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VT L[3K] 2014 - 2015 Thy Lm Phong

Gieo hnh vi gt thi quen. Gieo thi quen gt tnh cch. Gieo tnh cch gt s phn. ( Dick Lyles) 1

U

uO

M'2

M2

M'1

M1

-U U0

01

-U1Sng Sng

Tt

Tt

TNG HP KIN THC V DNG IN XOAY CHIU

Bi 1: I CNG V DNG IN XOAY CHIU

Nguyn tc to ra dng in xoaychiu:da trn hin tng cm ng in t (L hin tng c s bin thin ca t trngqua mt khung dy kn th trong khung xut

hin mt sut in ng cm ng sinh ramt dng in cm ng )

Cch to ra dng in xoay chiu:Cho khung dy dn din tch S, c N vngdy, quay u vi tn s gc trong in

trng u c cm ng t B (B trc

quay) th trong mch c dng in bin thin iu havi tn s gc gi l DXC.

T thngc phng trnh: = NBScos(t + ) Wb (V-be) = N ocos(t + ) Wb

(Trong o= BS l t thng cc i qua mi vng dy, S l din tch ca khung quay, N l s vngdy qun vo khung quay, l gc hp gia php tuyn ca khung v cm ng t B)

Sut in ng trong khung dy: e = - ' = NBSsin(t + ) = E ocos(t + -2

)

(Trong E o= NBSl sut in ng cc i quacc cun dy)

Khi nim v gi tr hiu dng, gi tr tc thi, gi tr cc i:

i: dng in tc thi, I: gi tr hiu dng, I o: gi tr cc i. Tng t ta c: I =I o

2, U =

U o

2, E =

E o

2

( E l sut in ng dng cho ni pht sinh dng in, U l hiu in th ni tiu th dng in )

Cc biu thc in p v dng in xoay chiu:

+ Biu thc in p tc thi v dng in tc thi:

u= U ocos(t + u)V v i= Iocos(t + i) A

Vi = u il lch pha ca uso vi i, c2 2

Cng thc tnh thi gian n hunh quang sng trong mt chu k:

Khi t in p u= U ocos(t + u) vo hai u bng n,bit n ch sng ln khi u U1.

t =

4

vicos =

U 1

U o, (0 < ZChay

1

LC

> 0 th unhanh pha hn i

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+ Khi ZL< ZChay1

LC < 0 th uchm pha hn i

+ Khi ZL= ZChay1

LC = 0 th ucng pha vi i.

Lc Max

UI =

Rgi l hin tng cng hng dng in

Bi 3:CNG SUT V H S CNG SUT CA MCH R - L - C

Cng sut tiu th trn mt mch in l: P = UICos = UIRZ

= U2RZ2

= RI2 .

( ty d kin bi m ta tnh cng sut ph hp, c th tm c gc = u- i)

Mch in ch tiu th cng sut khi c in tr R hoc r.

+ Nu mch ch c L hoc C th cng sut P = 0

H s cng sut: cos =PUI

=RZ

=U RU

c bit, nu mch c R-r-L-C th cos

= R + rZ

= UR+ U

r

U

+ Cng sut ph thuc vo cos, s dng hiu qu in nng tiu th th ta phi mc thm vomch nhng t in c in dung ln. Qui nh trong cc c s s dng in th cos 0,85.

Ch : nu mch in u,i c 2 thnh phn nh u = U 1+ U2cos(t) hay i = I

1+ I

2cos(t)th:

+ thnh phn U 1( hay I1) c xem l phn khng i ( dng in 1 chiu ),

+ thnh phn U 2( hay I2) c xem l thnh phn xoay chiu

c bit:+ nu mch l R-C th I 1v U

1khng tn ti do C khng dng in i qua P = RI

22.

+ nu mch l R-L th I 1v U1tn ti nhngZ

L= 0 i vi D1CP = RI

12 + RI

22 .

