Bai giang mach dien 1.pdf

ELECTRONIC CIRCUIT I

Page 1/94

LECTURE


Page 2/94

CHNG 1: CC KHI NIM C BN V MCH IN

1.Mch in v m hnh mch in

1.1.nh ngha mch in: gm tp hp cc thit b in, in t trong c s bin i nng

lng in sang cc dng nng lng khc.

Cu to mch in gm ngun in, ph ti, dy dn ngoi ra cn c cc phn t ph tr

khc

Ngun in: dng cung cp nng lng in hoc tn hiu in cho mch. Ngun

c bin i t cc dng nng lng khc sang in nng, v d my pht in (bin

i c nng thnh in nng), c quy (bin i ha nng sang in nng).

Ph ti: l thit b nhn nng lng in hay tn hiu in. Ph ti bin i nng lng

in sang cc dng nng lng khc, v d nh ng c in (bin i in nng thnh

c nng), n in (bin i in nng sang quang nng), bn l, bp in (bin i in

nng sang nhit nng) v.v.

Dy dn: lm nhim v truyn ti nng lng in t ngun n ni tiu th.

My pht in cquy

Ti

Ngun

I

-

+

E

Hnh 1.1


Page 3/94

Ngoi ra cn c cc phn t khc nh: phn t lm thay i p v dng trong cc

phn khc ca mch (nh my bin p, my bin dng), phn t lm gim hoc tng

cng cc thnh phn no ca tn hiu (cc b lc, b khuch i), v.v..

1.2.Cu trc ca mch in:

Nhnh: gm nhiu phn t ghp ni tip trong c cng mt dng in.

Nt: l im ni ca ba nhnh tr ln.

Vng: l tp hp nhiu nhnh to thnh vng kn, n c tnh cht l nu b i mt

nhnh th khng to thnh vng kn na.

Mc li : l vng m bn trong n khng cn vng no khc.

1.3. Cc hin tng in t

Gm hai hin tng l hin tng bin i nng lng v hin tng tch phng nng

lng in t.

Hin tng bin i nng lng gm hin tng ngun v hin tng tiu tn.

Hin tng ngun: l hin tng bin i t cc dng nng lng khc nh c

nng, ha nng, nhit nng thnh nng lng in t.

Hin tng tiu tn: l hin tng bin i nng lng in t thnh cc dng nng

lng khc nh nhit, c, quang, ha nng tiu tn i khng hon tr li trong

mch na.

Hin tng tch phng nng lng gm hin tng tch phng nng lng trong trng

in v trong trng t.

1.4.m hnh mch in

c dng trong l thuyt mch c xy dng t cc phn t l tng sau y:

Phn t in tr: l phn t c trng cho hin tng tiu tn nng lng in t,

quan h gia dng v p trn hai cc ca phn t in tr l: u = R.i. ( hnh 1.4 )

hnh 1.4

R i

E

F

D

C B

A

C

D A

A

B C

D

Hnh 1.2 Hnh 1.3


Page 4/94

Phn t in cm: l phn t c trng cho hin tng tch phng nng lng trng

t, quan h gia dng v p trn hai cc phn t in cm: u=dt

diL. ( hnh 1.5 )

Phn t in dung: l phn t c trng cho hin tng tch phng nng lng trng

in, quan h gia dng v p trn hai cc t in: i=dt

duC. thng s c bn ca mch

in, c trng cho qu trnh tch phng nng lng trng in. ( hnh 1.6 )

Phn t ngun: l phn t c trng cho hin tng ngun. phn t ngun gm phn

t ngun p v phn t ngun dng. ( hnh 4 ) v ( hnh 5 )

Phn t thc: phn t thc ca mch in c th c m hnh gn ng bi mt hay

nhiu phn t l tng c ghp vi nhau theo mt cch no m t gn ng

hot ng ca phn t thc t.

2. Cc khi nim c bn trong mch in.

2.1. Dng in v quy c chiu dng in:

Dng in l dng chuyn di hng ca cc in tch. Cng dng in ( gi tt l

dng in) l lng in tch chuyn qua mt b mt no ( tit din ngang ca dy dn, nu

l dng in chy trong dy dn ) trong mt n v thi gian.

Dng in k hiu l: I ( Ampe)

Quy c chiu dng in t cc dng sang cc m ca ngun (i>0), ngc li (i


Page 5/94

2.2.in p

in p gia hai im A v B l cng cn thit lm dch chuyn mt n v in tch (1

culong) t A n B.

in p k hiu l: U (vn)

V d: UAB: in p gia A v B

UBA: in p gia B v A

ta c : UAB = -UBA

2.3.Cng sut

Xt mch in chu tc ng 2 u mt in p u, qua n s c dng in i. Cng sut

tc thi c a vo mch in (c hp th bi mch in) l:

p(t) = u.i

n v cng sut l watt (w)

p(t) l mt i lng i s nn c th m hoc dng ti thi im t no Nu p > 0

th ti thi im t phn t thc s hp th nng lng vi cng sut l p, cn nu p <

0 th ti thi im t phn t thc s pht ra nng lng (tc nng lng c a t

phn t mch ra ngoi) vi cng sut l | p |.

3. Cc php bin i tng ng.

Trong thc t i khi cn lm n gin mt phn mch thnh mt mch tng ng n

gin hn. Vic bin i mch tng ng thng c lm cho mch c t phn t, t s

nt, t s vng v nhnh hn mch trc lm gim i s phng trnh phi gii.

Mch tng ng c nh ngha nh sau:

Hai mch c gi l tng ng nu quan h gia dng in v in p trn cc

cc ca hai phn t l nh nhau.

Mt php bin i tng ng s khng lm thay i dng in v in p trn

cc nhnh cc phn ca s khng tham gia vo php bin i.

Sau y l mt s php bin i tng ng thng dng :

3.1.Ngun sc in ng ghp ni tip

-

+

B

A

- + A B

U

5V

+

-

B

A

-5V

Hnh 1.10


Page 6/94

S tng ng vi mt ngun sc in ng duy nht c tr s bng tng tr s cc sc

in ng .

V d : e1= 3(v), e2= 5 (v), e3= 2(v) et= 3+5-2 = 6 (v).

Ngun in p c trng cho kh nng to nn v duy tr mt in p trn hai cc ca

ngun. k hiu: U(t)

Ngun p cn biu din bng s e(t).

e(t): chiu i t im c in th thp n im c in th cao.

u(t): chiu i t im c in th cao n im c in th thp.

3.2.Ngun dng in ghp song song

Ngun dng in mc song song s tng vi mt ngun dng duy nht c gi tr bng

tng i s cc ngun dng .

Jt=

n

kkj

1

V d : j1= 2 (A), j2= 3 (A), j3=1 (A) j = 2-3-1 = -2 (A)

Ngun dng in j(t) c trng cho kh nng ca ngun in to nn v duy tr mt dng

in cung cp cho mch ngoi.

3.3 in tr ghp ni tip v song song

in tr ghp ni tip s tng ng vi mt phn t in tr duy nht c tr s bng

tng cc in tr cc phn t .

Rt=RK

b

e1 e2 e3

a

et=e1+e2-e3

b a

et=

n

kke

1

u(t) e

j3 j2 j1 jtd= j1-j2-j3

i i

Rn R3 R2 R1 Rt

Hnh 1.11

Hnh 1.12

Hnh 1.13

Hnh 1.14


Page 7/94

V d : R1= 3 (), R2= 2 (), R3= 5 () Rt = 3+2+5 = 10 ()

in tr ghp song song s tng ng vi mt phn t in tr duy nht c in dn

bng tng in dn cc phn t . ( vi g = R

1: gi l in dn )

Gt=

n

KKG

1

ntd RRRR

1111

21

V d : R1= 2 (), R2= 2 (), R3= 5 () ntd RRRR

1111

21

= 30

31

5

1

3

1

2

1 ( )

3.4. Bin i -Y, Y-

3.4.1.bin i Y-:

R12=R1+R2+3

21.

R

RR(1)

R23=R2+R3+1

3.2

R

RR(2)

R31=R3+R1+2

13.

R

RR(3)

3.4.2.bin i -Y:

R1=312312

1231.

RRR

RR

(1)

R2=312312

1223.

RRR

RR

(2)

R3=312312

3123.

RRR

RR

(3)

Rdt

i

Rn R1 R2 R3

i

i1

i2

i3

R1

R3 R2 i3 i2

i1

R31

R23

R12

Hnh 1.15

Hnh 1.16


Page 8/94

Cc quan h trn c chng minh nh sau: v hai mch tng ng nn cc quan h sau

y th bng nhau i vi hai mch.

Rt12= 031

12 ii

u; Rt23= 01

2

23 ii

u; Rt31= 02

3

31 ii

u

i vi mch (Y) ta c:

Rt12=R1+R2; Rt23=R2+R3; Rt31=R1+R3

i vi mch () ta c:

Rt12=R12//(R23+R31); Rt23=R23//(R31+R12); Rt31=R31//(R23+R12)

Do ta c cc phng trnh sau:

R1+R2=312312

312312 )(

RRR

RRR

(1)

R2+R3=312312

123123 )(

RRR

RRR

(2)

R3+R1=312312

231212 )(

RRR

RRR

(3)

Gii h phng trnh(1),(2),(3) ta tm c cc php bin i trn.

3.5. Bin i tng ng gia ngun p v ngun dng.

Ngun p mc ni tip vi mt in tr s tng ng vi mt ngun dng mc song

song vi in tr v ngc li.

mch (hnh 1) ta c quan h gia u v i nh sau:

u = e-r.i (1)

mch (hnh 2) ta c: j = i+i1 (vi i1=u/r)

u = r.j-r.i (2)

So snh (1)v(2) ta thy hai mch s tng ng nu:

e = r.j hoc j = e/r

hnh2 hnh1

i1 r j

i

r

e i

Hnh 1.17


Page 9/94

Cu hi :

1. Mch in gm nhng phn no? Nu cng dng ca chng.

2. nh ngha nt ? vng ? mc li? iu kin no trong mch in c nt.

3. c trng ca phn t in tr l g ? Phn t in dung ? Phn t in cm ?

4. nh ngha dng in ? nh ngha in p ?

5. Tnh hiu in th (in p) UAB trong cc trng hp sau :

a. in th ti im A(UA=5 (V) ), in th ti im B(UB= 3 (V) ).

b. in th ti im A(UA=2 (V) ), in th ti im B(UB= -3 (V) ).

c. in th ti im A(UA= -1 (V) ), in th ti im B(UB= -4 (V) ).

6. Cng sut p(t) c trng nhng hin tng no ca thit b.

7. Ti sao phi thc hin php bin i tng ng ? Php bin i tng ng c lm thay

i dng v p trong mch in khng.

8. V li mch in v tnh in tr tng ng trong cc trng hp sau:

a. (R1nt R2)//R3. Bit R1 = 2 (), R2 = 1 (), R3= 4 ()

b. (R1 nt R2)//(R3 nt R4) nt R5. Bit R1 = 2 (), R2 = 2 (), R3= 1 (), R4= 1, R5= 3 ().

c. (R1nt R2)//(R3 nt R4 nt R5)//R6. Bit R1 = 2 (), R2 = 4 (), R3= R4= R5= 2 (), R6= 6 ().

d. (R1// R2)nt(R3 // R4 // R5)ntR6. Bit R1 = 2 (), R2 = 4 (), R3= R4= R5= 2 (), R6= 6 ().

9. Cho mch in nh hnh v:

Tnh in tr RAB

Ti Liu Tham Kho :

- Gio trnh mch in tc gi : Phm Th C NXBGD-1996

- Gio trnh in k thut tc gi : L Vn o NXBKHKT-1997

- Gio trnh mch in tc gi : L Vn Bng NXBGD-2008

1 1

2 2

2 2

B A d

c

b

a


Page 10/94

CHNG 2: MCH IN MT CHIU

1.Cc nh lut c bn trong mch in mt chiu

1.1. nh ngha dng in mt chiu:

Dng in mt chiu l dng in c chiu v ln khng i theo thi gian.

