V() R L C 513 Elecrical Circuis Tools Used in Lab 13 Series Circuis Damped Vibraions: Energy Van der Pol Circui A series circui wih an inducor, resisor, and capacior can be represened by Lq + Rq + 1, a second-order linear differenial C q = V() equaion wih consan coefficiens. Look familiar? 1. Circui Laws We know he following wo facs abou our series circui (due o Kirchhoff): 1. The curren in every par of he circui is he same. 2. The sum of he volage drops around he circui mus be equal o he inpu volage V(). From hese facs, we can obain a differenial equaion where each erm on he lef represens a volage drop (in vols) across he designaed circui elemen and each consan L, R, and 1 C, can be viewed as a consan of proporionaliy in henries, ohms, or (farads) 1, respecively. Noe ha q represens he charge (on he capacior) as a dq funcion of ime and I= represens curren in he circui as a d funcion of ime. The equaion can have hree forms, based on convenience. The basic series circui equaion is 1 LI + RI + C q = V () (1) dq In erms of charge, q(), in coulombs (using he fac ha I= ): d Lq + Rq + 1 (2) C q = V() 97
98 Ineracive Differenial Equaions In erms of curren, I(), in amperes (by differeniaing every erm of he basic equaion): LI RI 1 + + C I = V () (3) For he mos par, we use Equaion (2) in erms of q(). Noe ha i looks surprisingly like he equaion for he spring: mx + bx + kx = F() In fac, wihin he linear operaion of he circui elemens, he analogy is very igh and quie useful in undersanding he behavior of circuis. 2. L-C Circuis 2.1 The L-C Circui: Lq + 1 C q = 0 Open he Series Circuis ool. Se R = 0 and A = 0 precisely. Wihou an inpu volage, he only energy in he circui is due o he iniial condiions. Sar wih (nonrivial) iniial condiions by clicking he mouse on he phase plane o se q(0) = q 0 and q ( 0) I( 0) = I0. Are you surprised o ge he equivalen of simple harmonic moion on a spring? a. Solve he linear differenial equaion for q() above in erms of L and C. where b. Wha is he naural frequency of oscillaion for he circui? Compare i o he naural frequency of he spring, ω 0 = k m. where 1/C corresponds o he spring consan k and L corre- is analogous o sponds o he mass m. The capacior sores elecrical energy and has a volage drop in proporion o is charge, so ha 1/C is analogous o k, he spring consan. The inducor sores magneic energy and builds a back volage in proporion o he change in he curren flowing hrough i, di/d, so he inducance L is analogous o he mass m, (i.e. he ineria or resisance o change in velociy) in he spring sysem. Noaion: for convenience, we use for is analogous o, so ha L m 1 C k
Lab 13 Elecrical Circuis 99 2.2 Energy in he L-C Circui Look a he energy graph on he Damped Vibraions: Energy ool. Use he analogy wih he massspring sysem o obain he expression of he magneic (kineic) energy in he inducor and he elecric (poenial) energy in he capacior. Noice ha wihou resisance here is no dissipaion of energy in he circui, so he oal energy remains consan. Remember ha q x, he displacemen, and I v, he velociy ẋ. Complee he senence for he energy in he circui: mass-spring 1 1 Eoal = Ekineic + Epoenial = mv + kx 2 2 2 2 L-C circui Eoal = Emagneic + Eelecric = 2.3 The Forced L-C Circui: Lq + 1 C q = A cos( ω ) Use he Series Circuis ool o deermine he response when he frequency of he inpu volage V ( ) = Acos( ω) is near or equal o he naural oscillaing frequency ω 0 of he circui. Make a skech of he resuling oscillaions when ω is close o bu unequal o ω 0. q
100 Ineracive Differenial Equaions 3. L-R-C Circuis 3.1 The L-R-C Circui wih : Lq + Rq + 1 V () 0 C q = 0 a. Wha does he resisance, R, in a series circui correspond o in he spring sysem? Damping consan b b. Now se A = 0 and se R o a small nonzero value on he sliding scale. Wha kind of moion do you observe? Give a rough skech below. Sae your iniial condiions for he charge q 0 and ( )= ( ) curren q 0 I 0. q c. Are he charge q and curren I underdamped, overdamped, or criically damped? Underdamped d. Using he analogy wih he spring for he criical damping, bcr = 4 mk, find he value of he resisance R cr ha gives criical damping. e. Now use Lq + Rq + 1, via he characerisic equaion Lλ + Rλ + = 0, o find again he C q = 0 C resisance R cr ha gives criical damping. 2 1 f. We purposely lef he sliders wihou unis so ha any consisen se of realisic unis could be used. Se L = 2 henries and C = 0.5 microfarads. Vary he resisance o find he resisance ha gives criical damping. Wih he unis we are using, he slider for he resisance is in kilo-ohms (ha is, 10 3 ohms) and he ime scale is in milliseconds (10 3 seconds). Is he criical resisance you discovered he same as he calculaed resisance R cr? If no, figure ou he problem. I should be he same. Wha is he resisance? 4 kilo-ohms g. For every value of R 0, wha is he long-erm behavior of q() and I()? They boh approach zero as increases.
