CHUYN 3 Trng Ph Thng Nng Khiu HQG Tp.H Ch Minh CHUYN VT L MCH PHI TUYN Gio vin hng dn: thy Phm V Kim Hong Nhm bin son: Lp 11 L ( kho 2008-2011) L i Nam Nguyn Hng Nhung L B Tin Trin I.Mch phi tuyn l g? Lu : khi nim mch in dng trong chuyn ny nu khng ni g thm th ta hiu l mch in c dng in mt chiu v ta khng xt n mch in c dng in xoay chiu. Mch tuyn tnh Trc khi n vi khi nim mch phi tuyn, ta nm khi nim mch tuyn tnh. Mch tuyn tnh l mch in ch cha cc phn t tuyn tnh. Phn t tuyn tnh l cc phn t nh cc in tr R,(c th l cun dy L hay t in C trong mch xoay chiu) c tr s khng i theo thi gian, hay i vi in tr th quan h gia hiu in th U gia 2 u in tr v I chy qua in tr tho mn nh lut Ohm tc l UI =R V sao li gi cc phn t trn l tuyn tnh? Bi v cc phn t trn c c tuyn vn ampe U(I) hay I(U) l mt ng thng hay quan h gia U v I ca cc phn t trn l tuyn tinh nn chng cho tn gi l phn t tuyn tnh. Trng hp ca cun dy c t cm L khng i hay t in c in dung C khng i th c tuynvn ampe ca chng l ng thng khi mch c dng in xoay chiu c tuyn vn ampe ca mt s phn t tuyn tnh Mt in tr c tr s khng iThng c lm t hp kim constantan thng gm 55% ng v 45% nikenhay cc hp kim t chu nh hng bi nhit nh manganin86 12 2Cu Mn Ni , U = IR hay UI =R nn c tuyn vn ampe l mt ng thng i qua gc to c h s gc l R c tuyn vn-ampe ca mt in tr khng i I. Mch phi tuyn Mch phi tuyn l mch in c cha cc phn t phi tuyn. Phn t phi tuyn l cc phn t nh cc in tr R,(c th l cun dy L hay t in C trong mch xoay chiu) c tr s thay i theo thi gian, hay i vi in tr th quan h gia hiu in th U gia 2 u in tr v I chy qua in tr khng mn nh lut Ohm tc lUIRGi cc phn t trn l phi tuyn l bi v c tuyn vn ampe U(I) hay I(U) khng phi l ng thng hay quan h U v I ca phn t khng tuyn tnh. i vi trng hp ca t v cun cm th trong mch in xoay chiu quan h U-I khng tho mn ng thcUIZvi Z c th l dung khng ca t hoc cm khng ca cun cm Tng qut hn Z c th l tng tr ca mt on mch in xoay chiu Trong k tht v i sng ta thng gp cc phn t phi tuyn nhiu hn bi v trn thc t mi phn t trong mch in u c tr s ph thuc vo nhit . Mt s phn t phi tuyn v c tuyn vn ampe: Mt s phn t phi tuyn thng gp nh: + nhit in tr thermistor Trong thermistor ngi ta thng quan tm n th nhit - in p nhiu hn l c tuyn vn - ampe c tuyn vn ampe ca 1 thermistorS ph thuc gia nhit v in p ca 1 thermistor + in tr phi tuyn varistor Thng c i = f(u) th hin qua th sau + diode in t hay n in t + diode bn dn Diode bn dn l tngDiode Zener ( n cn c 1 vng in p ngc hot ng rt tt) + trong mch phi tuyn c dng in xoay chiu th cun cm c li st th c t cm L bin i mnh hay hin tng in tr mt s cht in mi nh scnht NaK(C2H2O3)2.4H2O lm cho in dung ca mt t in C thay i i vi hin tng bin i t cm L th ngi ta quan tm n c tuyn t thng- dng in i vi hin tng bin i in dung ca t th ngi ta quan tm n c tuyn in tch hin in th + dy tc bng n cng c coi l mt in tr c tr s thay i do tc dng nhit ca dng in qua n Cc thng s c trng cho mt phn t phi tuyn Mt phn t tuyn tnh c cc tr s khng i v d nh in tr R , t cm L, in dung C, c gi l cc thng s tnh. Mt phn t phi tuyn th ngoi cc thng s tnh cn c cc thng s ng Cc thng s ng hay thng s tc thi ca phn t phi tuyn c trng cho phn t phi tuyn cn cc thng s tnh v thng s trung bnh thng khng mang nhiu ngha vt l ( t c s dng trong bi ton) in tr: in tr tnh tUR =I in trtrung bnh tbUR =I in tr ng ( hay in tr tc thi) 'd IUR = UI= Cun dy t cm tnh tFL =I t cm trung bnh tbL =I t cm ng'd IL =I= T in in dung tnh tQC =U in dung trung bnh tbQC =U in dung ng'd