Din n vt l Ninh Hi 1 L THUYT GII TON IN XOAY CHIU BNG S PHC 1. S tng quan gia in xoay chiu v s phc -Xt on mch R, L, C mc ni tip,u = Uo cos(t + )(V ) . Ta c gin vect nh sau: + Trc honh biu din R UL + Phn dng ca trc tung biu din L + Phn m ca trc tung biu din C UC U UL Uc UR +Vect u c ln l U0 v to vi trc honh mt gc l UC -Xt mt s phc bt k: x = a + bi. S phc ny c ghi di dng lng gic l V c biu din nh hnh bn: +Trc honh biu din phn thc (s a) +Trc tung biu din phn o (s b) x = X o Z X0 b b x +Vect x c ln l Xo v to vi trc honh mt gc l a -Nh vy ta c th xem R nh l mt s phc ch c phn thc a (v nm trn trc honh) L v C l s phc ch c phn o b (v nm trn trc tung). Nhng chng khc nhau l L nm phn dng nn c biu din l bi. C nm phn m nn c biu din l bi. u hoc i c xem nh l mt s phc x v c vit di dng lng gic VD: X o Z . Cc i lng trong in xoay chiu Biu din di dng s phc R=50 50 ZL=100 100i ZC=150 -150i u = 100 cos(100tt + t )(V ) 6 100Z t 6 i = 2 2cos(100tt t )( A) 4 2 2Z( t ) 4 2. Cng thc tnh ton c bn: Khi gii cc bi tp in xoay chiu bng s phc, cc bn xem on mch ny nhl on mch mt chiu vi cc phn t R, L, C mc ni tip. Chng ta ch s dng mt nh lut duy nht gii. l nh lut Ohm trong mch in mt chiu. nh lut ny chng ta hc nm lp 9, qu quen thuc ng khng no: I = U R hayU = I.RhayR = U I Din n vt l Ninh Hi 2 Trong R khng ch ring mi in tr, m ch chung tt c nhng vt c tr khng (nhng ci c n v l ^^. VD: R, ZL, ZC...). TrongchngtrnhhcPhthng,chngtachhconmchxoay chiumcnitip, cho nn trong on mch mt chiu gm R1, R2,..., Rn ni tip ta c: R = R1 + R2 + ... +Rn U = U1 + U2 + ... + Un I = I1 = I2= ... =In 3.Thao tc trn my: a) Nhng thao tc c bn thc hin tnh ton s phc trn my, chng ta phi vo mode CMPLX bng cch n [Mode] [2]. Trn mn hnh hin CMPLX. Trong mode CMPLX, nhp k hiu i ta nhn phm ENG nhp k hiu ngn cchZ , ta nhn [SHIFT] [(-)] Nh ta bit, s phc c hai cch ghi, l i s v lng gic. - Khi my tnh hin th dng i s (a+bi), th chng ta s bit c phn thc v phn o ca s phc. Din n vt l Ninh Hi 3 - Khi my tnh hin th dng lng gic ( X o Z ), th chng ta s bit c di (modul) v gc (argumen) ca s phc. Mc nh, my tnh s hin th kt qu di dng i s. chuyn sang dng lng gic, ta nhn: [SHIFT] [2], mn hnh hin th nh sau: chn [3], nhn [=]. Kt qu s c chuyn sang dng lng gic b) Nhng li thng gp -Khi ci t my ch n v o gc no th phi nhp n v o gc y. Trong mode (mn hnh hin ch D), cc bn phi nhp n v l . VD: 450, 600 Trong mode rad (mn hnh hin ch R), cc bn phi nhp n v l radian. VD: Cch ci t my: Nhn [SHIFT] [Mode] t , t 43 Nhn [3] ci t my n v l . Nhn [4] ci t my n v l radian. -TrnmyFx 570 ES, bm nhanh, ccbn thng n du chia thay chodu phn s. Chnh v vy trong qu trnh bm my thng xut hin nhng li nh sau: 1 Z t 24 1 Z t 24 Khc1 2Z t 4 Khc 1 Zt 4 2 Din n vt l Ninh Hi 4 3 2iKhc3 (2i) Cch khc phc: s dng du ngoc II. CC DNG BI TP: (nhn [Mode] [2] chuyn sang mode s phc, ci t my n v gc radian) 1. Tm biu thc hiu in th, cng dng in : Bi 1:Mtonmch in gm in tr R = 50Omcni tipvi cun thun cm L= 0,5/t (H). tvohai u on mchmthiu in th xoay chiu u = 1002 sin(100tt-t/4)(V). Biu thc ca cng dng in qua on mch l: A. i= 2sin(100tt- t/2) (A).B. i= 2 2 sin(100tt - t/4)(A). C. i= 2 2 sin100tt (A).D. i= 2sin100tt (A). Gii: Gi : U Ta dng nh lut OhmI= gii. R Cch lm: -Ta c:R=50O ZL=50O. -Suy raI= U . R + Z L -Nhn [SHIFT] [2] [3] chuyn sang dng lng gic: p n : A Din n vt l Ninh Hi 5 Bi 2: Khi t hiu in th khng i 30V vohai u on mch gm in trthun mcni tipvi cun cm thun c t cm 1 4t (H) th dng in trong mch l dng in 1 chiu c cng 1A. Nu t vo hai u on mch ny in p cng dng in trong mch l: A.i = 5 2 cos(120tt t ) (A) 4 C.i = 5 2 cos(120tt + t ) (A) 4 u = 150 2 cos120tt (V) th biu thc B.i = 5 cos(120tt + t ) (A) 4 D.i = 5 cos(120tt t ) (A) 4 Gi : Gii: (Trch thi tuyn sinh i hc 2009) U Tnh R, sau dng cng thcI= tnh. R Cch lm: -Khi t hiu in th khng i (hiu in th 1 chiu) th on mch ch cn c R. -R = U I = 30 = 30O 1 -Ta c R=30O. ZL=30O. -Suy raI= U . R + Z L -Chuyn sang dng lng gic: Din n vt l Ninh Hi 6 p n : D Bi3:HiuinthxoaychiutvohaiuonmchR,L,Cnitipcbiuthc: u = 220 2 cos(100tt th gia hai bn t l: t )(V ) . Bit R = 100, L = 0,318H v C = 15,9 F. Biu thc hiu in 12 A.u = 440 cos(100tt t ) V 3 B.u = 400 cos(100tt t ) V 4 C.u = 440 cos(100tt + t ) V 6 Gi : Tnh I sau dng cng thc UC = I . ZC Cch lm: -Ta c:R=100O ZL=100O. ZC=200O. U Gii: D.u = 440 cos(100tt + t ) V 12 -I=. R + Z L + ZC Nhp vo my: Nhn [=] : Din n vt l Ninh Hi 7 -C I ri, ta suy ra UC bng cng thc: UC = I . ZC -Chuyn sang dng lng gic: p n : A 2. Tm cc thnh phn (Bi ton hp en) Ta chia R, L, C thnh 2 nhm: + Nhm 1: in tr (R). + Nhm 2: Cun cm v t in (L v C). Ly u chia i, hin th di dng i s th kt qu s ri vo nhng dng nh sau: -a + bi : on mch c c nhm 1 v nhm 2 ( Trong a l gi tr ca in tr, b l tng tr ca nhm 2.Nunhm 2ch c1phnt th bltr khng caphn t ) -a : on mch ch c in tr. -bi : on mch ch c nhm 2. Din n vt l Ninh Hi 8 Bi 1: t hiu in th xoay chiuu = 120 2 cos(100tt + t )(V )vo hai u ca mt cun dy khng 