INDEX. Transient analysis 1 Initial Conditions 1

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1 INDEX Secion Page Transien analysis 1 Iniial Condiions 1 Please inform me of your opinion of he relaive emphasis of he review maerial by simply making commens on his page and sending i o me a: Frank Mera 4398 Groveland Road Universiy Heighs, Ohio Your informaion will be used o revise nex year's exam review course. I may be reached for quesions abou he review maerial a: (home) (Case Wesern Reserve) (Case Wesern Reserve FAX) flm@po.cwru.edu i

2 Transien analysis Iniial Condiions Before we can solve ransien problems involving inducors and capaciors we mus undersand he iniial condiions ha apply o he differenial equaions: Because no circui can supply infinie power: (1) he curren hrough an i nducor canno change insananeously (2) he volage across a capacior canno change insananeously These rules are he boundary condiions which apply o inducors and capaciors and, for purposes of wriing ime dependen eqpressions for volage and curren, can be wrien as: For an inducor i(0)=i(0) For a capacior v(0)=v(0) where i is assumed ha a swich has been opened or closed in he nework a =0. where i is assumed ha a swich has been opened or closed in he nework a =0. Basically, his means ha for inducors he curren is coninuous, whereas for capaciors he volage is coninuous. The volage curren relaionship for a capacior canno be wrien wihou using inegrals or derivaives. C i q q C v C Using he above definiions we can wrie "Ohm's Law" for a capacior as i C = dq d = Cdv d or, in inegral form, as v() = v( 0 ) 1 C i C() d v L L i For he inducor, "Ohm's Law" is v L = L di d or, in inegral form, i() = i( 0 ) 1 L 0 v L() d 0 A ransien is wha occurs whenever you open or close a swich in an elecrical nework. The volages and currens mus quickly readjus hemselves o he new nework. The presence of capaciors and inducors in he nework means ha ha he volage and curren readjusmens occur according o he differenial equaions describing hese circui componens. 1

3 Consider he simple series circui shown below: R E L The quesion is wha is he curren in he circui as a funcion of ime for 0, he swich closes a =0. This requires he use of differenial equaions. To describe his circui begin wih Kirchoff's volage law  V n E ir L di d = 0 This is a differenial equaion in i L di d ir = E The homogeneous (or ransien) soluion comes from solving he equaion wih he righ hand side of he equaion se o zero. L di d ir = 0 The soluion o his equaion is an exponenial of he form i() = Ae k where A and k are consans o be deermined. Subsiuing his soluion ino he above differenial equaion we can find k LAke k Ae k R = 0 Dividing hrough by Ae k LkR=0 or k=r/l The consan A mus come from he seadysae soluion of he original differenial equaion. In he seadysae all derivaives are zero since nohing is changing. Therefore, he differenial equaion for his soluion reduces o ir=e or i = R E The resuling oal soluion is he sum of he seadysae and ransien soluions i()=i seadysae i ransien or i() = R E Ae R L To find A we mus apply he boundary condiions on i, i.e. ha i(0)=i(0 ). Le i(0 )=i O. A =0 i 0 ª R E A A = i 0 E R 2

4 Therefore, he final ime dependen expression for i() is i() = R E i 0 R E e R L The facor L/R is called he ime consan of he circui since i deermines how rapidly he curren changes in he circui. i() The curren in a RL circui mus always be coninuous a =0. i 0 E R The slope mus always go o zero as Æ For a similar series RC circui, he objecive is o predic he volage across he capacior as a funcion of ime given ha he capacior has a volage v O across i a ime =0. R E C Using KVL E ir 1 C i(') d' = 0 0 We canno use i as he variable for capaciors we mus use volage. v c () = 1 i(') d' C 0 which is he inegral form of i=c dv c d The differenial equaion is hen ERC dv c v d c =0 or, rewriing he equaion in a more sandard form, RC dv c v d c = E As before his equaion has a homogeneous and a seadysae soluion. 3

5 For he homogeneous soluion: RC dv c v d c =0 Leing v c ()=Ae k RCAke k Ae k = 0 or, solving for k, k = RC 1 The seadysae soluion comes from seing all ime derivaives o zero o give v c = E The oal soluion is hen v c () = v seadysae v ransien v c () = E Ae RC A =0, v c ()=v O so v c (0) = v 0 =EA and A = v O E The final soluion is hen v c () = v 0 E e RC E which is shown graphically below v() The volage in an RC circui mus always be coninuous a =0. v 0 E The slope mus always go o zero as Æ 4

