THTR DIEN XOAY CHIEU.doc

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CC BI TON LIN QUAN N LCH PHA

T HC T RN THNG 9

TRNG THPT HNG VNG

T: L------------ o0o --------------CHUYN T HC T RN THNG 9

MT S DNG TON TIU BIU TRONG MCH IN XOAY CHIU

CC BI TON LIN QUAN N LCH PHATRONG MCH IN XOAY CHIU(GV: MAI TH L GIANG)t vn :

Bi tp v mch in xoay chiu cng l mt phn kh quan trng trong cc chuyn bi tp vt l. Trong cc thi H v C thng cho dng trc nghim lin quan n lch pha trong mch in xoay chiu. .Dng ton ny thng lm hc sinh cm thy phc tp, ri v cch gii ng s cho kt qu ng nhng ta cn b mt vn m phi quan tm n khi gii trc nghim l THI GIAN LM BI. Sau y ti xin cp mt s kinh nghim gii quyt cc bi ton ny qua cc v d sau:1.Phng php chung:

+ Hay Thng dng cng thc ny v c du ca (, + Hay ; cos( = ; Lu cng thc ny khng cho bit du ca (.

+ sin( ;

+ Kt hp vi cc cng thc nh lut m :

+ Lu : Xt on mch no th p dng cng thc cho on mch .

+ Nu 2 on mch cng pha:

+ Nu 2 on mch vung pha:

a.Xc nh cc i lng khi bit hai on mch c in p cng pha, vung pha.

Bi tp 1: Mt mch in xoay chiu RLC khng phn nhnh c R=100, L=H, t in c in dung C thay i c. t vo hai u on mch mt in p xoay chiu . Gi tr ca C v cng sut tiu th ca mch khi in p gia hai u R cng pha vi in p hai u on mch nhn cp gi tr no sau y:

Gii: Ta thy khi uR cng pha vi uAB ngha l uAB cng pha vi cng dng in i. Vy trong mch xy ra cng hng in: ZL=ZC => . Vi ZL=L= 200 => C=F

Lc ny cng sut P=Pmax= Bi tp 2: Cho mach in xoay chiu nh hinh.

R1 = 4(, , R2 = 100( , H , f = 50Hz.

Tim in dung C2, bit rng in ap uAE va uEB ng pha.

Bi gii: ;

V uAE v uEB ng pha nn

; (F)

Bi tp 3: Cho mch in nh hnh v. UAN = 150V, UMB = 200V, uAN v uMB vung pha vi nhau, cng dng in tc thi trong mch c biu thc (A). Bit cun dy l thun cm. Hy vit biu thc uAB.

Bi gii:Ta c: V (1)

V (2)

V uAN v uMB vung pha nhau nn: (Vi , )

(3)

T (1), (2) v (3), ta suy ra : UL = 160V , UC = 90V, UR = 120V

Ta c : V

rad. Vy (V)

Bi tp 4: Cho vo on mch hnh bn mt dng in xoay chiu c cng (A). Khi uMB v uAN vung pha nhau, v (V). Hy vit biu thc uAN v tm h s cng sut ca on mch MN.

Bi gii: Do pha ban u ca i bng 0 nn

rad

Da vo gin vec-t, ta c cc gi tr hiu dng ca UL, UR, UC l:

UR = UMB cos (MB = (V)

(V)

V uMB v uAN vung pha nhau nn:

Ta c: (V)

Ta c: (V)

Vy biu thc (V).

H s cng sut ton mch:

b.Xc nh cc i lng khi bit hai on mch c in p lch pha gc (.Bi tp 1: Cho mch in nh hnh v. Bit F,

H, (V). in p uAM chm pha so vi dng in qua mch v dng in qua mch chm pha so vi uMB. Tnh r v R? s. v .Gii : ZL= 50(; ZC = 100(; .

. Bi tp 2: Mt mch in khng phn nhnh gm in tr thun R = 75, cun cm c t cm L =H v t in c in dung C. Dng in xoay chiu qua mch: i = 2 cos 100t(A). lch pha gia in p v cng dng in l (/4.Tnh C.Vit biu thc in p gia hai u on mch trn.

