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TRN QUANG THANH-K15-CH L H- VINHPHNG PHP DNG GIN VC T ( U -UI)

GII BI TP IN XOAY CHIUt vn : Ta bit khi gii bi tp in xoay chiu choon mch R, L , C khng phn nhnh , th trong 1 s bi tpyu cu cn phi v c gin vc t mi tm c cc i

lng cha bit. Tuy nhin iu ny khng phi d nu chngta khng nm c c im , tnh cht ca tng phn t mctrong mch . C 2 phng php v gin vc t , lphng php v chung gc v phng php v u ui . . Khigii bi tp ch c 1 phn t R, L, C trong on mch th vchung gc l n gin. Tuy nhin nu trong onh mch cnhiu hn 2 phn t , R,L , C th cch v u ui li hay hnc . Bng phng php thc nghim trong ging dy ti thy as cc em hc sinh khi gp bi tp dng ny u rt ngi.Nhng mt khi cc em nn c phng php v chung gc

th bi tan tr nn n gin hn. Trong gi hn cho php tixin mnh dn trnh by phng php u - ui. Hy vng ccem v cc ng nghip thy hu ch v cho kin phn hi.Mi thc mc lin lc theo a chemail:[email protected] hoc 0904.727271. hoc0383.590194. Xin chn thnh cm n

C S L THUYT :1.Dng in xoay chiu trong mch ch c R , hoc L, hoc C.a.

Mch ch c R: UR v i cng pha vi nhau . Nn trn gin vc t chng cng nm trn 1 ng thng hoc song songvi nhau .

R

ui R=

R

UI R0

0=

v o= b.Mch ch c L :Th U lun nhanh pha hn i mt gc 2

hay 2

=L V trn gin vc t UL lun vung gc vi

trc i

R

L

IO

UL

IURO


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TRN QUANG THANH-K15-CH L H- VINH

L

L

Z

ui =

:L

OL

Z

UI =0

c.Mch ch c CU lun chm pha hn i mt gc 2

hay 2

=C trn gin

vc t UC lun vung gc vi trc i nhng hng xung

C

C

Z

ui =

C

OC

Z

UI =0

2.Dng in xoay chiu trong mch khng phn nhnh R, L, C

CLRNBMNAMAB UUUUUUUrrrrrrr

++=++=

Hay : ABCLAB ZIZZRIU .)(.22=+=

TH1: Mch c tnh cm khng : (ZL>ZC)

CHUNG GC

M N B

C

IO

UC

OABUr

OLUr

CL UUrr

+

I

CU

r

RUr

O


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U UI: ch : vi cch v u ui th ui ca phn t

ny l u ca phn t kia v cc ch ci BNMA nitip nhau . Cui cng ta ni AB li ta c U

AB, nh l nu trong

on AM v UR th on tip sau m c UR v UL th nn vUL trc cho thun tin .

TH2: Mch c tnh dung khng(ZL


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u ui

lch pha gia U v I l : R

ZZ

U

UU

tgCL

CL

=

=

H s cng sut :ABAB

R

Z

R

U

Uk === cos

3.on mch ch cha 2 phn t RL ; RC; LCL cc trng hp ring ca on mch R, L , C khi khng c 1trong cc phn t C, L, R trong mch . Khi gii cc loi onmch ny ta vn dng cc cng thc v gin vv t cho on

mch R.L.C nhng b i cc i lng v vc t tng ng vicc phn t b thiu. C th :a.on mch RL(thiu C)Tng t :

LAB ZRZ22

+=

LRAB UUU22

+=

RZtg L= v 2


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b. on mch R, C (thiu L) CAB ZRZ22

+=

CRAB UUU22

+= v RZtg C=

v0

2


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khi ZL>ZC khi Zl


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Bi gii:Chn trc I lm trc pha ta c gin vc t :Ch : UR=5 ; UL=9 ; UAB=13

AM=5 ; MN=9 ; AB=13

222 MBAMAB += = AM2 + (NB-NM)2Hay :

222)( CLRAB UUUU +=

Hay222)( CLRAB UUUU =

Thay s :222

)(513 CL UU = Vy UL-UC=12 hoc UL-UC=- 12 . Do mch c tnh dungkhng nn ZC>ZL hay UC>UL Suy ra ly UL-UC=- 12 Suy raUC=UL + 12 = 9+12=21(V)

Bi 2: Cho mch in xoay chiu : )100sin(290 tUAB = (V)Cc my o khng nh hng ng k n dng in trongmch. Vn k V1 ch 120(V) , Vn k V2 ch 150(V) . Cho

tg370=3/4. Tm lch pha ca UAB i vi I ?

A.037= B.

045= C.060= D.

090=

Bi gii : Nhn xt :Do Hiu in th hiu dngUAB=90(V) nnGa s cun dy thun cm (R=O)

A NM

12 3

NUL

UC

A

UAB

IUR M

B

A

1 2

MN B


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TRN QUANG THANH-K15-CH L H- VINHth :

CLAB UUU =

Nhng theo bi ra :15012090

Nn cun dy c R khc

O . Chn trc I lm trc pha ta c gin vc t :

Nhn vo hnh v ta dng nhl o pitago chng minhc rng tam gic AMBvung ti A suy ra =

(gc c cp cnh

tng ng vung gc)AM=120 ; MN=150AB=90

Vy : 4

3

120

90===

AM

ABtg

Suy ra0

37==

Bi 3: Cho mch nh hnh v : )100sin(225 tUAB = . Vn k V

ch 12(V) ; Vn k V2 ch 17(V) . Cho cos370=4/5.Tm lchpha ca UAB so vi I

A.0

37= B.0

45= C.0

60= D.0

90=

Bi gii :