6 thun cm thy dng in trongmch c biu thci = 2 cos(100tt bng: t )( A) . in tr thun r c gi tr 12 A. 60B.85C.100D. 120 Gii: -Chuyn u, i sang s phc: u :120 2Z t 6 i : 2Z t 12 -Ly u chia i: -Suy ra r = 60. p n : A Bi2:inpgiahaiucundyvcngdnginquacundyl: u = 80 cos(100tt + t )(V ) ;i = 8 2 cos(100tt t )( A) .in tr thun Rv tcm Lcacundy 8 l: A. 40 v 0,368 HB. 40 v 0,127 H C.40 2 v 0,127 HD.40 2 v 0,048 H Gii: -Chuyn u, i sang s phc: u : 80Z t 8 i :2Z t 8 -Ly u chia i: Din n vt l Ninh Hi 9 0 -Suy raR = 40. ZL = 40 -C ZL = 40 , suy ra L = 0,127H. p n : B Bi 3: Cho on mch xoay chiu nh hnh v. R0, L ChoR=50,C = 2 .104 F , t uAM C = 80 cos(100tt )(V ) ; R A M B uMB = 200 2 cos(100tt + t )(V ) . Gi tr ca R 2 v L l: Gi : A. 250 v 0,8 HB. 250 v 0,56 H C. 176,8 v 0,56 HD. 176,8 v 0, 8 H Gii: Tnh I, sau ly UMB chia cho I. Cch lm: -Ta c:R = 50O ZC = 50O. -Chuyn uAM, uMB sang s phc: uAM: 80 uMB : 200 2Z t 2 Din n vt l Ninh Hi 10 3 \.\. \ | U -Tnh I :I= AM = R + ZC 80 50 50i -Ly UMB chia I: U MB I 200 2Z t = 2 4 + 4 i 55 -Suy raR0 = 176,8. ZL = 176,8 => L = 0,56 H p n : C 3. Cng cc u Nh ta bit, trong on mch mt chiu, mun bit hiu in th on mch th ta ch cn cng cc hiu in th thnh phn c trong mch li vi nhau. Bi 1: onmch AC c in tr thun, cun dy thun cm v t inmc ni tip. B lmt im trn t AC vi uAB = sin100tt (V) v uBC =3sin(100tt - 2) (V). Tm biu thc hiu in th uAC. A.uAC = 2 2 sin(100tt) V B.u AC =2 sin |100tt + t | V . |t | |t | C.u AC = 2 sin

100tt + 3 | V Gi : Cng cc hiu in th thnh phn li vi nhau. Cch lm: D.u AC = 2 sin

100tt 3 | V Gii: Din n vt l Ninh Hi 11 -Chuyn uAB, uBC sang s phc: uAB :1 uBC : 3Z t 2 -Tnh UAC : u= u+ u = 1 + 3Z t ACABBC 2 -Chuyn sang dng lng gic: -Suy ra p n : D u AC = 2 sin(100tt t )(V ) 3 (Bi ny cng c th gii nhanh bng phng php gin vect) III. BI TP T LUYN Bi 1: Cho on mch nh hnh v: R = 100, L = 0,138H v C = 15,9 F uMB = 220 cos(100tt t )(V ) . Biu thc cng 3 A L C R M B dng in trong mch l: A.i =2 cos(100tt t ) (A) 6 B.i = 2 cos(100tt + t ) (A) 6 C.i = 2 cos(100tt t ) (A) 2 D.i =2 cos(100tt + t ) (A) 2 Din n vt l Ninh Hi 12 3 \.\. \ | Bi2:Chomchinxoay chiugmmtcundycintrthun 103 r = 20 O ,tcm 3 L =1 ( H ) 5t vt in cin dungC = ( F ) 4t mcni tip.Bitbiuthc in phai u cun dy lud= 100 2 cos100tt (V). in p hai u mch l: A.u =100 2 cos(100tt 2t ) (V) 3 C.u =100 cos(100tt + t) (V) B.u =100 cos(100tt + 2t ) (V) 3 D.u =100 cos(100tt t) (V) Bi 3:Mtonmch xoay chiugm Rmcni tipvimt tronghai phn t Choc cun