ORDINARY DIFFERENTIAL EQUATIONS POWERTOOLLesson 29 -- Application: Electric CircuitsProf. Douglas B. MeadeIndustrial Mathematics InstituteDepartment of MathematicsUniversity of South CarolinaColumbia, SC 29208
URL: http://www.math.sc.edu/~meade/E-mail: meade@math.sc.eduCopyright \251 2001 by Douglas B. MeadeAll rights reserved-------------------------------------------------------------------<Text-field layout="_pstyle5" style="_cstyle3">Outline of Lesson 29</Text-field>29.A Electric Circuit29.A-1 Model for a General RLC Circuit29.A-2 Solution for the General RLC Circuit29.A-3 Special Case #1: LC Circuit29.A-4 Special Case #2: RC Circuit29.A-5 Special Case #3: RL Circuit<Text-field layout="_pstyle5" style="_cstyle3">Initialization</Text-field>restart;with( DEtools ):with( plots ):<Text-field bookmark="29.A" layout="_pstyle5" style="_cstyle3">29.A Electrical Circuit</Text-field><Text-field bookmark="29.A-1" layout="_pstyle7" style="_cstyle6">29.A-1 Model for a General RLC Circuit</Text-field>Consider an RLC series circuit with resistance NiMlIlJH (ohm), inductance NiMlIkxH (henry), and capacitance NiMlIkNH (farad). Denote the electric charge by NiMlInFH (coulomb). The current in the circuit is the instantaneous rate of change of the charge, so thatcharge := i(t) = diff( q(t), t );The unit for current is ampere. One of Kirchoff's Laws states that the sum of the instantaneous voltage drops (changes in potential) around a closed circuit must be zero, so thatKirchoffLaw := E[R] + E[L] + E[C] - E[emf] = 0;where NiMmJSJFRzYjJSJSRw==, NiMmJSJFRzYjJSJMRw==, and NiMmJSJFRzYjJSJDRw== are the voltage drops across the resistor, inductor, and capacitor, respectively, and NiMmJSJFRzYjJSRlbWZH is the voltage drop produced by an attached electromotive force.According to Ohm's Law, the voltage drop across a resistor is proportional to the current with constant of proportionality NiMlIlJH, so thatE[R] := R*i(t);Faraday's Law states that the voltage drop across the inductor is proportional to the instantaneous rate of change of the current, with NiMlIkxH as the proportionality constant, so thatE[L] := L*diff(i(t),t);The voltage drop across a capacitor is proportional to the electric charge on the capacitor, with constant of proportionality NiMqJiIiIkYkJSJDRyEiIg== , givingE[C] := 1/C * q(t);<Text-field bookmark="29.A-2" layout="_pstyle7" style="_cstyle6">29.A-2 Solution for the General RLC Circuit</Text-field>When the modeling assumptions for the potential drop across each component in the circuit are inserted into Kirchoff's Law, the resulting ODE isKirchoffLaw;Put this into the form of a second-order ODE for the charge via the substitutionsodeQ := eval( KirchoffLaw, {charge,E[emf]=e(t)} );The corresponding initial conditions areicQ := q(0)=q0, D(q)(0)=i0;The solution to this generic IVP isinfolevel[dsolve] := 3:solQ := subs(_z1=u, dsolve( { odeQ, icQ }, q(t) ));infolevel[dsolve] := 0:The current can be found by differentiation:solI := eval( charge, solQ );The same expressions for the charge and current can be derived as the solution to the first-order system of ODEs formed by Kirchoff's Law and the constitutive relation between current and charge. The appropriate first-order IVP issysQI := eval(KirchoffLaw,E[emf]=e(t)), charge;