Bi gii: ZL= (L= 100(.=125( ;

lch pha gia u v i: tan(= tan= // 1=

Suy ra:=> =>

a) Trng hp C= , th Z =

Ta c: U0 = I0 .Z = 2.75=150V ; (=(/4 nn: u= 150cos(100(t+ (/4)(V)

b) Trng hp C= , th Z =

Ta c: U0 = I0 .Z = 2.75=150V ; (= -(/4 nn: u= 150cos(100(t- (/4)(V)

Bi tp 3: Cho mch xoay chiu nh hnh v:, f=50(Hz); Bit lch pha mt gc 1350 v i cng pha vi . Tnh gi tr ca R?

A. B.

C. D.

Bi gii: Theo gi thit u v i cng pha nn trong mch xy ra hin tng cng hng ta c: . Mt khc on EB cha t C nn

Suy ra : Hay : ; Vy . ( Chn CBi tp 4: t in p (U0 v khng i) vo hai u on mch mc ni tip gm in tr R, t in c in dung C, cun cm thun c t cm L thay i c. Khi L=L1 v L=L2 in p hiu dng hai u cun cm c cng gi tr; lch pha ca in p hai u on mch so vi cng dng in ln lt l 0,52rad v ,05rad. Khi L=L0 in p gia hai u cun cm t cc i; lch pha ca in p hai u on mch so vi cng dng in l . Gi tr ca gn gi tr no nht sau y:

A. 0,41rad B, 1,57rad C. 0,83rad. D. 0,26rad.

+ Khi ULmax th ZLo =

(1)

+ V:

(2)

+ t: tan(0,52) = a v tan(1,05) = b th ta c: a.b = 1

+ Ta c :

(3)

Thay (3) vo (1) v t x = R/Zc th ta c PT:

(a+b)X3 a.b.X2 (a+b).X + 1 = 0,785 rad

V a.b = 1 nn PT c nghim: X = 1 nn tan = 0,785 radBi tp 5: : on mch ni tip gm cun cm thun, on mch X v t in (hnh v). Khi t vo hai u A, B in p V ( khng i) th v , ng thi UAN sm pha so vi UMB. Gi tr ca U0 l:

A.

B.

C.

D.

Hng dn :

- Do UMB = 2UAN v uAN lch pha uMB gc 600 nn ta v c gin vc t nh trn.

Cch 2: (Cch ny hay hn cch trn)

Bi tp 6: t in p (U0 v khng i) vo hai u on mch gm cun dy khng thun cm mc ni tip vi t in c in dung C (thay i c). Khi C=C0 th cng dng in trong mch sm pha hn u l () v in p hiu dng hai u cun dy l 45V. Khi C=3C0 th cng dng in trong mch tr pha hn u l v in p hiu dng hai u cun dy l 135V. Gi tr ca U0 gn gi tr no nht sau y:

A. 130V B. 64V C.95V D. 75VHng dn :

BI TON V CNG SUT CA MCH IN XOAY CHIU KHNG PHN NHNH(GV: NGUYN TH HIN)I. C s l thuyt

Cng sut ca mch RLC ni tip:

Cng sut ca mch chnh l cng sut ta nhit trn in tr, cn cun dy thun cm(L) v t in khng tiu th cng sut.

H s cng sut:

Trng hp cun dy c in tr thun r th:

Cng sut ca mch:

Cng sut trn in tr R:

Cng sut trn cun dy: vi

II. Cc bi ton v cng sut

1. Khi R thay i trong mch RLC ni tip

Bi 1 : Mt mch in xoay chiu gm t in C, mt cun cm thun L v mt bin tr R c mc ni tip. Khi R thay i th cng sut ta nhit cc i l Pmax. Khi bin tr gi tr ln lt l th cng sut tiu th trn on mch l P1, P2, P3. Nu P1=P2 th

A.

B.

C.

D.

GII

Dng ton ny nn dng phng php th l nhanh nht.

Trc ht ta kho st hm s P v v th.

Xt: t lp c bng bin thin nh sau:

T phc ha c th:

Thc cht khi lm nhanh ch cn phc ha th v biu din s liu nhn ra kt qu nhanh nht. S dng th gip cho chng ta gii nhn xt c cc bi ton bin thin.Bi 2: Mt on mch xoay chiu mc ni tip gm cun dy c in tr thun 40, c cm khng 60, t in c dung khng 80 v mt bin tr R (). in p hai u on mch n nh 200V-50Hz. Khi thay i R th cng sut ta nhit trn ton mch t gi tr cc i l:A. 1000W

B. 144W

C. 800W

D. 125W

Gii

Cng sut ton mch chnh l cng sut ta nhit trn bin tr v in tr trn cun dy.