Lab 13 Elecrical Circuis 101 3.2 Energy in he L-R-C Circui Now ha we have resisance in he circui, here is hea loss. In fac, he loss due o hea dissipaed in he resisor is E = E ( E + E ). Keep he analogy wih he mass-spring sysem firmly in mind. Describe carefully wha happens over he long run o he available energy E magneic + E elecric dissipaed oal magneic elecric (or in he case of he spring, dissipaed? The available energy is los in he form of hea dissipaed in he resisor. E poenial + E kineic ). Wha happens o he energy ha is 3.3 The L-R-C Circui wih : Lq + Rq + 1 V ( ) Acos( ω) C q = A cos( ω ) Reopen he Series Circuis ool. Se R o some nonzero values and experimen wih he response of he circui o various values of ω and A 0. a. Describe wha happens on he phase plane and on he ime series as becomes large. On he phase plane, he rajecory exhibis a limi cycle, and he soluion graphed on he ime axis evenually becomes periodic wih frequency ω (in rad/sec). b. The ransien soluions for he charge (and curren) are analogous o he ransien soluions for he posiion (and velociy) in Lab 11, Forced Vibraions: An Inroducion and die ou as ime becomes large. Was ha your experience in Exercise 3.1(g)? c. The seady-sae soluion depends on he impressed volage as well as he circui elemens and gives he long-erm behavior of he sysem. Se A = 2 and ω = 1. Se C = 0.5 microfarads R = 1 ohm, and L = 1 henry. Now ry several values for L, 1/C, and ω. Noe ha he wiggle a he beginning of he soluion curve is due o he ransien behavior of he circui. Wha is he angular frequency of he seady-sae soluion? 1 rad/sec d. The ransien soluion corresponds o he homogeneous soluion (homogeneous or paricular?) of he differenial equaion for he circui wih applied volage. The seady sae soluion corresponds o he paricular soluion (homogeneous or paricular?).
102 Ineracive Differenial Equaions 4. Addiional Exercises 4.1 A Simplified Inducion Moor Suppose we have a new inducion moor circui wih a problem. The AC power supply overheaed a small frequencies, so we added a capacior of 1 microfarad in series, bu now he moor circui has undesirable resonances as we accelerae i hrough is naural frequency. There are already 500 ohms of resisance in he circui, bu i has been suggesed ha an addiional resisor be placed in series o damp ou he resonance. If he inducance L is 1 henry, find he leas resisance required o damp ou he resonance in he circui. R added = R cr R moor, so he added resisance should be R added = 2000 500 = 1500 ohms. 4.2 A Nonlinear Example Jus for fun, open he Van der Pol Circui ool. The resisor is nonlinear and supplies variable damping, ha is, posiive damping (which dissipaes energy) for par of he ime and negaive damping (which adds energy) for par of he ime. Look for a limi cycle! (Tha is, ry a variey of iniial condiions. Wha happens?) You don have o look very hard, because he limi cycle is a global aracor. Describe wha happens. Every rajecory for every se of iniial condiions approaches he limi cycle.
Lab 13 Elecrical Circuis 103 Lab 13: Tool Insrucions Series Circuis Tool Seing Iniial Condiions Click he mouse on he qq plane or he ime series graph o se he iniial condiions for a rajecory. Clicking in he qq plane or he ime series graph while a rajecory is being drawn will sar a new rajecory. Parameer Sliders Use he sliders o change he values for he parameers L, R, 1/C, A, and ω. The circui is auomaically redrawn when he slider values are se o zero. Press he mouse down on he slider knob for he parameer you wan o change and drag he mouse back and forh, or click he mouse in he slider channel a he desired value for he parameer. Time Series Buons The buons labeled [ ] charge [ ] curren oggle he ime series graphs on and off. Oher Buons Click he [Pause] buon o sop a rajecory wihou canceling i. Click he [Coninue] buon o resume he moion of a paused rajecory. Click he mouse on he [Clear] buon o remove all he rajecories from he graph. Damped Vibraions: Energy Tool Seing Iniial Condiions Click he [Sar] buon o sar a rajecory using prese iniial condiions. Clicking in he ime series graph will se an iniial value of x and sar a rajecory. Clicking in he ime series graph while a rajecory is being drawn will sar a new rajecory. Parameer Slider Use he slider o se he damping coefficien, b. Press he mouse down on he slider knob for he parameer you wan o change and drag he mouse back and forh, or click he mouse in he slider channel a he desired value for he parameer. Time Series Buons The buons labeled [ ] energy [ ] velociy [ ] Posiion oggle he ime series graphs on and off. Oher Buons Click he [Pause] buon o sop a rajecory wihou canceling i. Click he [Coninue] buon o resume he moion of a paused rajecory.
104 Ineracive Differenial Equaions Van der Pol Circui Tool Seing Iniial Condiions Click he mouse on he iv graphing plane o se he iniial condiions for a rajecory. Clicking in he iv plane while a rajecory is being drawn will sar a new rajecory. Parameer Sliders Use he slider o se he nonlinear resisance parameer e. Press he mouse down on he slider knob for he parameer you wan o change and drag he mouse back and forh, or click he mouse in he slider channel a he desired value for he parameer. Time Series Buons The buons labeled [ ] capacior volage [ ] curren [ ] nonlinear resisance oggle he ime series graphs on and off. Oher Buons Click he mouse on he [Clear] buon o remove all he rajecories from he graph.