6 Example: Mos of he ime you do no need any fancy mah as shown above o solve he problem. Simply remember he boundary condiions and he general form of he soluions. R=10kW 100 vols C=0.1µf Consider he above circui. The ime consan is simply compued as = RC = ( )( ) = 10 3 seconds Assume v 0 =0 since i was no given. The problem is o deermine v across he capacior a = seconds. The general form of he soluion was shown graphically above. Expressing his general soluion mahemaically v() =E v 0 E e RC which for he given consans becomes v() = e 10 3 and, a = seconds, is v( ) =100 1 e = e 1 2 v( ) ª vols Summary of imporan relaionships for solving ransien problems. Componen Iniial condiion =0 = Ohm's Law Capacior V(0)=V(0) shor open RC I = C dv d or V = 1 C I d Inducor I(0)=I(0) open shor L/R V = L di d The differenial equaion A di d BI = C has wo soluions. One soluion is he d.c. (also called seady sae soluion) which occurs when all he derivaives go o zero. This soluion is simply I=C/B. The oher soluion is he ransien soluion (also called he homogeneous soluion) and requires ha C=0, i.e. solve A di d BI = 0 This soluion is always of he form I() = Ae m and can be solved for by simply subsiuing his expression for I() ino A di d BI = 0 5

7 CIRCUITS 31 The swich S closes a T=0. The complee response for i() for >0 is (a) 2.5 6e 0.75 (b) e 0.75 (c) e 1.33 (d) 0 (e) e vols S 4 W 3 H We use KVL o wrie he differenial equaion for he circui using he correc expression for he impedance of he inducor. 10 4i 3 di d = 0 Rewriing his in convenional form wih he sources on he righ hand side of he equaion 3 di 4i = 10 d The dc (or homogeneous) soluion is obained by seing he derivaives equal o zero or, in his case 4i = 10 giving i=2.5 amps. The ransien soluion is always an exponenial in form. Subsiuing i()=ae k ino he differenial equaion and seing he source (he righ hand side of he equaion) equal o zero we obain 3 di d 4i = 0 3kAe k 4Ae k = 0 3k4 = 0 k=4/3 The oal soluion is hen i() = i ransien i homogeneous = Ae The coefficien A is solved for by using he boundary condiion ha i(0 )=i(0 )=0. This requires ha i(0 ) = A 2.5 = 0, or ha A=2.5. Then i() = e 1.33 and he correc answer is (e). 6

8 Circuis 32 The form of he ransien par of V() for >0 is (a) (A 1 A 2 )e 0.5 (b) A 1 cos(0.5 ) A 2 sin(0.5) (c) A 1 e cos(0.5 ) A 2 e sin(0.5) (d) A 1 e 1.71 A 2 e 0.29 (e) 0 2 A 4W 1 2 5cos A =0 he swich moves from 1 o 2. Soluion: For >0 he equaion of he circui is V() di d 2i 1 i(a)da V 2 C (0 ) = 5cos 0 where V c (0 ), he iniial volage on he capacior, is zero. Differeniaing he above equaion o remove he inegral d 2 i d 2di 2 d 1 i() = 5sin 2 The lef hand side of his equaion describes he ransien response. For he ransien d 2 i d 2di 2 d 1 2 i() = 0 If we le i()=ae m we ge m 2 Ae m 2mAe m 1 2 Aem = 0 which reduces o he characerisic equaion m 2 2m0.5= 0 This equaion can be solved using he quadraic formula m = 2 ± = 1.71 and The only answer wih hese exponens is (d). 2 W i 1 H 2 F 7

9 Circuis 36 i() is (a) e 0.6 (b) e 1.67 (c) e 1.67 (d) e 0.6 (e) 2e u() V 2 W 2 W 1 3 F i() 8 W 8 W Soluion: There are many ways o solve his problem bu, perhaps, he easies way is o Thevenize he lef hand side of he circui (he volage source and he wo 2W resisors) and replace he righ hand side of he circui (he wo 8W resisors in parallel) by is equivalen resisance. Thevenizing he lef hand side of he circui V T = 2 10 = 5 vols W and 10 V 2 W R T = = 1 W NOTE: This is a good echnique o use o ge rid of curren sources in problems. The 8 W resisors in parallel can be replaced by a 4 W resisor. Redrawing he original circui and replaceing he lef hand side by is Thevenin equivalen and replacing he wo 8W resisors by a single 4W resisance, we ge he following circui 1 3 F 5 V 1 W i() 4 W Since V C (0 )=V C (0 )=0 i C 0 = 5 vols 1W 4W = 1 amp The ime consan for he circui can be direcly compued as = RC = 5 W ( 1 F) = 1.67 seconds 3 The soluions should be of he form Ae 1.67 = Ae 0.6 The correc answer is (a). 8

