VIẾT BIỂU THỨC của u HOẶC i TRONG ĐXC

VIT BIU THC CA u HOC i TRONG XC I.ON MCH CH Cể PHN T: UR R a) on mch ch cú in tr thun: uR cựng pha vi i : I = C B A b) on mch ch cú t in C: UC ZC - L ụm: I = ; vi ZC = C uC tr pha so vi i gúc l dung khỏng ca t in u = U cos t -t in ỏp vo hai u mt t in thỡ cng dũng in qua nú cú giỏ tr hiu dng l I Ti thi im t, in ỏp hai u t in l u v cng dũng in qua nú l i H thc liờn h gia cỏc i lng l : i2 u2 i2 u2 + = + =1 I 02 U 02C I 2U C2 u i2 + =2 U I2 Ta cú: i = I cos(t + ) -Cng dũng in tc thi qua t: L A B c) on mch ch cú cun dõy thun cm L: uL sm pha hn i gúc UL ZL - L ụm: I = ; vi ZL = L l cm khỏng ca cun dõy u = U cos t -t in ỏp vo hai u mt cun cm thun thỡ cng dũng in qua nú cú giỏ tr hiu dng l I Ti thi im t, in ỏp hai u cun cm thun l u v cng dũng in qua nú l i H thc liờn h gia cỏc i lng l : i2 u2 i2 u2 + =1 + =1 I02 U 0L 2I 2U L2 Ta cú: u i2 + =2 U I2 C A B R L N M i = I cos(t ) -Cng dũng in tc thi qua cun dõy: d) on mch cú R, L, C khụng phõn nhỏnh: u = U cos(t + u ) +t in ỏp vo hai u mch + lch pha gia u v i xỏc nh theo biu thc: tan = L Z L ZC R C = u i R = ; Vi U Z + Cng hiu dng xỏc nh theo nh lut ễm: I = R + (Z L - Z C ) Vi Z = l tng tr ca on mch i = I cos(t + i ) = I cos(t + u ) Cng dũng in tc thi qua mch: + Cng hng in on mch RLC: Khi ZL = ZC hay = U R LC thỡ U R Imax = , Pmax = , u cựng pha vi i ( = 0) Khi ZL > ZC thỡ u nhanh pha hn i (on mch cú tớnh cm khỏng) Khi ZL < ZC thỡ u tr pha hn i (on mch cú tớnh dung khỏng) R tiờu th nng lng di dng to nhit, ZL v ZC khụng tiờu th nng lng in C A B R L,r N M e) on mch cú R, L,r, C khụng phõn nhỏnh: u = U cos(t + u ) +t in ỏp vo hai u mch + lch pha gia uAB v i xỏc nh theo biu thc: tan = Z L ZC R+r C R+r L = = u i Vi U Z + Cng hiu dng xỏc nh theo nh lut ễm: I = (R+r)2 + (Z L - Z C )2 Vi Z = l tng tr ca on mch i = I cos(t + i ) = I cos(t + u ) Cng dũng in tc thi qua mch: + Cỏch nhn bit cun dõy cú in tr thun r R + (Z L Z C ) -Xột ton mch, nu: Z U R2 + (U L U C ) ;U R Z hoc P I2R hoc cos thỡ cun dõy cú in tr thun r -Xột cun dõy, nu: Ud UL hoc Zd ZL hoc Pd hoc cosd hoc d thỡ cun dõy cú in tr thun r II PHNG PHP 1: (PHNG PHP TRUYN THNG): a) Mch in ch cha mt phn t ( hoc R, hoc L, hoc C) - Mch in ch cú in tr thun: u v i cựng pha: = u - i = Hay u = i i = I 2cos( t+i ) u = U R 2cos( t+i ) I= UR R + Ta cú: thỡ ; vi +Vớ d 1: in ỏp gia hai u mt on mch in xoay chiu ch cú in tr thun R= 100 cú biu 200 cos(100 t + thc u= 2 cos(100 t A i= )(V ) )( A) Biu thc ca cng dũng in mch l : 2 cos(100 t + C.i= 2 cos(100 t + )( A) 2cos(100 t )( A) )( A) B i= D.i= +Gii :Tớnh I0 hoc I = U /.R =200/100 =2A; i cựng pha vi u hai u R, nờn ta cú: i = u = /4 2 cos(100 t + Suy ra: i = -Mch in ch cú t in: ) ( A) => Chn C uC tr pha so vi i gúc -> = u - i =- Hay u = i - u = U 2cos(t- i = I 2cos( t) +Nu cho i = I 2cos(t+ u = U 2cos( t) +Nu cho ; i = u + ) I= ) thỡ vit: UC zC v L ễm: ZC = vi C thỡ vit: +Vớ d 2: in ỏp gia hai u mt on mch in xoay chiu ch cú t cú in dung C= 104 (F ) cú 200 cos(100 t )(V ) biu thc u= Biu thc ca cng dũng in mch l : 2 cos(100t + A i= 2 cos(100 t ) ( A) 2 cos(100 t + C.i= )( A) 2 cos(100t B i= )( A) ) ( A) D.i= ZC = Gii : Tớnh .C =100, Tớnh Io hoc I = U /.ZL =200/100 =2A; 2 cos(100 t + i sm pha gúc /2 so vi u hai u t in; Suy ra: i = -Mch in ch cú cun cm thun: uL sm pha hn i gúc -> = u - i = => Chn C Hay u = i + u = U 2cos( t+ i = I 2cos( t) +Nu cho thỡ vit: u = U 2cos( t) i = I 2cos( t- )( A) ; i = u - ) ) I= v L ễm: UL zL ZL = L vi Nu cho thỡ vit: Vớ d 3: Hiu in th gia hai u mt on mch in xoay chiu ch cú cun cm cú t cm L= (H ) 200 cos(100t + cú biu thc u= 2 cos(100t + A i= ) ( A) ) (V ) Biu thc cng