UQC = QU= Cc thng s R, L, C l hm theo cng dng in I hoc hiu in th U ca chng nn n c trng cho phn t phi tuyn ti mi im trn c tuyn vn ampe. Cng sut: i vi cc phn t phi tuyn th ch c s dng cng thc P = U.I hay trong mch in c dng xoay chiu th P = U.I.cos Cc cch biu din c trng ca mt phn t phi tuyn: Cch 1: cho hm s U = f(I) hay I = f(U) hoc u = f(i) hay i = f(u) Vi f khng phi l mt hm tuyn tnh Cch cho ny l cch cho thun li Cch 2: cho c tuyn vn ampe ca phn t Cch cho ny tng i khng thun li Cch 3: cho th hoc c tuyn vn ampe ca phn t v km theo s liu hoc to ca 1 s im Cch cho ny thun li hn cch 2 cch 2 v cch 3 th i khi chng ta phi dng n phng php ngoi suy trn th lm v thng mang tnh thc nghim nhiu hn Cch 4: cho bng ghi s liu U I y cng l mt cch cho mang tnh thc nghim nhiu hn Cc tnh cht ca mch phi tuyn: + Mch phi tuyn khng c tnh xp chng nghim hay khng p dng c nguyn l chng chp cc trng thi in + Mch phi tuyn c tnh cht to tn s V d: vi nhng phn t phi tuyn R, L ,Ctrong mch in c ngun xoay chiu tn s gc th dng qua mch c th c tn s gc l 0,, 2, 3, Nu hiu in th kch thch dng hnh sin th do quan h phi tuyn nn cng dng in trong mch c th khng c dng sin m c th phn tch thnh tng cc dao ng iu ho c tn s khc nhau +Cc nh lut Kirchhoff vn ng trong mch in phi tuyn mt chiu v xoay chiu Ni ring v Diode: Nh ta bit, Diode l mt linh kin in t hay mt phn t trong mch in ( mt chiu v xoay chiu) ch cho dng in i qua theo 1 chiu nht nh v diode l mt phn t phi tuyn Diode gm c 2 loi l diode in t hay n in t v diode bn dn. Trong chuyn ny khng quan tm n cu to ca diode m ch quan tm n tnh phi tuyn ca diode c tuyn vn ampe ca diode nh sau Tuy nhin do c tuyn vn ampe ca diode kh phc tp nn ngi ta thng l tng ho thnh cc dng c tuyn n gin Sau y l cc dng l tng ho c tuyn vn ampe ca diode theo mc tin li gn vi kt qu ca thc nghim L tng ho 1: diode ng m Nu U < U0 th khng c dng chy qua Diode Nu U = U0 th Diode l mt phn t phi tuyn c in tr ng l dUR = = 0I L tng ho 2: vn l diode ng - m Nu U < U0 th khng c dng chy qua Diode Nu U U0 th Diode coi nh l 1 phn t phi tuyn c in tr ng l 0dU-U UR = = =const 0I I L tng ho 3: khng cn dng Diode ng m Nu U < U0 th Diode coi nh l 1 phn t phi tuyn c in tr ng l dU UR = = =const 0I INu U U0 th Diode coi nh 1 phn t phi tuyn c in tr ng l0d0U-U UR = = =const 0I I-I L tng ho 4: a v dng ng cong vi hm s gii tch v d nh hnh bn I = kU2 Diode c in tr ng lKhi th R-I c biu din nh hnh bn: dIU 1 1 1kR = = =I I k 2 I 2 kI= i Thng thng trong cc thi thng ch xut hin dng l tng ho 1 ,2 ,3 ca diode II.Phng php gii cc bi ton v mch phi tuyn Thng thng da vo cch cho v cu hi ca bi th c 3 phng php sau tm nghim gn ng trong mt bi ton mch phi tuyn v cc phng php ny kt hp vi cc nh lut v phng php trong mch tuyn tnh ( tr cc phng php khng tho tnh cht ca mch phi tuyn ) . Phng php th, phng php s v phng php biu din gn ng cc c tuyn bng hm xp x a)Phng php th: Ni dung: t cc c tuyn ca cc phn t ta v c tuyn chung ca mch sau xc nh im lm vic ca mch theo cc d kin ca bi ton Do cc mch u c cuto t 2 loi mch c bn l song song v ni tip nn ta xt 2 trng hp c bn ca mch phi tuyn + trng hp 1: mch gm cc phn t ghp ni tip Nguyn tc: trong mch ni tip th da vo 2 nh lut Kirchhoff, ta c: 11 ,i jniI I I i j nU U= = = T nguyn tc trn ta c th v c c tuyn vn ampe ca mch ni tip bng cch cng cc t tuyn I(Ui) theo trc honh ( trc OU) nh hnh v sau. + trng hp 1: mch gm cc phn t ghp song song Nguyn tc: trong mch ni tip th da vo 2 nh lut Kirchhoff, ta