2.9 Modeling: Electric Circuits

Transcription

1 SE. 2.9 Modeling: Elecric ircuis Modeling: Elecric ircuis Designing good models is a ask he compuer canno do. Hence seing up models has become an imporan ask in modern applied mahemaics. The bes way o gain experience in successful modeling is o carefully examine he modeling process in various fields and applicaions. Accordingly, modeling elecric circuis will be profiable for all sudens, no jus for elecrical engineers and compuer scieniss. Figure 61 shows an -circui, as i occurs as a basic building block of large elecric neworks in compuers and elsewhere. An -circui is obained from an -circui by adding a capacior. ecall Example 2 on he -circui in Sec. 1.5: The model of he -circui is Ir I E(). I was obained by KV (Kirchhoff s Volage aw) 7 by equaing he volage drops across he resisor and he inducor o he EMF (elecromoive force). Hence we obain he model of he -circui simply by adding he volage drop Q> across he capacior. Here, F (farads) is he capaciance of he capacior. Q coulombs is he charge on he capacior, relaed o he curren by I() dq d, equivalenly Q() I() d. See also Fig. 62. Assuming a sinusoidal EMF as in Fig. 61, we hus have he model of he -circui E() = E 0 sin ω ω Fig circui Name Symbol Noaion Uni Volage Drop Ohm s esisor Inducor apacior Ohm s esisance Inducance apaciance ohms ( ) henrys (H) farads (F) I di d Q/ Fig. 62. Elemens in an -circui 7 GUSTAV OBET KIHHOFF ( ), German physicis. aer we shall also need Kirchhoff s urren aw (K): A any poin of a circui, he sum of he inflowing currens is equal o he sum of he ouflowing currens. The unis of measuremen of elecrical quaniies are named afer ANDÉ MAIE AMPÈE ( ), French physicis, HAES AUGUSTIN DE OUOMB ( ), French physicis and engineer, MIHAE FAADAY ( ), English physicis, JOSEPH HENY ( ), American physicis, GEOG SIMON OHM ( ), German physicis, and AESSANDO VOTA ( ), Ialian physicis.

2 94 HAP. 2 Second-Order inear ODEs (1r) Ir I 1 I d E() E 0 sin v. This is an inegro-differenial equaion. To ge rid of he inegral, we differeniae wih respec o, obaining (1r) (1) Is Ir 1 I Er() E 0v cos v. This shows ha he curren in an -circui is obained as he soluion of his nonhomogeneous second-order ODE (1) wih consan coefficiens. In connecion wih iniial value problems, we shall occasionally use (1s) Qs Qs 1 Q E(), obained from (1r) and I Qr. Solving he ODE (1) for he urren in an -ircui A general soluion of (1) is he sum I I h I p, where I h is a general soluion of he homogeneous ODE corresponding o (1) and I p is a paricular soluion of (1). We firs deermine I p by he mehod of undeermined coefficiens, proceeding as in he previous secion. We subsiue (2) I p a cos v b sin v I p r v( a sin v b cos v) I p s v 2 ( a cos v b sin v) ino (1). Then we collec he cosine erms and equae hem o E 0 v cos v on he righ, and we equae he sine erms o zero because here is no sine erm on he righ, v 2 ( a) vb a> E 0 v v 2 ( b) v( a) b> 0 (osine erms) (Sine erms). Before solving his sysem for a and b, we firs inroduce a combinaion of and, called he reacance (3) S v 1 v. Dividing he previous wo equaions by v, ordering hem, and subsiuing S gives Sa b E 0 a Sb 0.

3 SE. 2.9 Modeling: Elecric ircuis 95 We now eliminae b by muliplying he firs equaion by S and he second by, and adding. Then we eliminae a by muliplying he firs equaion by and he second by S, and adding. This gives (S 2 2 )a E 0 S, ( 2 S 2 )b E 0. We can solve for a and b, (4) a E 0 S 2 S 2, b E 0 2 S 2. Equaion (2) wih coefficiens a and b given by (4) is he desired paricular soluion I p of he nonhomogeneous ODE (1) governing he curren I in an -circui wih sinusoidal elecromoive force. Using (4), we can wrie I p in erms of physically visible quaniies, namely, ampliude and phase lag u of he curren behind he EMF, ha is, I 0 (5) I p () I 0 sin (v u) where [see (14) in App. A3.1] I 0 2a 2 b 2 E S 2, a an u b S. The quaniy 2 2 S 2 is called he impedance. Our formula shows ha he impedance equals he raio E 0 >I 0. This is somewha analogous o E>I (Ohm s law) and, because of his analogy, he impedance is also known as he apparen resisance. A general soluion of he homogeneous equaion corresponding o (1) is where and are he roos of he characerisic equaion l 1 l 2 I h c 1 e l 1 c 2 e l 2 l 2 l 1 0. We can wrie hese roos in he form l 1 a b and l 2 a b, where a 2, 2 b B B 2 4. Now in an acual circui, is never zero (hence 0). From his i follows ha I h approaches zero, heoreically as :, bu pracically afer a relaively shor ime. Hence he ransien curren I I h I p ends o he seady-sae curren I p, and afer some ime he oupu will pracically be a harmonic oscillaion, which is given by (5) and whose frequency is ha of he inpu (of he elecromoive force).