dũng in mch l : 2 cos(100t C.i= ) ( A) 2 cos(100t + B i= ) ( A) cos(100t D.i= ZL = L Gii : Tớnh ) ( A) = 100.1/ =100, Tớnh I0 hoc I = U /.ZL =200/100 =2A; i tr pha gúc /2 so vi u hai u cun cm thun, nờn ta cú: 2 cos(100t ) ( A) =- Suy ra: i = => Chn C Trc nghim dng: Cõu 1: in ỏp gia hai u mt on mch in xoay chiu ch cú in tr thun R= 200 cú biu 200 cos(100 t + )(V ) thc u= Biu thc ca cng dũng in mch l : 2 cos(100 t ) ( A) cos(100 t ) ( A) A i= cos(100 t + ) ( A) C.i= 2cos(100 t )( A) B i= D.i= Cõu 2: in ỏp gia hai u mt on mch in xoay chiu ch cú in tr thun R= 100 cú biu thc 200 cos(100 t + )(V ) u= Biu thc ca cng dũng in mch l : 2 cos(100 t A i= 2 cos(100 t + )( A) )( A) B i= 2 cos(100 t + C.i= 2cos(100 t )( A) )( A) D.i= Cõu 3: in ỏp gia hai u mt on mch in xoay chiu ch cú t cú in dung C= 200 cos(100 t )(V ) thc u= Biu thc ca cng dũng in mch l : 2 cos(100t + A i= 2 cos(100 t B i= ) ( A) )( A) 2 cos(100 t + C.i= cos(100t D.i= )( A) ) ( A) 104 (F ) cú biu Cõu 4: Cho in ỏp hai u t C l u = 100cos(100t- /2 )(V) Vit biu thc dũng in qua mch, bit C= 10 (F ) B i = 1cos(100t + )(A) D i = 1cos(100t /2)(A) A i = cos(100t) (A) C i = cos(100t + /2)(A) u = 200 2cos(100 t) Cõu 5: t in ỏp (V) vo hai u on mch ch cú t ờn cú C = 15,9àF = (Ly 0,318) thỡ cng dũng in qua mch l: i = 2cos(100 t+ ) A C (A) i = 2 cos100 t B (A) Cõu Xỏc nh ỏp ỏn ỳng D i = cos100 t i = cos100 t + Cng dũng in qua t in i = 4cos100 t (A) in dung l 31,8 in l: F.Hiu in th t hai u t B uc = 400 cos(100 t + C uc = 400 cos(100 t ) (V) D uc = 400 cos(100 t Cõu 7: Cho in ỏp gia hai u on mch xoay chiu ch cú cun thun cm 100 cos( 100 t ) (V) ) (V) L = (H ) )(V ) Biu thc cng dũng in mch l : cos( 100 t A i= cos( 100 t + B i= (A) A- uc = 400cos(100 t ) (V) (A) )( A ) )( A ) cos( 100 t C.i= cos(100t D.i= )( A ) ) ( A) l : u = 200 2cos(100 t+ ) Cõu 8: t in ỏp (V) vo hai u on mch ch cú cun thun cm thỡ cng dũng in qua mch l: L= A C (H ) i = 2 cos100 t + i = 2 cos100 t (A) B (A) D i = cos100 t (A) i = cos100 t + (A) u = 200 2cos(100 t) Cõu 9: t in ỏp (V) vo hai u on mch ch cú cun thun cm L= = 0,318(H) (Ly A 0,318) thỡ cng dũng in qua mch l: (A) B i = cos100 t i = 2 cos100 t + C i = 2 cos100 t (A) D (A) i = cos100 t + Cõu 10: t mt hiu in th xoay chiu vo hai u cun dõy ch cú t cm L= cos(100t+ B u=150 cos(100t- )(V) D u=100cos(100t+ II.MCH IN KHễNG PHN NHNH (R L C) a Phng phỏp truyn thng): thỡ cng )(A) Biu thc no sau õy l hiu in th )(V) 2 C.u=150 cos(100t+ A u=150cos(100t+ H dũng in qua cun dõy cú biu thc i=3 hai u on mch: (A) )(V) )(V) -Phng phỏp gii: Tỡm Z, I ( hoc I0 )v ZC = ZL = L Bc 1: Tớnh tng tr Z: Tớnh 1 = C fC ; Z = R + (Z L ZC )2 v U I= Z Uo Z Bc 2: nh lut ễm : U v I liờn h vi bi ; Io = tan = Z L ZC R Bc 3: Tớnh lch pha gia u hai u mch v i: Bc 4: Vit biu thc u hoc i ; u = U 2cos( t+ ) i = I 2cos( t) -Nu cho trc: Hay i = Iocost ; thỡ biu thc ca u l thỡ u = Uocos(t + ) i = I 2cos( t- ) u = U 2cos( t) -Nu cho trc: thỡ biu thc ca i l: Hay u = Uocost thỡ i = Iocos(t - ) * Khi: (u 0; i ) Ta cú : = u - i => u = i + ; i = u - u = U 2cos(t+i +) i = I 2cos(t+i ) -Nu cho trc Hay i = Iocos(t + i) thỡ biu thc ca u l: thỡ u = Uocos(t + i + ) i = I 2cos(t+u -) u = U 2cos(t+u ) -Nu cho trc Hay u = Uocos(t +u) thỡ biu thc ca i l: thỡ i = Iocos(t +u - ) Lu ý: Vi Mch in khụng phõn nhỏnh cú cun dõy khụng cm thun (R ,L,r, C) thỡ: tan = Z = ( R + r )2 + ( Z L Z C )2 Tng tr : v Z L ZC R+r ; Vớ d 1: Mch in xoay chiu gm mt in tr thun R = 50, mt cun thun cm cú h s t cm L = (H ) C= v mt t in cú in dung 2.104 i = 5cos100 t ( A ) cú dng Gii : mc ni tip Bit rng dũng in qua mch Vit biu thc in ỏp tc thi gia hai u mch in Z L = L = 100 Bc 1: Cm