Pmax khi v ch khi mu s min: p dng bt ng thc C si ta c:

Vy nn khi

V nn

Ta c th dng th kim tra:

*Trng hp bi ton hi: Tm cng sut cc i trn bin tr R.

Pmax khi v ch khi mu s min: p dng bt ng thc C si ta c:

Vy nn khi

Thay s vo ta c: v

*Trng hp bi ton hi: Tm cng sut cc i trn cun dy( Imax , Udmax , UCmax) . khi v ch khi mu s nhn gi tr min.

Khi R=0 v

Bi 3: Mt mch in xoay chiu tn s f gm t in C, mt cun cm thun L v mt bin tr R c mc ni tip. Khi bin tr c gi tr hoc th cng sut trn on mch l nh nhau. Xc nh h s cng sut tiu th ng vi cc gi tr ca R1.

A. 0,707

B. 0,8

C. 0,5

D. 0,6

Gii

Cng sut:

Theo h thc Viet: .

Do vy khi R=R1 th h s cng sut :

Bi 4: Mch in xoay chiu gm ba phn t, in tr thun R, cun cm thun L v t in C mc ni tip. in tr R thay i c. t vo hai u on mch in p xoay chiu u = 120cos(100(t) V. iu chnh R, khi R = R1 = 18 th cng sut trn mch l P1, khi R = R2 = 8 th cng sut P2, bit P1 = P2 v ZC > ZL. Khi R = R3 th cng sut tiu th trn mch t cc i. Biu thc cng dng in qua mch khi R = R3 l

A. i = 5cos(100t + /3) (A).

B. i = 4cos(100t + /3) (A).

C. i = 5cos(100t + /4) (A).

D. i = 10cos(100t + /4) (A).

GiiV hai gi tr ca R1 v R2 cho cng mt gi tr cng sut nn:

Theo h thc Viet ta c: .

Mt khc vi R=R3 th Pmax khi :

( v ZC > ZL tc l i sm pha hn u).

Biu thc ca cng dng in qua mch: i = 10cos(100t + /4) (A). 2. Dng bi ton c L hoc C thay iBi 5: Cho on mch xoay chiu sau:

(in tr thun) FF

L: t cm thay i c ca mt cun thun cm

Hiu in th gia hai u AB ca on mch c biu thc:

a.Tnh L cng sut tiu th ca on mch cc i.V phc ha dng th ca cng sut tiu th P ca on mch theo L.b. Khi L = L1 v L = L2 = L1/2 th cng sut tiu th trn on mch l nh nhau, nhng cng dng in vung pha nhau. Gi tr L1 v in dung C ln lt l Gii

a. Khi L thay i Pmax. y l 1 du hiu ca bi ton cng hng. Bi v:

Pmax khi .

Khi :

- Kho st s bin thin ca cng sut theo L.

khi :

Bng bin thin:

b. Vi 2 gi tr ca L cho cng mt cng sut ( hoc cng cng dng in, hoc cng tng tr Z).

Do:

Khi : .

Mt khc:

v

Bi 6: Mch in in tr thun, cun dy v t in ni tip, c C thay i c. in p hai u on mch Khi th mch tiu th cng sut mch t cc i Pmax = 93,75 W. Khi th in p hai u on mch RC v cun dy vung pha vi nhau, in p hiu dng hai u cun dy khi l:A. 90 V.B. 120 V.C. 75 VD. 75V.

Gii

D tnh c : . + Khi C=C1 th cng sut trn mch t cc i ( du hiu ca bi ton cng hng: 1 trong 3 i lng hoc C hoc thay i lm cho Imax , ).

Cng sut mch :

Mt khc: Khi C=C2 th:V nn cun dy c cha in tr r.

Khi :

Ta nhn thy ngay R = r = 120. Khi

. Bi 7: t in p xoay chiu n nh vo hai u on mch gm cun dy ni tip vi t in c in dung C1. Khi dng in trong mch l i1 v cng sut tiu th ca mch l P1. Ly mt t in khc C=4C1 mc song song vi t in C1 th dng in trong mch l i2 v cng