icQI := q(0)=q0, i(0)=i0;with solutionsolQI := subs(_z1=u, dsolve( { sysQI, icQI }, { q(t), i(t) } ));That the two forms of the solutions are the same is verified in Maple withsimplify( eval( q(t), solQ ) - eval( q(t), solQI ) );<Text-field bookmark="29.A-3" layout="_pstyle7" style="_cstyle6">29.A-3 Special Case #1: LC Circuit</Text-field>If the circuit does not have a resistor, NiMvJSJSRyIiIQ==, and the solutions "simplify" toLCsolQI := simplify( eval( solQI, R=0 ));Note that NiMqKCUiQ0ciIiIlIj5HRiUiIiFGJQ== and NiMqKCUiTEciIiIlIj5HRiUiIiFGJQ== imply that since the exponentials in this problem all involve imaginary exponents, the real-valued solutions will involve sine and cosine. In particular, there is no transient solution for an LC circuit. Given that the second-order ODE for the charge in an LC circuit reduces toeval( odeQ, R=0 );which is equivalent to an undamped oscillator, this is not surprising.<Text-field bookmark="29.A-4" layout="_pstyle7" style="_cstyle6">29.A-4 Special Case #2: RC Circuit</Text-field>If the circuit does not have an inductor, NiMvJSJMRyIiIQ==, and it is not possible to obtain the solutions by simply inserting NiMvJSJMRyIiIQ== into the general solution to the RLC circuit. Attempting this leads to the errorsimplify( eval( solQI, L=0 ) );The problem is that when NiMvJSJMRyIiIQ== the ODE for the charge is no longer of second order, as seen withRCodeQ := eval( odeQ, L=0 );The solution isRCsolQ := subs(_z1=u, expand( dsolve( { RCodeQ, q(0)=q0 }, q(t) )));
RCsolI := i(t) = subs(_z1=u, expand( diff( rhs(RCsolQ), t )));Here, because NiMqKiUiQ0ciIiIlIlJHRiUlIj5HRiUiIiFGJQ==, the denominators all tend to NiMlKWluZmluaXR5Rw== as NiMlInRHNiMsJCUiPkchIiI=NiMlKWluZmluaXR5Rw==. This means that the initial charge in the system has no bearing on the long-time behavior of the solution. If the exponentially growing term in the integrand for the particular solution survives in a form that exactly cancels the exponential growth in the denominator, then the steady-state solution for the charge will be nontrivial. For example, taking NiMvLSUiZUc2IyUidEciIiI= leads to the solutioncollect( value( eval( { RCsolQ, RCsolI }, e=1 ) ), exp );where NiMtJSJxRzYjJSJ0Rw== tends to NiMlIkNH while NiMtJSJpRzYjJSJ0Rw== tends to zero.<Text-field bookmark="29.A-5" layout="_pstyle7" style="_cstyle6">29.A-5 Special Case #3: RL Circuit</Text-field>If the circuit does not have a capacitor, NiMvKiYiIiJGJSUiQ0chIiIiIiE=, and it is possible to obtain the charge on the circuit by simply taking a limit of the general solution to the RLC circuit. This limit is obtained viaRLsolQ := simplify( map( limit, solQ, C=infinity )) assuming R>0;This substitution works because the ODE for the charge, namely,RLodeQ := limit( odeQ, C=infinity );is still of second-order, with two distinct eigenvalues. Note that while the formal substitution NiMvJSJDRyUpaW5maW5pdHlH yields the same ODE, that is,eval( odeQ, C=infinity );this substitution cannot be used directly on the solution, since that results ineval( solQ, C=infinity );However, the charge in an LR circuit can be obtained as the limit of the generic solution for an RLC circuit as NiMlIkNHNiMsJCUiPkchIiI=NiMlKWluZmluaXR5Rw==, as seen bysimplify( map( limit, solQ, C=infinity ) ) assuming R>0;The corresponding current isi(t) = expand( diff( rhs(RLsolQ), t ) ) assuming R>0;Careful inspection of the expression for the charge in the circuit verifies that this solution is obtained from a second-order ODE with eigenvalues NiMvJSdsYW1iZGFHIiIh and NiMvJSdsYW1iZGFHLCQqJiUiUkciIiIlIkxHISIiRio=. The presence of a zero eigenvalue implies that the initial charge and current in the circuit impact the solution for all time. Moreover, because NiMqJiUiUkciIiIlIkxHISIi > 0, only the portion of the homogeneous solution contributed by the negative exponential is guaranteed to be transient.[Back to ODE Powertool Table of Contents]