8. Basic RL and RC Circuits

Transcription

1 8. Basic L and C Circuis This chaper deals wih he soluions of he responses of L and C circuis The analysis of C and L circuis leads o a linear differenial equaion This chaper covers he following opics The Source-Free L Circui The Source-Free C Circui The Uni-Sep Funcion Driven L Circui Driven C Circui 1

2 Firs Order Linear Differenial Equaions A firs order homogeneous linear differenial equaion is one of he form dy py=0 d where "Firs Order'' indicaes ha boh dy/d and y occur o he firs power and "homogeneous'' refers o he zero on he righ hand side of he equaion In circui analysis, y can eiher be he volage v or he curren i of he circui The soluion of a homogeneous linear differenial equaion is called a complemenary funcion In circui analysis, we refer o he soluion of he circui as a naural response or a ransien response

3 A Direc Approach Since he variables can be separaed, he differenial equaion can be rewrien as dy = p d y We inegrae boh sides as dy y = p d Then, we have ln y = p A The consan of inegraion mus be seleced o saisfy he iniial condiion y(0)=y0 ln Y 0 =A Thus, we obain ln y = p ln Y 0 p y =Y 0 e for 0 ln y ln Y 0 = p ln y = p Y0

4 A General Soluion Approach We assume a general soluion of y() in exponenial form y =Aes where A and s are consan o be deermined Subsiue he exponenial soluion ino he differenial equaion dy py=0 d s s As e p Ae =0 where (s +p) is he characerisic equaion Deermine he value of s s p =0 s= p s s p Ae =0 The characerisic equaion dy py=0 d sy py= s p y=0 Invoke he iniial condiion o deermine he remaining consan A The final form is y 0 =Y 0 =Aes0 y =Y 0 e p A=Y 0 for 0

5 5 8.1 The Source-Free L Circui We assume a series L circui for which i() o be deermined Apply KVL i vl =0 di i=0 d L - We will solve he naural response + i =Aes The characerisic equaion s i=0 L s= L Using he iniial condiion i(0)=i0, we have i 0 =I 0 =Aes 0 A=I 0 The naural response is given as i =I 0 e L for 0

6 Example: L wih a swich We have wo differen circuis: one wih he swich closed and one wih he swich open We are asked o find v(0.2) for he circui shown in Figure (c) From Figure (b), we compue he curren il 24 il = =2.4 A 10 his curren is used as iniial condiions when he il swich is opened Wrie he differenial equaion of he circui (c) di 40 il 10 il 5 L =0 d dil 10 il =0 d The general soluion of i() in exponenial form i =Aes Since il(0)=2.4 A, he soluion is i =2.4e 10 for 0 6

7 7 The Exponenial esponse The expression for he curren in an L series circui describes he naural response of he inducor i =I 0 e L + The curren decreases exponenially wih ime The L/ erm in he above equaion is known commonly as he ime consan, τ, of he L series circui i =I 0 e Power in The L Series Circui The power being dissipaed in he resisor is p =v i = i =I e The oal energy is found as w= 0 p d=i 1 = LI e d The ime consan is τ=l/

8 8 8.2 The Source-Free C Circui We assume a series L circui for which i() o be deermined Apply KCL dv v = d dv v =0 d C ic C We will solve he naural response v =Aes The characerisic equaion 1 s v=0 C s= 1 C Using he iniial condiion v(0)=v0, we have v 0 =V 0 =Aes 0 A=V 0 The naural response is given as v =V 0 e 1 C for 0

9 9 The Exponenial esponse The expression for he curren in an C series circui describes he naural response of he inducor v =V 0 e 1 C The volage decreases exponenially wih ime The 1/C erm in he above equaion is known commonly as he ime consan, τ, of he C series circui v =V 0 e Power in The C Series Circui The power being dissipaed in he resisor is 2 v2 V 0 2 C p =vi = = e The oal energy is found as V 20 2 C w= 0 p d= 0 e d 1 = CV 20 2 The ime consan is τ=c

10 Deermine he inducor volage v in he circui for > 0. For <0, l0 V appears across he 4 Ω resisor, so a dc curren of il= l0/4 = 2.5 A flows hrough he inducor (which acs as a shor circui) For >0, he baery is removed so we wrie he simple KVL equaion: dil dil 6 il 4 il 5 =0 2 il =0 d d Thus, we can represen he circui wih he equaion i =il 0 e 2 =2.5 e 2 A Finally, he volage is v =L d i = e 2 = 25 e 2 V d 10

11 Noing carefully how he circui changes once he swich in he circui is hrown, deermine v() a = 0 and a = 160 μs Before he swich is hrown, he 80Ω resisor is conneced only by one of is erminals and herefore may be ignored (i=0) Wih no curren flow permied hrough he capacior, we know v(0)=50 V since he capacior volage canno change in zero ime Afer he swich is hrown, he only remaining circui is a simple source-free C circui. Wih τ = C =160 μs v =v 0 e =50e 6250 V Finally, he volage is v 160 s =50 e 1 =18.39 V 11

12 8.3 A More General Perspecive General L Circuis The ime consan of a single-inducor circui will be τ=l/eq where eq is he resisance seen by he inducor Example: eq=3+4+12/ (1+2) General C Circuis The ime consan of a single-capacior circui will be τ=eqc, where eq is he resisance seen by he capacior Example: eq=2+13/ (1+3) 12

13 13 A =0.15 s in he circui, find he value of (a) il; (b) i1; (c) i2 For < 0, (he swich is open) i2 0 =0 2 =0.4 A il 0 =2 =1.6 A 2 8 For >0, 100% of he 2-A source conribues o i2, The 8-Ω resisor is shored ou so i1=0 i1 0 =2 Thus i2 =2 il where il =il 0 e Finally, he currens are, = L 0.4 = =0.2 s, and il 0 =il 0 =il 0 =1.6 A eq il 0.15 =1.6 e =755.6 ma i =2 il 0.15 =1.244 A