c: 11 ,i jniU U U i j nI I= = = T nguyn tc trn ta c th v c c tuyn vn ampe ca mch ni tip bng cch cng cc t tuyn I(Ui) theo trc tung ( trc OI) nh hnh v sau. Nh ni trn, mch phc tp c cu to t cc mch n gin nn ta c th nhm cc on mch ni tip v song song v c tuyn chung cho on mch u im ca phng php ny l gii quyt c vi nhng bi ton cho th hay c tuyn vn-ampe Ngoi ra c tuyn vn-ampe ca on mch cn cho bit tnh phi tuyn ca on mch , kt hp gia th v php ngoi suy c th gii mt s bi ton mt cch d dng Nhc im ca php trn l khi gii quyt cc mch phc tp th tnh chnh xc khng cao, gii bng th tr nn kh khn v km chnh xc, gp sai s ln. b)Phng php s: khc phc nhng khim khuyt ca phng php th cho mch phc tp, ngi ta dng n cc phngphp s.Chng ta a bi ton v dng i s v gii phng trnh i s bng cc phng php s. Phng php s c tin cy v chnh xc cao l phng php lp. Phng php lp c bn: C s ton hc: xt phng trnh g(x) = 0 (1) Ta a phng trnh v dng x = f(x) (2) sao cho f(x) l hm c tp xc nh l R Chn x0 l mt nghim gn ng ca (1) Ta c x1 = f(x0); x2 = f(x1); x3 = f(x2),, xn+1 = f(xn) Ta lp i lp li n khi xn+1 = xn = x (*)th x l nghim ca (1) Chng minh: d dng thy rng khi xy ra iu kin (*) th x tho (2) do x l nghim ca (1) Phng php lp Newton: C s ton hc: xt phng trnh g(x) = 0. Chn nghim ban u tng i gn ng l x0 Ta c: nghim gn ng bc 1 l 01 00( )'( )g xx xg x= Nghim gn ng bc n+1 l 1( )'( )nn nng xx xg x+= Khin th( ) 0ng x Chng minh : Theo nh ngha o hm th( ) ( ') ( )'( ) '( )'( ') ( )''( )g x g x g xg x g xx x xg x g xx xg x= = = +vi x v x l 2 nghim gn ng bc lin tip nhau Thay x = xn+1 v x = xn th do xn+1 l nghim ca phng trnh g(x) = 0 th g(xn+1) = 0 Do nn+1 nng(x )x =x -g'(x ) Php lp cng ln th hiu xn+1 xn cng nh do biu thc tng qut ca o hm cng ng cho x cng nh. Phng php lp c bn v phng php lp Newton thng c dng trong cc bi ton v mch phi tuyn khi phi gii cc phng trnh phi tuyn tnh Kt hp vi phng php th nhm tm ra x0 gn ng nht th ch sau vi bc lp ta c th xc nh c nghim gn ng chnh xc ca phng trnh. Phng php lp Newton tuy khng chnh xc bng phng php lp c bn nhng li nhanh hn i vi nhng hm tng i phc tp. Do khi phi s dng n phng php lp th ta nndng phng php lp Newton gii quyt bi ton mt cch nhanh nht c)Phng php biu din gn ng cc c tuyn bng hm xp x : i khigp cc bi ton cho th hoc bng s liu m khi v th ta ra nhng dng th gn ng vi nhng hm gii tch quen thuc v ta biu din I = f(U) hay U = f(I) cho bi cc hm gii tch . Phng php trn c gi l biu din gn ng c tuyn bng hm xp x Lu l tng chnh xc, sau khi tm ra hm xp x th ta nn v mt th biu din quan h I f(U) (hayU f(I)) vi I = kf(U) ( hay U = kf(I))v k l mt hng s Ta xem th th I f(U) c gn ng l ng thng hay khng v s c li hn khi dng ng thng ngoi suy trn th Mt s th ca cc hm gii tch c bn 1.Hm s x ayx b+=+ Hnh bn tri l khi a > b cn hnh bn phi l khi a < b. 2.Hm s 2y ax =Hm ny tng i quen thuc vi chng ta Hnh bn tri l th 2y ax =trong trc to Oxy Cn Hnh bn phi l biu din quan h gia y x2 l mt ng thng 3.Hm s xy a =Hnh bn tri khi a > 1Cn hnh bn phi khi a < 1 4.Hm s xy e = 5.Hm s y = lgax 6.Hm s y = ln x 7.Hm y = sin x 8.Hm y = cos x 9.Hm y = tan x Lu : Khng phi ch c 3 phng php trn gii mch in phi tuyn. 3 phng php trn c dng gii quyt bi ton mt cch gn ng v thun tin cho tng trng hp. i lc ch cn p dng 2 nh lut Kirchhoff vit ra h phng trnh in th - nt cng c th gii quyt c bi ton. d)V d v ng dng cc phng php trn : Sau y l mt bi v d n gin v mch in phi tuyn Vi mi cch cho khc nhau th ta li chn nhng phng php ph hp gii bi ton