1.2.nh lut omh:

Cng dng in trong mt on mch t l thun vi hiu in th hai u on

mch t l nghch vi in tr ca on mch.

I = R

U (A)

1.2.nh lut Joule- Lenx:

Nhit lng ta ra trong mt vt dn t l thun vi in tr ca vt dn vi bnh phng

cng dng in v vi thi gian dng in i qua.

Q = R.I2.t ( J )

1.3.nh lut Faraday:

Khi lng m ca cht c gii phng ra in cc t l vi ng lng ha hc A/n

ca cht v vi in lng q i qua dung dch in phn.

m = K.n

A.q = tI

n

A

F...

1

A: nguyn t lng

n : ha tr

1/k = F =9,65.107C/kg ( s Faraday )

I : cng dng in qua bnh in phn

t : thi gian dng in chy qua

1.4.nh lut kichop:

nh lut kichop 1 v 2 l hai nh lut c bn nguyn cu v tnh ton mch in.

Hnh 2.2

I(t)

t

i

0

B A

UAB

R

Hnh 2.1


Page 11/94

a. nh lut kichop 1: ni ln mi quan h gia cc dng in ti mt nt.

Tng i s cc dng in ti mt nt th bng khng.

01

K

n

K

i

Vi mch hnh bn: i1 - i2 - i3 = 0

hoc - i1 + i2 + i3= 0

Trong nu ta quy c cc dng in i ti nt mang du dng th cc dng in ri

khi nt mang du m v ngc li.

b.nh lut kichop 2: ch r cc mi lin h gia in p trong mt vng kn.

i theo mt vng kn vi chiu ty , tng i s in p ri trn cc nhnh bng khng.

01

K

n

K

U

nh lut kichop 2 pht biu li nh sau:

i theo mt vng kn vi chiu ty , tng i s cc in p ri trn cc nhnh bng

tng i s cc s c trong vng, trong cc s v dng in no c chiu trng vi chiu

i ca vng s mang du dng ngc li mang du m.

2. Cc phng php gii mch in mt chiu.

2.1.Phng php bin i in tr: (phng php ny ch yu s dng nh lut omh)

Cc in tr mc ni tip: trong on mch mc ni tip dng in qua cc phn t l

nh nhau. (I1=I2=I3==In)

I = Rt

U

RnRRR

U

...321

Trong : Rt=R1+R2+R3++Rn

V d : R1= 1 (), R2= 3 (), R3= 4 (), UAB= 10 (V).

i3

i2

i1

vng 1: I1.R1+I3.R3=e1 (1) vng 2: I2.R2+I3.R3=e2 (2)

Rn R3 R2 R1 R

+

- -

+

e2 e1

I3

I2 I1

R3

R2 R1

I II

Hnh 2.3

Hnh 2.4

Hnh 2.5


Page 12/94

Tnh RAB, IAB

Cc in tr mc song song: trong on mch mc song song in p hai u mi

mch nhnh bng nhau v bng in p hai u on mch.

U1=U2=U3==Un

I = Rt

U

RnRRRRt

1...

3

1

2

1

1

11

V d : R1= 1 (), R2= 3 (), R3= 4 (), UAB= 10 (V).

Tnh RAB, IAB

2.2.Phng php xp chng dng in

Phng php :

Bc 1 :

Cn c mch in ban u c bao nhiu ngun( ngun in p v ngun dng in ) ta phn

tch thnh by nhiu hnh v p dng quy tc sau:

Nu l ngun p: loi b ngun p v ni tc ngun p

Nu l ngun dng: loi b ngun dng v ni tt li

Bc 2 :

Da vo nh lut m xc nh dng in trn cc mch va phn tch.

Bc 3 :

Tnh dng in trn cc nhnh nh sau:

Dng in qua nhnh ban u bng tng i s cc dng in cng i qua nhnh y trn

cc mch in mi v p dng quy tc sau, nu dng in no cng chiu vi dng in

trn mch chnh s mang du dng (+), ngc li mang du m (-).

Vi mch in hnh trn ta c: I1=I/1- I1

//

I2= I2/- I2

/

I3= I3/+ I3

//

Rdt

i

Rn R1 R2 R3

i

+

I1

e2

I2

+

-

I3

I2

R3

R2 R1

+

- -

+

e2 e1

I3

I2 I1

R3

R2 R1

-

+

e1

I3

I1

R3

R2 R1

Hnh 2.8 Hnh 2.9

Hnh 2.6

Hnh 2.7


Page 13/94

V d: E1=10(v), E2 = 4(v), R1= 4(), R2= 2(), R3= 4(). Tnh I1, I2, I3 bng phng php

xp chng ca mch in trn.

Gii

Hnh 1 ta c:

R1 nt (R2//R3) R123= R1+R23= R1+ 32

32

RR

RR

= 4+

3

16

3

4 ()

8

15

16

30

3

16

10

123

1'1

R

eI (A)

4

5

42

4

8

15

32

3'1

'2

RR

RII (A)

8

5

42

2

8

15

32

2'1

'3

RR

RII (A)

Hnh 2 ta c:

R2 nt (R1//R3) R213= R2+R13= R2+ 31

31

RR

RR

= 2+ 4

8

16 ()

14

4

213

2''2

R

eI (A)

2

1

44

41

31

3''2

''1

RR

RII (A)

2

1

44

41

31

1''2

''3

RR

RII (A)

Vy : I1=I/1- I1

// = 8

11

2

1

8

15 (A)

I2= -I2/+ I2

//= 4

11

4

5 (A)

I3= I3/+ I3

//= 8

9

2

1

8

5 (A)

2.3. Phng php dng in nhnh

n s bi ton l dng in nhnh.

B

A

+

- -

+

e2 e1

I3

I2 I1

R3

R2 R1

I II

Hnh 2.10


Page 14/94

Bc 1: Ty v chiu dng in trong cc nhnh, chn chiu i ca vng.

Bc 2: Xc nh s nt, s nhnh v s vng dc lp (mc li ), nu gi n l s nt, m l s

nhnh s phng trnh cn phi vit l:

Vit (n-1) phng trnh k1. Khng cn vit cho nt th n v c th suy ra t (n-1)

phng trnh vit.

Vit m-(n-1) = (m+1-n) phng trnh k2. Vy ta phi chn (m = 1-n) mc li.

Ti nt A: I1+I2-I3=0 (1)

Vng I : I1.R1+I3.R3=E1 (2)

Vng II : I2.R2+I3.R3= E2 (3)

Bc 3: Gii h phng trnh (1), (2) v (3) tm I1,I2, I3.

V d 1: Cho mch in nh hnh v.

Cho E1= 8(v), E2= 5(v), R1=1(), R2=3(),

R3=5().Tnh dng in trn cc nhnh.

Gii

Chn chiu dng in v chiu i ca vng nh hnh v.

p dng nh lut K1 ti nt A ta c: I1+I2-I3= 0 (1)

p dng nh lut K2 cho vng 1 v vng 2 ta c:

Vng 1: I1.R1+I3.R3=E1 (2)

Vng 2: I2.R2+I3.R3= E2 (3)

Gii h phng trnh (1), (2), (3)

I1+I2-I3= 0 (1) I1+I2-I3= 0 (1)

I1.R1+I3.R3=E1 (2) I1+5I3 = 8 (2)

I2.R2+I3.R3= E2 (3) 3I2+5I3 = 5 (3)

T phng trnh (1) I1= -I2+I3 (4)

Th phng trnh (4) vo phng trnh (2) (-I2+I3)+5I3= 8 -I2+6I3= 8 (5)

Gii h phng trnh (3) v (5)

3I2+5I3 = 5 (3) 3I2+5I3 = 5 (3)

3 x -I2+6I3 = 8 (5) -3I2+18I3 = 24 (6)

Nhn 2 v phng trnh (5) vi 3. Ly pt (3) + pt (6)

23I3= 29 I3= 23

29(A).

B

A

+

- -

+

e2 e1

I3

I2 I1

R3

R2 R1

I II

Hnh 2.11


Page 15/94

Th I3 vo pt (3) 3I2+ 5. 23

29=5 I2=

23

10

323

29.55

(A).

Th I3 vo pt (2) I1+5. 23

29= 8 I1=

23

39

23

145184

23

29.58

(A).

Th li:

Th cc gi tr I1, I2, I3 vo phng trnh (1) I1+I2-I3 = 023

29

23

10

23

39

Ch : Nu gii ra dng in no c gi tr m ta kt lun chiu dng in i trong mch

ngc vi chiu ta chn. Vy chiu I2 i trong mch ngc vi chiu chn.

V d 2 : Cho mch in mt chiu nh hnh v.

E1= 5 (v), E2= 4 (v), E3= 7 (v), R1= 2 (), R2= 3 (), R3= 4 ().

Tnh I1, I2, I3 bng phng php dng in nhnh.

Gii

Chn chiu dng in v chiu i ca vng nh hnh v.

p dng nh lut K1 ti nt A ta c: I1-I2-I3= 0 (1)

p dng nh lut K2 cho vng 1 v vng 2 ta c:

Vng 1: I1.R1+I3.R3=E1-E3 (2)

Vng 2: I2.R2-I3.R3= -E2 +E3 (3)

Gii h phng trnh (1), (2), (3)

I1-I2-I3= 0 (1) I1-I2-I3 = 0 (1)

I1.R1+I3.R3=E1-E3 (2) 2I1+4I3 = -2 (2)

I2.R2-I3.R3= -E2+ E3 (3) 3I2-4I3 = 3 (3)

T phng trnh (1) I1= I2+I3 (4).

Th phng trnh (4) vo pt (2) 2(I2+I3)+4I3= -2 2I2+6I3= -2 (5)

Gii h phng trnh (3) v (5) ta c:

3I2-4I3 = 3 (3)

-

+

-

e3

B

A

+

-

+

e2 e1

I3

I2 I1

R3

R2 R1

I II

Hnh 2.12


Page 16/94

2I2+6I3= -2 (5)

p dng phng php gii h phng trnh bc nht hai n s gii h pt (3) v (5) nh sau:

= = 3.6 (-4).2 = 18+8 = 26

x = = 3.6 (-4).(-2) = 18-8 = 10

y = = 3.(-2) 3.2 = -12

I2 = 26

10

x (A), I3=26

12

y (A)

Th I3= 26

12(A) vo phng trnh (2) 2I1 + 4.

26

12 = -2 I1=

26

2

226

482

(A)

Th li:

Th cc gi tr I1, I2, I3 vo phng trnh (1) I1-I2-I3 = 026

12

26

10

26

2

Vy chiu I1, I3 i trong mch ngc vi chiu chn.

2.4.Phng php dng in vng.

Phng php:n s ca h phng trnh l dng in vng

Gi m l s nhnh n l s nt s vng c lp cn phi chn l m-n+1.Mi vng s c mt

dng in vng chy khp kn trong vng y

Dng in chy khp kn trong vng a gi l dng Ia.

Dng in chy khp kn trong vng b gi l dng Ib.

Cc dng in Ia, Ib l n s ca h phng trnh.

Bc 1: chn chiu cc dng in vng Ia,Ib.

Bc 2: vit h phng trnh k2 cho(m-n+1)vng.

3 -4 2 6

3 -4 -2 6

3 3 2 -2

e2 e1

+

- -

+ I3

I2 I1

R3

R2 R1

Ia Ib

Hnh 2.13


Page 17/94

(tng i s in p ri trn cc nhnh ca vng do cc dng in vng gy ra bng tng i s

cc sc in ng c trong vng, trong cc s, cc dng in vng c chiu trng vi

chiu i ca vng s mang du dng ngc li mang du m).

