Chuong3_ptrinh Mach Dien

7/30/2019 Chuong3_ptrinh Mach Dien

1/19

_______________________________________________Chng 3 Phng trnh mchin - 1

Chng3PHNG TRNH MCH IN

KHI NIM V TOPO Mt snh ngha

nh l v topo mch PHNG TRNH NT Mch cha ngun dng in Mch cha ngun hiu th

PHNG TRNH VNG Mch cha ngun hiu th

Mch cha ngun dng in BIN I V CHUYN V NGUN

Bin i ngun Chuyn v ngun

__________________________________________________________________________________________

Trong chng ny, chng ta gii thiu mt phng php tng qut gii cc mchin tng i phc tp. l cc h phng trnh nt v phng trnh vng. Chng ta cng cp mt cch slc cc khi nim cbn v Topo mch, phn ny gip cho vic thit lpcc h phng trnh mt cch c hiu qu.

3.1 Khi nim v Topo MCH

Trong mt mch, n s chnh l dng in v hiu th ca cc nhnh. Nu mch c Bnhnh ta c 2B n s v do cn 2B phng trnh c lp gii. Lm th no vit v

gii 2B phng trnh ny mt cch c h thng v t c kt qu chnh xc v nhanh nht, l mc ch ca phn Topo mch.Topo mch ch n cch ni nhau ca cc phn t trong mch m khng n

bn cht ca chng.

3.1.1. Mt snh ngha

Gin thng

v gin thng tng ng ca mt mch ta thay cc nhnh ca mch bi ccon thng (hoc cong) v cc nt bi cc du chm.

(a) (b)

(H 3.1)

___________________________________________________________________________Nguyn Trung Lp L THUYTMCH


2/19


Trong gin cc nhnh v nt c t tn hoc nh s th t. Nu cc nhnh cnh hng (thng ta ly chiu dng in trong nhnh nh hng cho gin ), ta c gin hu hng.

(H 3.1b) l gin nh hng tng ng ca mch (H 3.1a). Gin con

Tp hp con ca tp hp cc nhnh v nt ca gin . Vng

Gin con khp kn. Mi nt trong mt vng phi ni vi hai nhnh trong vng .Ta gi tn cc vng bng tp hp cc nhnh to thnh vng hoc tp hp cc nt thuc vng.

Th d:(H 3.2a): Vng (4,5,6) hoc (a,b,o,a).(H 3.2b): Vng (1,6,4,3) hoc ( a,b,o,c,a).

(a) (b)(H 3.2)

Cy

Gin con cha tt c cc nt ca gin nhng khng cha vng.

Mt gin c th c nhiu cy.Th d:(H 3.3a): Cy 3,5,6 ;(H 3.3b): Cy 3,4,5 . . ..

(a) (b)(H 3.3)

* Cch v mt cy: Nhnh th nht c chn ni vi 2 nt, nhnh th hai ni 1trong hai nt ny vi nt th 3 v nhnh theo sau li ni mt nt na vo cc nt trc. Nhvy khi ni N nt, cy cha N-1 nhnh.

Th d v cy ca (H 3.3b) ta ln lt lm tng bc theo (H 3.4).

___________________________________________________________________________(H 3.4)

Nguyn Trung Lp L THUYTMCH


3/19


phn bit nhnh ca cy vi cc nhnh khc trong gin , ngi ta gi nhnh cacy l cnh v cc nhnh cn li gi l nhnh ni. Cnh v nhnh ni ch c ngha sau khi chn cy.

Gi L l s nhnh ni ta c:

B = (N - 1) + L

Hay L = B - N +1 (3.1)Trong B l s nhnh ca gin , N l s nt.Trong gin trn hnh 3.1 : B = 6, N = 4 vy L = 6 - 4 + 1 = 3

Nhn thy, mt cy nu thm mt nhnh ni vo s to thnh mt vng c lp ( lvng cha t nht mt nhnh khng thuc vng khc ).

Vy s vng c lp ca mt gin chnh l s nhnh ni L.