mt cch chnh xc Cho mch in nh hnh v: on mch nh hnh v c ngun khng i E = U = 10V, in tr R l mt in tr phi tuyn. ng kho K. Tm s ch Ampe k.in tr ca Ampe k, ca ngun v ca dy dn tng cng l R1 = 500 Cho bit R c c trng vn ampe c cho nh sau a/ th i (mA) u(V) cho nh hnh sau b/ Cho bng s liu nh sau: U(V)1,02,03,04,0 I(mA)4,05,76,98,0 Lu : cc cu a,b c lp vi nhau Gii: a/ do bi cho th nn ta s dng phung php th cho tin li Theo 2 nh lut Kirchhoff, ta c U = i.R1 + u1U uiR=Thay U = 10V , R1 =500 =0.5k Ta c20 2 i u = V th hm s20 2 i u = , ta c nh hnh sau Giao im ca 2 th chnh l to m mch ang lm vic Theo th th im mch ang lm vic l I = 9.3 mA v hiu in th 2 u in tr R l u = 5.4V b/ Da vo bng s liu, ta c th gn ng nh sau: Ta phng on rng c tuyn U-I trn th c dng ca mt parabol. kim tra li d on ta v mt th khc l th U-I2v nu n l ng thng th d on ca ta l ng Do tr s I2 tng i ln nn ta v th U-0.25I2 cho thun tin ( hnh bn phi) Ta thy rng 4 im trn nm gn ng trn mt ng thng c h s gc k = 0.25 u = 0.0625i2 . Hoc ti y ta c th U-I ta c th dng phng php th gii nh cu a hoc s dng phng php s gii ti dy ta li c nh sau: U = i.R1 + uu = U iR1 v u = 0.0625i2 0.0625i2 + iR1 U = 0 thay U = 10V v R1 = 0.5k.0.0625i2 +0.5i 10 = 0 vi i o bng mA C th s dng cng thc gii phng trnh bc 2 v tm ra gn ng i = 9.3mA ( con s chnh xc l 9.266499161 ) Hc ta c th dng phng php lp c bn li =4 10 0.5i Chn gi tr ban u i = 10, ta bt u lp v c bng sau: I 4 10 0.5i 108.94427191 8.944271919.404564037 9.4045640379.206708842 9.2067088429.292272556 9.2922725569.255367067 9.2553670679.271303223 9.2713032239.264425196 9.2644251969.267394371 9.2673943719.266112725 9.2661127259.26666597 9.266665979.266427156 9.2664271569.266530243 9.2665302439.266485745 9.2664857459.266504953 9.2665049539.266496661 9.2664966619.266500241 9.2665002419.266498696 9.2664986969.266499362 9.2664993629.266499075 9.2664990759.266499199 9.2664991999.266499145 9.2664991459.266499168 9.2664991689.266499158 9.2664991589.266499163 9.2664991639.266499161 9.2664991619.266499162 9.2664991629.266499161 Tht ra chng ta khng cn lm t m nh bng ny bi v chng ta mun chnh xc n phn thp phn th my th thy phn c lp li l chnh xc. Th d ta mun thy chnh xc n ch s th ba sau du phy th ch mt 10 bc l c th tnh cra kt qu l I = 9.266mA. Hoc nu ch cn 2 ch s sau du phy th ch cn 8 bc l c th ra kt qu l I = 9.26 mA Phng php lp thng dng gii cc phng trnh tng i phc tp hoc chng ta khng c cng thc tng qut cho chng nh phng trnh x = sin x hay x = cos x chng hn, Phng php lp Newton Ta th xem phng php lp Newton trong bi ton ny nh th no y g(i) = 0.0625i2 +0.5i 10 do g(i) = 0.125i + 0.5 ta c21( ) 0.0625 0.5 10'( ) 0.125 0.5n nn n nn ng x i ii i ig x i++ = = + Chn i0 = 10 mA in+1 20.0625 0.5 100.125 0.5nnni iii+ + 109.285714285 9.2857142859.266513057 9.2665130579.266499161 9.2664991619.266499161 Ta thy rng ch sau 4 bc l cho kt qu chnh xc Ta lu rng do cc kt qu cui cng thng c o bng thc nghim nn ta phi chn chnh xc cho ph hp vi thc th. V d: I = 9.3mA l ph hp vi thc t cn I = 9.266 mA khng c ph hp cho lm III. Cc dng bi ton v mch phi tuyn. Mt s bi ton v mch phi tuyn Theo kinh nghim t cc bi tp v mch phi tuyn th c th chia ra cc dng bi v mch phi tuyn nh sau: Nhng bi mch cu c phn t phi tuyn , v nhng bi mch phi tuyn xoay chiu ( c th l mch nn dng t - diode,) v mt s dng bi khc 1.Bi ton:mch cu c phn t phi tuyn i vi cc bi ton mch cu c phn t phi tuyn th ta s dng phng php in th - nt nh trong mch tuyn tnh bnh thng. Tuy nhin vi mi phn t phi tuyn th li c mt cch gii ring gii quyt bi ton. Bi 1: Cho mch in nh hnh v: VA > VB vi A1 v A2 l 2 ampe k l tng ln lt ch IA1 = 30mA v IA2 = 5mA. R1 = 1 k ;R2 = 2 k ; R3 = 3 k; R4 = 4 kX l mt phn t phi tuyn. Hiliu X c th l 1 trong cc phn t sau y hay khng v nu c thUAB bng bao nhiu v cng sut tiu th trn X: a/ X l mt varistor c c trng vn ampe l i = ku2 k o bng mA/V2