Bi 4: Cch S Dng Gin Vect Trt Gii Cc Bi Ton R-L-C v Hp en.PHNG PHP GIN VECT TRT( Gii ton in bng hnh hc )Chn ngang l trc dng in.(Chuyn do thy Chu Vn Bin bin son )Chn im u mch (A) lm gc.V ln lt cc vc-t biu din cc in p, ln lt t A sang B ni ui

nhau theo nguyn tc:+ L - ln.+ Cxung.+ Rngang.

di cc vc-t t l vi cc gi tr hiu dng tng ng. *Ni cc im trn gin c lin quan n d kin ca biton.* Biu din cc s liu ln gin .* Da vo cc h thc lng trong tam gic tm cc in p hoc gc cha

bit.

Gii thiu mt s gin thng dng.+ Gin R-L-C:Cho mch in gm cun cm thun c t cm L,

in tr thun R, t in c in dung C mc ni tip.

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+ Gin R-Lr:Cho mch in gm in tr thun R, cun dy c t cm L v in tr thun r mcni tip.

+ Gin Lr-R-C:Cho mch in gm cun dy khng thun cm, in tr thun R v t in c indung C mc ni tip.

+ Gin R-C-L:Cho mch in gm cun in tr thun R, t in c in dung C v cun dy c t cm L mc ni tip.

+ Gin R-C-Lr:Cho mch in gm cun in tr thun R, t in c in dung C v cun dykhng thun cm c in tr r mc ni tip.

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+ Gin C-R-Lr:Cho mch in gm t in c in dung C, in tr thun R v cun dy khngthun cm c in tr r mc ni tip.

+ Gin R-Lr-C:Cho mch in gm intr thun R v cun dy khng thun cm c in tr r vt in c in dung C mc ni tip.

+ Kinh nghimcho thy khi trong bi ton c lin quan n lch phahoc qu nhiu s liuth nngii bng phng php gin vc t s c li gii ngn gn hn gii bng phng php i s.

Bi 5A: BI TON CNG HNG CA MCH R - L - CKhi trong mch c hin tng Cng Hng in l khi: = u-

i= 0

Khi = 0 cos= 1 cos ln nhtv khi R = Z U R= UKhi = 0 tan= 0 Z L- Z C= 0 Z L= Z CU L= U C

Khi = 0 P max= UI =U2R

Khi Z L= ZCZ

min= R m I =

UZ

I max=UR

hay LC2 = 1=1LC

Khi u, i cng pha nhau( V gin vect th chng trng ln nhau )Khi ulch pha so vi U

Cmt gc 90

o .

V rt nhiu trng hp khc a v cc trng hp trn u c xem l Cng hng.

Bi 5B: Mch R - L - C - f - C Cc Phn T Thay i.Mch R Thay i.

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+ Chnh R = R

1

R = R2th

P1= P

2

I 1= I2R1+ R

2=

U2P

v R1R2= (Z

L- Z

C)

2 ( Tha mn PT Vi-et).

Khi tng ng ta c 1+ 2= 90

o hay sin

1= cos

2

+ Chnh R = Roth P

ton mchCc i Ro= |Z

L- Z

C| = R

1R

2( bng in tr nhm cn li).

Khi P max=U22Ro

( Vi Rong vi gi tr trn )

V H s cng sut cos=R

Z=R

R2 + (ZL- Z

C)

2=

R

R2 + R2

=1

2

+ Nu mch R-rL-C, chnh R = Roth P

trn RCc i Ro= r

2 + (Z

L- Z

C)

2

+ Nu mch R-rL-C, chnh R = Roth P

ton mchCc i Ro+ r = |Z

L- Z

C|

Mch L Thay i.