Vng a: Ia.R1+Ia.R3+Ib.R3=E1 (1)

Vng b: Ib.R3+Ib.R2+Ia.R3=E2 (2)

Bc 3: Gii h phng trnh tm Ia, Ib.

Bc 4: Tnh dng in nhnh nh sau:

Dng in trn mt nhnh bng tng i s cc dng in vng i qua nhnh y, trong

dng in vng no c chiu trng vi chiu dng in nhnh s mang du dng

ngc li mang du m.

I1=Ia, I2=Ib, I3=Ia+Ib

V d 1 : Cho E1= 8(V), E2= 6(V), R1 = 2(), R2 = 3( ), R3 = 4().

Tnh dng in qua cc nhnh bng phng php dng in vng ca mch in trn.

Gii

Chn chiu i ca dng in vng nh hnh v

Vng a: Ia.R1+Ia.R3+Ib.R3= E1 (1)

Vng b: Ib.R2+Ib.R3+Ia.R3= E2 (2)

Gii h phng trnh (3), (4)

2.Ia+ 4.Ia+ 4.Ib= 8 (1) 6Ia+ 4Ib= 8 (3)

3.Ib+ 4.Ib+ 4.Ia= 6 (2) 4Ia+ 7Ib= 6 (4)

= = 6.7 4.4 = 42 - 16 = 26

x = = 8.7 6.4 = 56 - 24 = 32

y = = 6.6 4.8 = 36 - 32 = 4

Ia = 13

16

26

32

x (A), Ib=13

2

26

4

y (A)

I1=Ia= 13

16(A), I2 = Ib=

13

2(A), I3 =

13

18(A).

V d 2 : Cho mch in nh hnh v

8 4 6 7

6 8 4 6

6 4 4 7


Page 18/94

E1= 7(v), E2= 4(v), R1= 2(), R2= 5(), R3= 4().

Tnh I1, I2, I3 bng phng php dng in vng

v PR1, PR2, PR3.

Gii

Chn chiu i ca dng in vng nh hnh v

Vng a: Ia.R1+ Ia.R3 - Ib.R3= E1 - E2 (1)

Vng b: Ib.R2 + Ib.R3- Ia.R3= E2 (2)

Gii h phng trnh (3), (4)

2Ia + 4Ia-4Ib= 7 (1) 6Ia + 4Ib= 8 (3)

5Ib + 4Ib+4Ia= 6 (2) 4Ia + 7Ib= 6 (4)

= = 6.7 4.4 = 42 - 16 = 26

x = = 8.7 6.4 = 56 - 24 = 32

y = = 6.6 4.8 = 36 - 32 = 4

Ia = 13

16

26

32

x (A), Ib=13

2

26

4

y (A)

I1=Ia=13

16(A), I2=Ib=

13

2(A), I3=

13

18(A).

2.5.Phng php nt.

Phng php: (p dng nh lut k1)

Tng i s cc dng din chy vo mt nt th bng 0.

I1+I2+I3=0

Kho st mch in (hnh 2.16).

Bc1: chn mt nt lm chun(chn 0 lm chun) Hnh 2.16

I3I2

I1

8 4 6 7

6 8 4 6

6 4 4 7

I1

I3 Ib I2 Ia

Ic R2

R3 R1

0

B A

I1 I2

e1 +

-

-

+

e2

I3

I2 I1

R3

R2 R1

Ia Ib

Hnh 2.14

Hnh 2.15


Page 19/94

V0=0

VA=VAO

VB=VBO

Bc2: kho st cc nt

kho st nt A: Ia+Ib+Ic=0

Ia = 1R

VaVo , Ib =

2R

VV AB , Ic = I1

1R

VA

2R

VV AB +I1=0

1111 1221

IR

VRR

V BA

kho st nt B: I1+ I2

+ I3= 0 (I1

=I2)

I2=

2R

VV BA , I3=

3

0

R

VV B

32 R

V

R

VV BBA

+I1=0

VB 211 2232

IR

V

RRA

H phng trnh nt: 1111 1221

IR

VRR

V BA

VB 211 2232

IR

V

RRA

Bc 3: Gii h phng trnh tm c VA,VB dng in qua cc nhnh.

Nhn xt:

21

11

RR : Tng in dn ni ti nt A

32

11

RR: Tng in dn ni ti nt B

2

1

R: in dn chung gia nt Av B

I1: l gi tr ngun dng ni ti nt A, mang du (+) nu ngun dng chy vo nt A v

mang du (-) nu ngun dng chy ra t nt A.


Page 20/94

I2: l gi tr ngun dng ni ti nt B, mang du (+) nu ngun dng chy vo nt B v

mang du (-) nu ngun dng chy ra t nt B.

V d 1: Cho I1= 2(A), I2 = 3(A), R1= 3(V), R2 = 4(V), R3 = 5(V).

Tnh in th ti cc nt mch in (hnh 2.16)

Gii

Chn mt nt lm chun(chn 0 lm chun)

V0=0

VA=VAO

VB=VBO

Kho st cc nt

kho st nt A: Ia+Ib+Ic=0

Ia=1R

VaVo , Ib=

2R

VV AB , Ic=I1= 2 (A)

1R

VA

2R

VV AB +I1=0

1111 1221

IR

VRR

V BA

kho st nt B: I1+I2

+I3=0 (I1

=I2 = 3 (A))

I2=

2R

VV BA , I3=

3

0

R

VV B

32 R

V

R

VV BBA

+I1=0

VB 211 2232

IR

V

RRA

H phng trnh nt: 1221

111I

RV

RRV BA

(1) 2

4

1

4

1

3

1

BA VV (3)

VB 2232

11I

R

V

RRA

(2) VB 3

45

1

4

1

A

V (4)

= = )4

1).(

4

1(

20

9

12

7 =

5

1

240

48

240

1563

16

1

240

63

12

7 -

4

1

-4

1

20

9


Page 21/94

x = = )4

1.(3

20

92 =

20

33

20

1518

4

3

20

18

y = = 20

9.23

12

7 =

20

17

120

102

120

108210

20

18

12

21

VA = 4

335.

20

33

5

120

33

x (V), VB = 4

17

5

120

17

y (V)

V d 2: Cho mch in nh hnh 2.17

Tnh in th ti cc nt mch in trn. Bng phng php nt

Gii

Chn mt nt lm chun(chn 0 lm chun)

V0=0

VA=VAO

VB=VBO

Kho st cc nt

kho st nt A: I1+I2+I3+I4=0

I2=1R

VaVo , I4=

2R

VV AB , I1= 4 (A), I3= - 2(A)

1R

VA

2R

VV AB + 4-2 = 0

24111

221

RV

RRV BA (A)

22

1

2

1

4

1

BA VV (1)

kho st nt B: I3+I4

+I5 = 0 (I3 = 2 (A))

12

7 2

20

9 3

2 -4

1

3 20

9

I5

I4

I3

I2

4

2

0

2A

4A

B A I1

4

Hnh 2.17


Page 22/94

I4=

2R

VV BA , I5=3

0

R

VV B

32 R

V

R

VV BBA

+I3= 0

VB 211

232

R

V

RRA

VB 224

1

2

1

A

V ( 2 )

Gii h phng trnh (1) v (2).

22

1

2

1

4

1

BA VV (1) 3VA- 2VB = 8 (3)

VB 224

1

2

1

A

V (2) - 2VA+ 3VB = 8 (4)

= = 3.3 (-2).(-2) = 9 - 4 = 5

x = = 8.3 (-2).8 = 24 + 16 = 40

y = = 3.8 8.(-2) = 24 + 16 = 40

VA = 85

40

x (V), VB = 85

40

y (V)

2.6. Phng php in p hai nt.

Phng php: (p dng cho nhng mch c nhiu nhnh song song nhng ch c hai nt).

Bc 1: Chn chiu dng ca in p v dng in trong cc nhnh.

Bc 2 : Xc nh in p hai nt theo cng thc :

8 -2 8 3

3 8 -2 8

3 -2 -2 3

UAB

-

A

B

e1

R1

I3

I2

+

- -

+ e2

I1

R3 R2

Hnh 2.18


Page 23/94

UAB=

n

KK

n

KKK

G

GE

1

1

.

Trong : EK, GK l sc in ng v dng in trn nhnh th k (gk=R

1)

Nu EK c chiu trng vi chiu dng gi thit ca in p th tch EK.GK mang du

m ngc li mang du dng.

Mch no khng c ngun sc in ng (EK=0) th EK.GK=0

Bc 3: p dng nh lut omh tm dng in trong cc nhnh.

V d 1: Cho E1=120(V), E2=119(V), R1=5, R2= 3, R3= 22 .

Tnh dng in qua cc nhnh bng phng php in p hai nt mch in (hnh 2.18)

Gii

Chn chiu dng in p v chiu dng in i trn cc nhnh nh hnh v

UAB=

n

KK

n

KKK

G

GE

1

1

.

= 321

2211 ..

ggg

gEgE

=

191

330

15

955

330

15

330

110

330

6615

595

15

360

22

1

3

1

5

13

1.119

5

1.120

= 110(V)

p dng nh lut omh cho cc nhnh

I1= 25

110120

1

1

R

UE AB (A)

I2= 33

110119

2

2

R

UE AB (A)

I3= 522

110

3

R

UAB (A)

Th li: I1+I2+I3=0 2 + 3 5 = 0

V d : Cho mch in nh hnh 2.19

E1= 5 (v), E2= 4 (v), E3= 7 (v), R1= 2 (), R2= 3 (), R3= 4 ().

Tnh dng in qua cc nhnh bng phng php in p hai nt mch in trn.

UAB

-

+ I1

-

+

-

e3

B

A

+

-

+

e2 e1

I3 I2 R3 R2

R1

Hnh 2.19


Page 24/94

Chn chiu dng in p v chiu dng in i trong cc nhnh nh hnh v

UAB=

n

KK

n

KKK

G

GE

1

1

.

= 321

332211 ...

ggg

gEgEgE

=

13

67

13

12

12

67

12

34612

211630

4

1

3

1

2

14

17

3

1.4

2

1.5

(V)

p dng nh lut omh cho cc nhnh

I1= 13

1

13.2

6765

213

675

1

1

R

UE AB (A)

I2= 13

5

13.3

15

13.3

6752

313

674

2

2

R

UE AB (A)

I3= 13

6

13.4

24

13.4

6791

413

677

3

3

R

EUAB (A)

Th li: I1- I2- I3= 0 -13

6

13

5

13

1 = 0

Vy dng I1, I2, I3 i trong mch ngc vi chiu chn

3.Cng v cng xut ca dng in mt chiu

Trong mt mch kn bao gi cng c hai s chuyn ha nng lng l bn trong ngun

in v bn ngoi ngun in.

Trong ngun in: c mt dng nng lng no (ha nng, c nng, ni nng)

chuyn ha thnh in nng.

Bn ngoi ngun in: in nng c chuyn ha thnh nhng dng nng lng khc

(ni nng, ha nng, c nng).

S o nng lng y biu th cng ca dng in.

3.1.Cng ca dng in.

Cng ca dng din sinh ra trong mt on mch bng tch ca ht gia hai u on

mch vi cng dng in v thi gian dng in i qua.

A = q.U=U.I.t (jun)

3.2.Cng sut ca dng in.

Cng sut ca dng in l i lng c trng cho tc sinh cng ca dng in. N c

ln bng cng ca dng in sinh ra trong mt giy.

P = IUt

A. (W)


Page 25/94

P = RI2 (W)

Cng sut ca dng in trong mt on mch bng tch hiu in th gia hai u on

mch vi cng dng in trong on mch.

3.3.o cng v cng sut.

Mun o cng v cng sut trn mt on mch ta dng ampe k o cng dng in

qua on mch v vn k o hiu in th hai u on mch P=U.I

o cng ca dng in tc in nng tiu th trn on mch ta dng cng t in.