b/ X l mt n in t ng m. NuVD > VB th n cho dng chy qua vi I5 < 5mA khng i v t D n B cn nu ngc li th khng c dng in qua X c/ X l mt Diode c c trng vn ampe qua th sau Phn tch: bi ny cha cho chiu ca ngun nn c th nhn nghim m phi dng nhng iu kin vt l loi b nghim trong bi. Gii a/ta c cc phng trnh in th nt sau: thun tin ta gii bi ton vi cc dng in o bng mA Gi s dng in trong mch chnh i t A n B 2 3 1 21 2 3 5 5 13 1 41 3 41 3 2 51 4 2301 2 31 3 4AABu u ui i i i ku Iu u ui i iu u u u Uu u u+ = + => + = + = == + => = ++ = + =+ =1 2 212 2 35 3 52 2 2 45 5 2 42 2 2 25 2 5 5 52230 301 2 230 90 3330 30 4 2 20( )2 4330 90 3 3 1202 230 202u u uuuku u kuu uku ku u u Vu uku u u ku uuu+ = = + = = = + = = + = + + + = = Nu 22542 2 25 5 5 5 5100( )365( )20 33 13 120 5( ) . 108 108.33( )2 355( )ABu Vku mAuuku u u V P u ku WU V=== + + = = = = = = Nu 225 542 2 25 5 5 5 520( )320( )3203 23 120 130( ) . 866 866.66( )2 3410( )3ABu Vi ku mAuuku u u V P u ku WU V== == + + = = = = = = Tuy nhin khi u5 > 0 m i5 < 0 th v l i vi varistor Vy ta c kt qu sau: UAB = 55V v Px = 108.33W cu a liu c phc tp hn nu i v tr R4 v varistor X hay khng??? b/ gi s VD > VB

i1 + i2 = i3 + i5 = 30i1 = i3+ i4 v i2 +i4 = i5 v i4 < i5 nn i4 = -5mA => i1= i3 -5 v i2 -5 = i5 li c u1 +u4= u2 => i1 - 20 = 2i2 i3 -5 20 = 2(I5 + 5) => i3 = 2I5 + 35 i1 = 20+2(I5 +5) = 30 + 2I5 i2 = I5 + 5 I5 = - 0.6 mA (loi v yu cu bi) Vy VD VB v khi th UAB < 0 loi v VA > VB Do iu kin bi khng ph hp vi thc t c/ u tin ta phi tm c trng ca Diode khi U < 5V th khng c dng chy qua diode khi U 5V th c dng chy qua diode v in tr ng ca Diode l 13dR k = Ta xt nu UDB < 5V th khng c dng chy qua diode, mch lc ny ch cn (R1 // (R2 nt R4))nt R3 Khi ta tnh c dng qua R4 v R2 l IA2 = 5mA v qua R3 l IA1 = 30mA . Do UDB = R4IA2 + R3IA1 nn suy ra UDB = 20+90 = 110V > 5V.VyUDB > 5V v c dng in chy qua n Lc ny ta li c 5 5153u i = +3 5 1 21 2 3 5 13 1 41 3 41 3 2 51 4 253011 2 331 3 4AABu u u ui i i i Iu u ui i iu u u u Uu u u + = + => + = + = == + => = ++ = + = + ==> 213 52 25 52 25 5 2 52230245 330 15 5 15 52 2330 45 3 4 752 230 202uuu uu uu uu uu u u uuu= = = = + = + + = = Chia ra 2 trng hp u4 = 20 V v u4 = -20V Gii ra tng trng hp ta c kt qu sau U4 = 20 V th25510032541103u Vuu====> loi v u5 nhn 2 gi tr khc nhau U4 = -20V th255203654403u Vuu====> loi v u5 nhn 2 gi tr khc nhau Nhn xt: bi ny ta phi lu mt s iu 1.v mt n v, ta thy rng 1mA . 1k = 1V ta c th quy c v n v thun tin khi gii 2.trong qu trnh gii, cc kt qu trung gian khng nn lm trn m gia nguyn cn, phn s, 3.sau khi ra kt qu phi bin lun bi ton vi thc t, v du v v ln Bi 2 : Trong mch cu hnh c cc in tr R1= 2 ; R2= 4; R3= 1; X l mt varistor c i=kU2. a.V ng c tuyn Vn-Ampe U= f(i) ca varistor. Gi dURdi=l in tr tc thi ca varistor. C th ni g v in tr ny khi i bin thin t 0 n +. b.Bit k= 0,25 (A/V2) nu i o bng Ampe, U bng Vn. Ngi ta iu chnh hiu in th U0= UAD cu cn bng. Tnh cng sut in P tiu th trn varistor, tnh cc dng i1, i2 qua 2 nhnh v hiu in th U c.R1, R2, R3 v k c gi tr bt