mach mot chieu 2015b mk

Mch mt chiu

C s l thuyt mch in

Nguyn Cng Phng

Mch mt chiu - sites.google.com/site/ncpdhbkhn 2

Ni dungI. Thng s mchII. Phn t mchIII. Mch mt chiuIV. Mch xoay chiuV. Mng hai caVI. Mch ba phaVII.Qu trnh qu VIII.Khuch i thut ton


Mch mt chiu L mch in ch c ngun mt chiu Cun dy (nu c) b ngn mch T in (nu c) b h mch Ni dung:

Cc nh lut c bn Cc phng php phn tch Cc nh l mch Phn tch mch in bng my tnh


Mch mt chiu1. Cc nh lut c bn

a) nh lut Ohmb) Nt, nhnh & vngc) nh lut Kirchhoff

2. Cc phng php phn tch3. Cc nh l mch4. Phn tch mch in bng my tnh


nh lut Ohm

Lin h gia dng & p ca mt phn t Nu c nhiu phn t tr ln th nh lut Ohm cha Cc nh lut Kirchhoff

Riu

Rui

u

i R


Nt, nhnh & vng (1) Nhng khi nim xut hin khi kt ni cc phn t mch Cn lm r trc khi ni v cc nh lut Kirchhoff Nhnh: biu din 1 phn t mch n nht (v d 1

ngun p hoc 1 in tr) Nhnh c th dng biu din mi phn t c 2 cc


Nt, nhnh & vng (2) Nt: im ni ca t nht 2 nhnh Biu din bng 1 du chm Nu 2 nt ni vi nhau bng dy dn, chng to thnh 1

nt

a b

c

a b

c


Nt, nhnh & vng (3) Vng: mt ng khp kn trong mt mch ng khp kn: xut pht 1 im, i qua mt s im khc, mi

im ch i qua mt ln, ri quay tr li im xut pht

Vng c lp: cha mt nhnh, nhnh ny khng c mt trong cc vng khc

Mt mch in c d nt, n nhnh, v vng c lp s tho mn h thc:

v = n d + 1 (3 = 5 3 + 1)


nh lut Kirchhoff (1) 2: nh lut v dng in & nh lut v in p nh lut v dng in vit tt l KD KD da trn lut bo ton in tch (tng i s in tch

ca mt h bo ton) KD: tng i s cc dng i vo mt nt bng khng

N: tng s nhnh ni vo nt in: dng th n i vo (hoc ra khi) nt

N

nni

10


nh lut Kirchhoff (2) KD: tng i s cc dng i vo mt nt bng khng

Quy c: Dng i vo mang du dng (+), dng i ra mang du m () Hoc ngc li

N

nni

10

i1i2

i3 i4

i5i1 i2 i3 + i4 i5 = 0

Hoc: i1 + i2 + i3 i4 + i5 = 0


nh lut Kirchhoff (3) Mt cch pht biu khc ca KD:

Tng cc dng i vo mt nt bng tng cc dng i ra khi nt KD c th m rng cho mt mt kn:

Tng i s cc dng i vo mt mt kn bng khng

C th coi nt l mt mt kn co li

i1

i2

i3 i4

i5

i1 i2 i3 + i4 i5 = 0

nh lut Kirchhoff (4)


VD1i1 = 3A, i2 = 2A, tm i3?

1 2 3 0i i i 3 1 2i i i 3 2 1A

VD2i1 = 3A, i2 = 2A, tm i3?

1 2 3 0i i i 3 2 1i i i 2 3 1A


nh lut Kirchhoff (5) nh lut th nht l KD nh lut th hai l v in p, vit tt KA KA da trn nh lut bo ton nng lng KA: tng i s cc in p trong mt vng kn bng khng

M: s lng in p trong vng kn, hoc s lng nhnh ca vng kn

um: in p th m ca vng kn

M

mmu

10


nh lut Kirchhoff (6) KA: tng i s cc in p trong mt vng kn bng khng

M

mmu

10

u1 + u2 + u3 u4 u5 = 0 u1 u2 u3 + u4 + u5 = 0



100V

20

VD3Tnh dng in ca in tr?

i

u

100 0u 20 100 0i 100 5A20

i

E

1R

2R

VD4Tnh dng in & in p ca in tr?

i

1u

1u1 2 0u u E 1 2 0R i R i E 1 2

EiR R

1 1 1

1 2

;Eu R i RR R

2 2 2 1 2Eu R i R

R R



1R

2R

i1i

2i

VD5Bit i, tnh dng in ca cc in tr?

1u

2u

1 2

1 2

00

i i iu u

1 2

1 1 2 2

0i i iR i R i

1 21 2

2 11 2

ii RR R

ii RR R


nh lut Kirchhoff (9)u1 u3

u2

u1 + u2 30 = 0u3 u2 = 0u1 = 8i1u2 = 3i2u3 = 6i3

8i1 + 3i2 30 = 0

6i3 3i2 = 0

i1 i2 i3 = 0

8i1 + 3i2 30 = 0

6i3 3i2 = 0

i1 i2 i3 = 0

Tnh cc dng & p

VD6


nh lut Kirchhoff (10)u1 u3

u2

8i1 + 3i2 30 = 0

6i3 3i2 = 0

i1 i2 i3 = 0

Tnh cc dng & p

i2 = 2 A

i3 = 1 A

i1 = 3 A

VD6



8i1 + 3i2 30 = 0

6i3 3i2 = 0

i1 i2 i3 = 0

8i1 + 6i3 30 = 0

i1 + i2 + i3 = 0

8i1 + 3i2 30 = 0

6i3 3i2 = 0

i1 i2 i3 = 0 H 5 phng trnh 3 n s tha 2 phng trnh ch cn 3 phng trnh

H ny c 3 p/tr c lp & 2 p/tr ph thuc

VD6



8i1 + 6i3 30 = 0

i1 + i2 + i3 = 0

8i1 + 3i2 30 = 0

6i3 3i2 = 0

i1 i2 i3 = 0

H trn c 3 p/tr c lp & 2 p/tr ph thucChn 3 p/tr no?