A= P.t = Kw.h

Cu hi:

1. nh ngha dng in mt chiu ? Trnh by nh lut omh, nh lut k1 v nh lut k2.

2. Cho mch in nh hnh v

a. R1=R2=R3=R4=2(), R5=R7=1(),

R6=3(),UAB=10(V).

Tnh: RAB, I, IR3, UR6. (hnh 2.20)

b. R1=R7= 2(), R2=R3=R6=1(),

R4=R5=3(), UAB= 12(V).

Tnh: RAB, I, IR4, UR3, P. (hnh 2.21)

c. R1=R2= R3= 1(), R4=R5=1(),

R7=R8=4, R6 =R9=3(), UAE= 5(V).

Tnh: RA, IAB, UBC, UCD, IR6, IR8,

PR8, P. (hnh 2.22)

d. UAB= 10(V). Tnh RAB, IAB. (Hnh 2.23)

I B A R7 R6 R5

R4 R3 R2 R1

R3

I B A R7

R6 R5

R4

R2

R1

I E D C A

R2

R3

R1

R5

B R9

R8 R7

R6

R4

IAB A

B

1

2

2

4 4

6 6

c d

e

b

a

-

+

-

e3

B

A

+

-

+

e2 e1

I3

I2 I1

R3

R2 R1

Hnh 2.20

Hnh 2.21

Hnh 2.22

Hnh 2.23

Hnh 2.24


Page 26/94

3.Cho mch in nh hnh 2.24. Vi R1=1(), R2=3(), R3=6(),E1= 10(V), E2=4(),

E3=6().

Tnh: I1, I2, I3 PR1, PR2, PR3,bng phng php xp chng.

4. Cho mch in nh hnh 2.25 R1=3(), R2=4(),

R3=8(), E1=6(V), E2=8(V).

Tnh: I1, I2, I3 bng phng php xp chng.

5. Tnh: I1, I2, I3, I4 bng phng php xp chng.

6. Tnh: I1, I2, I3, bng phng php xp chng.

7. Tnh dng in qua cc nhnh bng phng php xp chng.

-

+

-

e2

B

A

+

e1

I2

I3 I1

R2

R3 R1

-

+

4

I4

I3 I2

2

2

2A

12V

B A I1

4

+ -

C

I2

-

+

24V 4A

12V

2

I3

6

3 B A I1

3 6

6 12

36V

- + 6A

Hnh 2.25

Hnh 2.26

Hnh 2.27

Hnh 2.28


Page 27/94



10. Cho mch in nh hnh 2.31

E1= 10(V), E2= 6(V), E3= 2(V).

R1= 4(), R2= 4(), R3= 4().


11. Cho mch in nh hnh 2.32

E1= 10(V), E2= 4(V), R1= 4().

R2= 4(), R3= 4().


12. Cho E1= 6(V), E2= 8(V), E3= 10(V).

R1= 2(), R2= 4(), R3= 6().

Tnh I1, I2, I3 bng phng php dng in vng.

(hnh 2.33).

E2 E1 R3

+ - -

+

I3

I2 I1 R2 R1

E3 - +

E2

E3

E1 R3 +

-

+ - -

+

I3

I2 I1 R2 R1

6A 2 2

6

3 + -

8A

3V

E2 E1 R3

+ - -

+

I3

I2 I1 R2 R1

2A

4A

2

2

4

Hnh 2.29

Hnh 2.30

Hnh 2.31

Hnh 2.32

Hnh 2.33


Page 28/94

13. Cho E1= 10(V), E2= 4(V), R1= 2(), R2= 4().

R3= 2, R4= 4(), R5= 2, R6= 2()

Tnh dng in qua cc nhnh bng phng php

dng in vng.

14. Cho E1= 8(V), E2= 4(V), R1= 1(), R2= 2().

R3= 4, R4= 4(), R5= 2().

Tnh dng in qua cc nhnh bng phng php

dng in vng.

15. Cho E1= 8(V), E3= 4(V), R1= 2(), R2= 4().

R3= 1().

Tnh dng in I1, I2, I3 bng phng php

dng in nhnh.

16. Cho E1= 10(V), E2= 6(V), R1= 2(), R2= 6().

R3= 4().


dng in nhnh.

17. Cho E1= 10(V), E2= 6(V), E3= 5(V)

R1= 2(), R2= 6(), R3= 4(), R1=1().


dng in nhnh.

18. p dng phng php nt tnh V1, V2, V3

R1 R1

E2

E3

E1 R3 +

-

+ - -

+

I3

I2 I1 R2

E3

E1 R3 +

- - +

I3

I2 I1 R2 R1

E1 R5 R4

R3 R2 R1

- +

- + E2

E2 E1 R3

+ - -

+

I3

I2 I1 R2 R1

V3

V2

V1

+

+

A

B

0

3A4

5A4

2

4A

R6

R5 E2

- + E1 - +

R3 R4

R1 R2

Hnh 2.34

Hnh 2.35

Hnh 2.36

Hnh 2.37

Hnh 2.38

Hnh 2.39


Page 29/94

19. p dng phng php nt tnh I

20. p dng phng php nt tnh V.

21. p dng phng php nt tnh V.

22. Cho E1= 6(V), E2= 2(V), R1= 3(), R2= 4(), R3= 2().

Tnh dng in I1, I2, I3 bng phng php in p hai nt.

23. Cho E1= 10(V), E2= 6(V), R1= 2(), R2= 6(), R3= 4().

Tnh dng in I1, I2, I3 bng phng php in p hai nt.

24. Cho E1= 6(V), E2= 5(V), E3= 4(V), R1= 2(),

R2= 6(), R3= 4(), R1=1().


in p hai nt.

Hnh 2.42

Hnh 2.41

E3

E1 R3 +

- - +

I3

I2 I1 R2 R1

0

I4 I3 I1 I2 V

c b a

12 4 2

6 - +

8V

28V - +

0

I B A

2

8 4 V1 AV

31 2A

-

2A6A

I4

0

I3 I1 I2

b

V

a

6 2

+

12V

R1 R1

E2

E3

E1 R3 +

-

+ - -

+

I3

I2 I1 R2

E2 E1 R3

+ - -

+

I3

I2 I1 R2 R1

Hnh 2.40

Hnh 2.43

Hnh 2.44

Hnh 2.45


Page 30/94

25. Hai bng n cng sut nh mc ca mi bng ln lt l 25W v 100W u lm vic

bnh thng ht 110V hi.

a. Cng dng in qua bng no sng hn ?

b. in tr ca bng no ln hn.

c. C th mc nt hai bng n ny vo mng in c hiu in th 220V c khng ? ti sao?

26. Mt bng n c ghi Um= 220V, Pm= 100W nu mc bng n vo ht 110V th cng

sut tng hay gim.

27. bng n loi 120V-60W sng bnh thng mng in c hiu in th l 220V

ngi ta mc ni tip vi n mt in tr ph R. Tm in tr ph .

Ti Liu Tham Kho :




- Gio trnh mch in Trng HSPKT-TPHCM lu hnh ni b.


Page 31/94

CHNG 3: MCH IN XOAY CHIU

Trong i sng v trong k thut dng in xoay chiu c s dng rng ri v n c cc

u im sau y:

D truyn ti i xa.

D dng thay i cp in p nh my bin p.

MF in, ng c in xoay chiu lm vic tin cy, vn hnh n gin.

ch s kinh t v k thut cao.

Trng hp cn thit dng in xoay chiu c bin i sang dng in mt

chiu nh thit b nn dng ( chnh lu ).

1.Khi nim v dng in xoay chiu.

1.1.Dng in xoay chiu

L dng in thay i chiu v ln theo thi gian.

Dng in hoc sc in ng c tri s bin i tun hon theo quy lut ca mt hm

hnh singi l sc in ng hay dng in hnh sin.

f(t)=Fm.sin( ) t

f(t) c th l dng in i(t), in p u(t), sc in ng e(t) hoc tr s ca dng in j(t).

Fm>0: bin

>0: tn s gc, n v o l rad/s (radian/giy)

t : gc pha ti thi im t, n v o l radian hoc .

: gc pha ban u, n v o l radian hoc (0 3600)

1.2.Chu k, tn s

+ Chu k: l khong thi gian ngn nht sc in ng (e) hoc l dng in (i) tr v gi

tr c.

K hiu : T=

2 (giy)

i

0 t

Hnh 3.1


Page 32/94

+ Tn s: l s chu k trong mt n v thi gian (1giy).

k hiu: f, n v o l hec (hz)

f=

2

1

T (hz)

1.3.Pha v s lch pha.

+ Pha: l trng thi bin i ca sc in ng (hay dng in) theo thi gian (tng ln hay

gim xung qua tr s khng v cc i) gi l pha ca sc in ng hoc dng din.

+ S lch pha: nu hai dng in hoc hai sc in ng hnh sin c tr s bin i ng thi

( cng tng ln cng gim xung qua tr s 0 v cng cc i, cng i chiu ) th gi l hai

dng in (hoc s) cng pha. Tri li l s lch pha.

1.4.Tr hiu dng.

o v nh gi c cc gi tr ca dng in xoay chiu nh cng dng in, hiu

in th, sc in ng. Ngi ta da vo gi tr hiu dng ca dng in xoay chiu.

Tr hiu dng I ca mt dng in i(t) bin thin tun hon trong mt chu k T bng vi

dng in khng i gy ra cng mt cng sut tiu tn trung bnh trn mt in tr R.

2

.1

..1

0

2

0

22 m

TT Idtti

TIRIdtiR

T

Vi i(t)=Im.sin( ) t , T

2

T

2

T

i e

0 t

T

i e

0 t

e

i

Hnh 3.2

Hnh 3.3


Page 33/94

Tng t ta tnh c: U=2

mU , E=2

mE , J=2

mJ

1.5.Biu din lng hnh sin bng th vect.

Mt lng hnh sin, c th biu th bng mt biu thc hoc bng mt vect.

* phng php:

di ca vect bng tr s cc i ca lng hnh sin.

Gc pha u l gc hp bi vect vi trc honh thi im ban u.

Tc gc quay ca vect bng tc gc ca lng hnh sin.

Chiu quay ca vect ngc vi chiu kim dng h.

Hnh chiu ca vect trn trc tung l tr s tc thi ca lng hnh sin.

V d: e=Em.sin( )300t

1.6.Cng v tr bng th vect

e1=Em.sin( )1 t

e1=Em.sin( )2 t

21 mmm EEE

ln:

21212

22

12 ...2 mmmmmmm ECosEEEEEE

Gc pha:

Tg2211

2211

cos.cos.

sin.sin.

mm

mm

EE

EE

Khi cng cc vect cng pha vi nhau th tr s cc i ca vect tng hp bng tng cc

tr s cc i ca cc vect thnh phn.

Em=Em1+Em2

Php tr cng nh php cng ch vic m cng lng b tr bng tr s lng m, ngha l:

0

e

t+

mE

mE

x

y

e

e2 e1

mE

0

t+1

2mE

1mE

x

y

t+2

Hnh 3.4

Hnh 3.5


Page 34/94

)( 2121 mmmmm EEEEE

2.Gii mch in xoay chiu

2.1.Mch xoay chiu thun tr

Xt on mch: UAB=Um.sin t. (v). Trong khong thi gian t v cng nh coi dng

in l khng i:

i= tR

U

R

u m sin. (A)

V Um, R l khng i, t Im=R

Um i=Im.sin t mch AC thun tr in p v dng in

cng pha nhau.

V d : t hai u in tr R=50() hiu in th xoay chiu U=220(V), f=50(hz). Tnh

dng in hiu dng I, vit biu thc cng dng in i qua mch.