khỏng: (F ) = 100 ZC = ; Dung khỏng: = C = 50 2.10 100 Z = R + ( Z L Z C ) = 502 + ( 100 50 ) = 50 2 Tng tr: Bc 2: nh lut ễm : Vi Uo= IoZ = 5.50 = 250 V; tan = Bc 3: Tớnh lch pha gia u hai u mch v i: (rad) Bc 4: Biu thc in ỏp tc thi gia hai u mch in: Z L Z C 100 50 = =1 = R 50 u = 250 cos 100 t + ữ Vớ d 2: Mt mch in xoay chiu RLC khụng phõn nhỏnh cú R = 100 cng dũng in qua mch cú dng: i = 2cos100 mch v hai u mi phn t mch in Hng dn : Z L = L. = -Cm khỏng : 100 = 200 ; C= 104 F ; L= (V) H t (A) Vit biu thc tc thi in ỏp ca hai u ZC = = C ; Dung khỏng : 104 100 = 100 R + ( Z L Z C )2 = 1002 + ( 200 100 )2 = 100 -Tng tr: Z = 100 -HT cc i :U0 = I0.Z = tan = V =200 V Z L Z C 200 100 = = = rad R 100 - lch pha: u = i + = 0+ ;Pha ban u ca HT: U cos(t + u ) = 200 cos(100t + =>Biu thc HT : u = (V) (t + u R ) -HT hai u R :uR = U0Rcos ; Vi : U0R = I0.R = 2.100 = 200 V; (t + u R ) Trong on mch ch cha R : uR cựng pha i: uR = U0Rcos (t + u L ) -HT hai u L :uL = U0Lcos ) = 200cos 100t V Vi : U0L = I0.ZL = 2.200 = 400 V; Trong on mch ch cha L: uL nhanh pha hn cd uL = i + : = 0+ = 2 rad = 4 (100t + (t + u R ) => uL = U0Lcos ) = 400cos V (t + uC ) -HT hai u C :uC = U0Ccos Vi : U0C = I0.ZC = 2.100 = 200V; Trong on mch ch cha C : uC chm pha hn cd : (100t (t + uC ) => uC = U0Ccos uL = i =0 = 2 rad ) = 200cos V Vớ d 3: Mch in xoay chiu gm mt in tr thun R = 40, mt cun thun cm cú h s t cm L= ,8 (H ) C= v mt t in cú in dung 104 F i = 3cos(100 t )( A) mc ni tip Bit rng dũng in qua mch cú dng a Tớnh cm khỏng ca cun cm, dung khỏng ca t in v tng tr ton mch b Vit biu thc in ỏp tc thi gia hai u in tr, gia hai u cun cm, gia hai u t in, gia hai u mch in Hng dn: Z L = L = 100 a Cm khỏng: 0,8 = 80 ZC = = C ; Dung khỏng: = 50 2.10 100 Z = R + ( Z L ZC ) = 402 + ( 80 50 ) = 50 Tng tr: b Vỡ uR cựng pha vi i nờn : u R = U oR cos100 t ; Vi UoR = IoR = 3.40 = 120V Vỡ uL nhanh pha hn i gúc nờn: Vi UoL = IoZL = 3.80 = 240V; Vỡ uC chm pha hn i gúc nờn: Vy u = 120cos100 t (V) uL = U oL cos 100 t + ữ Vy u L = 240cos 100 t + ữ uC = U oC cos 100 t ữ (V) b Biu thc cng dũng in: tan d = Z C 173 = = d = R 100 - Khi K úng: lch pha : id = u d = d = rad Pha ban u ca dũng in: Vy id = 0,25 cos 100 t + ữ tan m = - Khi K m: (A) Z L Z C 346 173 = = m = R 100 lch pha: im = u m = m = Pha ban u ca dũng in: R C L N M A B Vy im = 0,25 cos 100 t ữ (A) Vớ d 8: Cho mch in nh hỡnh v : UAN =150V ,UMB =200V lch pha UAM v UMB l / Dũng in tc thi mch l : i=I0 cos 100t (A) , cun dõy thun cm Hóy vit biu thc UAB Hng dn: Ta cú : (1) U AN = U C + U R U AN = U C2 + U R2 = 150V (2) U MB = U L + U R U MB = U + U L R = 200V Vỡ UAN v UMB lch pha / nờn tg1 tg = U L U C =1 U R U R hay U2R = UL.UC (3) T (1),(2),(3) ta cú UL=160V , UC = 90V , U R = 120V ; U AB = U + (U L U C ) = 139V R tg = U L U C = = 0,53rad / s UR 12 vy uAB = 1392 cos(100t +0,53) V , cun dõy thun cm L v t in C =10 - Vớ d 9: Cho mch in khụng phõn nhỏnh gm R = 100 /2 (F) t vo u mch in mt hiu in th u = 100 cos 100 t Bit hiu in th ULC = 50V ,dũng in nhanh pha hn hiu in th.Hóy tớnh L v vit biu thc cng dũng in i mch Hng dn: Ta cú = 100 rad/s ,U = 100V, ZC = = 200 C Hiu in th u in tr thun l: U R = U U LC = 50 3V cng dũng in U I = R = 0,5 A R v Z LC = U LC = 100 I Vỡ dũng in nhanh pha hn hiu in th,m trờn gin Frexnen,dũng in c biờ din trờn trc honh vy hiu in th c biu din di trc honh ngha l Z L< ZC Do ú ZC-ZL =100ZL =ZC -100 =100 suy L= ZL = 0,318 H i = 0,5 2cos(100 t + )( A) lch pha gia u v i : Z ZC tg = L = = R ; vy TRC NGHIM: Cõu 11: Cho mch in xoay chiu cú R=30 cos100 , L= (H), C= t (V), thỡ cng dũng in mch l 10 0.7 (F); in ỏp u mch l u=120 A i = cos 100 t + ữ( A) i = 4cos(100 