k, Tnh U0 cu cn bng, tnh dng I trong mch chnh. Thay X bng mt bin tr R ta c cu Uytston, hy nu s ging v khc nhau gia cu nghin cu trong bi v cu Uytston. ( Trch thi hc sinh gii Quc gianm hc 1990-1991) Gii: a.ta c i=kU2 iu=kv R = du 1=di 2 ki khi I bin thin t 0 n + th R bin thin t + v 0 b.khi cu cn bng th ta c i1 = 01 2UR +Rv u3 = u1 = i1R1 = 0 11 2U RR +R; u4 = i1R2 = 0 21 2U RR +R li c 22 3 0 1 0 2 1 1 24 0 23 3 1 2 1 2 3 2u U R U R R (R +R )=ku =k U =R R (R +R ) R +R kR R| | |\ = 3V T rt ra c i1 = 0.5A v i2 = 1A v P = 2W c. 1 1 20 23 2R (R +R )U =kR R nu thay X bng mt bin tr R th cu cn bng R = 2 31R RR Cu Uytxton khc vi cu trong bi ny ch bin tr c mt gi tr xc nh cu cn bng vi U bt k cn varistor th cu cn bng ta phi t U = U0 xc nh th cu mi cn bng Ging nhau l nu 1 trong 3 in tr cn li khng bit gi tr th vn tnh c nu bit 2 in tr cn li v tr s ca varistor hoc bin tr do ng dng trong vic o chnh xc in tr cha bit Bi 3: cho mch in nh hnh v R1 = 1k; R2 = 2k; R3 = 3k;R4 = 4k Q l mt n quang in c ant ni vi im C, catt ni vi im D Nu in th ant cao hn in th catt th n m i0 = 10 mA i qua, ngc li th n ng, khng c dng i qua. Hiu in th t gia A,B l UAB = 100 V 1.a. n Q ng hay m? b.Tnh hiu in th gia 2 cc ca n 2.Gi nguyn cc in tr v hiu in th gia A v B nhng thay n Q bng mt Diode Kch cho dng i qua theo chiu t C n D. Diode c ng c trng Vn-Ampe v hnh. a.Nu cc c in ca diode v mc dn in. b.Nu diode m, tnh dng qua diode. ( Trch thi hc sinh gii quc gia nmhc 1987-1988) Gii 1) a) gi s bng n ng, khi mch in cho in tr tng ng l R = 2,4 k Khi UCB = 75 V v UDB = 33,3 V => VC > VB nn n m Vy n in t m b) n m . ta xt mch ACB c i1 R1 + i3 R3 = UAB hay i1 + 3(i1 10) = 100 hay i1 = 32,5 mA i3 = 22,5 mA tng t vi mch ADB. Ta cng c i4 = 13,3 mA v i2 = 23,3 mA UCD = 20,9 V 2) a) nu UCD > 20 V th Diode m v c in tr l 1k cn khi UCD < 20 V th Diode ng v xem nh c in tr v cng ln b) tng t nh cu 1 ta c cc phng trnh sau 11 324 1 3 431 1 3 44303 10021, 24 2( ) 10023, 24 014, 2i mAi ii mAi i i ii mAi i i ii mA= + = = + + = = + == Dng qua Diode l i1 i3 = 7 mA v UCD = 27,2 V > 20V 2.Bi ton v th Bi 4: Trong hnh 1.1 cho ng c trng vn ampe ca bng n pin. Bng n c mc trong mch in nh trn hnh 1.2 a)Hy tm cng dng in qua n bng th. b)Vi v tr no ca con chy bin tr th hiu in th gia hai im A v B bng khng? c)Vi v tr no ca con chy bin tr, hiu in th gia im A v B hu nh s khng thay i khi bin i khng nhiu sut in ng ca pin? B qua in tr trong ca pin. Gii: a)ta c U = e Ire - UI = rv th e-UI = rv da vo th ta xc nh c I = 0,24 A v U = 1,6V. Lu : ng thng e-UI = r gi l ng ti b) UAB = 0 th ta c Ur = 2,4 V v U = 1,6VUAB = 0 th U1 = Ur = 2,4 V v U2 = U = 1,6V 1 112 221 2R U 3= = R =24R U 2R =16R +R =40 c)khi sut in ng bin i khng nhiu 1 lng e th in p 2 u n bin i 1 lng U = e rI in tr ng ( cn gi l in tr vi phn ) ca n ln cn vng hot ng l cotgdUrI= = v theo th trn th rd = 12,5 vitanl h s gc ca tip tuyn T suy ra U = e rI =e rdUr =e 0.8U do suy ra 1,8eU =Hiu in th phn di ca bin tr l 2 2 22 2eR R R40U U e eR R= = = hiu in th A, B hu nh khng i th U2 = U hay R2 = 18 Khi UAB xp x 0,6 V l e thay i trong khong t -1V n 1V th UAB khng thay i qu 0,03 V Dng c th ny c th coi l mt b n p cho cc in p khng ln 3.Bi ton v bng s liu ca bng n Bi 5: cho cc bng n ging nhau c bng s liu nh sau: U(V)00,60,711,52,53,54 I(A)00,10,140,150,220.260,270,28 Bit rng cc s liu trn o c trong qa trnh kho st bi ton ny nn ta c th coi gn ng vi sai s khng qu 5% Hnh