Mt mch in c nKD p/tr c lp vit theo KD & c nKA p/tr c lp vit theo KA

nKD = s_nt 1

nKA = s_nhnh s_nt + 1

VD6


Mch mt chiu1. Cc nh lut c bn2. Cc phng php phn tch

a) Dng nhnhb) Th ntc) Dng vngd) Bin i tng nge) Ma trn

3. Cc nh l mch4. Phn tch mch in bng my tnh


Dng nhnh (1) n s l cc dng in ca cc nhnh S lng n s = s lng nhnh (tr ngun dng) ca

mch p dng trc tip KD & KA Lp h phng trnh bng cch

p dng KD cho nKD nt, v p dng KA cho nKA vng


Dng nhnh (2)

nKD = s_nt 1 = 3 1 = 2 vit 2 p/tr theo KDa: i1 + i2 i3 = 0b: i3 i4 + j = 0

nKA = s_nhnh s_nt + 1 = 4 3 + 1 = 2 vit 2 p/tr theo KA

A: u1 u2 + e2 e1 = 0 R1i1 R2i2 + e2 e1 = 0B: u2 + u3 + u4 e2 = 0 R2i2 + R3i3 + R4i4 e2 = 0

A B

VD1


Dng nhnh (3)

i1 + i2 i3 = 0

i3 i4 + j = 0

R1i1 R2i2 + e2 e1 = 0

R2i2 + R3i3 + R4i4 e2 = 0

i1 + i2 i3 = 0

i3 i4 = j

R1i1 R2i2 = e1 e2R2i2 + R3i3 + R4i4 = e2

i1i2i3i4

A B

VD1


Dng nhnh (4)

1. Tnh nKD & nKA2. Chn nKD nt & vit nKD

phng trnh KD cho cc nt

3. Chn nKA vng & chiuca chng

4. Vit nKA phng trnh KA cho nKA vng

A B

nKD = 3 1 = 2; nKA = 4 3 + 1 = 2

a: i1 + i2 i3 = 0

b: i3 i4 + j = 0

A: R1i1 R2i2 = e1 e2B: R2i2 + R3i3 + R4i4 = e2

VD1


Dng nhnh (5)VD2 nKD = s_nt 1 = 4 1 = 3

nKA = s_nhnh s_nt + 1 = 6 4 + 1 = 3

a: i1 + i2 i6 = 0b: i1 i5 + i3 + j = 0c: i3 i4 + i6 j = 0A: R1i1 + R5i5 + R2i2 = e1B: R3i3 + R5i5 R4i4 = 0C: R2i2 + R6i6 + R4i4 = e6

BA

C

1. Tnh nKD & nKA2. Chn nKD nt & vit nKD phng trnh

KD cho cc nt 3. Chn nKA vng & chiu ca chng4. Vit nKA phng trnh KA cho nKA vng

Dng nhnh (6)


R1 R2

R3R4

R5

R6

E6

E3

E1 J

d

a

bc

VD3

A

BC2 3 6: 0b i i i

4 3 5: 0c i i i

i1

i5

i2

i6

i3i41 4: 0d i i J

1 1 5 5 4 4 1:A R i R i R i E 3 3 6 6 5 5 3 6:B R i R i R i E E 6 6 2 2 6:C R i R i E

Dng nhnh (7)


VD4R1 = 10, R2 = 20, R3 = 15, e1 = 30V, e3 = 45V,j = 2A. Tnh cc dng in trong mch?

1 2 3 0i i i j 1 1 2 2 1R i R i e 2 2 3 3 3R i R i e

1 2 3

1 2 3

1 2 3

1 1 1 210 20 0 30

0 20 15 45

i i ii i ii i i

1 2 3

1 2 3; ;i i i

1 1 110 20 0 ;0 20 15

1

2 1 130 20 0 ;45 20 15

2

1 2 110 30 0 ;0 45 15

3

1 1 210 20 30 ;0 20 45

Dng nhnh (8)


R1 = 10, R2 = 20, R3 = 15, e1 = 30V, e3 = 45V,j = 2A. Tnh cc dng in trong mch?

1 2 3 0i i i j 1 1 2 2 1R i R i e 2 2 3 3 3R i R i e

1 2 3

1 2

2 3

210 20 30

20 15 45

i i ii i

i i

1 1 1

10 20 00 20 15

20 0 1 1 1 11 10 0

20 15 20 15 20 0

1( 20.15 20.0) 10[1.15 20( 1)] 0[1.0 ( 20)( 1)] 650

VD4

Dng nhnh (9)



1 2 3 0i i i j 1 1 2 2 1R i R i e 2 2 3 3 3R i R i e

1 2 3

1 2

2 3

210 20 30

20 15 45

i i ii i

i i

1 2 3

1 2 3; ;i i i

1 2 3650; 1350; 300; 2350

1

2

3

1350 2,08 A650

300 0,46 A6502350 3,62 A650

i

i

i

VD4

Dng nhnh (10)



VD4

1 2 3

1 2

2 3

210 20 30

20 15 45

i i ii i

i i

1

2

3

2,08 A0,46 A

3,62 A

iii

1 2 3

1 2

2 3

210 20 30

20 15 45

i i ii i

i i

1

2

3

2,08 A0,46 A3,62 A

iii


Dng nhnh (11) Khi lng tnh ton gii h 4 phng trnh 4 bin

= 5 nh thc bc 4= 5 x 4 nh thc bc 3= 5 x 4 x 3 nh thc bc 2= 60 nh thc bc 2

Khi lng tnh ton gii h 3 phng trnh 3 bin:= 4 nh thc bc 3= 4 x 3 nh thc bc 2= 12 nh thc bc 2

Khi lng tnh ton gii h 10 phng trnh 10 bin ?