t )( A) B i = 2cos(100 t )( A) C i = 2cos(100 t + )( A) D 0,3 / Cõu 12:Cho on mch gm R, L, C mc ni tip; R = 10 ;L= u = 100 cos ( 100 t ) hai u on mch mt hiu in th a) Vit biu thc cng dũng in mch (H); C = i = 2cos ( 100 t + / ) (A) i = 5cos ( 100 t / ) (F) t vo (V) i = 2cos ( 100 t / ) A 103 / B C (A) b) Vit biu thc hiu in th hai u mi phn t R; L; C i = 5cos ( 100 t + / ) D (A) u R = 86,5 cos ( 100 t + / ) A (A) u L = 150 cos ( 100 t + / 3) ; uC = 100 cos ( 100 t / 3) ; uR = 86,5 cos ( 100 t / ) u L = 150 cos ( 100 t + / 3) uC = 100 cos ( 100 t / ) B A C ; ; u R = 86,5 cos ( 100 t / ) A ; uC = 100 cos ( 100 t / 3) ; u R = 86,5 cos ( 100 t + / ) D A u L = 150 cos ( 100 t + / 3) ; uC = 100 cos ( 100 t + / 3) Cõu 13: Cho mch xoay chiu cú R, L, C mc ni tip cú R=30 hiu in th u mch l U=100 ú l: Z L = 117,3(), i = A uL = 150 cos ( 100 t + / 3) cos100 ; , C= 104 (F) , L thay i c cho t (V) , u nhanh pha hn i gúc cos(100 t )( A) rad thỡ ZL v i Z L = 100(), i = 2cos(100 t )( A) B Z L = 117,3(), i = cos(100 t + )( A) Z L = 100(), i = 2cos(100 t + )( A) C C Cõu 14: Mt mch gm cun dõy thun cm cú cm khỏng bng 10 C= 10 F mc ni tip vi t in cú in i = 2 cos100 t + )A dung Dũng in qua mch cú biu thc th ca hai u on mch l: u = 80 2co s(100 t ) A u = 120 2co s(100 t ) C u = 80 cos(100 t + (V) B Cõu 15: Mch in xoay chiu gm in tr i= u = 80co s100 t R = 40 i = 2co s(100 t ) D (V) ghộp ni tip vi cun cm L Hiu in th tc UL v in ỏp hiu dng hai u cun cm co s(100 t ) A A ) (V) u = 80 2co s(100 t + (V) thi hai u on mch qua mch l: Biu thc hiu in i= co s(100 t + ) A B )A =40V Biu thc i i = 2co s(100 t + )A C D Cõu 16: Mt on mch in gm in tr R = 50 mc ni tip vi cun thun cm L = 0,5/ (H) t vo hai u on mch mt in ỏp xoay chiu u = 100 dũng in qua on mch l: cos(100t - /4) (V) Biu thc ca cng A i = 2cos(100t - /2) (A) B i = cos(100t - /4) (A) C i = cos100t (A) D i = 2cos100t (A) Cõu 17: Khi t in ỏp khụng i 30V vo hai u on mch gm in tr thun mc ni tip vi cun cm thun cú t cm (H) thỡ dũng in on mch l dũng in mt chiu cú cng A Nu t vo hai u on mch ny in ỏp in on mch l u = 150 cos120t i = cos(120t ) A (A) (V) thỡ biu thc ca cng dũng i = 5cos(120t + ) B (A) i = cos(120t + ) i = 5cos(120t ) C (A) D (A) Cõu 18: Cho on mch xoay chiu LRC mc ni tip hai u AB, L mc vo AM, R mc vo MN, C mc vo NB Biu thc dũng in mch i = I cos 100 u AN = 100 2cos ( 100 t + / 3) t (A) in ỏp trờn on AN cú dng (V) v lch pha 900 so vi in ỏp ca on mch MB Vit biu thc uMB ? uMB = 100 cos 100 t ữ uMB = 100cos ( 100 t ) A B, uMB = uMB = 100cos 100 t ữ 100 cos 100 t + ữ C D Cõu 19: t in ỏp xoay chiu u = Uocos(100t + ) (V) vo hai u mt cun cm thun cú t 2 cm L= (H) thi im in ỏp gia hai u cun cm l 100 cun cm l A Biu thc ca cng dũng in qua cun cm l A i = cos(100t + C i = cos(100t + V thỡ cng dũng in qua ) (A) B i = cos(100t - ) (A) D i = cos(100t - ) (A) ) (A) C= Cõu 20: Xột on mch gm mt in tr hot ng bng 100, mt t in cú in dung v mt cun cm thun cú t cm u = 200cos100 t H mc ni tip Nu t vo hai u mt in ỏp (V) thỡ in ỏp gia hai u in tr hot ng cú biu thc uR = 200cos(100 t ) A 50 àF u R = 200cos(100 t + ) u R = 100 cos(100 t ) (V) B (V) uR = 100 cos(100 t ) C (V) D (V) Cõu 21: Cho on mch in AB khụng phõn nhỏnh gm cun cm thun, t in cú in dung thay i c, mt in tr hot ng 100 Gia A, B cú mt in ỏp xoay chiu n nh u = 110cos(120 t ) 125 àF (V) Cho C thay i Khi C = thỡ in ỏp hiu dng gia hai u cun cú giỏ tr ln nht Biu thc ca in ỏp gia hai u cun cm l uL = 220 cos(120 t + ) A uL = 110 cos(120 t + ) (V) B (V) A M N B C R L u L = 220cos(120 t + ) C (V) Cõu 22: Cho mch in nh hỡnh v: uL = 110 cos(120 t + ) D U AN = 150V ,U MB = 200V i = I sin(100 t )( A) (V) lch pha