+ Chnh L = L

1

L = L 2th

P1= P

2

I 1= I2Z C=

Z L1+ Z

L2

2 v u=

i1+

i2

2

+ Chnh

L = L1

L = L 2 th

P1= P

2

I 1= I 2 v chnh L = L

3th U

Lmax2

L 3=

1

L 1+

1

L 2

+ Chnh L U LmaxZL=

R2 + ZC

2

Z C, U Lmax=

U

R R2 + Z

C

2 =

U

U RU R

2 + U

C

2

Khi U Lmax2 = U

2 + U

RC

2 U

Lmax

2 = U

2 + U

R

2 + U

C

2

U RCU tan

RC. tan= -1(1)U L.U

C= U

R

2 + U

C

2 (2)U

L

2 = U

R

2 + U

C

2 + U

2

(3)U L= U. 1 +U CU R

2

(4)1

U2+

1

U R2 + U

C

2

=1

U R2

+ Chnh L U RLmax2 2

42

C C

L

Z R ZZ th ax2 2

2 R

4RLM

C C

UU

R Z Z

(R-L mc ni tip)

+ Chnh L P max, I

max, U

Cmax, U

Rmax, v.v... ( Nhng phn t khc MAX ngoi U

Lmax)

CNG HNG ( Xem bi 5)

+ Mch L-RC c R v L thay i, Chnh L U RCkhng ph thuc vo R

khi U RC= U v ZL= 2Z

CLC

2 = 2

Mch C Thay i. ( tng t L )

+ Chnh C = C 1C = C 2

th P 1= P 2I 1= I

2

Z L=Z C1+ Z C2

2 v u=

i1+ i22

+ Chnh C = C

1

C = C 2th

P1= P

2

I 1= I2

v chnh C = C 3th U

CmaxC3=

1

2(C 1+ C

2)

+ Chnh C U CmaxZC=

R2 + ZL

2

Z L, U Cmax=

U

RR2 + Z

L

2 =

U

U RU R

2 + U

L

2

Khi U Cmax2 = U

2 + U

RL

2 U

Cmax

2 = U

2 + U

R

2 + U

L

2

U RLU tan

RL. tan= -1

+ Chnh C U RCmax2 2

42

L LC

Z R ZZ

th ax

2 22 R

4RCM

L L

UU

R Z Z

( R v C mc ni tip)

+ ChnhC P max, I

max, U

Lmax, U

Rmax, v.v... ( Nhng phn t khc MAX ngoi U

Cmax)

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CNG HNG ( Xem bi 5)

+ Mch C-RL c C v R thay i, Chnh C U RLkhng ph thuc vo R

khi U RL= U v ZC= 2Z

LLC

2 = 0,5

Mch thay i. ( tng t vi tn s f )

+ Chnh

= 1

=

2

th

P1= P

2

I

1= I

2

, nu chnh = 3th I

maxhoc P

max32 =

1.

2

+ Chnh = 1= 2

th I 1= I2=

I maxn

(n > 1) th khi R =L( 1-

2)

n2 - 1 vi ( 1>

2)

Hoc R =| 1-

2|

C 12 n

2 - 1

+ Chnh U Lmaxth 2 =

2

2LC - R2C2v ax

2 2

2 .

4LM

U LU

R LC R C

=U

1 -Z C

2

Z L2

+ Chnh U

Cmaxth

2

=

2LC - R2C2

2L2C2 v ax 2 2

2 .

4CM

U LU

R LC R C

=

U

1 -Z L2Z C

2

+ Chnh= 1th U

Lmax

= 2th U

Cmax

= 3th U

Rmax

th 32 =

1

2.

+ Chnh = 1= 2

th U C1= U

C2.Nu chnh = 3th U

Cmax

32 =

12( 1

2 +

22)

+ Chnh = 1= 2

th U L1= U

L2. Nu chnh = 3th U

Lmax

1

32=

12(

1

1

2

+1

2

2)

+ Chnh = 1v = 2= n

1th mch tiu th cng h s cng sut ngha l cos

2= cos

1

vi L = CR2 . Khi : tan1=

1

2-

2

1=

f1f2

-f2f1

(cng thc ny ch p dng khi L = CR2)