3.1.2. nh l v Topo mch

Nhc li, mt mch gm B nhnh cn 2B phng trnh c lp gii, trong Bphng trnh l h thc v - i ca cc nhnh, vy cn li B phng trnh phi c thit lp t

nh lut Kirchhoff .nh l 1:

Gin c N nt, c (N -1) phng trnh c lp do nh lut KCL vit cho (N-1) ntca gin .

Tht vy, phng trnh vit cho nt th N c th suy t (N-1) phng trnh kia.nh l 2

Hiu th ca cc nhnh (tc gia 2 nt) ca gin c th vit theo (N-1) hiu thclp nhnh lut KVL.

Tht vy, mt cy ni tt c cc nt ca gin , gia hai nt bt k lun c mt

ng ni ch gm cc cnh ca cy, do hiu th gia hai nt c th vit theo hiu th cacc cnh ca cy. Mt cy c (N - 1) cnh, vy hiu th ca mt nhnh no ca gin cngc th vit theo (N-1) hiu thc lp ca cc cnh.

Trong th d ca (H 3.1), cy gm 3 nhnh 3, 4, 5 c bit quan trng v cc cnh can ni vi mt nt chung O, O gi l nt chun. Hiu th ca cc cnh l hiu th gia ccnt a, b, c (so vi nt chun). Tp hp (N - 1) hiu th ny c gi l hiu th nt.

Nu mch khng c c tnh nh trn th ta c th chn mt nt bt k lm nt chun.nh l 3

Ta c L = B - N +1 vng hay mt li c lp vi nhau, trong ta c th vit phngtrnh tnh lut KVL.

nh l 4Mi dng in trong cc nhnh c thc vit theo L = B - N +1 dng in c lpnhnh lut KCL.

Cc vng c lp c c bng cch chn mt cy ca gin , xong c thm 1 nhnhni vo ta c 1 vng. Vng ny cha nhnh ni mi thm vo m nhnh ny khng thucmt vng no khc. Vy ta c L = B - N + 1 vng c lp. Cc dng in chy trong ccnhnh ni hp thnh mt tp hp cc dng in c lp trong mch tng ng .

Th d: Trong gin (H 3.1b), nu ta chn cy gm cc nhnh 3,4,5 th ta c ccvng c lp sau y:



4/19


(H 3.5)Mt phng php khc xc nh vng c lp l ta chn cc mt li trong mt

gin phng (gin m cc nhnh ch ct nhau ti cc nt). Mt li l mt vng khngcha vng no khc. Trong gin (H 3.1b) mt li l cc vng gm cc nhnh: (4,5,6),(2,3,4) & (1,2,6).

Mt mt li lun lun cha mt nhnh khng thuc mt li khc nn n l mtvng c lp v s mt li cng l L.

Cc nh l trn cho ta B phng trnh gii mch :Gm (N-1) phng trnh nt v (L = B - N + 1) phng trnh vng.

V tng s phng trnh l:(N-1) + L = N - 1 + B - N + 1 = B

3.2 Phng trnh Nt

3.2.1 Mch ch cha in trv ngun dng in

Trong trng hp ngoi in trra, mch ch cha ngun dng in th vit phngtrnh nt cho mch l bin php d dng nht gii mch. Chng ta lun c th vit phngtrnh mt cch trc quan, tuy nhin nu trong mch c ngun dng in ph thuc th ta cn

c thm cc h thc din t quan h gia cc ngun ny vi cc n s ca phng trnh miiu kin gii mch. Ngun dng in c lp:

Nu mi ngun trong mch u l ngun dng in c lp, tt c dng in cha bitc th tnh theo (N - 1) in th nt. Ap dng nh lut KCL ti (N - 1) nt, tr nt chun, tac (N - 1) phng trnh c lp. Gii h phng trnh ny tm hiu th nt. T suy racc hiu th khc.Th d 3.1:

Tm hiu th ngang qua mi ngun dng in trong mch (H 3.6)

(H 3.6)Mch c 3 nt 1, 2, O; N = 3 vy N - 1 = 2, ta c 2 phng trnh c lp.