gia uAN v uMB l Dũng in tc thi mch l , cun dõy thun cm Biu thc ca uAB l u AB = 139 sin(100 t + 0,53)V u AB = 612 sin(100 t + 0,53)V A B u AB = 139sin(100 t + 0,53)V C u AB = 139 sin(100 t + 0,12)V R = 60 D Cõu 23: Cho ba linh kin gm in tr thun , cun cm thun L v t in C Ln lt t in ỏp xoay chiu cú giỏ tr hiu dng U vo hai u on mch ni tip RL hoc RC thỡ biu thc cng i1 = 2cos(100 t )( A) 12 i1 = 2cos(100 t + )( A) 12 dũng in mch ln lt l v t in ỏp trờn vo hai u on mch RLC ni tip thỡ dũng in mch cú biu thc: i = 2cos(100 t + )( A) A Nu i = 2cos(100 t + )( A) B i = 2cos(100 t + )( A) i = 2cos(100 t + )( A) C D Cõu 24: Cho on mch in xoay chiu gụm cun dõy thun cm va tu iờn mc nụi tiờp iờn ap hiờu dung hai õu cuụn cam la 150V, gia hai õu tu iờn la 100V.Dong iờn mach co biờu thc i =I0cos(t + /6)((A) Biờu thc iờn ap hai õu oan mach la u = 50 cos(100t + / 2)V u = 50 cos(100t / 2) A V B u = 50 cos(100t / 3) C u = 50 cos(100t + / 3) V D V Cõu 25: t in ỏp u = 120cos(100t + ) (V) vo hai u mt on mch gm cun cm thun mc ni tip in tr thun R= 30 thỡ in ỏp hiu dng hai u cun cm l 60 V Dũng in tc thi qua on mch l i = 2 cos(100t + A ) 12 i = 2 cos(100t ) C i = cos(100t + ) (A) B (A) D i = 2 cos(100t + ) (A) (A) 4.Trc nghim vit biu thc u hoc i nõng cao Cõu 26 Cho linh kin gm in tr thun R= 60, cun cm thun L v t in C Ln lt t in ỏp xoay chiu cú giỏ tr hiu dng U vo hai u on mch ni tip RL hoc RC thỡ biu thc cng dũng in nch ln lt l i1=cos(100-)(A) v i2=cos(100+)(A) nu t in ỏp trờn vo hai u on mch RLC ni tip thỡ dũng in mch cú biu thc: A 2cos(100t+)(A) B cos(100t+)(A) C 2cos(100t+)(A) D 2cos(100t+)(A) ỡù =- ( 1) I 01 = I 02 ị Z RL = Z RC ị ùớ ùù Z L = Z C ợ HD: Theo u - i1 =1 ( 2) ỹ ùù ( 1) i + i2 = ( 3) ýị u = u - i2 = ùùỵ Mt khỏc ( 2) , ( )ị = Z ị L 3= R ị Z60 ) L =3 ( T ị U = I 01Z RL = 120 ( V ) U0 R RLC nt đ cng hng: i= Khi u cos(100t+ )= 2cos(100t+)(A) Cõu 27: t vo hai u mch in xoay chiu gm mt cun dõy v mt t in mc ni tip mt in 100 cos(100 t + )(V ) ỏp xoay chiu n nh cú biu thc u = Dựng vụn k cú in tr rt ln ln lt o in ỏp gia hai u cun cm v hai bn t in thỡ thy chỳng cú giỏ tr ln lt l 100V v 200V Biu thc in ỏp gia hai u cun dõy l: ud = 200 cos(100 t + )(V ) ud = 100 cos(100 t + )(V ) A B ud = 100 cos(100 t + )(V ) ud = 200 cos(100 t + )(V ) C D Cõu 28: Cho on mch gm R, L, C mc theo th t trờn vo on mch AB M l im gia L v C; Biu thc hiu in th tc thi gia hai im A v M l u AM = uRL = 200 cos100 t(V) Vit biu thc uAB? u AB = 200 cos ( 100 t ) u AB = 200 cos ( 100 t ) A (V) B u AB = 200 cos ( 100 t / ) (V) u AB = 200 cos ( 100 t + / ) C (V) D (V) Cõu 29: Cho on mch in AB gm R, L, C mc ni tip vi R l bin tr Gia AB cú mt in ỏp u = U cos(t + ) luụn n nh Cho R thay i, R = 42,25 hoc R = 29,16 thỡ cụng sut tiờu th ca on mch nh nhau; R = R0 thỡ cụng sut tiờu th ca on mch t giỏ tr ln nht, v i = cos(100 t + ) 12 cng dũng in qua mch u = 140, 2cos(100 t + )(V ) 12 A u = 140, 2cos(100 t )(V ) C u = 70, 2cos(100 t B )(V ) 12 u = 70, 2cos(100 t + )(V ) D R1 R2 Gii: R0 = (A) in ỏp u cú th cú biu thc =35,1 R0 = Z L Z C ú thỡ u , t ú tớnh c U0 v tan bn s tỡm c Cõu 30: Cho on mch in xoay chiu AB khụng phõn nhỏnh gm mt cun cm thun, mt t in cú in dung C thay i c, mt in tr hot ng 100 Gia AB cú mt in ỏp xoay chiu luụn n u=110cos(120t- ) 125 F nh (V) Cho C thay i, C = giỏ tr ln nht Biu thc ca in ỏp gia hai u cun cm l A u L =110 2cos(120t+ ) u L =220cos(120t+ ) (V) B u L =220cos(120t+ ) C (V) u L =110 2cos(120t+ ) (V) Z L = ZC Gii: thay i c ULmax thỡ D (V) ,tự ú sua U0L=I0R=220V uL = + M ú thỡ u,i cựng pha ,t ú suy Cõu 31: t in ỏp xoay chiu u=U 0cos 120t+ ữV L= H thỡ in ỏp gia hai u cun cm cú = vo hai u mt cun cm thun