Gii

Gi s biu thc hiu in th tc thi hai u in tr c dng

u=Um.Sint (V) = 220 2 .sin100t (V)

I= 50

220

R

U4,4(A)

Biu thc cng dng in qua mch

i=Im.sint=4,4. 2 .sin100t(A)

2.2.Mch in xoay chiu thun cm

Nu qua phn t in cm c dng in

iL(t) = ILm.sin t

Trn n s xut hin in p: uL(t)=L. )(.)( , tiL

dt

tdiL

L

uL(t) = L.ILm. .cos t =ULm.cos )2

sin(.

tUt Lm

t ULm = L.Im. ).(. LZZI LLm

Trong mch in xoay chiu thun cm in p uL(t) nhanh pha hn iL(t) gc 2

uL(t)

IL(t)

mI mU 0

i R

u

uL(t)

iL(t) L

Hnh 3.6

Hnh 3.7

Hnh 3.8

Hnh 3.9


Page 35/94

V d 1 : Mt cun dy c h s t cm L= 0,1(H), cho dng in xoay chiu I= 0,5(A) i qua

mch. Tnh hiu in th hiu dng hai u on mch, vit biu thc hiu in th tc thi.

Gii

Gi s biu thc cng dng in tc thi qua mch c dng

i=Im.Sint (V) = 0,5. 2 .sin100t (A)

ZL=.L=100.0,1=31,4()

U=I.ZL=0,5.31.4=15,7(V)

Biu thc hiu in th hai u on mch

u=Um.sin(t+2

)=15,7. 2 .sin(100t+

2

)(V)

V d 2: Mt cun dy c h s t cm L=

1(H). Tnh dng in quan mch trong cc trng

hp sau:

a. t hai u cun dy mt hiu in th mt chiu U=110(V).

b. t hai u cun dy mt hiu in th xoay chiu U= 100(V), f=50(hz).

Gii

a. V f = 0ZL= 0 I= 0

110

Z

U

b.V f=50(hz) ZL=.L=100.

1=100() I= 1

100

100

R

U(A)

2.3.Mch xoay chiu thun dung

t in p uc(t)=UCm.sin t hai u t in C, s c dng in

ic(t)=C. tUCdt

tduCm

c cos...)(

)2

sin(.)(

tIti Cmc (A)

ICm=C.

.

.C

IUU CmCmCm

t Zc= Ccmcm ZIUC

..

1

Trong mch in xoay chiu thun cm in p uL(t) nhanh pha hn iL(t) gc 2

IC(t)

UC(t)

+ - uC

iC C

Hnh 3.10

Hnh 3.11


Page 36/94

V d 1 : Mt t in c in dung C=25(F), t hai u t in hiu in th xoay chiu

U=220(V), f=50(hz). Tnh dng in hiu dng qua mch, biu thc tc thi ca dng in.

Gii

ZC= 40100.10.

250

1

.

1

6

C

()

I= 40

220

CZ

U4,4(A)

I=Im.sin(t+2

) = 4,4. 2 (sin100t+

2

) (A)

V d 2: Mt t in c in dung C=

10(F). Tnh hiu in th hai u t in trong cc

trng hp sau:

a. Cho dng in mt chiu I=0,5(A) qua mch.

b. Cho dng in xoay chiu I=0,2(A) qua mch, f=50(hz).

Gii

a. V f = 0ZC = U= I.ZC=

b.V f=50(hz) ZC=

100.10.

10

11

6

C

=1000() U=I.ZC=0,2.1000=200 (V)

2.4.Mch xoay chiu R-L-C mc ni tip

t hai u mch R-L-C hiu in th xoay chiu. gi s cng dng in qua mch

AB tai thi im t l:

i(t)=Im.sin t (A)

uR(t)=IRm.sin t (

uL(t)=ILm.Sin( t + )2

(V)

uc(t)=Icm.Sin( t - )2

(V)

u(t)=uR(t)+uL(t)+uc(t)=Um.sin( ) t

URm=Im.R , ULm=Im.ZL , UCm=Im.ZC

Um= 22

.

22 . CLmCmLmRm ZZRIUUU

t Z= 2

2CL ZZR : tng tr mch R,L,C mc ni tip. Um=Im.ZU=I.Z

L C i R

u

UmL-UmC Im

UmL

UmC

UmL

Um

UmR

Hnh 3.12

Hnh 3.13


Page 37/94

tg =R

ZZ CL ht hai du on mch R-L-C lch pha so vi I gc .

Nu: ZL>ZCht nhanh pha hn cng gc .

Nu: ZL


Page 38/94

Z= (13)56,124(10)( 2222 CL ZZR )

Tg =

85,010

56,124

R

ZZ CL = - 040

I 9,1613

220

Z

U(A)

)40100sin(.2.9,16 0 ti (A)

3.Mch R-L-C phn nhnh

t hai u on mch hiu in th xoay chiu.

u=Um.sin t (V)

Dng in qua cc phn t R-L-C

iR=IRm.sin t (A)

iL=ILm.sin( t -2

) (A)

iR=ICm.sin( t +2

) (A)

i=iR+iL+iC=Im.sin( ) t

Im= 22

mLmCmR III

ImR=R

Um , ImC=C

m

Z

U , ImL=

L

m

Z

U

2

2

111

LC

mmZZR

UI

LCmR

mlmC

ZZR

I

IITg

11

V d : Cho R= 20(), C= )(4

1F

, L=

1(H). Tnh dng in qua cc nhnh v dng in qua

mch mch in trn, bit u=50.sin2 t (V).

Gii

Dng in qua cc nhnh ln lt l

ZC= )(22.

4

1

1

.

1

c

iC iL iR

i

C L R

u

ImC - ImL

Um

UmL

ImL

ImC

Im

ImR

Hnh 3.16

Hnh 3.17


Page 39/94

ZL=.L=2 .

24()

iR= )(2sin.5,220

2sin.50At

t

R

u

iL= )2

2sin(.

tZ

U

L

m = ))(2

2sin(.4

2.50At

iC= ))(2

2sin(.2

2.50)

22sin(. Att

Z

U

C

m

i=iR+iL+iC Im= 22

mLmCmR III =

2

2

4

2.50

2

2.505,2

= 6,45(A)

)(56,4414,1

45,6

2A

II m

14,142105,2

4

2.50

2

2.50

mR

mlmC

I

IITg 850

i=6,45.sin(2 )850t (A)

4.Gii mch in xoay chiu bng s phc

4.1.khi nim v s phc:

S phc k hiu ibaC

(v i l k hiu tr s dng in tc thi nn ta thay i bng j).

Vy ta c s phc jbaC

trong :

a : thnh phn thc

b : thnh phn o

Trong a v b l cc s thc ( j= 1 hay j2= -1 l n v o).

MP phc l mt phng c hai trc trc giao nhau, trong trc thc 1 l trc honh, trc o

j l trc tung.

Mt s phc c th vit mt trong hai dng sau:

Dng i s: jbaC

0 -1 +1

-j

+j

Hnh 3.18


Page 40/94

Dng m (dng cc):

CeCC j.

C l moun s phc (|C|= 22 ba )

l argumen ca s phc ( =arctga

b)

(0 3600)

Ta c th biu th s phc C trn mt phng phc .

sin.,cos.,22 CbCabaC

Mt s phc bt k c th nm mt trong bn mt phng phc.

Ni chung mt s phc bt k c k hiu bng ch in hoa c du chm trn u.

V d: i s phc sau t dng i s sang dng s m.

1C 4+j3,

2C = 1+j2.

Chuyn mt s phc t dng s m sang dng i s ta s dng cng thc euler.

sincos je j

1C 4+j3=0374

322 .5.34 j

jarctg

ee

2C = 1+j2=0631

222 .5.21 j

jarctg

ee

Mt i lng hnh sin c th biu din di dng phc v ngc li. Khi biu din moun

s phc tng ng tr hiu dng v argumen tng ng pha ban u ca lng hnh sin.

V d : Chuyn cc biu thc sau sang dng s phc

i= t100sin.2 (A), u=220.sin(100 030t )(V), e=110.sin(100 060t )(V)

Gii

00 00 .1.2

2. jjj eeeII

(A)

C |C|

0

b

a

+j

+1 0 +1

C b

-a

+j

jbaC

jbaC

jbaC

-1 +1

-j

+j

C

|C|

-b

-a

jbaC

a +1

-j

+j

C

|C|

-b

|C|

0

Trc o

Trc thc

b C

a

+j

+1

Hnh 3.19

Hnh 3.20


Page 41/94

00 3030 .2.110.2

220. jjj eeeUU

(V)

00 6060 .2.55.2

110.

jjj eeeEE (V)

4.2.Cp s phc lin hp

S phc C2 gi l lin hp vi s phc C1 nu chng c phn thc bng nhau, phn o

bng nhau nhng tri du.

V d: 43,43 21 jCjC

4.3.Php cng- php tr

Gp trng hp phi cng (tr) cc s phc, trc tin ta bin i v dng i s ri

cng (tr) phn thc vi phn thc phn o vi phn o.

V d: 211 jC

, 322 jC

045

1 .2jeC

, 090

2 .1jeC

Tnh

21 CC .

21 CCC 1+j2+2+j3 = 3+j5

045

1 .2jeC

=2.(cos450+jsin450) = 2.( )2

2

2

2j = 22 j

090

2 .1jeC

=1.(cos900+jsin900) = 1.(0+j1) = j1

)12(212221

jjjCCC

Tnh

21 CC

)32()21(21 jjCCC -1-j

)12(21)22(21

jjjCCC

4.4.Php nhn-php chia:

Khi nhn (chia) hai s phc vi nhau u tin ta da v dng s m sau nhn (chia)

moun vi moun cn argumen cng hoc tr cho nhau.

V d: 045

1 .2jeC

, 090

2 .1jeC

211 jC

, 322 jC


Page 42/94

Tnh:

21 .CC ,

2

1

C

C

0000 135)9045(9045

21 .2.2.1..2.jjjj eeeeCCC

000

0

0

45)9045(

90

45

2

1 .2.2.1

.2

jjj

j

eee

e

C

CC

1C = 1+j2 =0631

222 .5.21 j

jarctg

ee

322 jC

=0562

322 .13.32 j

jarctg

ee

00000 119)5663(5663

21 .65.65.13..5.jjjj eeeeCCC

000

0

0

7)5663(

56

63

2

1 .13

65.

13

65

.13

.5 jjj

j

eee

e

C

CC

4.5.Gii mch in hnh sin bng s phc

Quy tc:

Nu: u = Um.sin( UeUUt j.).

Nu: e = Em.sin( EeEEt j.).

Tng tr phc c xc nh:

j

j eZeI

U

I

UZ iu .)(

jXRjZeZZ j sincos.

Phn thc l in tr R

Phn o l in khng X

V d 1: Cho mch in xoay chiu. Bit R = 3(), L = 40(mH), u = 10.cos(100t + 200)(V).

L i jL

I

I i C .C

j

R i

R

I

Hnh 3.21


Page 43/94

Tnh

I , vit biu thc i.

Gii

u= 10.cos(100t +200) = 020.

2

10 je

)(410.40.100... 3

jLjZL

LZRZ = 3 + j4 () = 3

442 .43

jarctg

e = 5.053je ()

000

0

0

33)5320(

53

20

.2.2.5

.2

10

jjj

j

eee

e

Z

UI (A)

)33100sin(.2 0 ti (A)

V d 2: Cho mch in xoay chiu. Bit R = 4(), C = F8

1, u = 8.sin2t(V).

Tnh

I , vit biu thc i.

Gii

u = 8.sin2t = 00.

2

8 je (V)

)(42.