10 Example afernoon problem (his is acually ougher han mos afernoon problems): K a L g R R 1 2 _ e S f C b R 3 i C d i S i L You are given ha e S ()=EE 1 sin(500)e 2 sin(1000), L=10 millihenries, C=200 microfarads, R 1 =10 ohms, R 2 =5.0 ohms and R 3 =5.0 ohms in he above circui. For quesions assume ha swich K is closed a =0 and answer he quesions for he insan immediaely afer he swich is closed, i.e. for ime = If E=30V, E 1 =40V and E 2 =20V, he curren i C is mos nearly (A) 0.0 amperes (B) 1.5 amperes (C) 2.8 amperes (D) 3.0 amperes (E) 6.0 amperes This problem is mos easily solved by recalling he iniial condiions for capaciors which require ha V(0) = V(0). The iniial volage on he capacior is 0 vols so he volage across he capacior immediaely afer he swich is closed mus also be 0 vols. The applied volage e S (=0) ª 30 since sin(0) ª 0. A =0 e S appears enirely across R 1 and he resuling curren (which is equal o i C since R 1 and C are in series) mus be given by i C = e S = 30 vols = 3.0 Amperes R 1 10W The correc answer is (D). 12. If E=30V, E 1 =40V and E 2 =20V, he magniude of he volage beween poins a and b is mos nearly (A) 5.0 vols (B) 7.5 vols (C) 15 vols (D) 22 vols (E) 30 vols The volage being referred o is across he series combinaion of he inducor and R 2 as shown in he diagram below. 9

11 a L g Volmeer e S _ f b R 2 R 3 d As in quesion #11 he volage e S (0 ) ª 30 vols, i L (0 )=i L (0 ) since he curren hrough inducors is coninuous, and V C (0 )=V C (0 ) since he volage across capaciors is coninuous. Since here is no curren flow hrough L a =0 he inducor represens an open circui. The poenial a a is 30 vols; he poenial a b is zero since i is conneced o ground hrough R 3 and no curren is flowing hrough R 3. The poenial v ab is hen 30 vols. The correc answer is (E) 13. If E, E 1 and E 2 are magniudes such ha i C a =0 is 2.0 amperes, he rae of change of he volage beween poins f and d is mos nearly (A) 0.0 vols/second (B) 20 vols/second (C) vols/second (D) vols/second (E) vols/second This quesion is a lo simpler han i sounds and is a direc applicaion of of he definiion of capciance. By definiion, i C = C dv d which includes a direc expression of he rae of change of volage. Evaluaing his expression for =0, i C 0 = C dv d 0 which can be solved for he ime rae of change of volage a =0 dv d = i C 0 = 0 C = 2 Amperes Farads The closes answer is (E). = 104 vols second 14. If E, E 1 and E 2 are magniudes such ha he volages beween poins a and g is 40 vols a =0, he rae of change of i L is mos nearly (A) 0.0 amps/second (B) amps/second (C) amps/second 10

12 (D) amps/second (E) amps/second The soluion of his problem is almos idenical o ha of problem #13 wih he excepion ha he volage expression mus be ha for an inducor, i.e. V L = L di d Evaluaing his expression a =0 di d = V L 0 = 0 = 40 vols amps = 4000 L Henrys second The answer is (C). 15. If E=30V, E 1 =40V and E 2 =20V, he magniude of he curren i C is mos nearly (A) 0.0 amperes (B) 1.5 amperes (C) 2.8 amperes (D) 3.0 amperes (E) 6.0 amperes e S ()=EE 1 sin(500)e 2 sin(1000) i C (0 )=0 and i L (0 )=0 since here are no volage sources for <0. Using our rules for boundary condiions on inducors and capaciors i L (0 )=i L (0 ) bu i C (0 )πi C (0 ). A =0 e S (0 )=3040sin(500 0 )20sin( ) ª 30 vols since he sine of a small number is approximaely zero. Since V C (0 )=V C (0 ) (remember ha he volage across a capacior is coninuous) he curren i C hrough R 1 a =0 mus hen be given by i C = 30vols/10ohms = 3 amperes. The answer is (D). 11