C)^2 - 4*L*C))/(2*L*C)% in Rads/sec Sminus = (-(R1+R2)*C - sqrt(((R1+R2)*C)^2 - 4*L*C))/(2*L*C)% in Rads/sec tauPlus = 1000/Splus % in mS tauMinus = 1000/Sminus % in mS % % use Ax = b to solve for K1 & K2 A = [1 1; 8.39 1.18] % unitless coefficients b = [-8; 0]; % units are volts K = A\b % in K1 = K(1) K2 = K(2) % % Define function for the total solution vtot = @(y) K1*exp(y/tauMinus) + K2*exp(y/tauPlus) + 22.5 t = linspace(0, 60, 500); % in mS vc = vtot(t); % % graph Solution plot(t,vc), grid, xlabel('time (ms)'), ylabel('vc (V)') % disp('Showing FULL-time plot. Hit AnyKey to show the 2nd-Order IC CHECK-Plot') disp(' ‘)

  4. MATLAB Code - 2 pause % % make the [dvc/dt](0+) =0 check plot tdv = linspace(0, 0.5); % in mS dvcdt = vtot(tdv); plot(tdv,dvcdt), grid, xlabel('time (ms)'), ylabel('vc (V)') % disp('Showing SHORT-time plot. Hit AnyKey to show the FULL-TIME io(t) Plot') disp(' ') pause % % the solution for io(t) Kio = 36.61; % in mA io = @(t) Kio*(exp(Splus*t) - exp(Sminus*t)); tio = linspace(0, 0.04, 1000); % in SECONDS ioplot = io(tio); plot(tio*1000, ioplot), xlabel('time (mS)'), ylabel('io (mA)'), title('MTE2 2nd-Order Problem Exam Review'), grid % % % Find ioMax and tioMax using MAX command [ioMax,ItioMax] = max(ioplot); disp('Maximum value for io in mA = ') disp(ioMax) disp(' ') tioMax = tio(ItioMax); disp('TIME for Maximum value for io in mS = ') disp(1000*tioMax) % disp('Showing FULL-time plot. Hit AnyKey to show the Short-TIME io(t) Plot') disp(' ') pause tio = linspace(0, 0.008, 1000); % in SECONDS ioplot = io(tio); plot(tio*1000, ioplot, 1000*tioMax, ioMax, 'h'), xlabel('time (mS)'), ylabel('io (mA)'), title('MTE2 2nd-Order Problem Exam Review'), grid

  5. vC(t) MATLAB Plot

  6. CHK: dvC(t)/dt|t=0+ = 0 • Slope at t=0 looks FLAT; thus 1st Order IC is CONFIRMED

  7. io(t) MATLAB Plot • Note io,max at about 3 mS

  8. io,max(2.683 mS) = 22.82 mA

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