8

1.

jj

C

jZC

045122 .2.4.4444

jjarctgC eejZRZ ()

.c

j C

i

u

-

+

R

I

U -

+

R

i

u

-

+ L

R

I

U -

+ jL

R

Hnh 3.24

Hnh 3.22 Hnh 3.23

Hnh 3.25


Page 44/94

000

0

0

45)450(

45

0

.1.1.24

.2

8

jj

j

j

eee

e

Z

UI

(A)

))(452sin(.2 0 Ati

V d 2: Cho mch in xoay chiu. Bit R = 4(), C = F2

1, L = 2(H), u = 8.cos2t(V).

Tnh

I ,

CV ,vit biu thc i.

Gii

u = 8.cos2t = 00.

2

8 je (V)

)(

2.2

1.

jj

C

jZC

)(42.2..

jjLjZL

CL ZZRZ 4 +j4 j = 4 j3() = 5.037 ()

000

0

37)370(

37

0

.10

2.8.

10

2.8

.5

.2

8

jjj

j

eee

e

Z

UI (A)

00 1279037 .10

2.8.1..

10

2.8.

jjjCC eeeZIV (V)

)372cos(.10

16 0 ti

5.Cng sut, h s cng sut (cos ) v bin php nng cao cos .

5.1.Cng sut tc dng

Cng xut tc dng tc thi c xc nh nh sau:

P(t) = u(t).i(t)

jL

.c

j C

i

u

-

+

R

I

U -

+

R

Hnh 3.26 Hnh 3.27


Page 45/94

Cng xut trung bnh trong mt chu k:

P= T n

nn IRUIdttpT

0 1

2.cos.).(1

(W)

U,I ht v dng in hiu dng.

cos : h s cng sut

Rn,In: in tr v dng in trn cc nhnh.

P: c trng cho hin tng bin i in nng sang c nng v nhit nng.

5.2.Cng sut phn khng:

Q = U.I.sin = In2(ZLn - ZCn) (VAR)

Q c trng cho cng trong qu trnh trao i nng lng in t trng

Trong : ZLn,ZCn,In,ln lt l in khng, in dung, dng in ca mi nhnh.

5.3.Cng sut biu kin:

S=U.I= 22 QP (VA)

5.4.Cng sut phc:

- thun li cho vic tnh ton cng sut, ngi ta nh ngha khi nim cng sut phc nh

sau:

jQPS

5.5.Phng php nng cao cos :

a. ngha ca cos :

H s cos l ch tiu k thut rt quan trng n c ngha rt ln v mt kinh t:

- Mi mt my in c ch to vi mt cng xut biu kin nht nh Sm, t n c th

cung cp mt cng xut tc dng P = Sm.cos

Nu cos =1 P = Sml cng xut ln nht m my c th cung cp.

Nu cos gim, kh nng pht cng xut P gim. Do vy mun tn dng kh nng lm

vic ca thit b th cos phi ln,v P l hng sI=cos.U

P, cos , I , dn

n tc hi

Dng in ln phi dng dy dn ln, lm tng kim loi mu v vn du t xy dng

ng dy .

Tn hao nng lng trn dy dn t l vi bnh phng dng in

AItRIA 2 .


Page 46/94

b.Bin php nng cao cos :

Ta c: cos = 22 QP

Pmun nng cao cos phi tm cch gim Q = I2(ZL+ZC)

ZL: ti in cm (trong sinh hot v cng nghip ZL l ch yu)

ZC: ti in dung

Mun Q, ZL v ZC:

Mun ZL, khng s dng cc thit b c tnh cht cm khng lm vic ch khng

ti hoc non ti.

Mun ZC dng t in mc ss ti (bin php b)

V d:

Khi cha mc t in dng in trn ng dy I = dng in qua ti I1 h s cng xut ca

mch l cos 1 . Khi c t b (mc t // mch ) dng in trn ng dy s l:

CIII 1 , ta

thy I, cos .

I


Page 47/94

cos 1 =0,8 =36,80 tg1= 0,72

cos = 0,95 =180 tg= 0,32

p dng cng thc: )(.

12

tgtg

U

PC = )32.072,0(

220.100

10.102

3

= 26010.6,2 4 (F).

Cu hi :

1. nh ngha tn s? chu k? Pha v s lch pha.

2. Cho bit s lch pha gia dng in v in p trong mch in xoay chiu thun tr?

Thun cm? Thun in dung.

3. nh ngha s phc? Ti sao s phc c ng dng gii cc mch in xoay chiu hnh

sin.

4. Cho cc s phc:

a.

1C = 1+j4,

2C 2+j3. Tnh

21 CC ,

21 CC ,

21 .CC ,

2

1

C

C.

b. 321 jZ

, 532 jZ

. Chuyn cc s phc v dng s m v tnh

21 .ZZ ,

2

1

Z

Z.

c. 030

1 .2jeZ

, 060

2 .3

jeZ . Chuyn cc s phc v dng i s v tnh

21 ZZ ,

21 ZZ .

5. Chuyn cc biu thc sau y v dng s phc

i = 4.sin(2t + 100) (A)

u = 10.cos(5t+150) (V)

e = 5.sin(10t 200) (V)

6. Gii cc mch in xoay chiu sau y:

a. R=3(), L = 10

1(H), i = 2.sin20 t (A).

Tnh Z, U, vit biu thc u, Vit biu thc in p hai u cun dy uL, cng sut P.

b. R=3(), C=16

1(F), u = 10.sin4 t (V).

Tnh Z, I, vit biu thc i, Vit biu thc in p hai u t in uC, cng sut P.

7. Mt bng n loi 110V- 60w mc ni tip vi 1 cun dy c h s L= )(1

H

, cun dy c

in tr RL=10(). t hai u cun dy mt hiu in th xoay chiu U=220(V), f=50(hz).

Tnh dng in qua mch, Vit biu thc hiu in th tc thi hai u cun dy.

i

L R

i

C R

Hnh 3.30

Hnh 3.31


Page 48/94


t hai u on mch mt hiu in th xoay chiu u = 12.sin4t(V), R=5(), L=1(H).

a. Khi cng tc v tr 1,bit C=8

1(F). Tnh Z, I, ULC, vit biu thc uLC.

b. Khi cng tc v tr 2. Tnh gi tr in dung C1 dng in qua mch t cc i v tnh

cng sut ca mch in trong trng hp ny.


u = 5.cos3t, R = 1(), R1= 3(), L=1(H)

C=9

1(F). Tnh i, iC, P.


u = 18.cos8t(V), R= 6(), R1= 4(), C = 32

1(F)

L = 4

1(H). Tnh

I ,

CV .


u = 10.cos8t(V), R = 20(), R1=10(), C = 40

1(F)

Tnh

I ,

CV , P.


u = 10.cos(t+100)(V), R =1(), R1=1(), L =1(H)

iC i1

i

R

L

C

R1 u

iC i1

i

R L

C

R1 u

iC i1

i

R

C

R1 u

L

2

1

i

C

R

C1

Hnh 3.32

Hnh 3.33

Hnh 3.34

Hnh 3.35


Page 49/94

Tnh

I ,

LV , P.


u = 16.sin(2t + 200)(V), L = 2(H), L1=1(H)

R= 4(), R1= 2(). Tnh

I ,

1LV , P.


R1=10(), R2= 5(), L =100(mH), C = 50(F)

u = 2

100.sin100t(V). Tnh i1, i2

Ti Liu Tham Kho :





i R2

I2

L

i1 R1

u

C

L R1

i1

L1

i1

i

R u

iL

L

i1

i

R

R1 u

Hnh 3.36

Hnh 3.37

Hnh 3.38


Page 50/94

CHNG 4: MCH IN BA PHA

1.Nguyn l pht sinh h thng dng in xoay chiu 3 pha.

1.1.nh ngha mch in 3 pha:

- Mch in xoay chiu 3 pha gm ngun in 3 pha, ng dy truyn ti v ph ti 3 pha

1.2.Nguyn l to ra ngun 3pha.

- to ra ngun ba pha ngi ta dng my pht in ng b ba pha.

a.Cu to: gm hai phn chnh l stato v roto.

Stato: bao gm li thp v dy qun

- Li thp ghp bng cc l thp k thut in c dng hnh tr rng bn trong c dp rnh t

dy qun.

Dy qun gm ba b dy ging nh nhau t lch nhau

gc 3

2 trong khng gian, mi b dy l mt pha..

Dy qun pha A: (A,X)

Dy qun pha B: (B,Y)

Dy qun pha C: (C,Z)

Roto: l mt nam chm in.

b.nguyn l hot ng

- Khi quay roto, t trng ln lt qut qua cc dy qun trn stato v cm ng vo dy qun

cc sc in ng hnh sin cng bin , cng tn s v lch pha nhau gc 3

2.

Nu chn pha u ca sc in ng eA ca dy qun AX bng khng th biu thc s tc

thi ca pha l:

Hnh 4.1

Stato

roto

eA eC eB e

t 120

0 240

0 360

0

Hnh 4.2


Page 51/94

eA=E tsin.2 (V)

eB=E )3

2sin(.2

t (V)

eC=E )3

2sin(.2

t (V)

Biu din bng s phc:

0

0

0

120

120

0

.

.

.

jC

jB

jA

eEE

eEE

eEE

Ngun in gm 3 s hnh sin cng bin , cng tn s, lch nhau v pha gc 3

2 gi

l ngun in ba pha i xng: 0 CBA EEE .

Sc in ng, in p, dng in mi pha ca ngun (ti) gi l s pha (EP), in p

pha (UP), dng in pha (IP).

Dng in chy trn ng dy pha t ngun n ti gi l dng in dy (Id), in p

gia hai dy pha gi l in p dy (Ud).

Thng thng trc khi u vi ti ngun ba pha c u li, c hai cch u l u

sao v u tam gic.

2.S u dy trong mang ba pha.

2.1.Ni hnh sao(Y)

0

Ip

Ip

Ip

Id

Id

Ud

Id

UP

C B

A

Ud Up

Ip

Ip Ip

Id

Id

Id

C B

A

C B

A

Ip

0 0

Ip

Ip

Ip

Id

Id

Ud

Id

UP

C B

A

Ip

Ip

Hnh 4.3 Hnh 4.4

Hnh 4.5


Page 52/94

- Ni 3 im cui ca ba pha vi nhau to thnh im chung gi l im trung tnh.

Ngun: ni 3 im cui X,Y,Zim trung tnh O

Ti: ni 3 im cui X,Y,Zim trung tnh ti O

OO gi l dy trung tnh

a.quan h gia dng in dy v dng in pha.

b.quan h gia in p dy v in p pha.

BAAB UUU (in p gia pha Av pha B)

CBBC UUU (in p gia pha Bv pha C)

ACCA UUU (in p gia pha Cv pha A)

Xt tam gic OAB:

AB=2.OA.cos300=2.OA. OA.32

3

AB=UAB=Ud, OA=Up Ud= pU.3

2.2.ni tam gic ()

- Mun ni tam gic ta ni u pha ny vi cui pha kia.

V d: ni A vi Z, B ni vi X, C ni vi Y.

a. Quan h gia Idv If

p dng nh lut kichop 1 ti cc nt.

Ti nt A: CAABA III

Ti nt B: ABBCB III

Ti nt C: BCCAC III

Xt tam gic OAB: OB=2.OA.cos300=2.OA.2

3

Hnh 4.6

IP=Id

ABU

CAU

BCU

ABU

300 H B A

O

0

AI

CAI

ACI

ABI

CAI

Zp

Zp

Zp

C

B

A

Ud Up

Ip

Ip Ip

Id

Id

Id

C B

A

Hnh 4.7

Hnh 4.8


Page 53/94

OB=Id, OA=Ip

Id= PI.3

b.Quan h gia Ud v UP

Vy v pha dng in Id chm sau IP gc 300

3.cng xut mch in ba pha.

3.1.cng xut tc dng:

- Gi PA,PB,PC tng ng l cng xut tc dng ca pha A,B,C.