14 1s Order esponse Observaions The volage on a capacior or he curren hrough an inducor is he same prior o and afer a swich a =0 esisor curren (or volage) prior o he swich i(0-) can be differen from he volage afer he swich i(0+) All volages and currens in an C or L circui follow he same naural response e-/τ 14

15 The Uni-Sep Funcion So far, we have been sudying he naural response of L and C circuis (when no sources or forcing funcions were presen) In oher words, we have been solving problems in which energy sources are suddenly removed from he circui We shall consider ha ype of response which resuls when energy sources are suddenly applied o a circui The uni-sep funcion u() is a convenien noaion o represen change: { u = { u 0 =

16 Swiches and Seps In order o obain an exac equivalen for he volage-sep forcing funcion, we may provide a single-pole double-hrow swich The exac equivalen for he curren-sep forcing funcion, we may replace his circui by a dc source in series wih a swich 16

17 Modeling Pulses Using u() By manipulaing he uni-sep forcing funcion, we can generae many useful forcing funcions A recangular volage pulse { 0 v = V The wo uni seps u( 0) and u( 1) are needed o obain he recangular volage pulse A pulsed sinewave v =V m u 0 u 1 sin 2 f 17

18 8.5 Driven L Circui Now, we consider he behavior of a simple L nework o he sudden applicaion of a dc source The shown circuis represen an L circui subjecs o a volage-sep forcing funcion V0u() Applying Kirchhoff s volage law di i L =V 0 u d Since i()=0 for <0, we sudy he soluion for >0 i L di =V 0 d Then, we have di L =d V 0 i 0 0 Inegraing boh sides L ln V 0 i = K 18

19 19 invoke he iniial condiion o find K (i()=0 for <0) L ln V 0 =K and hence L ln V 0 i ln V 0 = earranging L Which can be rewrien as ln V 0 i ln V 0 = or V 0 V 0 L i = e V0 i = 1 e L u ln V 0 i = V0 L 0

20 20 The expression for he volage in an L series circui describes he energizing characerisics of he inducor Forced response i = V 0 Naural response V e 0 L The Naural response: The exponenial erm has he funcional form of he naural response of he L circui; i is a negaive exponenial, i approaches zero as ime increases, and i is characerized by he ime consan L/ The Forced response: I is he response ha is presen a long ime afer he swich is closed

21 21 A General Soluion Approach The soluion of any linear differenial equaion may be expressed as he sum of wo pars: he complemenary soluion (naural response) and he paricular soluion (forced response) The inhomogeneous linear differenial equaion has he form of V dy di p y=f i= 0 u d d L L or dy p y d=f d Forced response The soluion is given as paricular soluion y=e f e d p p complemenary soluion Ae p i = V 0 Naural response V e 0 We noe ha, when f() is zero (a source-free circui), he soluion is he naural response p y n =Ae Since f()=f, he paricular soluion leads o he following forced response F y f =e p F e p d = p L

22 22 Example: L wih a forcing funcion Applying KVL yields V0 di i= u d L L Firs, we evaluae he naural response by solving di i=0 d L s We assume in =Ae, where A and s1 are consan o be deermined 1 The characerisic equaion s i=0 s1 = L L Then, we have in =Ae L Nex, we deermine he forced response if = Finally, he complee soluion is i()=in()+if() V0 i = Ae L where V0 V0 i 0 =0= A A= V0

23 Deermine i() for all values of ime in he circui We noe ha he circui conains a dc volage source as well as a sep-volage source Using superposiion, we solve he circui for each source alone We compue he curren due o only a dc volage source idc = 50 =25 A 2 23

24 24 We compue he curren due o only a a sep-volage source isep volage =if in 2 =25 25 e The complee response 2 i =idc isep volage =50 25e

25 25 Find he curren response in a simple series L circui when he forcing funcion is a recangular volage pulse of ampliude V0 and duraion 0 We represen he forcing funcion as he sum of wo sepvolage sources V0u() and -V0u(-0) Using superposiion, assume i1() designae he par of i() due o V0u() acing alone, and i2() represens ha par due o -V0u(-0) acing alone i =i1 i2 We solve he response i1() as V0 L i1 = 1 e We solve he response i2() as 0 V 0 i2 = 1 e L 0 0 We now add he wo soluions, bu do so carefully, since each is valid over a differen inerval of ime 0

26 { i = i1 i1 i

27 8.6 Driven C Circui Similar o L circuis, he complee response of an C circui consiss of he naural and he forced response Since he procedure is virually idenical o wha we have already discussed in deail for L circuis, we consider a relevan example for a driven C circui Find he capacior volage vc() and he curren i() in he 200 Ω resisor for all ime 27

28 For <0, he circui has wo separae loops The capacior acs as open circuis 50 vc 0 =120 =100 V The curren in he 200 Ω resisor 1 i 0 =50 =192 ma For >0 The complee response of he capacior is vc =vcf vcn The naural response vcn() vcn =Ae eq C where eq = =24 The forced response is vcf =50 =20 V

29 29 The complee response is vc =20 Ae 1.2 We use vc 0 =vc 0 =vc 0 =100 V, hen 100=20 A A=80 Thus, we have vc =20 80 e and vc =100 0 The curren in he 200 Ω resisor vc i = = e 1.2 ma 200 i =192 ma 0 0

30 Homework Assignmen 7 P8.7, P8.10, P8.20, P8.22, P8.24, P8.27, P8.29, P8.35, P8.38, P8.40, P8.44, P8.46, P8.48, P8.52, P8.54, P8.57, P8.61 and