Hn 200 php tnh (cng, nhn, chia)

1

2

3

4

2 3 106 5 4

7 93 4 5

ii

ii

Di 8 php tnh (cng & chia)

NG THI

KHNG NG THI

?

i1 + i2 i3 = 0i3 i4 + j = 0R1i1 R2i2 = e1 e2R2i2 + R3i3 + R3i3 = e2

gim khi lng tnh ton th cn phi thay h phng trnh ng thi bng h phng trnh khng ng thi


gim khi lng tnh ton th cn phi thay h phng trnh ng thi bng h phng trnh khng ng thi

C 2 cch thay th:1. i bin s

Phng php th nt Phng php dng vng

2. Phn r mch in (ln lt tnh ton thng s ca tng phn ca mch in)

Bin i tng ng Mng mt ca (s hc trong Cc nh l mch)


Th nt (1)

1 2 3

3 4

: 0: 0

a i i ib i i j

(h 2 phng trnh 4 n s)

i1 = f1(a, b)i2 = f2(a, b)i3 = f3(a, b)i4 = f4(a, b)

A11a + A12b = B1A21a + A22b = B2

(h 2 phng trnh 2 n s)

VD1


Th nt (2) n s l in th ca cc nt Cn gi l th nh Dng KA i n s dng in nhnh thnh n s in th

nt

(60 nh thc bc 2)


i1 = f1(a, b)i2 = f2(a, b)i3 = f3(a, b)i4 = f4(a, b)

A11a + A12b = B1A21a + A22b = B2

(3 nh thc bc 2 + 4 hm f )


Th nt (3)

11 1 1 1

1( ) a ca c

eR i e iR 1

11

aeiR

Nu t c = 0

1 ( , , ) ?a b ci f

Th nt (4)


a

bR

ei

( )a bRi e a beiR

0 abei

R

a

bR

ei

( )a bRi e a bei

R

0 abei

R

Th nt (5)


a

bR

ei

( )a bRi e a bei

R

0 abei

R

a

bR

ei

( )a bRi e a beiR

0 abei

R

Th nt (6)


a

bRi

( ) 0a bRi a biR

0 ab i R

a

bRi

( ) 0a bRi a bi

R

0 ab i R


Th nt (7)

Rei

1

11 R

ei a

2

22 R

ei a

33 R

i ba

0: 321 iiia

032

2

1

1 RR

eR

e baaa

t c = 0

VD1


Th nt (8)

44 R

i b3

3 Ri ba

3 4: 0b i i j 0

43

jRR

bba

t c = 0

VD1

Rei


Th nt (9)

032

2

1

1 RR

eR

e baaa

t c = 0

043

jRR

bba

1 2

1 2 3 3 1 2

3 3 4

1 1 1 1

1 1 1

a b

a b

e eR R R R R R

jR R R

VD1


Th nt (10)

b

a

t c = 0

1 2

1 2 3 3 1 2

3 3 4

1 1 1 1

1 1 1

a b

a b

e eR R R R R R

jR R R

1

11 R

ei a

2

22 R

ei a

33 R

i ba

44 R

i b

VD1


Th nt (11)

032

2

1

1 RR

eR

e baaa

t c = 0

043

jRR

bba

1 2

1 2 3 3 1 2

3 3 4

1 1 1 1

1 1 1

a b

a b

e eR R R R R R

jR R R

VD1


Th nt (12)

1 2

1 2 3 3 1 2

3 3 4

1 1 1 1:

1 1 1:

a b

a b

e eaR R R R R R

b jR R R

t c = 0

VD1


Th nt (13)

::

a a ab b a

ab a b b b

a G G jb G G j

Tng dn ring ca nt a

Tng dn tng h gia nt a& nt b

Tng dn ring ca nt b

Ngun dng chy vo nt a

Ngun dng chy vo nt b

t c = 0


Th nt (14)

1 2

1 2 3 3 1 2

3 3 4

1 1 1 1

1 1 1

a b

a b

e eR R R R R R

jR R R

Tng dn ring ca mt nt: tng ca in dn ca tt c cc nhnh ni TRC TIP vi nt

Tng dn tng h gia 2 nt: tng ca in dn ca tt c cc nhnh ni TRC TIP 2 nt

t c = 0


Th nt (15)

1. Chn mt nt lm gc

2. Tnh cc tng dn ring v cc tng dn tng h

3. Tnh cc ngun dng vo nKD nt

4. Lp h phng trnh

5. Gii h phng trnh tm cc th nt

t c = 0

1 2

1 2 3 3 1 2

3 3 4

1 1 1 1

1 1 1

a b

a b

e eR R R R R R

jR R R


Th nt (16)

1 2

1 2 3 3 1 2

3 3 4

1 1 1 1:

1 1 1:

a b

a b

e eaR R R R R R

b jR R R

t c = 0

VD1


Th nt (17)

3 4 4

1 2

4 1 2 4 1 2

1 1 1:

1 1 1 1:

b c

b c

b jR R R

e ec jR R R R R R

t a = 0

VD1


Th nt (18)

1 2

1 2 3 1 2 1 2

1 2

1 2 1 2 4 1 2

1 1 1 1 1:

1 1 1 1 1:

a c

a c

e eaR R R R R R R

e ec jR R R R R R R

t b = 0

VD1


Th nt (19)VD2

t d = 0

1 6

1 2 6 1 6 1 6

1

1 1 3 5 3 1

6

6 3 3 4 6 6

1 1 1 1 1:

1 1 1 1 1:

1 1 1 1 1:

a b c

a b c

a b c

e eaR R R R R R R

eb jR R R R R R

ec jR R R R R R

nKD = s_nt 1 = 4 1 = 3


Th nt (20)VD2

t c = 0

61

1 2 6 1 2 1 6

1

1 1 3 5 5 1

2 5 2 4 5

1 1 1 1 1:

1 1 1 1 1:

1 1 1 1 1: 0

a b d

a b d

a b d

eeaR R R R R R R

eb jR R R R R R

dR R R R R

nKD = s_nt 1 = 4 1 = 3

Th nt (21)


R1 R2

R3R4

R5

R6

E6

E3

E1 J

d

a

bc

VD3

3 6

2 3 6 3 3 6

3

3 3 4 5 4 3

1

4 1 4 1

1 1 1 1: 0

1 1 1 1 1:

1 1 1: 0

b c d

b c d

b c d

E EbR R R R R R

EcR R R R R R

Ed JR R R R

Th nt (22)


R1 R2

R3R4

R5

R6

E6

E3

E1 J

d

a

bc

VD3

61

1 2 5 6 2 6 1 1 6

3 6

2 6 2 3 6 3 6

1

1 1 4 1

1 1 1 1 1 1 1:

1 1 1 1 1: 0

1 1 1: 0

a b d

a b d

a b d

EEa JR R R R R R R R R

E EbR R R R R R R

Ed JR R R R

Th nt (23)



1 1 1 30 45210 20 15 10 15a

9,23 Va

1

2

3

30 9,23 2,08 A10

9,23 0,46 A20

45 9,23 3,62 A15

i

i

i


Th nt (24)VD5

t d = 0

1 2 3 41 2 3 4

; ; ;a b a c c bi i i iR R R R

1 2 3 4i i i i 1 2 3 4

a b a c c bR R R R

1a E 2b c E

11 4 2 3 1 2

2

1 1 1 1 1 1b c

b c

ER R R R R R

E


Dng vng (1)

A: R1i1 R2i2 = e1 e2 (h 2 phng trnh 4 n)B: R2i2 + R3i3 + R4i4 = e2

(h 2 phng trnh 2 n)

i1 = f1(iA, iB)i2 = f2(iA, iB)i3 = f3(iA, iB)i4 = f4(iA, iB)

A11iA + A12iB = B1A21iA + A22iB = B2

A B

VD1

Dng vng (2)


5A 3A

2A

5A 3A

Dng vngDng nhnh


Dng vng (3)

6A 3A

5A2A

8A

?6A ?3A

?8A

?5A?2A

?6 (3) = 9A

?3 8 = 11A

?6 2 = 8A

??

?

VD2


Dng vng (4) n s l dng in chy trong mt vng Dng vng l i lng khng c thc, nhng tin li cho vic

phn tch mch in Dng KD i n s dng in nhnh thnh nKA n s dng

in vng

(60 nh thc bc 2)


i1 = f1(iA, iB)i2 = f2(iA, iB)i3 = f3(iA, iB)i4 = f4(iA, iB)

A11iA + A12iB = B1A21iA + A22iB = B2

(3 nh thc bc 2 + 4 hm f )


Dng vng (5)

A: R1i1 R2i2 = e1 e2i1 = iAi2 = iB iA

R1iA R2(iB iA) = e1 e2

B: R2i2 + R3i3 + R4i4 = e2i3 = iBi4 = iB + j

R2(iB iA) + R3iB + R4(iB + j) = e2

iA iBj

VD1


Dng vng (6)

A: R1iA + R2(iA iB) = e1 e2B: R2(iB iA) + R3iB + R4(iB + j) = e2

A: (R1+ R2)iA R2iB = e1 e2B: R2 iA + (R2 + R3 + R4)iB = e2 R4j

iA iBj

VD1


Dng vng (7)

iA iB

A: R1(iA ) + R2(iA iB) = e1 e2B: R2(iB iA) + R3(iB ) + R4(iB + j) = e2

j

VD1


Dng vng (8)

(R1+ R2)iA R2iB = e1 e2 R2 iA + (R2 + R3 + R4)iB = e2 R4j

iAiB

i1 = iAi2 = iB iAi3 = iBi4 = iB + j

Ch : chiu ca cc dng nhnh khng nh hng n h p/trnh dng vng

iA iBj

VD1


Dng vng (9)

iA iB j

1( )AR i 2( )A BR i i j 1 2e e 2( )B AR i i j 3( )BR i j 4( )BR i 2e

VD1


Dng vng (10)

iA iB j

1( )AR i j 2( )A BR i i 1 2e e 2( )B AR i i 3( )BR i j 4( )BR i 2e

VD1


Dng vng (11)VD3

nKA = 6 4 + 1 = 3 cn chn 3 dng vng

A: R1(iA ) + R5(iA iB) + R2(iA iC) = e1B: R3(iB + j) + R4(iB iC) + R5(iB iA) = 0

C: R2(iC iA) + R4(iC iB) + R6(iC ) = e6

i1 = iA; i2 = iA iC; i6 = iCi3 = iB j; i4 = iB iC; i5 = iA iB;

iA

iC

iB

iA iB

iC

j

iA iB

iC

j


Dng vng (12)VD3

iA iB

iC

j

1( )AR i 3( )A BR i i 6( )A CR i i 1 6e e 3( )B AR i i 4( )B CR i i j 5( )BR i j 02( )CR i 4( )C BR i i j 6( )C AR i i 6e