cú t cm Ti thi im in ỏp gia hai u cun cm l thỡ cng dũng in qua cun 40 V cm l 1A Biu thc ca cng dũng in qua cun cm l A B i=3 2cos 120t- ữA C i=3cos 120t- ữA D i=2 2cos 120t- ữA u2 U 02 + i2 I 02 =1 u2 Z 2L + i = I 02 i=2cos 120t+ ữA = p dng cụng thc c lp : I0 = 3A i = Chon ỏp ỏn B Cõu 32: t dũng in ỏp xoay chiu vo hai u mch gm in tr thuõn R mc ni tip mt t in C thỡ biu thc dũng in cú dang: i1=I0 cos(t+ )(A).mc ni tip thờm vo mch iin cun dõy thun cm L ri mc vo in ỏp núi trờn thỡ biu thc dũng in cú dng i 2=I0 cos(t- )(A) Biu thc hai u on mch cú dng: A:u=U0 cos(t +)(V) B: u=U cos(t +)(V) C: u=U0 cos(t -)(V) D: u=U0 cos(t -)(V) Gii: Gi s u = U0 cos(t + ) Gi 1; gúc lch pha gia u v i1; i2 ZC R Z L ZC R Ta cú: tan1= = tan( - /6); tan2= = tan( + /3); Mt khỏc cng dũng in cc i hai trng hp nh nhau, nờn Z = Z2 Z L ZC R ZC R ZC2 = (ZL ZC)2 ; ZL = 2ZC Vỡ vy: tan2= = = tan( + /3); tan( - /6) = - tan( +/3) tan( - /6) + tan( +/3) = => sin( - /6 + +/3) = => - /6 + +/3 = => = - /12 => u=U0 cos(t -)(V) Chn C Cõu 33: Mt on mch gm cun cm cú t cm L v in tr thun r mc ni tip vi t in cú in dung C thay i c t vo hai u mch mt hiu in th xoay chiu cú giỏ tr hiu dng U v f tn s khụng i Khi iu chnh in dung ca t in cú giỏ tr C=C thỡ in ỏp hiu dng gia hai u t in v hai u cun cm cú cựng giỏ tr v bng U, cng dũng in mch ú cú i1 = 6cos 100 t + ữ( A) biu thc Khi iu chnh in dung ca t in cú giỏ tr C=C thỡ in ỏp hiu dng gia hai bn t in t giỏ tr cc i Cng dũng in tc thi mch ú cú biu thc l A i2 = 3cos 100 t + ữ( A) 12 B i2 = 2cos 100 t + ữ( A) C D Gii: Khi C = C1 UD = UC = U => Zd = ZC1 = Z1 r + ( Z L Z C1 ) Zd = Z1 => r +Z 2 L C1 3Z Zd = ZC1 => r2 +ZL2 = ZC!2 => r2 = 3Z tan1 = => = - Z C1 (1) C1 => r = Z C1 Z C1 Z L Z C1 = = r 3 Z C1 (2) Z2 r +Z = C1 = Z C ZL Z C1 2 r + (Z L Z C ) = Khi ú Z2 = i2 = 3cos 100 t + ữ( A) => ZL ZC1 = ZL => ZL = = Khi C = C2 UC = UCmax ZC2 = i2 = 2cos 100 t + ữ( A) 12 L Zc Z C1 + ( Z C1 ) = 3Z C21 = 3Z C1 Z L ZC2 r Z C1 Z C1 = = 3 Z C1 tan2 = => = - Z1 I = = =2 Z2 3 U = I1Z1 = I2Z2 => I2 = I1 (A) cos(100t + + ) Cng dũng in qua mch: i2 = I2 B cos(100t + ) 12 =2 (A) Chn Cõu 34( H -2009): t in ỏp xoay chiu cú giỏ tr hiu dng 60 V vo hai u on mch R, L, C I cos(100t + ) mc ni tip thỡ cng dũng in qua on mch l i = (A) Nu ngt b t in C i = I cos(100t ) 12 thỡ cng dũng in qua on mch l u = 60 cos(100t A u = 60 cos(100t + (A) in ỏp hai u on mch l ) 12 u = 60 cos(100t ) (V) ) 12 B C (V) Gii: Gi biu thc ca u = Uocos(100t + ) u = 60 cos(100t + ) (V) D (V) Z L ZC = Z L Ta thy : I1 = I2 suy Z1 = Z2 hay Z ZC Z tan = L = L R R Lỳc u: Z L = ZC/2 i1 = Io cos(100t + + ) + = /4 tan = ZL R Lỳc sau: i2 = Io cos(100t + - ) - = - /12; u = 60 cos(100t + = M = /12 Vy Gii 2: Ta thy I1 = I2 => (ZL ZC)2 = ZL2 => ZC = 2ZL Z L ZC R tan1 = ZL R =- ZL R (*) tan1 = ) 12 (V).Chn C (**) => + = = u - ; = u + 12 12 => 2u - u = 60 cos(100t + ) 12 + = => u = 12 Do ú , Chn C Cõu 35 Cho linh kin gm in tr thun R= 60, cun cm thun L v t in C Ln lt t in ỏp xoay chiu cú giỏ tr hiu dng U vo hai u on mch ni tip RL hoc RC thỡ biu thc cng dũng in nch ln lt l i1=cos(100-)(A) v i2=cos(100+)(A) nu t in ỏp trờn vo hai u on mch RLC ni tip thỡ dũng in mch cú biu thc: A 2cos(100t+)(A) B cos(100t+)(A) C 2cos(100t+)(A) D 2cos(100t+)(A) ỡù =- ( 1) I 01 = I 02 ị Z RL = Z RC ị ùớ ùù Z L = Z C ợ Gii 1: Theo Mt khỏc u - i1 =1 ( 2) ỹ ùù ( 1) i + i2 = ( 3) ýị u = u - i2 = ùùỵ ( 2) , ( )ị = Z ị L 3= R ị Z60 ) ị U = I Z = 120 ( V ) L =3 ( 01 RL T U0 R RLC nt đ u Khi cng hng: i= cos(100t+ )= 2cos(100t+)(A) Chn C Gii 2: Ta thy cng hiu dng on mch RL v RC bng suy ZL = ZC lch pha gia u v i1 v gia u v i2 i tan1= - tan2 Gi s in ỏp t vo cỏc on mch