bn l mt mch n i vo hnh lang ca mt cng ty. Khi i qua hnh lang, ngi bo v ln lt ng cc kho t trong ra thy ng. Tuy nhin do s sut trong vic lp t nn trong mch n c 1 n ni tip nh hnh v. H thng n c t vo mt in p 5V, bit rng in p nh mc ca n l 3,5 V v nu in p vt qu 4V th n s chy. in p ca n di 1V th n sng rt yu v ngi bo v c th tng lm l n b chy Hi liu khi i qua dy hnh lang th ngi bo v c thy n no chy khng? i qua ht dy hnh lang m ngi bo v khng thy n no b chy th in p t vo mch phi l bao nhiu?nu khng c th in p t vo s n sng l nhiu nht? Khng tho b cc n m ch lp thm, liu c cch no khc ph s c trn hay khng? Gii:Ta xt khi i qua dy n th nht, khi mch n ch c 2 n mc ni tip nn do in p mi n l 2,5 V Khi i qua dy hnh lang th hai th lc ny ta c do n c in p khng qu 4V th mi khng chy nn n ni tip khng chy th in p n khng qu 4V, do m cng dng in qua n khng qu 0,28 A => cng dng in qua 2 mch nhnh khng qu 0,14A hay in p ca 2 n cn li khng qu 0,7 V. Nh vy ngi bo v vn thy c n b chy d in p ca 2 n cn li di 1 V khng c n no b chy th in p ln nht c th t vo mch gn ng l 4,6 V v n ni tip c in p khng qu 4V ko theo cng dng in ca n khng qu 0,28A hay khi ng ht kho th cng dng in quami bng khng qu 0,07A => in p ca chng khng qu 0,6V Do 4,6 V < 5V nn chc chn l ngi bo v vn nhn thy c n chy. Do khng th t in p sao cho ngi bo v khng thy bng no chy. khc phc ta c th mc thm cc n song song vi n ni tip. gi n l s n mc thm Ta c khi ng kho K u tin th dng qua mch chnh khng qu 0,28A m khng bnhn l chy th cc n khc phi c dng khng di 0,15A => s n ti a l 1 n Do ta khng ch mc thm n song song m cc nhnh song song phi ni vi cc cng tc iu chnh cng dng in trong mch 4.Bi ton v nhit in tr: Bi 6: Trn hnh v l s biu din s ph thuc ca in tr ca 1 linh kin vo nhit . Khi nung nng linh kin n 100oC th in tr tng t R1 = 50 n R2 = 100. Cn khi gim nhit linh kin t 100oC th in tr n gim t ngt t R1 =100 n R2= 50. t vo hai u linh kin ni trn 1 hiu in th khng i U =80V th trong mch hnh thnh 1 dng dao ng. Hy tnh chu k T, Imax, Imin. Cho bit: Nhit mi trng l t0 = 20oC, cng sut to nhit trn b mt linh kin t l thun vi chnh lch nhit gia linh kin v mi trng vi h s t l l k = 1.2WK. Nhit dung ca linh kin C = 3JK. Gii: Xt ti nhit t ti thi im trong qu trnh dao ng khi linh kin nng dn 2013( ) 128 1, 2( 20) 3 (152 1, 2 ) 3152 1, 2U dtd k t t d Cdt d t d dt t d dt dR t = + = + = = Tch phn 2 v ta c 110012, 5ln (152 1, 2 )tt = Khi linh kin t t = 100oC th 2023( ) 64 1, 2( 20) 3 (88 1, 2 ) 31, 2 88U dtd k t t d Cdt d t d dt t d dt dR t = + = + = = Tch phn 2 v ta c 12 1002, 5ln (88 1, 2 )tt = t1 l nhit thp nht trong qu trnh dao ng. Imax = 1801, 650UAR= =v Imin = 1800, 8100UAR= = 5.Bi ton v tm cng sut cc i trong mch phi tuyn Bi 7: Mt mng lng cc c c trng vn ampe c m t bi cng thc 210 U I = , trong cng dng in c o bng ampe cn hiu in th c o bng vn. C 2 mng lng cc nh vy c mc ni tip vi nhau v mc vo mt ngun in l tng vi sut din ng E =10V. Mc thm mt in tr song song vi mt trong hai mng lng cc. in tr ny l bao nhiu cng sut nhit gii phng trn n t cc i? Gii Gi U l hiu in th trn in tr mc song song vi mt mng lng cc. Khi hiu in th trn mng lng cc kia s l E- U.Dng qua in tr:10 10RU UI= E Vy cng sut to nhit trn R l: ( )10 10R RU UU I U= = EP (1) o hm 2 v (1) theo U ta c 32 2d U U UUdU= = P EP E (2) Pcc i th (2) phi bng khng

2 2302 218 21 4 0U U UUU U = + =EEE E(3) Gii (3) ta c 2 nghim:

(21 153)2.436(21 153)9.336U VU V= += EE