Dng vng (13)


R1 R2

R3R4

R5

R6

E6

E3

E1 J

d

a

bc

VD4

A BC

i1

i5

i2

i6

i3i4

J

1 5 4 1

3 6 5 3 6

2 6 6

( ) ( ) ( )( ) ( ) ( )( ) ( )

A A B A

B B C B A

C C B

R i J R i i R i ER i R i i R i i E ER i R i i E

A

B

C

iii

1 ;Ai i J 2 ;Ci i 3 ;Bi i 4 ;Ai i 5 ;A Bi i i 6 C Bi i i

Dng vng (14)


R1 R2

R3R4

R5

R6

E6

E3

E1 J

d

a

bc

VD4

AB

CJ

1 5 4 1

3 6 5 3 6

2 6 6

( ) ( ) ( )( ) ( ) ( )( ) ( )

A A B A

B B C B A

C C B

R i R i i J R i J ER i R i i R i i J E ER i R i i E

Dng vng (15)



iA iB

10 20( 2) 3020( 2) 15 45

A A B

B A B

i i ii i i

30 20 1020 35 85

A B

A B

i ii i

2,08 A3,62 A

A

B

ii

1

2

3

2,08 A0,46 A

3,62 A

A

A B

B

i ii i i ji i

j

Dng vng (16)


VD5

10 20( 2) 3020( 2) 15 45

A A B

B A B

i i ii i i 2,08 A3,62 A

A

B

ii

1

2

3

2,08 A0,46 A

3,62 A

A

A B

B

i ii i i ji i

iA iBj iA iBj

10( ) 20( 2) 3010( ) 15 30 45

A B A

B A B

i i ii i i 1,54 A

3,62 AA

B

ii

1

2

3

2,08 A0,46 A

3,62 A

A B

A

B

i i ii i ji i


Dng vng (17)

ia ib

R1ia + (R3 + R4)ib = ej = i4 i1 = ib ia

VD6

R1ia + (R3 + R4)ib = e ia + ib = j


i vi mt mch in c n nhnh, p/p dng nhnh s dnn vic gii ng thi h n phng trnh n n

Rt t khi dng phng php dng nhnh Hai p/p dng vng & th nt gim s lng phng trnh

& s lng n Nn dng hai p/p dng vng & th nt khi gii mch in Cho mt mch in, chn p/p th nt hay dng vng? La chn:

Chn p/p no c t n s hn P/p th nt rt thch hp cho mch in ch c 2 nt C mt s kiu mch in kh dng p/p th nt C mt s kiu mch in kh dng p/p dng vng


VD7 Tnh i7 ?

Phng php dng nhnh c my n?

Phng php th nt c my n?

Phng php dng vng c my n?

Bin i tng ng


Bin i tng ng (1) Hai phn t mch c gi l tng ng

nhau nu chng c quan h gia dng & pging nhau

Dng phn r mch in gim khilng tnh ton

Cc php bin i tng ng: Ngun p ni tip Ngun dng song song in tr ni tip in tr song song Y (ngun p ni tip in tr) (ngun dng song

song in tr) Millman

5A

10V 2

5A

10V 2


Bin i tng ng (2) Ngun p ni tip (hai phn t gi l ni tip nu chng c chung t nht 1 u & c cng mt dng in chy qua)

= e1 + e2 e3

N ktd ee1

Bin i tng ng (3)


VD1

1 4e e

VD2

1 4e e


Bin i tng ng (4) Ngun dng song song (Hai phn t gi l song song nu chng c chung 2 u)

= j1 + j2 j3

N ktd jj1

Bin i tng ng (5)


VD3

1 2j j


Bin i tng ng (6) in tr ni tip:

Rtd = R1 + R2 + R3

1R 2R 3R tdR

1R 2Ra b 1 1 1 2

2 21 2

abR

abR

uu RR R

uu RR R


Bin i tng ng (7) in tr song song:

321

1111RRRRtd

1 2

1 2td

R RRR R

1R

2R

tdR

1R

2R

i 1i

2i

21

1 2

12

1 2

Ri iR R

Ri iR R

Bin i tng ng (8)


VD4

3ad ab cdR R R R 1 2

1 2ab

R RRR R

4 5 6

4 5 6 4 5 5 6 6 4

1 1 1 1cd

cd

R R RRR R R R R R R R R R

1 2 4 5 6

31 2 4 5 5 6 6 4

adR R R R RR R

R R R R R R R R


Bin i tng ng (9)

(4 6)84 6 8

1 2 31 2 3( )R R R

R R R

4, 44

1 2 3( ) //tdR R R R R1 = 4 ; R2 = 6 ; R3 = 8 ; Tnh Rtd?

VD5

R1 = 4 ; R2 = 6 ; R3 = 2 ; R4 = 10 ; Tnh Rtd ?