Ch : nu c thm r = R hay L = CR2 = Cr2 th tan1=

12

1

2-

2

1

Khi tnh c tanta dng 1 + tan2 =1

cos2

+ t in p xoay chiu u = U 2cost (U khng i, thay i c) vo hai u mch c R, L ,C

mc ni tip. Khi = 1th cm khng v dung khng ca on mch ln lt l Z L1v Z C1. Khi = 2th

trong on mch xy rahin tng cng hng. H thc ng l: 1= 2

Z L1Z C1

Bi 6: DNG IN XOAY CHIU BA PHA Dng in xoay chiu ba phal h thng ba dng in xoay chiu, gy bi ba sut in ng xoay chiu

cng tn s, cng bin nhng lch pha tng i mt l2

3

( Da trn hin tng ng in t )

Cu to: Phn ngl ba cun dy ging nhau gn c nh trn mt ng trn tm O ti ba v tr tcch nhau mt gc 120o . Phn Cml mt nam chm c th quay quanh trc O vi tc gc khng i.

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1 0

2 0

3 0

os( )

2os( )

3

2os( )

3

e E c t

e E c t

e E c t

trong trng hp ti i xng th

1 0

2 0

3 0

os( )

2os( )

3

2os( )

3

i I c t

i I c t

i I c t

Bi 7: HIU SUT TRUYN TI IN NNG.

Cng sut tiu th: P = UIcosv P hao ph= RI2 P

hao ph = R

P2U2cos

2

U tng n ln th P hao ph gim n2Trong :P l cng sut truyn i ni cung cp

U l in p ni cung cpcosl h s cng sut ca dy ti in

lR

S

l in tr tng cng ca dy ti in (lu :dn in bng 2 dy)

+ gim in p trn ng dy ti in: U = IR

+ Hiu sut ti in: H =P - P

P.100 %

S l tit din trn ca dy.Do ta c S = r2 = d24

(d = 2r : l ng knh ca dy )

T cc mi quan h t l thun - nghch ta c:P 1

P 2=

U 2U 1

2

=S 2S 1

=

r 2r 1

2

trong P = 100 - H

Trong qu trnh truyn ti in i xa, gim in p hiu dngtrn ng dy ti in mt pha bng n

ln (n < 1)in p cui ng dy ny. Coi rng cng dng in lun cng pha vi in p. cngsut hao phtrn ng dy gim m ln (m > 1)nhng vn m bo cng sut tiu th nhn c khng

i.Cn phitng in p a vo truyn ti :n + mm(n +1)

Bi 8: MY BIN PHot ng: da trn hin tng Cm ng in T.Tc dng: bin i in p ( v cng dng in ) ca dng xoay chiu m vn gi nguyn tn s,khng c tc dng bin i nng lng.

Cng thcquan trng nht ca my bin p:1 1 2 1

2 2 1 2

U E I N

U E I N = k

+ Nu k > 1 N 1> N2U

1> U

2: My h p

+ Nu k < 1 N 1< N2U

1< U

2: My tng p

i vi bi ton ny khi thay i s vng dy cc cun s cp (N 1) hay thay i cun th cp (N2) u

nh hng n U 1v U2.

Hiu sut my bin p: H =P 2P 1

=U 2I

2cos

2

U 1I1cos

1

Bi 9 : NHNG LU KHI GII BI TP IN XOAY CHIU

Hai on mch AM gm R1L1C1ni tip v on mch MB gm R2L2C2ni tip mc ni tip vi nhau cUAB= UAM+ UMB uAB; uAMv uMBcng pha tanuAB= tanuAM= tanuMB

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Hai on mch R1L1C1v R2L2C2cng uhoc cng ic pha lch nhau

+ Vi 1 11

1

tan L C

Z Z

R

v 2 22

2

tan L C

Z Z

R

(gi s 1> 2)

C 12= 1 21 2

tan tantan

1 tan tan

+ Trng hp c bit = /2 (vung pha nhau) th tan1tan2= -1.

Mch in xoay chiu mt pha cha cc phn t R-L-C+ = ocos(t + ) (vi

o= NBS : t thng cc i v l gc hp gia php tuyn n v cm ng t B)

+ c bit sut in ng to ra in p xoay chiu l e = - ' = NBS.cos(t + ) ( vi E o= NBS)

TH1: Mch ch c L: khi : I =E

Z L=

NBSL

=NBS

L= hng s

D c thay i tc quay n th no ? th I khng i.