P=PA+PB+PC

=UA.IA.cos A + UB.IB.cos B + UC.IC.cos C

+ Khi mch in ba pha i xng:

UA=UB=UC

IA=I =IC

cos A = cos B = cos c = cos

P=3.UP.IP.cos =3.RP. cos...32

ddP UII

3.2.cng xut phn khng

- Cng xut phn khng Q ca ba pha:

Q = QA+QB+QC

Q = UA.IA.sin A + UB.IB.sin B + UC.IC.sin C

- Khi ti i xng:

Q = 3.UP.IP.sin =3.XP. sin...32

ddP UII

3.3.cng xut biu kin

S = ddPP IUIUQP .3..322

4.Gii mch in ba pha i xng.

- i vi mch in ba pha i, dng in, in p trn cc pha c tr s bng nhau v lch pha

nhau mt gc 3

2v vy khi gii mch in ba pha ta ch cn tch mt pha ra gii.

4.1. Ngun ni sao i xng.

- Theo hnh v ta c O l im trung tnh ca ngun, nu ti ni sao, O l im trung tnh

UP=Ud


Page 54/94

ca ti. Cc dy t ngun n ti AA, BB, CC gi l dy pha. Dy OO gi l dy trung tnh.

Mch in c dy trung tnh gi l mch in ba pha bn dy, mch in khng c dy trung

tnh gi l mch in ba pha ba dy. i vi mch i xng ta lun lun c quan h:

00 CBA IIII

v th dy trung tnh khng c tc dng, c th b qua dy trung tnh. in th im trung tnh

ca ti i xng lun lun trng vi in th ca trung tnh ngun.

Nu gi sc in ng ngun l Ep th:

in p dy Ud v in p pha UP cua mch in ba pha l:

in p pha pha u ngun l: Up=Ep

in p dy pha u ngun l: Ud= 3 Ep

4.2. Ngun ni tam gic i xng

in p pha pha u ngun l: Up=Ep

in p dy pha u ngun l: Ud=Up= Ep

T gi tr in p dy (hoc in p pha) ca mch in ba pha, ta xc nh in p pha

ca ti.

4.3. Gii mch in ba pha ti ni sao i xng

4.3.1. Khi khng xt tng tr ng dy pha.

UP=3

dU (in p t ln mi pha ca ti)

- Tng tr pha ti: ZP=22PP XR

Trong RP, XP in tr, in khng mi pha ti

Dng in pha ca ti:

IP = 22.3 PP

d

P

P

XR

U

Z

U

Gc lch pha gia in p pha v dng in pha.

C B

A

Ip

0 0

Ip

Ip

Ip

Id

Id

Ud

Id

UP

C B

A

Ip

Ip

Id=If Zp

Ud

C

B

A

Hnh 4.9

Hnh 4.10


Page 55/94

P

P

R

Xarctg

V ti ni sao nn dng in dy bng dng in pha

Id=If

th vect nh hnh v

4.3.2. Khi xt tng tr ng dy pha.

Khi xt n tng tr ng dy:

IP = 22.3 PdPd

d

XXRR

U

Rd, Xd in tr in khng ng dy.

V d: C ba cun dy ging nhau in tr v in khng ca mi cun ln lt l R=3(),

X=4(), in p nh mc ca mi cun dy l Up=220(V). Hi ba cun dy phi mc th no

s dng c ngun in xoay chiu 3 pha c Ud=380(V). Tnh Ip, Id, P3pha, Q3pha, S3pha.

Gii

Ba cun dy trn phi u hnh sao v: Ud= 3 Ud=220. 3 =380(V)

Zp=22PP XR =

22 43 5()

IP=5

220

p

p

Z

U= 44(A)

=arctg p

p

R

Xarctg

3

4= 530 (dng in chm pha so vi in p)

P3pha= 3.R.2pI = 3.3.44

2= 17424(W)

Q3pha= 3.Xp.2pI = 3.4.44

2=23232(VAR)

S3p=3.Up.Ip=3.220.44=29040(VA)

4.4.Gii mch in ba pha ti ni tam gic i xng

4.4.1.Khi khng xt tng tr ng dy pha.

IP = 22PP

d

P

P

XR

U

Z

U

P

P

R

Xarctg

Id= PI.3

If

Uf

Xp Rp Xd Id=If Rd

Ud

C

B

A

380V 380V

Id

Ud

A

B

C

Zp

Zp

Zp

C

B

A

Hnh 4.11

Hnh 4.12

Hnh 4.13

Hnh 4.14


Page 56/94

V d: C ba cun dy ging nhau in tr v in khng ca mi cun ln lt l R=6(),

X=8(), in p nh mc ca mi cun dy l Up=220(V). Hi ba cun dy phi mc th no

s dng c ngun in xoay chiu 3 pha c Ud=220(V). Tnh Ip, Id, P3pha, Q3pha, S3pha.

Gii

Ba cun dy trn phi u hnh tam gic: Up=Ud=220(V)

Zp=22PP XR = 1086

22 ()

IP=10

220

p

p

Z

U= 22(A)

Id= 3 Ip= 3 .22= 31,1(A)

=arctg p

p

R

Xarctg

6

8= 530 (dng in chm pha so vi in p)

P3pha= 3.R.2pI = 3.6.22

2= 8712(W)

Q3pha= 3.Xp.2pI = 3.8.22

2=11616(VAR)

S3p=3.Up.Ip=3.220.22=14520(VA)

4.4.2. Khi xt tng tr ng dy pha.

Tng tr mi pha khi u tam gic: PP XjRZ .

Bin i sang hnh sao: 333PP

Y

Xj

RZZ

Id=

3

33.3

22

Yd

P

Pd

Pd

d

II

XX

RR

U

5.Gii mch in ba pha khng i xng.

Rd Xd Id

Ud

A

B

C

Zp

Zp

Zp

C

B

A

220v 220v

Z

C

Y

B A

X

Hnh 4.15

Hnh 4.16


Page 57/94

- Mch ba pha khng i xng (ZAZBZC), do vy dng in v in p trn cc pha ti s

khng i xng xem mch in nh mt mch phc tp.

CBA IIII 0 0

5.1.Ti ni sao dy trung tnh c tng tr Z0.

- gii mch in trn ta dng phng php in p hai nt.

U00=0

..

YYYY

YUYUYU

CBA

CCBBCA

5.1.1.Khi khng xt n tng tr ng dy.

Trong : A

AZ

Y1

,B

BZ

Y1

,C

CZ

Y1

,0

0

1

ZY

-Ngun i xng: 000 1201200 .,.,. jPC

JPB

jPA eUUeUUeUU

0

000

'00

'00

'00

0

120120

'00

'

00

...

Z

UI

Z

UIUUU

Z

UIUUU

Z

UIUUU

YYYY

eYeYYUU

C

CCCC

B

BBBB

A

AAAA

CBA

jC

jBA

P

V d: Ba cun dy A, B, C ni sao nh hnh v c cp in t mt ng dy 3 pha ba dy

(khng c dy trung tnh) vi in p dy Ud=346V v tn s f=50hz. Bit in tr ca cc

cun dy A, B, C ln lt l 03010

AZ (), )(453,130

BZ , )(3720 0

CZ .

Tnh dng in trong mi cun dy v in p gia hai u mi cun dy.

Gii

IO ZO O O

ZC

ZB

Id=If ZA

Ud

C

B

A

C B

A

Ip

0 0

Ip

Ip

Ip

Id

Id

Id

C B

A

Ip

Hnh 4.17

Hnh 4.18


Page 58/94

p dng phng php in p hai nt

Ta c: EA = EB = EC = 3

346200(V)

Khi chn

AE lm gc pha, ta c

AE = EA= 200(V); 0120200

BE = 200(-0,5 j0,86); 0120200

CE = 200(-0,5 + j0,86)

YA = 0

0301,0

3010

11

AZ= 0,086 j0,05

YB = 0

045075,0

453,13

11

BZ= 0,053 j0,053

YC = 0

03705,0

3720

11

CZ= 0,04 j0,03

CBA

CCBBAAOO

YYY

YEYEYEU

...

'

03,004,0053,0053,005,0086,0

)86,05,0)(03,004,0()86,05,0)(053,0053,0()05,0086,0(200'

jjj

jjjjjU OO

= 3,775,27)(73,147,2861,36223,0

34,51032,0200

133,0179,0

)025,002,0(2000

0

jVj

j

(V)

in p gia hai u mi cun dy l

OOAA UEU ,

= 200 27,75 + j7,3

= 172,25 + j7,3 = 172,4 42,20 (V)

OOBB UEU ,

= -100 j173 27,75 + j7,3

= - 127,75 j165,7 = 209,22 63,1270 (V)

OOCC UEU ,

= -100 + j173 27,75 + j7,3

= - 127,75 + j180,3 = 221 32,1250 (V)

Dng in tronh cc cun dy s l

58,2724,173010

42,24,172 00

0

A

AA

Z

UI (A)

63,17273,15453,13

63,12722,209 00

0

B

BB

Z

UI (A)


Page 59/94

32,8805,113720

32,125221 00

0

C

CC

Z

UI (A)

5.1.2.Khi xt n tng tr ng dy.

dC

C

dB

B

dA

AZZ

YZZ

YZZ

Y

1

,1

,1

5.2.Khi tng dn dy trung tnh 00 Z

- im trung tnh 0 trng vi 0 v in p trn cc pha ti bng in p pha tng ng ca

ngun. R rng nh c dy trung tnh in p pha trn ti vn i xng.

- Tnh dng in trong cc pha, ta p dng nh lut m cho tng pha ring r:

A

AA

A

AA

Z

UI

Z

UI

B

BB

B

BB

Z

UI

Z

UI

C

CA

C

CC

Z

UI

Z

UI

5.3. Khi dy trung tnh b t hoc khng c dy trung tnh )0,( 00 YZ .

- in p U00c th rt ln, do in p trn cc pha ti khc in p pha ngun rt nhiu c

th gy nn qu in p mt pha no .

V d:

- Ti 3 pha khng i xng. Pha A l mt t in thun dung, hai pha B v C l hai bng n.

Tng dn phc ca pha A l:

jbjC

YA

1

Hai pha B v C l hai bng n c tng dn: gR

YY CB 1

- Ngun 3 pha i xng, c in p l Up. Tnh in p t

Ln mi bng n.

Gii

p dng phng php in p hai nt

ggjb

egegjbUU

jj

p

00

,

120120

00..

IC

IB

IA

ZC

ZB

ZA

Ud

C

B

A

IC

IB

IA

Zd

Zd

ZA

IO ZO O O

ZC

ZB

Zd

Ud

C

B

A

B

C

A

C

Hnh 4.19

Hnh 4.20

Hnh 4.21


Page 60/94

)120sin()120cos( 001200

je j = -0,5 j0,866

0120 120cos0

je +jsin1200 = - 0,5 + j0,866

AU Up

0120. jpB eUU

= Up(-0,5 j0,866)

0120. jpC eUU

= Up(-0,5 + j0,866)

Thay vo cng thc trn ta c:

ggjb

jUjgjbUU pp

)866,05,0()866,05,0(.00'

Cho g = b )6,03,0(00' jUU p

- Ta suy ra in p t ln bng n pha B:

)6,02,0()866,05,0(00' ' jUjUUUU ppBB

= Up(-0,3-j1,466)

V tr s th: UB= Up 22 466,13,0 = 1,5Up

- Tng t in p t ln bng n pha C:

)6,02,0()866,05,0(00' ' jUjUUUU ppCC

= Up(-0,3+j0,266)

V tr s th: UC= Up 22 266,03,0 = 0,4Up

- Ta nhn thy in p t ln bng n pha B ln hn in p t ln bng n pha C. Ta c

th dng thit b lm ci ch th t pha. Mun bit th t pha ca mt h thng no , m

ci ch th t pha ni vo h thng in p . Nu gi pha ni vo nhnh in dung l pha A

th pha ni vo bng n sng r l pha B v pha ni vo bng n ti s l pha C.