VD6

43

43

21

21

RRRR

RRRRRtd 07,4102

10.264

6.4


Bin i tng ng (10)

6.86 8

2 3

2 3

R RR R

3,43

2 3//tdR R RR1 = 4 ; R2 = 6 ; R3 = 8 ; Tnh Rtd?

VD7

R1 = 4 ; R2 = 6 ; R3 = 2 ; Tnh Rtd ?

VD8

1 2

1 2td

R RRR R

4.6 2,40

4 6


Bin i tng ng (11)

a

b

10 1

3

1

4 5

6

12

VD9 Tnh Rab

VD10 Tnh Rab


Bin i tng ng (12)

cba

cabac RRR

RRRRRR )(31

cba

bacab RRR

RRRRRR )(21

a b

c

R1 R2

R3

a b

c

Ra

Rc

Rb

Rac(Y) = R1 + R3 Rac() = Rb // (Ra + Rc)=

cba

cbabc RRR

RRRRRR )(32

Tng t:


Bin i tng ng (13)

cba

cabac RRR

RRRRRR )(31

cba

bacab RRR

RRRRRR )(21

cba

cbabc RRR

RRRRRR )(32

a b

c

R1 R2

R3

a b

c

Ra

Rc

Rb

cba

cb

RRRRRR 1

cba

ac

RRRRRR 2

cba

ba

RRRRRR 3


Bin i tng ng (14)

cba

cba

RRRRRR

2133221 )()(

cba

cbacba

RRRRRRRRRRRRRRR

a b

c

R1 R2

R3

a b

c

Ra

Rc

Rb

cba

cb

RRRRRR 1

cba

ac

RRRRRR 2

cba

ba

RRRRRR 3

x R2

x R3

x R1

(+)


Bin i tng ng (15)

1

133221

RRRRRRRRa

cbacba

RRRRRRRRRRRR 133221

a b

c

R1 R2

R3

a b

c

Ra

Rc

Rb

cba

cb

RRRRRR 1

(:)

2

133221

RRRRRRRRb

3

133221

RRRRRRRRc

Tng t:


Bin i tng ng (16)a b

c

RaRb

RcR1 R2

R3

cba

cb

RRRRRR 1

cba

ac

RRRRRR 2

cba

ba

RRRRRR 3

1

133221

RRRRRRRRa

2

133221

RRRRRRRRb

3

133221

RRRRRRRRc


Bin i tng ng (17)

hoc

13

15 35


Bin i tng ng (18) Hai phn t mch c gi l tng ng nhau nu chng c quan

h gia dng & p ging nhau Cc php bin i tng ng:

Ngun p ni tip Ngun dng song song in tr ni tip in tr song song Y (ngun p ni tip in tr) (ngun dng song song in tr) Millman


Bin i tng ng (19) (Ngun p ni tip in tr) (ngun dng song song in tr)

Rje

ejR

Riue e ui R R


Bin i tng ng (20)

Rje

ejR

11

1

ejR

66

6

ejR

VD11 Tnh dng qua R3?

1 212

1 2

R RRR R

4 6

464 6

R RRR R

12 12 1e R j

46 46 6e R j

12 463

12 3 46 5

e eiR R R R


Bin i tng ng (21)

4 4 10.2 20 Ve R j

43

12 3 4

202,4 2 10

1,39 A

eiR R R

4,2646.4

21

2112 RR

RRR

j = 2 A; R1 = 4 ; R2 = 6 ; R3 = 2 ; R4 = 10 ; Tnh i3 ?

VD12


Bin i tng ng (22) Hai phn t mch c gi l tng ng nhau nu chng c quan

h gia dng & p ging nhau Cc php bin i tng ng:

Ngun p ni tip Ngun dng song song in tr ni tip in tr song song Y (ngun p ni tip in tr) (ngun dng song song in tr) Millman


Bin i tng ng (23) Bin i Millman

1 2 3

11 1 1tdR

R R R

1 2 3

1 2 3

1 2 3

1 1 1td

e e eR R Re

R R R


Bin i tng ng (24)VD13 Tnh dng qua R3

1 2 3

11 1 1tdR

R R R

1 2 3

1 2 3

1 2 3

1 1 1td

e e eR R Re

R R R

012

0 1 2

11 1 1R

R R R

0 0 1 1012

0 1 2

/ /1 1 1

e R e Re

R R R

5678

5 6 7 8

11 1 1 1R

R R R R

6 6 7 7 8 85678

5 6 7 8

/ / /1 1 1 1

e R e R e Re

R R R R

012 56783

012 3 4 5678

e eiR R R R


Ma trn (1) Xy dng phng trnh:

Ax = b x: vct dng nhnh hoc th nt hoc dng vng Nghim:

x = A-1b


Ma trn (2)

i1 + i2 i3 = 0i3 i4 = j

R1i1 R2i2 = e1 e2R2i2 + R3i3 + R4i4 = e2

Ai = b

2

21

4

3

2

1

432

21

0

0001100

0111

eeej

iiii

RRRRR

iA iB


Ma trn (3)

i1 i2 i3 i4a

b

A

B

a

b

A

B

2

21

4

3

2

1

432

21

0

0001100

0111

eeej

iiii

RRRRR

iA iB


Ma trn (4)

i1 i2 i3 i4 i5 i6abcABC

VD1 nKD = s_nt 1 = 4 1 = 3nKA = s_nhnh s_nt + 1 = 6 4 + 1 = 3

6

5

4

3

2

1

iiiiii

i = A =

abcABC

b =

1 1 0 0 0 11 0 1 0

10

0 0 1 10 1

R1 R2 R500 0R3 R5 R40 0 0

R2 R4 R60 0 0

0 jj

e10e6

Ai = b

BA

C


1 2 2 1 2

2 2 3 4 2 4

A

B

R R R i e eR R R R i e R j

Ma trn (5)