dy thun cm L. in p gia hai u mch in v cng dng in qua mch c biu thc: u = 100 2 cos(100tt )(V ) ,i = 2 cos(100tt t )( A) . on mch gm 4 A. R v C c R = 30, ZC = 30B. R v L c R = 40, ZL = 30 C. R v C c R = 50, ZC = 50D. R v L c R = 50, ZL = 50 Bi 4: on mch AC cin tr thun, cun dy thun cm v t inmcni tip.B lmt t im trn AC vi uAB = sin100tt (V) v uBC =3sin(100tt - 2) (V). Tm biu thc hiu in th uAC. A.uAC = 2 2 sin(100tt) V B.u AC =2 sin |100tt + t | V . |t | |t | C.u AC = 2 sin

100tt + 3 | V D.u AC = 2 sin

100tt 3 | V Bi5:tmthiuin thu=2002 .sin(100 tt+ t/6)(V)vohai ucamtcun dy thun cm c t cm L = 2/t (H). Biu thc ca cng dng in chy trong cun dy l A. i=2sin (100tt + 2t/3 ) (A).B. i= 2 sin ( 100tt + t/3 ) (A). C. i=2sin (100tt - t/3 )(A).D. i=2sin (100tt - 2t/3 ) (A). Bi 6: Cho mt on mch in xoay chiu gm in tr thun R v t in c in dung C mc ni tip. in pt vo hai u onmch lu = 100 2 cos100tt (V ) ,b quain tr dy ni.Bit cng dng in trong mch c gi tr hiu dng l3 A v lch pha Gi tr ca R v C l: t so vi in p hai u mch. 3 A.R = 50 OvC = 3 104 F t B.R = 50 OvC = 3 103 F 5t C.R = 50 3OvC = 104 F t D.R = 50 3OvC = 103 F 5t Din n vt l Ninh Hi 13 Bi 7: Cho mt on mch in xoay chiu gm hai trong trong phn t: in tr thun R, cun dy thuncmL,tCmcni tip.Hiuinthgiahaiumchvcngdngin trong mch c biu thc: u=220 2sin (100tt- t/3 )(V) i=2 2sin (100tt+ t/6) (A) Hai phn t l hai phn t no? A. R v L.B. R v C C. L v C.D. R v L hoc L v C. Bi 8 : tmt hiu in thxoay chiu u = 60sin100tt (V) vohai uon mch gm cun thun cmL = 1/tHv t C = 50/tFmcni tip.Biu thc ng ca cng dng in chy trong mch l A. i = 0,2sin(100tt + t/2) (A).B. i = 0,2sin(100tt - t/2) (A). C. i = 0,6sin(100t + /2) (A).D. i = 0,6sin(100tt - t/2) (A). Bi9 :Choonmchnhhnhv,R=50,L=1/(H),C=2.10-4/(F),bit u MB = 1002 sin(100t 3)(V ) . Tm biu thc hiu in th uAB? A. 100 C. 100 2 sin(100t 2 sin(100t+ 6)(V ) 4)(V ) B. 100 D. 100 2 sin(100t+ 2 sin(100t+ 6)(V ) 3)(V ) Bi 10: Mch in ni tip R, L, C trong cun dy thun cm (ZL < ZC). t vo hai u on mch mt in p xoay chiuu = 200 2 cos(100tt + t )V. Khi R = 50 cng sut mch t gi 4 tr cc i. Biu thc dng in qua mch lc : A.i = 4 cos(100tt + t ) (A) 2 C.i = 4 2 cos(100tt + t ) (A) 4 B.i = 4 cos(100tt + t ) (A) 4 D.i = 4 2 cos(100tt) (A) Gi : Khi R = 50 cng sut mch t gi tr cc i, suy ra Cc dng ton cc tr trong dng in xoay chiu). Mt khc ZC > ZL nn trong s phc ta c: ZL + ZC = -50i 200 2Z t Z L ZC = R = 50 (Xem thm chuyn Suy ra:i = u = 4 = 4Z t R + ZL + ZC 50 50i2