cú dng: u = U cos(100t + ) (V) Khi ú = (- /12) = + /12 = 7/12 tan1 = tan( + /12) = - tan2 = - tan( 7/12) tan( + /12) + tan( 7/12) = sin( + /12 + 7/12) = Suy = /4 - tan1 = tan( + /12) = tan(/4 + /12) = tan /3 = ZL/R R + Z L2 = RI1 = 120 ZL = R v U = I1 (V) Mch RLC cú ZL = ZC => cú s cng hng I = U/R = 120/60 = (A) v i cựng pha vi u: u=U cos(100t + /4) Vy i = cos(100t + /4) (A) Chn C R = 60 Cõu 36: Cho ba linh kin: in tr thun , cun cm thun L v t in C Ln lt t in ỏp xoay chiu cú giỏ tr hiu dng U vo hai u on mch ni tip RL hoc RC thỡ biu thc cng dũng in i1 = cos(100 t /12)( A) i2 = cos(100 t + /12)( A) mch ln lt l v Nu t in ỏp trờn vo hai u on mch RLC ni tip thỡ dũng in mch cú biu thc: i = 2 cos(100 t + / 3) ( A) i = 2cos(100 t + / 3) ( A) A B i = 2cos(100 t + / 4) ( A) D i = 2 cos(100 t + / 4) ( A) C = C L = I0 = I 01 cos Gii: Pha ban u ca i: => = Ta cú th m rng bi toỏn ny nh sau: Mc mch RL vo hiu in th u thỡ dũng in l i = I cos(t + ) Mc mch RC vo hiu in th u thỡ dũng in l i = I cos(t + ) chn A I 0' Mc mch RLC vo hiu in th u thỡ dũng in l i = cos(t + ) Ta luụn cú mi quan h:(v gin hoc s dng cụng thc tan ta d dng chng minh c): C L Z L = Z C = R tan = I 0' = I0 cos Vy bi toỏn ny mch RLC ta cú th tớnh v vit c biu thc ca: R,L,C,u,i,P Cõu 37: t in ỏp u = U cos t ữ vo hai u on mch cha mt in tr thun v mt t i = I0 cos t ữ in mc ni tip Khi ú, dũng in mch cú biu thc Mc ni tip vo mch t th hai cú cựng in dung vi t ó cho Khi ú, biu thc dũng in qua mch l i = 0, 63I0 cos ( t 0,147 ) (A) A i = 1, 26I0 cos ( t 0,147 ) (A) C i = 0, 63I0 cos ( t 0,352 ) (A) B i = 1, 26I0 cos ( t 0,352 ) (A) D u = U cos t ữ i = I0 cos t ữ R = ZC ZC mc thờm t na thỡ I0 I 02 = = 2Z C tan = ỏp ỏn A 0,3 / Cõu 38: Cho mch in xoay chiu RLC Cun dõy thun cm L = u = 200 cos ( 100 t ) bin tr t mch vo hiu in th a) Vit biu thc uR cụng sut ca mch t cc i uR = 100 cos ( 100 t / ) C (F); R l V uR = 200 cos ( 100 t / ) A (H), C = 4.104 / u R = 200 cos ( 100 t + / ) V B V D uR = 100 cos ( 100 t / ) V V b) Cho R = 20 , Hi phi ghộp vi C mt t C1 nh th no v bng bao nhiờu cụng sut tiờu th ca mch t cc i; Vit biu thc hiu in th gia hai u cun cm ú A mc song song C1 = 0,637 mF B mc ni tip C1 = 0,637 mF à C mc song song C1 = 0,637 F D mc ni tip C1 = 0,637 F Cõu 39: oan mach AC co iờn tr thuõn, cuụn dõy thuõn cam va tu iờn mc nụi tiờp B la mụt iờm trờn AC vi uAB = cos100t (V) va uBC = cos (100t - ) (V) Tim biờu thc hiờu iờn thờ uAC u AC = 2cos 100t ữV u AC = 2cos(100 t) V A B u AC = 2cos 100t + ữV u AC = 2cos 100t ữV C D L= Cõu 40: t in ỏp xoay chiu vo vo hai u mt cun cm thun cú t cm H thỡ cng dũng in qua cun cm cú biu thc i = I 0cos(100t - ) (V) Ti thi im cng tc thi ca dũng in qua cun cm cú giỏ tr 1,5 A thỡ in ỏp tc thi hai u cun cm l 100 V in ỏp hai u cun cm cú biu thc 2 A u =100 cos(100t + C u = 75 cos(100t + ) V B u = 125cos(100t + ) V D u = 150cos(100t + ) V ) V Cõu 41: t vo hai u AMNB ca on mch RLC gm ni tip M l im ni gia t in v cun dõy thun cm, N l im ni gia cun dõy v in tr thun Khi ú biu thc in ỏp ca hai u on mch NB l uNB = 60 cos(100t - hai u on mch AB mt gúc cos(100t - C u = 40 Biu thc ca in ỏp hai u on mch AB l 6 A u = 60 cos(100t + ) V v in ỏp gia hai u on mch AN sm pha hn in ỏp 6 ) V B u = 40 ) V D u = 60 cos(100t - cos(100t + ) V ) V [...]