(R1+ R2)iA R2iB = e1 e2 R2 iA + (R2 + R3 + R4)iB = e2 R4j

Tt c cc in tr c mt trn ng i ca iA

Tt c cc in tr c mt trn ng i ca iB

Tt c cc in tr chung ca iA & iB; nu cng chiu th (+), ngc chiu th ()

Tt c cc ngun p c mt trn ng i ca dng vng:

-ngun p e: cng chiu th (+), ngc chiu th ()

-ngun p Rj: cng chiu th (), ngc chiu th (+)

iA iBj


Ma trn (6)VD3


A A B A C A A

B A B B C B B

C A C B C C C

R R R i eR R R i eR R R i e

RA = ? RA = R1 + R5 + R2RB = ? RB = R3 + R4 + R5RC = ? RC = R2 + R4 + R6

RA-B = ? RA-B = R5 = RB-ARA-C = ? RA-C = R2 = RC-ARB-C = ? RB-C = R4 = RC-B

iA iB

iC

j


Ma trn (7)VD3


A A B A C A A

B A B B C B B

C A C B C C C


eA = ? eA = e1eB = ? eB = R3jeC = ? eC = e6

iA iB

iC

j


Ma trn (8)VD3


A A B A C A A

B A B B C B B

C A C B C C C


RA = R1 + R5 + R2RB = R3 + R4 + R5RC = R2 + R4 + R6

RA-B = R5 = RB-ARA-C = R2 = RC-ARB-C = R4 = RC-B

eA = e1eB = R3jeC = e6

iA iB

iC

j


Mch mt chiu1. Cc nh lut c bn2. Cc phng php phn tch3. Cc nh l mch4. Phn tch mch in bng my tnh


Cc nh l mch Nu mch in phc tp th cc phng php phn tch mch

hc s mt nhiu thi gian tnh ton Cc nh l mch gip cho vic phn tch mch tr nn n gin

hn Dng phn r mch in gim khi lng tnh ton Cc nh l ny p dng cho mch in tuyn tnh Ni dung:

Mch in tuyn tnh Nguyn l xp chng nh l Thevenin nh l Norton Truyn cng sut cc i


Mch in tuyn tnh Cc nh l mch ch p dng cho mch in tuyn tnh Mch in tuyn tnh: ch gm cc phn t th ng

tuyn tnh Phn t tuyn tnh: u ra (p ng) t l thun vi u

vo (kch thch) C 2 tnh cht:

1. Nu [u = Ri & k = const] th [ku = kRi]2. Nu [u1 = Ri1 & u2 = Ri2] th [u = (i1 + i2)R = Ri1 + Ri2 = u1 + u2]


Cc nh l mcha) Nguyn l xp chngb) nh l Theveninc) nh l Nortond) Truyn cng sut cc i


Xp chng (1) p dng cho mch in c t 2 ngun tr ln tng: ln lt tnh thng s ca mch khi cho ln lt tng

ngun c mt trong mch in, sau cng cc thng s Nguyn l: in p (hoc dng in) ca mt phn t ca mt

mch in tuyn tnh l tng i s ca cc in p (hoc ccdng in) do tng ngun gy ra

Ch : 1. Khi xt tc dng ca mt ngun, phi trit tiu tt c cc ngun khc2. Khng p dng nguyn l ny cho cng sut

Li ch: vic p dng nguyn l ny c th lm cho cu trc mchtr nn n gin hn d phn tch hn


Xp chng (2)

1 2 3

1 2 3

1 1,

ng ngu i

2 2,

ng ngu i 3 3,ng ngu i

1 2 3 1 2 3;

ng ng ng ng ng ngu u u u i i i i


Xp chng (3) Khi xt tc dng ca mt ngun, phi trit tiu tt c cc

ngun khc

Phn cn li ca mch in

Phn cn li ca mch in

Trit tiu ngun p

Phn cn li ca mch in

Phn cn li ca mch in

Trit tiu ngun dng

Xp chng (4)


Trit tiu ngun p

Trit tiu ngun dng

Xp chng (5)


R1 = 10, R2 = 20, e = 30V, j = 2A.Tnh dng in chy qua R2?

VD1

e j

1R2R

1. Trit tiu j, tnh i2|e

e

2 ei1R

2R

21 2

30 1A10 20

e

eiR R

2. Trit tiu e, tnh i2|j

2 ji

j

1R

2R

2 ji

1R j

1R

2R10.2

10 200,67 A

12

1 2j

R jiR R

3. Tnh i2|e + i2|j

2 2 2

1 0,67

1,67 A

e ji i i


Xp chng (6)VD2 e1 = 16 V; e2 = 9 V; j = 2 A; R1 = 4 ;

R2 = 6 ; R3 = 2 ; R4 = 10 ; Tnh i2

1. Trit tiu e2 & j, tnh i2|e12. Trit tiu e1 & j, tnh i2|e23. Trit tiu e1 & e2, tnh i2|j4. Tnh i2|e1 + i2|e2 + i2|j


Xp chng (7)

1. Trit tiu e2 & j, tnh i2|e1

0c

2 12

8 1,33A6

ae

iR

8Va

1

1 2 3 4 1

1 1 1a

eR R R R R

VD2 e1 = 16 V; e2 = 9 V; j = 2 A; R1 = 4 ; R2 = 6 ; R3 = 2 ; R4 = 10 ; Tnh i2


Xp chng (8)