Chn nt O lm chun, 2 nt cn li l 1 v 2 .v

1 vv

2 chnh l hiu th cn tm.Vit KCL cho nt 1 v 2.



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Nt 1: 024

5 211 =

++vvv

(1)

Nt 2: 026322212 =+++

vvvv(2)

Thu gn:

521

21

41 21 =

+ vv (3)

26

1

3

1

2

1

2

121 =

+++ vv (4)

Gii h thng (3) v (4), ta c :v1 = 8 (V) v v2 = 2 (V)

Thit lp phng trnh nt cho trng hp tng qutXt mch ch gm in trR v ngun dng in c lp, c N nt. Nu khng k

ngun dng in ni gia hai nt j v k, tng s dng in ri nt j n nt k lun c dng:

Gjk(vj - vk) (3.2)Gjkl tng in dn ni trc tip gia hai nt j , k ( j k ) gi l in dn chung gia hai nt j, k ; ta c:

Gjk= Gkj (3.3)

Gi ij l tng i s cc ngun dng in ni vi nt j.nh lut KCL p dng cho nt j:

( ) =k

jkjjkG ivv (ij > 0 khi i vo nt j )

Hay jk k

kjkjkj GG ivv = ( j k ) ( 3.4)

Gjkk

: L tng in dn ca cc nhnh c mt u ti nt j. Ta gi chng l in

dn ring ca nt j v k hiu:

(3.5)=k

jkjj GG

Phng trnh (3.4) vit li:

(3.6)( kjGG jk

kjkjjj = ivv )

Vit phng trnh (3.6) cho (N - 1) nt ( j = 1, ..., N - 1 ), ta c h thng phng trnhNt 1: G11v1 - G12v2 - G13v3 . . . - G1(.N-1)vN-1 = i1Nt 2: - G21 v1 + G22 v2 - G23 v3 . . . - G2.(N-1) vN-1 = i2:::

Nt N -1: - G(N-1).1 v1 - G(N-1).2 v2 . . . +G(N-1)(.N-1) vN-1 = iN-1

Di dng ma trn:



6/19


=

1N

2

1

1N

2

1

11.NN1.2N1.1N

12.N2221

11.N1211

:

:

:

:

:

:

G...............GG-

:::

:::

:::

G...............GG-

G...............GG

i

i

i

v

v

v

Hay

[G][V] = [I] (3.7)[G]: Gi l ma trn in dn cc nhnh, ma trn ny c cc phn ti xng qua ng chochnh v cc phn t c th vit mt cch trc quan t mch in .[V]: Ma trn hiu th nt, phn t l cc hiu th nt.[I]: Ma trn ngun dng in c lp, phn t l cc ngun dng in ni vi cc nt, c gitr dng khi i vo nt.

Trli th d 3.1:

G11 =2

1

4

1+ ; G22 =

6

1

3

1

2

1++ ; G12 =

2

1

i1 = 5A v i2 = - 2AH phng trnh thnh:

=

++

+

2

5

6

1

3

1

2

1

2

1

2

1

2

1

4

1

2

1

v

v

Ta c kt qu nh trn. Ngun dng in ph thuc :

Phng php vn nh trn nhng khi vit h phng trnh nt tr s ca ngun dng in nyphi c vit theo hiu th nt gii hn sn s vn l N-1. Trong trng hp ny matrn in dn ca cc nhnh mt tnh i xng.

Th d: 3.2Tn hiu th ngang qua cc ngun trong mch (H 3.7).

(H 3.7)Ta c th vit phng trnh nt mt cch trc quan:



7/19


=

+++

=

+

321

21

36

1

3

1

2

1

2

1

52

1

2

1

4

1

ivv

vv

(1)

H thng c 3 n s, nh vy phi vit i3 theo v1 v v2.

221

3

vvi

= (2)

Thay (2) vo (1) v sp xp li

52

1

4

321 = vv & 0

2

121 = vv

v1 = - 20 (V) v v2 = - 40 (V)

Th d 3.3Tnh v2 trong mch (H 3.8).