4 96 HAP. 2 Second-Order inear ODEs EXAMPE 1 -ircui Find he curren I() in an -circui wih 11 (ohms), 0.1 H (henry), 10 2 F (farad), which is conneced o a source of EMF E() 110 sin (60 # 2p) 110 sin 377 (hence 60 Hz 60 cycles> sec, he usual in he U.S. and anada; in Europe i would be 220 V and 50 Hz). Assume ha curren and capacior charge are 0 when 0. Soluion. Sep 1. General soluion of he homogeneous ODE. Subsiuing,, and he derivaive Er() ino (1), we obain Hence he homogeneous ODE is 0.1Is 11Ir 100I 0. Is characerisic equaion is The roos are l 1 10 and l The corresponding general soluion of he homogeneous ODE is Sep 2. Paricular soluion I p of (1). We calculae he reacance S and he seady-sae curren wih coefficiens obained from (4) (and rounded) 0.1Is 11Ir 100I 110 # 377 cos l 2 11l I h () c 1 e 10 c 2 e 100. I p () a cos 377 b sin 377 a 110 # , b 110 # Hence in our presen case, a general soluion of he nonhomogeneous ODE (1) is (6) I() c 1 e 10 c 2 e cos sin 377. Sep 3. Paricular soluion saisfying he iniial condiions. How o use Q(0) 0? We finally deermine c 1 and c 2 from he in iniial condiions I(0) 0 and Q(0) 0. From he firs condiion and (6) we have (7) I(0) c 1 c , hence c c 1. We urn o Q(0) 0. The inegral in (1r) equals I d Q(); see near he beginning of his secion. Hence for 0, Eq. (1r) becomes Ir(0) # 0 0, Differeniaing (6) and seing 0, we hus obain so ha Ir(0) 0. Ir(0) 10c 1 100c # 377 0, hence by (7), 10c 1 100(2.71 c 1 ) The soluion of his and (7) is c , c Hence he answer is I() 0.323e e cos sin 377. You may ge slighly differen values depending on he rounding. Figure 63 shows I() as well as I p (), which pracically coincide, excep for a very shor ime near 0 because he exponenial erms go o zero very rapidly. Thus afer a very shor ime he curren will pracically execue harmonic oscillaions of he inpu frequency 60 Hz 60 cycles> sec. Is maximum ampliude and phase lag can be seen from (5), which here akes he form I p () sin ( ).

5 SE. 2.9 Modeling: Elecric ircuis 97 y I() Fig. 63. Transien (upper curve) and seady-sae currens in Example 1 Analogy of Elecrical and Mechanical Quaniies Enirely differen physical or oher sysems may have he same mahemaical model. For insance, we have seen his from he various applicaions of he ODE yr ky in hap. 1. Anoher impressive demonsraion of his unifying power of mahemaics is given by he ODE (1) for an elecric -circui and he ODE (2) in he las secion for a mass spring sysem. Boh equaions Is Ir 1 I E 0v cos v and mys cyr ky F 0 cos v are of he same form. Table 2.2 shows he analogy beween he various quaniies involved. The inducance corresponds o he mass m and, indeed, an inducor opposes a change in curren, having an ineria effec similar o ha of a mass. The resisance corresponds o he damping consan c, and a resisor causes loss of energy, jus as a damping dashpo does. And so on. This analogy is sricly quaniaive in he sense ha o a given mechanical sysem we can consruc an elecric circui whose curren will give he exac values of he displacemen in he mechanical sysem when suiable scale facors are inroduced. The pracical imporance of his analogy is almos obvious. The analogy may be used for consrucing an elecrical model of a given mechanical model, resuling in subsanial savings of ime and money because elecric circuis are easy o assemble, and elecric quaniies can be measured much more quickly and accuraely han mechanical ones. Table 2.2 Analogy of Elecrical and Mechanical Quaniies Elecrical Sysem Mechanical Sysem Inducance Mass m esisance Damping consan c eciprocal 1> of capaciance Spring modulus k Derivaive E 0 v cos v of } elecromoive force Driving force F 0 cos v urren I() Displacemen y()

6 98 HAP. 2 Second-Order inear ODEs POBEM SET 2.9 elaed o his analogy are ransducers, devices ha conver changes in a mechanical quaniy (for insance, in a displacemen) ino changes in an elecrical quaniy ha can be moniored; see ef. [Genef11] in App IUITS: SPEIA ASES 1. -ircui. Model he -circui in Fig. 64. Find he curren due o a consan E. E() 4. -ircui. Solve Prob. 3 when E E 0 sin v and,, E 0, and are arbirary. Skech a ypical soluion. urren I() Fig circui π 8π 12π c urren I() Fig. 68. Typical curren I e 0.1 sin ( 1 4 p) in Problem 4 Fig. 65. urren 1 in Problem ircui. This is an -circui wih negligibly small (analog of an undamped mass spring sysem). Find he curren when 0.5 H, F, and E sin V, assuming zero iniial curren and charge. 2. -ircui. Solve Prob. 1 when E E 0 sin v and,, E 0, and v are arbirary. 3. -ircui. Model he -circui in Fig. 66. Find a general soluion when,, E are any consans. Graph or skech soluions when 0.25 H, 10, and E 48 V. E() E() Fig circui 6. -ircui. Find he curren when 0.5 H, F, E 2 2 V, and iniial curren and charge zero urren I() Fig circui Fig. 67. urrens in Problem GENEA -IUITS 7. Tuning. In uning a sereo sysem o a radio saion, we adjus he uning conrol (urn a knob) ha changes (or perhaps ) in an -circui so ha he ampliude of he seady-sae curren (5) becomes maximum. For wha will his happen? 8 14 Find he seady-sae curren in he -circui in Fig. 61 for he given daa. Show he deails of your work. 8. 4, 0.5 H, 0.1 F, E 500 sin 2 V 9. 4, 0.1 H, 0.05 F, E 110 V 10. 2, 1 H, 1 20 F, E 157 sin 3 V