V hiu in th trn on song song phi nh hn mt na hiu in th 2 cc ngun nn ta chn2.4 U V Vy R 6.27 6.Bi ton Diode nn dng Bi 8:Trong mch in nh hnh sau. l it l tng, t in c in dung l C, hai cun dy L1 v L2 c t cm ln lt l L1 = L, L2 = 2L. in tr ca cc cun dy v cc dy ni khng ng k. Lc u kho K1 v K2 u m. 1.u tin ng kho K1, khi dng qua cun dy L1 c gi tr I1 th ng thi m kho K1 v ng kho K2. Chn thi im ny lm mc tnh thi gian t. a)Tnh chu k ca dao ng in t trong mch. b)Lp biu thc ca cng dng in qua mi cun dy theo t. 2.Sau , vo thi im dng chy qua cun dy L1 bng khng v hiu in th UAB c gi tr m th m kho K2. a)M t hin tng in t xy ra trong mch. Lp biu thc v v phc th biu din cng dng in qua cun dy L1 theo thi gian tnh t lc m kho K2. (trch thi chn HSG quc gia mn vt l, nm 2002-2003) Gii: 1. a) Khi ng thi m K1 v ng K2 Qui c chiu dng cc dng nh hnh Gi q l in tch bn t ni vi B.Lp h phng trnh: iC = i1 + i2(1) Li1 2Li2 = 0 (2) Li1 = qC (3) i = - q (4) o hm 2 v (1), (2), (3) theo thi gian, ta c: iC = i1 +i2 (1) L i1 - 2Li2 = 0 (2) Li1 = - iC/ C(3) i1 = 23 iC ; iC = 32LCiC

Phng trnh trn chng t iC dao ng iu ho vi tn s 32LC =b)iC = I0sin( ) t + (5) T (2) suy ra : (Li1 2Li2) = const i1 2i2 = const Ti t = 0 th i1 = I1 i1 2i2 = I1(6) i1 + i2 = iC = I0sin( ) t +Gii h ta c:

1 111 12012 3sin3 3 23sin3 3 22' cos( )3CABI Ii tLCI Ii tLCI qu Li LC tC = += + = = = + Ti thi im t = 0 : i1 = I1 ; i2 = 0 ; uAB = 0. Gii h ta c : I0C = I1 ; 2=p s:1 111 122 3cos3 3 23cos3 3 2I Ii tLCI Ii tLC= += + 2. thi im t1, m K2 : a) M t : V i1 =0, VA < VB nn khng c dng i qua , ch c dao ng trong mch L2C vi T = 2 2LC . n thi im t2 tip theo th uAB bng 0 v i sang du dng. T thi im ny c dng i qua c hai cun dy, trong mch c dao ng in t vi T = 223LC . Ta s chng minh c t thi im t2 lun c dng i qua it. b) V i1 = 0 v uAB < 0 nn t (6) suy ra i2 = 0.5I1 ; Ngy sau thi im t1 = 0 khng c dng i qua it. Trong mch dao ng L2C c dao ng vi chu k T1 = 2 2LC vi nng lng 212IL . iu kin ban ui2 = 0.5I1. Bin dng in l I0 : 2 20 1 1 10 22 sin( )2 2 2 2 2I I I I tL L I iLC = = = +122Li' 2 cos( )2 2ABI tu LLC LC = = + . Ti t1 = 0, i2 = 0.5I1 , uAB < 0 nn / 4 = . Tip theo, khi 4 2 2tLC + =hay khi 224LCt t= =th 122ii = , uAB = 0 v i du. Vy t sau thi im t2 c dng qua it. Tng t nh cu 1, ta c: 322 223LCT LC== = 1 2 01 2 C 0 20 01 20 12 21 0 2i 2iIiiiI' sin{ ( ) }2 2 'sin{ ( ) }3 3' 2sin{ ( ) }3 3' 2 ' cos{ ( ) }CCCAB Ct t tI Ii t t tI Ii t t tqu Li I LC t t tC =+ = = + = + += + + = = = + Vi iu kin u l t = 0 th i1 = 0 ; u = 0 2 =; 0 0I' IC=1 11 22 2 3 3{1 cos ( )} {1 cos( )}3 3 2 4I Ii t t t tLC = = 0 nn sau thi im t2 lun c dng qua it. Kt lun : Vi 0 < t < 24LC th i1 = 0 Vi 24LCtth 112 3 3{1 cos( )}3 2 4Ii tLC = Ta c th nh sau 7.Bi ton mch phi tuyn xoay chiu Bi 9 Mch R L C phi tuyn Cho u = U0 cost , R nt L nt C Tnh i(t) = ?, cng sut tiu th ca mch bit a)L v C l 2 phn t tuyn tnh v R l mt varistor c t trng vn ampe l i = ku2 b)R v C l 2 phn t tuyn tnh, L l cun cm c li st nn do hin tng t tr, t cm ca n bin i mnh, ta coi gn ngi = a2 + b c)R v L l 2 phn t tuyn tnh, C l t c in mi bin i theo in tr, in dung n bit i mnh, ta coi gn ng u = cq2 + d Gii Bi ny chng ta c th s dng s phc gii quyt bi ton nya)u = U0 cost => u*= U0 li c ZLi* + ZCi* + *ik=u* = U0 T y gii ra phng trnh bc 2 gii ra i* = I0 ( cos+jsin) => i = I0 cos (t+) Cu b v c cng tng t, ta c i vi cun cm th: 'i add d dibu idt di dt di = = = i vi t th'u dddq dq duci udt du dt du= = = nn biton tr nn tong i phc tp hn