(H 3.8)Chn nt chun O, v1 & v2 nh trong (H 3.8)H phng trnh nt l:

=

++

+=

+

321

321

9

11

412

1

ivv

ivv

(1)

Vi i3 = 5v1 (2)Ta c :

=+

=

09

104

42

7

21

21

vv

vv

(3)

Suy ra :v2 = - 114 (V)

3.2.2 Mch ch cha in trv ngun hiu th

Ngun hiu thc lp



8/19


Nu mt nhnh ca mch l 1 ngun hiu th c lp, dng in trong nhnh khng th tnh d dng theo hiu th nt nh trc. V hiu th ca ngun khng cn l n snn ch cn (N-2) thay v (N-1) hiu th cha bit, do ta ch cn (N-2) phng trnh nt,vit nhnh lut KCL gii bi ton. c (N-2) phng trnh ny ta trnh 2 nt ni vingun hiu th th dng in chy qua ngun ny khng xut hin.

Cui cng, tm dng in chy trong ngun hiu th, ta p dng nh lut KCL ti

nt lin h vi dng in cn li ny, sau khi tnh c cc dng in trong cc nhnh ti ntny.

Th d 3.4Tnh v4 v in trtng ng nhn t 2 u ca ngun hiu thv1 trong (H 3.9).

(H 3.9)Mch c N = 4 nt v mt ngun hiu thc lp. Chn nt chun O v nt v1 ni vi

ngun v1 = 6 V nn ta ch cn vit hai phng trnh cho nt v2 v v3.Vit KCL ti nt 2 v 3.

=+

+

=

++

024

6

1

0121

6

3323

3222

vvvv

vvvv

(1)

Thu gn:

=+

=

2

3

4

7

62

5

32

32

vv

vv

(2)

Gii h thng (2):

v2 =9

32V v v3 =

9

26V

v4 = v2 - v3 =3

2 V

Dng i1 l tng cc dng qua in tr1 v 4.

9

29

9

7

9

22

4

6

1

6 321 =+=

+

=

vvi A

in trtng ng:

Rt =29

54

9

29

6=

Rt =29

54



9/19


Chng ta cha tm c mt phng php tng qut vit thng cc phng trnhnt trong nhng mch c cha ngun hiu th.

Trong thc t ngun hiu th thng c mc ni tip vi mt in tr(chnh l nitrca ngun) nn ta c th bin i thnh ngun dng in mc song song vi in tr(bin i Thevenin, Norton).

Nu ngun hiu th khng mc ni tip vi in ta phi dng phng php chuyn v

ngun trc khi bin i ( cp trong mt phn sau ).Sau cc bin i, mch n gin hn v ch cha ngun dng in v ta c th vit hphng trnh mt cch trc quan nh trong phn 3.2.1.

Trong th d 3.3 trn, mch (H 3.9) c th v li nh (H 3.10a) m khng c gthay i v bin cc ngun hiu th ni tip vi in trthnh cc ngun dng song song viin trta c (H 3.10b).

(H 3.10)V phng trnh nt:

612

11 32 =

++ vv

- v2 + 1,512

1

4

13 =

++ v

Gii h thng ta tm li c kt qu trn. Ngun hiu th ph thuc :

Ta cn mt phng trnh ph bng cch vit hiu th ca ngun ph thuc theo hiuth nt.

Th d 3.5Tm hiu thv1 trong mch (H 3.11)

(H 3.11)Mch c 4 nt v cha 2 ngun hiu th nn ta ch cn vit 1 phng trnh nt cho nt

b. Chn nt O lm chun, phng trnh cho nt b l:04

3

2

1

24 1bb =

+ vvv

(1)



10/19


Vi phng trnh ph l quan h gia ngun ph thuc v cc hiu th nt:

1b -24 vv = (2)Thay (2) vo (1), sau khi n gin:

v1=2 (V)

3.3 Phng trnh Vng

Mch c B nhnh, N nt c th vit L = B - N + 1 phng trnh vng c lp . Midng in c th tnh theo L dng in c lp ny.