TH2: Mch ch c C: khi I =E

Z C=

NBS1

C

= NBSC2

Nu tc quay tng n ln th I tng n2 ln

TH3: Mch gmL-R. v I =E

Z LR=

E o

2 R2 + ZL

2

=NBS

2 R2 + ZL

2

, t =NBS

2= const I =

R2 + Z

L

2

Mi lin h nm ch tc quay rto t l vi , t l vi ZLtheon 1n 2

=f1f2

=

1

2=

Z L1Z L2

=Z C2Z C1

( Cc trng hp cn li tng t)

Ni hai cc ca mt my pht in xoay chiu mt phavo hai u on mch ngoi RLC ni tip. Bqua in tr dy ni, coi t thng cc i gi qua cc cun dy ca my pht khng i. Khi rto ca my

pht quay vi tc n1 vng/phtv n2vng/pht th cng sut tiu th mch ngoi c cng mt gitr. Khi rto ca my pht quay vi tc n vng/pht th cng sut tiu th mch ngoi t cc i.

Khi :1

1

2+

1

22=

2

o

2

hay1

f12+

1f2

2

=2

fo2

hay1

n 12

+1

n 22

=2

n o2

Cho mch in xoay chiu gm cun cm (L,r) ni tip vi t in, c cm khng v dung khng ln lt lZLv ZC. Bit in p ga hai u cun dy vung pha vi hai in p hai u mch. H s cng sut mch

c tnh:Cos =Z LZ C

S Vung Phaca 2 thnh phn bt k trong mch R-L-C.

TH1:Nu mch ch c C. Khi u Ci i2I o

2+

u2U o

2

= 1

Chng minh: Gi s u = U ocost (1). Do i sm pha hn u 1 gc /2 i= /2

i = I ocos(t + /2) = - Iosint (2).

T (1) u2

U o2= cos2t ,

i2I o

2

= sin2t. Cng v ta ci2I o

2

+u2

U o2= 1.

TH2:Nu mch ch c L. Chng minh tng t ta ci2I o

2

+u2

U o2

= 1

TH3: Nu u RCu u

RC

2

U oRC

2

+ u

2

U o

2= 1

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Gieo hnh vi gt thi quen. Gieo thi quen gt tnh cch. Gieo tnh cch gt s phn. ( Dick Lyles)10

TH4: Nu u RLu u RL

2

U oRL2+

u2U o

2

= 1 ( C nh vy ta m rng ra cc trng hp c s vung gc )

Bi ton lin quan n lch pha vung gc:+ TH1: 1+

2= 90

o tan

1.tan

2= 1

+ TH2: 1- 2= 90

o tan

1.tan

2= -1

+ TH3: | 1| + |2| = 90

o tan

1.tan

2= 1

Cc xc nh phn ttrong bi ton hp enX.+ Mch in n gin:

a. Nu on NB cng pha vi i X cha Ro

b. Nu on NB sm pha vi i mt gc /2 X cha cun cm L.c. Nu on NB tr phavi i mt gc /2 X chat in C.

+ Mch in phc tp:a. Mch 1

NuU ABcng pha vii X chchaL

NuU

ANU

NBX chaR

Vy X cha R v L

b. Mch 2

NuU ABcng pha vii X chaCNuU ANU

NBX chaR

Vy X cha R v C

+ Mch inc cun dy cha bit thun cm hay cha thun cm ?Nu cun dy lch pha vi i 1 gc = /2 cun dy ch c L

Nu cun dy lch pha vi 1 gc vi 0 < < /2 Cun dy c L v r( C th dng gin vecto trt chng minh. Thng ta gi s L thun cm ri bin lun )

CHC CC EM N TP HIU QU V T KT QU CAO NHT TRONG KTHI TUYN SINH I HC 2014- 2015

[email protected] - [email protected]

R

C

XA N B

R

L C

XA N B

R

L

XA N B

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