5.3. Gii mch in ba pha ti ni tam gic khng i xng.

- Nu khng xt n tng tr cc dy dn pha, in p t ln cc pha ca ti l in p ngun.

AB

ABAB

AB

ABAB

Z

UI

Z

UI

BC

BCBC

BC

BCBC

Z

UI

Z

UI

ABI

CAI

ABI

CI

AI

BI

CAZ

ABZ

BCZ

A

ABZ

C

B

Hnh 4.22


Page 61/94

CA

CACA

CA

CACA

Z

UI

Z

UI

- p dng nh lut K1 tm cc dng in nh sau:

CAABA III

ABBCB III

BCCAC III

V d:

Ba cun dy A, B, C ni tam gic nh hnh v c cp in t mt ng dy 3 pha ba dy

vi in p dy Ud=200V v tn s f=50hz. Bit in tr ca cc cun dy A, B, C ln lt l

03010

AZ (), )(453,130

BZ , )(3720 0

CZ .

- Tnh dng in trong mi dy pha t ngun n cc cun dy.

Gii

Ngun ba pha i xng:

00 .200. jjpA eeUU

, 00 120120 .200.

jjpB eeUU , 00 120120 .200. jjpC eeUU

0

0

0

30

30

0

.20.10

.200

jj

j

AB

AAB e

e

e

Z

UI (A) =

0

0

0

75

45

120

.04,14.3,13

.200

jj

j

BC

BBC e

e

e

Z

UI (A)

0

0

0

83

37

120

.10.20

.200 jj

j

CA

CACA e

e

e

Z

UI

(A)

CAABA III = 20.

00 8330 .10 jj ee = 20(cos(-300) + jsin(-300)) 10(cos(830) + jsin(830))

= 20( )2

1

2

3j - 10(0,12 + j0,99)

= 20(0,866 j0,5) 1,2 j9,9

CI

BI

AI

CAI

BCI

ABI

C

B

A

ZCA

ZBC

ZAB

C

B

A

Hnh 4.23


Page 62/94

= 17,32 j10 1,2 j9,9 = 16,12 j19,9 = 25,61.051je (A)

ABBCB III = 14,04.

075je - 20030je = 14,04(cos(-750) + sin(-750)) 17,32 - j10

= 14,04(0,26 j0,96) 17,32 j10

= 3,64 j13,46 17,32 j10

= - 13,68 j23,46 = 27,15.060je (A)

BCCAC III = 10.

00 7583 .04,14 jj ee = 10(cos(830) + jsin(830)) - 14,04(cos(-750) + sin(-750))

= 10(0,12 + j0,99) - 14,04(0,26 j0,96)

= 1,2 + j9,9 - 3,64 + j13,46 = -2,44 + j23,3

= 23,48.084je (A)

- Nu c tng tr ng dy Zd ta nn bin i tng ng ti ni tam gic thnh hnh sao.

6.cch ni ngun v ti trong mch ba pha

- Ngun v ti u c th ni sao hoc ni tam gic, ty theo iu kin c th nh in p quy

nh ca thit b, in p ca mng in v mt s yu cu k thut khc.

6.1.cch ni ngun in

- Cc ngun in trong sinh hot thng ni sao c dy trung tnh. Ni nh vy c u im l

c th cung cp hai cp in p khc nhau l in p pha v in p dy.

+ Mng in 380v/220v (Ud=380v,Uf=220v)

+ Mng in 220v/127v (Ud=220v,Uf=127v)

6.2.Cch ni ti ba pha.

-Ti ba pha thng l cc ng c in ba pha, gm ba b dy ging nhau. Khi thit k ngi

ta quy nh in p cho mi dy qun, lc lm vic yu cu in p phi ng vi quy nh.

380v/220v

UAN=Up

N

UAB=Ud

C

B

A

220v/1270v

UAN=Up

N

UAB=Ud

C

B

A

A B C N

Hnh 4.24

Hnh 4.25


Page 63/94

6.3.Cch ni ti mt pha.

- Khi chn cc thit b trong sinh hot (thit b mt pha), ta cn chn in p thit b bng in

p pha, nh vy ta s dng mt dy pha v dy trung tnh, in p t ln cc thit b l in

p pha. Nh c dy trung tnh in p t ln cc thit b khng vt qu in p pha.

Cu hi:

1. nh ngha mch in 3 pha ? Nguyn l to ra ngun in xoay chiu 3 pha.

2. Chng minh trong mch in xoay chiu 3 pha ni (Y), in p dy Ud= 3 Up.

3. Chng minh trong mch in xoay chiu 3 pha ni (), in p dy Id= 3 Ip.

4. C su bng n loi 110V- 60w, ngi ta ni vo ngun in xoay chiu 3 pha 3 dy c

in p dy Ud= 220V c c khng ? V hnh.

5. Mt phn xng c cp in t mt ng dy 3 pha voi in p dy Ud=120V v tn s

f=50hz. Phn xng bao gm cc loi ti nh sau:

CD

A B C N

N

A B C

Hnh 4.26

Hnh 4.27


Page 64/94

a. 150 bng n loi 100w ni gia cc pha v dy trung tnh, mi pha gm 50 bng n.

b. Mt ng c ba pha 14,9kw vi hiu sut =0,9 v cos = 0,85.

c. Mt b t in gm 3 nhm ni tam gic, mi nhm gm ba t in ni song song, in

dung mi t l C = 20F.

d. Ba cun dy ni tam gic, mi cun dy c in tr R = 5 v in cm L = 0,01H.

Xc nh dng in trong dy pha t ngun n phn xng v h s cng sut cos ca

phn xng.

6. Mt ng dy 3 pha in p dy 200V tn s 50hz cp in cho mt phn xng nh hnh

v.

Phn xng bao gm cc ti in nh sau:

a. Mt ng c M cng sut 3kw, hiu sut = 0,8 v cos = 0,82.

b. Ba n cng sut mi n 500w c ni gia mi dy pha v dy trung tnh.

c. Ba cun dy mc tam gic, mi cun dy c in tr R = 10 v in cm L = 0,02H. Xc

nh dng in i trong mi dy pha t ngun n phn xng v cos ca phn xng.

7. Mt ng dy ba pha vi in p dy 200v v tn s 50hz cp in cho ba cun dy nh

nhau c ni tam gic nh hnh v. Mi cun dy c in tr R = 1,6 v in cm L =

0,00328H. Mi dy pha ca ng dy c in tr Rd = 0,02 v in khng Xd= 0,04. Hy

xc nh :

a. Dng in trong mi dy pha.

b. in p gia cc dy pha u ng dy.

c. H s cng sut cos u ng dy.

N

A B C

M

R, L

C

B

A

Rd

Rd Xd

Xd

Xd Rd

Hnh 4.28

Hnh 4.29


Page 65/94

8. Mt ng dy 3 pha 3 dy vi in p dy 240V cp in cho mt ti ba pha ni tam gic

nh hnh v. Bit tng tr cc pha ca ti l 0010ABZ (), 030BCZ (),

03015 CAZ (A).

- Xc nh cc dng in qua cc pha ca ti v trong cc dy pha cung cp cho ti.

9. Mt ng dy 3 pha 4 dy vi in p dy 220V cung cp in cho mt ti 3 pha ni sao

nh hnh v.

Bit tng tr cc pha ca ti l AZ = 6 00 , 0306BZ , 0455CZ . Xc nh dng

in i trong dy pha v dy trung tnh ca ng dy cp in cho ti trong hai trng hp.

a. B qua in tr dy trung tnh.

b. t dy trung tnh.

10. Mt ng dy 3 pha 3 dy vi in p dy 500V cp in cho mt ti ba pha ni tam gic

nh hnh v.

- Bit tng tr cc pha l )(3010 0 ABZ , )(025 0 BCZ , )(3020 0 CAZ .

Xc nh dng in i trn cc dy pha ca ng dy cung cp v cng sut tiu th ti.

CI

BI

AI

CAI

BCI

ABI

C

B

A

ZCA

ZBC

ZAB

C

B

A

I0

IC

IB

IA

ZC

ZB

ZA

N

C

B

A

Hnh 4.30

Hnh 4.31

Hnh 4.32

CI

BI

AI

CAI

BCI

ABI

C

B

A

ZCA

ZBC

ZAB

C

B

A


Page 66/94

11. Mt ti ba pha ni sao vi tng tr cc pha )(010 0 AZ , )(6010 0 BZ ,

)(6010 0 CZ c cp in t mt ng dy 3 pha 3 dy vi in p dy 200V.

- Xc nh cc in p pha ph ti UAO, UBO, UCO.

Ti Liu Tham Kho :




- Gio trnh k thut in tc gi : Trng tr Ng NXBXD-2004


CHNG 5: GII MCH IN NNG CAO

1.nh lut omh, kichop dng phc

ZC

ZB

ZA

C B

A

Ip

0 0

EA

EB EC

IC

IB

IA

Ip

Ip

Hnh 4.33

Hnh 4.34


Page 67/94

1.1.nh lut omh.

ZIU .

- Cc trng hp ring:

hai cc l phn t in tr: RIU .

hai cc l phn t in cm: ILjULjZL .....

hai cc l phn t in dung: IC

jU

C

jZC

.

1.2.nh lut kichop dng phc

a.nh lut k1:

0 KI

b.nh lut k2:

0

KU

2.Gii mch in nng cao.

2.1.Gii mch in AC bng phng php dng in nhnh.

V d: Cho mch in nh hnh v vi 01 0100

E (V), 02 30100

E , Z1=Z1= 50 + j30(),

Z3= 100(). Vit phng trnh dng in nhnh v gii h phng trnh .

Gii

Chn chiu dng in v chiu i ca vng nh hnh v

p dng nh lut K1 i nt 1: 0321

III (1)

p dng nh lut K2 cho vng 1 v 2:

Z1

1331 EIZI (2)

23322 EIZIZ (3)

(50+j30) 031 0100100

II (2)

II I 3

I

2

I

Z3

Z2 Z1

2

E 1

E

1

I

Hnh 5.1


Page 68/94

(50+j30) 032 30100100

II (3)

T phng trnh (1) suy ra:

213 III (4)

Thay phng trnh (4) vo (2) v (3) ta c:

(150+j30) 021 0100100

II (2)

100 021 30100)30150(

IjI (3)

30150

100

100

30150

30150

100

30100

1000

1

j

j

jI x

= 08,136952,0900011600

80006340

j

j(A)

00

2 32,854528,0900011600

49024490

30150

100

100

30150

30100

100

100

30150

j

j

j

j

j

I y (A)

0213 217672,0)2855,07122,0(

jIII (A)

2.2.Gii mch in AC bng phng php dng in vng.

V d: Cho mch in nh hnh v vi 01 0100

E (V), 02 30100

E , Z1=Z1= 50 + j30(),

Z3= 100(). Vit phng trnh dng in vng v gii h phng trnh .

Gii

Chn chiu i cc dng in vng nh hnh v

Vng I: (Z1+Z3).

13 EIZI III (1)

Vng II: (Z2+Z3)

23 EIZI III (2)

Gii h phng trnh (1) v (2):

100100)30150(

III IIj (1)

II I 3

I

2

I

Z3

Z2 Z1

2

E 1

E

1

I

Hnh 5.2


Page 69/94

030100)30150(100

III IjI (2)

30150

100

100

30150

30150

100

30100

1000

j

j

jI xI

= 08,136952,0900011600

80006340

j

j(A)

00

32,854528,0900011600

49024490

30150

100

100

30150

30100

100

100

30150

j

j

j

j

j

I yII (A)

)(217672,0)028557122,0()4153,003695,0()1658,06752,0(

)(32,864528,0

)(8,136952,0

0213

02

01

AjjjIII

AII

AII

II

I

2.3.Gii mch in AC bng phng php in p hai

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