3.3.1 Mch ch cha in trv ngun hiu th

Ngun hiu thc lp :

Nu mch ch cha ngun hiu thc lp, cc hiu th cha bit u c th tnh theoL dng in c lp.

p dng KVL cho L vng c lp (hay L mt li) ta c L phng trnh gi l hphng trnh vng. Gii h phng trnh ta c cc dng in vng ri suy ra cc hiu thnhnh t h thc v - i.Th d 3.6: Tm cc dng in trong mch (H 3.12a).

(a) (b) (c)

(H 3.12)Mch c N = 5 v B = 6Vy L = B - N + 1 = 2

Chn cy gm cc ng lin nt (H 3.12b). Cc vng c c bng cch thm ccnhnh ni 1 v 2 vo cy.

Dng in i1 v i2 trong cc nhnh ni to thnh tp hp cc dng in c lp. Ccdng in khc trong mch c th tnh theo i1 v i2.

Mt khc, thay v ch r dng in trong mi nhnh, ta c th dng khi nim dng

in vng. l dng in trong nhnh ni ta tng tng nh chy trong c vng c lpto bi cc cnh ca cy v nhnh ni (H 3.12c).Vit KVL cho mi vng:

(1)

0=24+4+)-6(+2

0=60-3+)-6(

2122

121

iiii

iii

Thu gn:

(2)( )

=+++ 246426-

60=6-3)+6(

21

21

ii

ii

Gii h thng ta c :

i1 = 8A v i2 = 2ADng qua in tr6: i1 - i2 = 6 (A)



11/19

_______________________________________________Chng 3 Phng trnh mchin - 11 Thit lp phng trnh vng cho trng hp tng qut

Coi mch ch cha in trv ngun hiu thc lp , c L vng.Gi ij, ik...l dng in vng ca vng j, vng k ...Tng hiu th ngang qua cc in trchungca vng j v k lun c dng:

Rjk( ijik) (3.8)Du (+) khi ij v ikcng chiu v ngc li.Rjkl tng in trchung ca vng j v vng k. Ta lun lun c:

Rjk= Rkjvj l tng i s cc ngun trong vng j, cc ngun ny c gi tr (+) khi to ra dng in cngchiu ij ( chiu ca vng ).p dng KVL cho vng j:

( ) =k

jkjjkR vii (3.10)

Hay (3.11)jvii = kk

jkk

jkj RR

Rjkk chnh l tng in trchung ca vng j vi tt c cc vng khc tc l tng in tr

c trong vng j.

t = RRjkk

jj v vi qui c Rjkc tr dng khi ij v ikcng chiu v m khi ngc li,

ta vit li (3.11) nh sau:

Rjjij + (3.12)jkk

jkR vi =i vi mch c L vng c lp :Vng 1 : R11i1 + R12i2 + . . . . R1LiL = v1

Vng 2 : R21i1 + R22i2 + . . . . R2LiL = v2: : : : :: : : : :

Vng L: RL1i1 + RL2i2 + . . . . RLLiL = vLDi dng ma trn

=

L

2

1

2

1

.2L.1

2.2221

1.1211

:

::

:

::

.....R..........RR

:::

::::::

.....R..........RR

.....R..........RR

v

v

v

i

i

i

LLLL

L

L

H phng trnh vng vit di dng vn tt:[R] .[I] = [V] (3.13)

[R]: Gi l ma trn in tr vng c lp. Cc phn t trn ng cho chnh lun lundng, cc phn t khc c tr dng khi 2 dng in vng chy trn n cng chiu, c tr mkhi 2 dng in vng ngc chiu. Cc phn t ny i xng qua ng cho chnh.[I] : Ma trn dng in vng[V]: Ma trn hiu th vng



12/19

_______________________________________________Chng 3 Phng trnh mchin - 12Trli th d 3.6 ta c th vit h phng trnh vng mt cch trc quan vi cc s liu sau:

R11 = 3 + 6 = 9 ,

R22 = 2 + 4 + 6 = 12 ,

R21 = R12 = - 6 ,

v1 = 60 Vv

v2 = - 24 (V) Ngun hiu th ph thuc

Nu mch c cha ngun hiu th ph thuc, tr s ca ngun ny phi c tnh theocc dng in vng. Trong trng hp ny ma trn in th mt tnh i xng.