Nhn xtAM=12

MB=17 ; AB= 25Chn trc I lm trc phata c gin vc t( ch : sau im M ta nnv tip UL ch khng nnv tiP UR2)

p dng nh l hm s cosin

N

UR

UL

M

UC

UAB

B

I

1

R1

MA B

R2, L

21


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TRN QUANG THANH-K15-CH L H- VINHcho tam gic nhn ABMta c :

BM2= AM2+AB2-2.AM.AB. cos(MAB)

Hay : cos...2 12

12

22

UUUUU += Thay s :

5

4

25.12.2

172512

..2cos

222

12

222

12

=+

=+

=

UU

UUU

Suy ra0

37=

Bi 4: Cho 2 cun dy (R1; L1) v (R2; L2) mc ni tip . Tmmi lin h gia R1;L1; R2 ; L2 tng tr on mch AB tha

mn : ZAB=Z1+Z2 ( Z1, v Z2 l tng tr ca cun dy 1 v 2)

A.2

1

2

1

L

L

R

R=

B.1

2

2

1

L

L

R

R=

C. 2.12

1 .LLR

R=

D. 2.121 .. LLRR =

Bi gii : Ta c :

ZAB=Z1+Z HayIO.ZAB=I0.Z1+I0.Z2

Tng ng :

U0AB=U01+U02

c th cng bin cc hiu in th th cc thnh phn U1v U2 phi cng pha . C ngha l trn gin vc t chngphi cng nm trn mt ng thng. Chn trc I lm trc phata c gin vc t :

B

I

ULUAB

UR2

UMB=17

A

UR1

M

M

R2,L2R1.L1


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1

Trn hnh v 3 im A,M, B thng hnghay ni cch khc U1; U2; v UAB cng phatam gic AHM ng dng tam gic MKB nn tac cc t s ng dng sau:

BK

MK

MH

AH=

Hay2

1

2

1

L

L

R

R

U

U

U

U=

Hay2

1

2

1

L

L

R

R=

Bi 5: Cho mch nh hnh v : )100sin(2 tUuAB = (V)Vn k V1 ch 40(V) ; Vn k V2 ch 90(V) ; Vn k V3 ch120(V) . Tm s ch vn k V?A. 50(V) B. 70(V) C.100(V) D.200(V)Bi gii :

V1 ch UR=40 ; V2 ch UL=90 ; V3 ch UC=120 ; V ch UAB=?Chn trc I lm trc pha ta c gin vc t :

AM= 40; MN=90; NB= 120

Xt tam gic AMB c :

AB2=AM2+BM2

Hay : U2AB=U2

R+(UL-UC)2

Thay s U2AB=402+(90-120)2

H

MK

I

UR1

UL1

UR2

UL2

U1

U2

NM

12 3

B

A MRU

r

N

UL

UC

UAB

B


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1

Vy UAB= 50(V)

Bi 6: Cho mch nh hnh v : f=50(Hz) Vn k V1 ch 70 (V)V2 ch 100(V). Hiu in th U2 hai u cun dy lch pha450 so vi cng dng in trong mch ,. Tnh hiu in thhiu dng UAB ?

A. 50(V) B. 70(V) C.158(V) D.200(V)

Bi gii : Chn trc Ilm trc phata c gin vc t :

AM=70= 250 ; BM=100

Xt tam gic AMB dng nh lhm s cosin ta c :

22

2

222 co...2)cos(...2 BMAMBMAMBMAMBMAMAB ++=+=

Do gc )( == AMB

Thay s : Vi0

2 45= Do U2 sm pha hn I mt gc 450

022

245cos.100.250.2100250 ++=OABU

Hay : UOAB=158(V)Bi 7: Cho vn k V1 ch 120 (V) , Vn k V2 ch 150(V) , v

U1 lch pha 530 so vi dng in. Tm s ch ca vn k V ? (cho tg530=4/3)?A. 50(V) B. 90(V) C.158(V) D.200(V)

R1MA B

R2, L

ULUAB

UR2

2 UR1 M

1 2

AM

N BR,L


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1

Bi gii : Chn trc I lm trc pha ta c gin vc t :

p dng nh l hm s cosin

cho tam gic AMB ta c :

022237cos...2 BMAMBMAMAB +=

Hay :0

2122

122

37cos...2 UUUUU AB +=

Thay s : )(9037cos.150.120.21501200222

VUU ABAB =+=

Bi 8: Cho mch nh hnh v : )100sin(2100 tuAB = , Vn k Vch 100(V), vn k V2 ch 100(V). ampe k ch 2(A) . Vit biu

thc cng dng in .

A. )100sin(22 ti = B.)

6100sin(22

+= ti

C. )100sin(2 ti = D.)

6100sin(22

= ti

Bi gii: nhn xt : do CLAB UUU nn trong cun dy c

cha in tr R .AM=MB=AB=100Chn trc I lm trc phata c gin vc t : nhn vogin vt t ta thy I nhanh pha

hn UAB mt gc 6

(Do tam gic AMB u ) Suy ra

URA

B

M

UC

UAB

ULU1

30370

I

MA B

12


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1

6

=

. Vy biu thc)

6100sin(22

+= ti

Bi 9: Cho mch nh hnh v : )100sin(2100 tuAB = , Vn k Vch 100(V) , Hiu in th UAM v UMB vung pha nhau. Vit

biu thc UAM v UMB ?Bi gii : Ga s cun dy

thun cm(R=0) th

CLAB UUU = iu ny c ngha lUAM v UMB cng phngngc chiu nhau( tri vi gi thit l 2 U nyvung pha nhau).Vy cun dy c R khc O . Chn trc I lm trc pha ta c gin vc t . Vi AM=100; AB=100

Chn )100sin(2100 tuAB = lm trc pha gc : lch pha gia

UAM v I l R

Ztg L=

1

20

1


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