... sin( - /6 + +/3) = 0 => - /6 + +/3 = 0 => = - /12 => u= U0 cos(t -)(V) Chn C C u 33: Mt on mch gm cun cm cú t cm L v in tr thun r mc ni tip vi t in cú in dung C thay i c t vo hai u mch mt hiu in th xoay chiu cú giỏ tr hiu dng U v f tn s khụng i Khi iu chnh in dung ca t in cú giỏ tr C=C 1 thỡ in ỏp hiu dng gia hai u t in v hai u cun cm cú cựng giỏ tr v bng U, cng dũng in trong mch khi ú cú i1 ... Z L = ZC Gii: khi thay i c ULmax thỡ D (V) ,tự ú sua ra U0 L =I0 R=220V uL = + 3 2 M khi ú thỡ u, i cựng pha ,t ú suy ra C u 31: t in ỏp xoay chiu u= U 0cos 120t+ ữV 3 1 L= H 6 thỡ in ỏp gia hai u cun cm cú 6 = vo hai u mt cun cm thun cú t cm Ti thi im in ỏp gia hai u cun cm l thỡ cng dũng in qua cun 40 2 V cm l 1A Biu thc ca cng dũng in qua cun cm l A B i= 3 2cos 120t- ữA 6 C i= 3cos 120t-... im cng tc thi ca dũng in qua cun cm cú giỏ tr 1,5 A thỡ in ỏp tc thi hai u cun cm l 100 V in ỏp hai u cun cm cú biu thc 2 2 A u =100 cos(100t + 2 C u = 75 cos(100t + 3 3 ) V B u = 125cos(100t + ) V D u = 150cos(100t + 3 ) V ) V C u 41: t vo hai u AMNB ca on mch RLC gm ni tip M l im ni gia t in v cun dõy thun cm, N l im ni gia cun dõy v in tr thun Khi ú biu thc in ỏp ca hai u on 3 2 mch NB l uNB... D i= 2 2cos 120t- ữA 6 u2 U 02 + i2 I 02 =1 u2 Z 2L + i 2 = I 02 i= 2cos 120t+ ữA 6 = 3 2 6 p dng cụng thc c lp : I0 = 3A i = Chon ỏp ỏn B C u 32: khi t dũng in ỏp xoay chiu vo hai u mch gm in tr thuõn R mc ni tip mt t in C thỡ biu thc dũng in cú dang: i1 =I0 cos(t+ )(A).mc ni tip thờm vo mch iin cun dõy thun cm L ri mc vo in ỏp n i trờn thỡ biu thc dũng in cú dng i 2 =I0 cos(t- )(A) Biu thc... gm in tr i= u = 80co s100 t R = 40 i = 2co s(100 t 2 ) 3 D (V) ghộp ni tip vi cun cm L Hiu in th tc UL v in ỏp hiu dng hai u cun cm 2 co s(100 t ) A 2 4 A ) 6 (V) u = 80 2co s(100 t + (V) thi hai u on mch qua mch l: Biu thc hiu in i= 2 co s(100 t + ) A 2 4 B )A 4 =40V Biu thc i i = 2co s(100 t + )A 4 C D C u 16: Mt on mch in gm in tr R = 50 mc ni tip vi cun thun cm L = 0,5/ (H) t 2 vo hai u. .. minh c): C L 2 Z L = Z C = R tan = I 0' = I0 cos Vy bi toỏn ny trong mch RLC ta cú th tớnh v vit c biu thc ca: R,L,C ,u, i, P C u 37: t in ỏp u = U 0 cos t ữ 2 vo hai u on mch cha mt in tr thun v mt t i = I0 cos t ữ 4 in mc ni tip Khi ú, dũng in trong mch cú biu thc Mc ni tip vo mch t th hai cú cựng in dung vi t ó cho Khi ú, biu thc dũng in qua mch l i = 0, 6 3I0 cos ( t 0,147 ) (A) A i. .. pha vi u: 2 u= U 2 cos(100t + /4) Vy i = 2 cos(100t + /4) (A) Chn C R = 60 C u 36: Cho ba linh kin: in tr thun , cun cm thun L v t in C Ln lt t in ỏp xoay chiu cú giỏ tr hiu dng U vo hai u on mch ni tip RL hoc RC thỡ biu thc cng dũng in i1 = 2 cos(100 t /12)( A) i2 = 2 cos(100 t + 7 /12)( A) trong mch ln lt l v Nu t in ỏp trờn vo hai u on mch RLC ni tip thỡ dũng in trong mch cú biu thc: i = 2... MB = 200V i = I 0 sin(100 t )( A) (V) lch pha gia uAN v uMB l Dũng in tc thi trong mch l , cun dõy thun cm Biu thc ca uAB l u AB = 139 2 sin(100 t + 0,53)V u AB = 612 2 sin(100 t + 0,53)V A B u AB = 139sin(100 t + 0,53)V C u AB = 139 2 sin(100 t + 0,12)V R = 60 D C u 23: Cho ba linh kin gm in tr thun , cun cm thun L v t in C Ln lt t in ỏp xoay chiu cú giỏ tr hiu dng U vo hai u on mch ni tip RL hoc... Cho mch in nh hỡnh v : UAN =150V ,UMB =200V lch pha UAM v UMB l / 2 Dũng in tc thi trong mch l : i= I0 cos 100t (A) , cun dõy thun cm Hóy vit biu thc UAB Hng dn: Ta cú : (1) U AN = U C + U R U AN = U C2 + U R2 = 150V (2) U MB = U L + U R U MB = U + U 2 L 2 R = 200V Vỡ UAN v UMB lch pha nhau / 2 nờn tg1 tg 2 = 1 U L U C =1 U R U R hay U2 R = UL.UC (3) T (1),(2),(3) ta cú UL=160V , UC = 90V , U R =... 150V, gia hai u tu i n la 100V.Dong i n trong mach co bi u thc i =I0 cos(t + /6)((A) Bi u thc i n ap hai u oan mach la u = 50 2 cos(100t + / 2)V u = 50 2 cos(100t / 2) A V B u = 50 2 cos(100t 2 / 3) C u = 50 2 cos(100t + 2 / 3) V 3 D V C u 25: t in ỏp u = 120cos(100t + ) (V) vo hai u mt on mch gm cun cm thun mc ni tip in tr thun R= 30 thỡ in ỏp hiu dng hai u cun cm l 60 V Dũng in tc thi qua on