Th d 3.7 Tnh i trong mch (H 3.13)

(H 3.13)Vit phng trnh vng cho cc vng trong mch

6i1- 2 i+ 4i2=15 (1)

4i1+ 2 i+ 6i2= 2 i (2)-2i1+ 8 i+ 2i2=0 (3)

(2) cho 212

3ii = (4)

(3) cho4

21 iii

= (5)

Thay (5) vo (1)11i1+ 9i2=30 (6)

Thay (4) vo (6) ta ci2=- 4 Ai1= 6 A

V i = 2,5 (A)

3.3.2. Mch cha ngun dng in

Ngun dng in c lpNu mt nhnh ca mch l mt ngun dng in c lp, hiu th ca nhnh ny kh

c th tnh theo dng in vng nh trc. Tuy nhin nu mt dng in vng duy nht cv qua ngun dng in th n c tr s ca ngun ny v ch cn (L-1) n s thay v L (bngcch khng chn nhnh c cha ngun dng lm cnh ca cy).

Th d 3.8: Tnh dng in qua in tr2 trong mch (H3.14a)



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(a) (H 3.14) (b)Mch c B = 8, N = 5, cy c 4 nhnh v 4 vng c lp .Chn cy nh (H 3.14b) (nt lin), cnh ca cy khng l nhnh c cha ngun dng c lp.Ta c:

i3 = 10 A v i4 = 12 AVit phng trnh vng cho hai vng cn li.Vng 1: ( 4 + 6 + 2 )i1 - 6i2 - 4i4 = 0 (1)

Vng 2: - 6i1 + 18i2 + 3i3 - 8i4 = 0 (2)Thay i3 = 10 A v i4 = 12 A vo (1) v (2)

12i1 - 6i2 = 48

- 6i1 + 18i2 = 66Suy ra i1 = 7 (A)

Th d trn cho thy ta vn c th vit c h phng trnh vng cho mch changun dng in c lp. Tuy nhin ta cng c th bin i v chuyn v ngun (nu cn) c mch cha ngun hiu th v nh vy vic vit phng trnh mt cch trc quan d dnghn.

Mch (H 3.14a) c th chuyn di v bin i ngun c mch (H 3.15) di

y.

(a) (H 3.15) (b)Vi mch (H 3.15b), ta vit h phng trnh vng.Vng 1: 12i1 - 6i2 = 48Vng 2: - 6i1 + 18i2 = 96 - 30Ta c li kt qu trc. Ngun dng in ph thucTm v1 trong mch (H 3.16)



14/19


(a) (b) (c)

(H 3.16)Mch c B = 5, N = 3 cy c hai cnh v 3 vng c lp .Chn cy l ng lin nt ca (H 3.16b). Cc ngun dng in nhnh ni

Vit phng trnh cho vng 326i3 + 20i2 + 24i1 = 0 (1)

Vi i1 = 7A v i2= 31

4

1

8i

v= (2)

Thay (2) vo (1)26i3 - 5i3 + 168 = 0 i3 = - 8 (A) v v1= 16 (V)

3.4 Bin i v chuyn v ngun

Cc phng php bin i v chuyn v ngun nhm mc ch sa son mch chovic phn gii c d dng. Mch sau khi bin i hoc phi n gin hn hoc thun tinhn trong vic p dng cc phng trnh mch in .

3.4.1. Bin i ngun

Ngun hiu th ni tip v ngun dng in song song (H 3.17).

(H 3.17) Ngun hiu th song song v ngun dng in ni tip.Ta phi c: v1 = v2 v i1 = i2.

(H 3.18)Ngun hiu th song song vi in trv ngun dng in ni tip in tr: C th bintrm khng nh hng n mch ngoi.



