MEC702_Wk_7_Lect_2 - MEC702 Wk 7 Lect 2 Modeling Electrical...

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Unformatted text preview: MEC702 Wk 7 Lect 2 Modeling: Electrical Circuits. Designing good models is a task the computer cannot do. Hence setting up models has become an important task in modern applied mathematics. The best way to gain experience in successful modeling is to carefully examine the modeling process in various elds and applications. Accordingly, modeling electric circuits will be protable for all students, not just for electrical engineers and computer scientists. Setting up the Model We try to model the RLC -circuit given below and nd the total current I(t) for the circuit. Note the physical laws will be taking into consideration: • Current is dened as the time rate change of charge at a particular point. This gives the formula: d [Q(t)] dt ˆ Q(t) = I(t) dt I(t) = • Dierential formula Integral formula. Name Notation Unit Formula for Voltage Drop Resistor R L C ohms (Ω) henrys (H ) ∆V = IR ∆V = LI 0 ∆V = Q C Inductor Capacitor Farads (F ) 1 • Kircho 's Voltage Law (KVL): The sum of all the voltage drop across the crcuits equals the electromotive force provided by the cell. • Kircho  Current Law (KCL): At any point in the circuit, the amount of inowing current at that point is equal to the amount of outowing current. Assuming that R, L C, and by KVL, we have ∆Vind + ∆Vres + ∆Vcap = ∆Vcell 1 LI 0 + RI + (Q) = V0 sin ωt ˆ C  1 0 Idt = V0 sin ωt LI + RI + C We have the function for Current as the unknown function I(t) with respect to time t, I(t). Dierentiating we have 00  0 LI + RI + 1 C  I = V0 ω cos ωt This is an non-homogeneous linear second order dierential equation with constant coecients. Solving the ODE for I(t) The general solution of the ODE is the sum I(t) = Ih + Ip . For yh , we have the characteristic equation λ2 + R 1 λ+ =0 L LC and the solution Ih (t) = c1 eλ1 t + c2 eλ2 t where λ1 = −α + β and λ2 = −α − β , α= For yp , we notice R 2L , r(x) = V0 ω cos ωt, and β= 1 2L q R2 − 4L C . then using Method of Undetermined Coef- cients, we choose Ip (t) = A cos ωt + B sin ωt Ip0 (t) Ip00 (t) = ω(−A sin ωt + B cos ωt) = −ω 2 (A cos ωt + B sin ωt). 2 Then substituting into LIp00 + RIp0  + 1 C  Ip = V0 ω cos ωt and collecting the cosine and sine terms together, we have two simultaneous equations A = V0 ω C B Lω 2 (−B) + Rω(−A) + =0 C Lω 2 (−A) + RωB + (Cosine terms) (Sine terms). Introducing a few other notions RL = ωL 1 RC = ωC S = RL − RC Z= (Reactance) p p R2 + (RL − RC )2 = R2 + S 2 (Impedance) we are able to simplify the simultaneous equations into −SA + RB = V0 −RA − SB = 0. Solving, we have V0 S + S2 V0 R . R2 + S 2 A = − B = R2 Hence, Ip (t) = A cos ωt + B sin ωt or in its alternative form, Ip (t) = I0 sin(ωt − θ) where I0 = tan θ = V0 Z RL − RC . R 3 Therefore the general solution for the non-homogeneous linear ODE that we were modeling is I(t) = c1 eλ1 t + c2 eλ2 t + I0 sin(ωt − θ) Notice that as time and e λ2 t → 0. t increases (the experiment continues), the terms This is since λ1 = −α + β and λ2 = −α − β are both negative 0. Thus, we obtain the values and exponentials of large negative values tend to steady-state solution of the model I(t) = I0 sin(ωt − θ) 4 eλ1 t → 0 ...
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