15/19


(H 3.19) Ngun hiu th mc ni tip vi in trhay ngun dng mc song song vi in tr. Ta cth dng bin i Thevenin Norton bin i ngun hiu th thnh ngun dng in hayngc li cho ph hp vi h phng trnh sp phi vit.

(H 3.20)

3.4.2. Chuyn v ngun :

Khi gp 1 ngun hiu th khng c in trni tip km theo hoc 1 ngun dng inkhng c in trsong song km theo, ta c th chuyn v ngun trc khi bin i chng.Trong khi chuyn v, cc nh lut KCL v KVL khng c vi phm. Chuyn v ngun hiu th :

(H 3.21) cho ta thy mt cch chuyn v ngun hiu th . Ta c th chuyn mt ngunhiu th " xuyn qua mt nt " ti cc nhnh khc ni vi nt v ni tt nhnh c changun ban u m khng lm thay i phn b dng in ca mch, mc d c s thay i v

phn bin th nhng nh lut KVL vit cho cc vng ca mch khng thay i. Hai mchhnh 3.21a v 3.21b tng ng vi nhau.

(a) (b)

(H 3.21)Th d 3.9: Ba mch in ca hnh 3.22 tng ng nhau:

(H 3.22) Chuyn v ngun dng in:

Ngun dng in i mc song song vi R1 v R2 ni tip trong mch hnh 3.23a cchuyn v thnh hai ngun song song vi R1 v R2 hnh 3.23b.

(H 3.23)



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nh lut KCL cc nt a, b, c ca cc mch (H 3.23) cho kt qu ging nhau.Hoc mt hnh thc chuyn v khc thc hin nh(H 3.24a) v (H 3.24b). nh lut

KCL cc nt ca hai mch cng ging nhau, mc d s phn b dng in c thay inhng hai mch vn tng ng .

(a) (H 3.24) (b)

Th d 3.10: Tm hiu th gia a b ca cc mch hnh 3.25a

(a) (b) (c)

(H 3.25)

Suy ra vab =8

55

3

11

8

15= V

vab =8

55V



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Tm li, khi gii mch bng cc phng trnh vng hoc nt chng ta nn sa soncc mch nh sau:

- Nu gii bng phng trnh nt, bin i ch c cc ngun dng in trong mch.- Nu gii bng phng trnh vng, bin i ch c cc ngun hiu th trong mch.

BI TP

--o0o--1. Dng phng trnh nt, tm v1 v v2 ca mch (H P3.1)2. Dng phng trnh nt , tm i trong mch (H P3.2).

(H P3.1) (H P3.2)

3. Dng phng trnh nt tm vv i trong mch (H P3.3).4. Dng phng trnh nt, tm v trong mch (H P3.4)

(H P3.3) (H P3.4)

5. Dng phng trnh nt, tm vv v1 trong mch (H P3.5)6. Cho vg = 8cos3t (V), tm vo trong mch (H P3.6)

(H P3.5) (H P3.6)7. Tm v trong mch (H P3.7), dng phng trnh vng hay nt sao cho c t phng trnhnht.



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(H P3.7)

8. Tm Rin theo cc R, R2, R3 mch (H P3.8).Cho R1 = R3 = 2K. Tm R2 sao cho Rin = 6K v Rin = 1K

(H P3.8)

9. Cho mch khuch i vi sai (H P3.9)

- Tmv

o theov

1,v

2, R1, R2, R3, R4.- Tm lin h gia cc in trsao cho: vo = ( )12

1

2

R

Rvv

10. Tm hiu thvngang qua ngun dng in trong mch (H P3.10) bng cch dng phngtrnh vng ri phng trnh nt.

(H P3.9) (H P3.10)

11. Tnh li dng ini

0

i

ica mch (H P.11) trong 2 trng hp.

a. R2 = 0b. R2 = 1

12. Tm ix trong mch (H P.12)



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(H P.11) (H P.12)

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Chuong3_ptrinh Mach Dien