2. Trit tiu e1 & j, tnh i2|e2

0c

3Va

2

1 2 3 4 2

1 1 1a

eR R R R R


22 2

2

9 3 1A6

ae

eiR


Xp chng (9)

3. Trit tiu e1 & e2, tnh i2|j

4 4 10.2 20 Ve R j

22

3,33 0,56A6

aj

iR

3,33Va

4

1 2 3 4 3 4

1 1 10;c ae

R R R R R R



2 11,33A

ei

2 21A

ei

2 0,56Aji

i2 = i2|e1 + i2|e2 + i2|j

= 1,33 + 1 0,56

= 0,89 A

VD2

Xp chng (10)


Xp chng (11) p dng cho mch in c t 2 ngun tr ln Ch :

1. Khi xt tc dng ca mt ngun, phi trit tiu tt c ccngun khc

2. Khng p dng nguyn l ny cho cng sut Li ch: vic p dng nguyn l ny c th lm cho cu

trc mch tr nn n gin hn d phn tch hn c bit tin li khi phn tch mch in c nhiu

ngun c tn s khc nhau (s cp trong phn Mchxoay chiu)


Thevenin (1)

Rtet


Thevenin (2) Mt mng tuyn tnh 2 cc c th c thay th bng mt mch tng ng gm c ngun p etd& in tr Rtd, trong : etd: ngun p h mch trn 2 cc Rtd: in tr trn hai cc sau khi trit

tiu cc ngun

ttd

tdt RR

ei

Mng tuyn tnh 2 cc

it

Rt

Rt

it


Thevenin (3)

Mng tuyn tnh 2 cc

Mng tuyn tnh 2 cc trit tiu ngun c lp

Rtd

Mng tuyn tnh 2 cc

etd


Thevenin (4)


Rtd


uvo

ivoMng tuyn tnh 2 cc trit tiu ngun c lp

uvo ivotd

uRi

vovo


Thevenin (5)

1 1 2 2

1 1 2 2

......

o m m

m m m n n

u A j A e A e A eA j A j A j

0 0u A j B

0 1 1 2 2

1 1 2 2

......

m m

m m m n n

B A e A e A eA j A j A j

Mng tuyn tnh 2 cc

a

b

ju

Gi s mng tuyn tnh 2 cc c m ngun p& n ngun dng, theo tnh cht xp chng:

0 0u A j B 0j 0 0jB u

0 0B 00

0B

uAj

= etd (in p h mch)

= Rtd (in tr vo khi trit tiu ngunbn trong mng tuyn tnh 2 ca)

td tdu R j e


Thevenin (6)Mng tuyn tnh 2 cc

a

b

ju

a

b

ju

td tdu R j e td tdu R j e

Mt mng tuyn tnh 2 cc c th c thay th bng mt mch tng ng gm c ngun p etd & in tr Rtd, trong : etd: ngun p h mch trn 2 cc Rtd: in tr trn hai cc sau khi trit tiu cc ngun

Thevenin (7)


R1 = 10, R2 = 20, e = 30V. Tnh dng in chyqua R3 vi cc gi tr R3 ln lt l 30, 60, 100?

VD1

e2R

1R3R

tde

tdR3R

e2R

1R tde

etd: ngun p h mch trn 2 cc

2Ru 2 2R i2

1 2

eRR R

20V

Rtd: in tr trn hai cc sau khi trit tiu cc ngun

2R

1R tdR 1 21 2

R RR R

6,67

Thevenin (8)



VD1

e2R

1R3R

tde

tdR3R

3 33

2030 0,55A6,67 30

td

td

eR iR R

20V; 6,67td tde R

3 33

2060 0,30A6,67 60

td

td

eR iR R

3 33

20100 0,19A6,67 100

td

td

eR iR R

Thevenin (9)



VD1

e2R

1R3R

2R

1R 1 21 2

6,67tdR RR

R R

11 0,1A

10i

21 0,05A20

i 0,1 0,050,15A

i

vo

2R

1R

1V

1i 2i ivo

1 6,670,15td

uRi

vovo

2R

1R

1A

1i 2i uvo2101 0,33A

10 20i

20.0,33 6,67 Vu vo

6,67 6,671td

uRi

vovo

Thevenin (10)


R1 = 10, R2 = 20, e = 30V, j = 2A.Tnh dng in chy qua R2? e j

1R2R

VD2

tde

tdR2R

22

td

td

eiR R


e j

1Rtde

1i1 1

1

tdR i e ei j

1tde e R j

50VRtd: in tr trn hai cc sau khi trit tiu cc ngun

1R tdR 1R 10 5010 20

1,67 A


Thevenin (11)

ttd

tdt RR

ei

VD3e = 16 V; j = 2 A; R1 = 4 ; R2 = 6 ; R3 = 8 ; Rt = 5 ; Tnh it?



Thevenin (12)


1 2 3 1

1 1a

e jR R R R

1 1 16 24 6 8 4a

t c = 0e = 16 V; j = 2 A; R1 = 4 ; R2 = 6 ; R3 = 8 ; Rt = 5 ; Tnh it?

etd = u3

18,67 Va

32 3

18,67 1,33 A6 8

aiR R

3 3 3 8.1,33 10,67 Vu R i 3 10,67 Vtde u

VD3


Thevenin (13)

ttd

tdt RR

ei


Rtd: in tr trn hai cc saukhi trit tiu cc ngun

e = 16 V; j = 2 A; R1 = 4 ; R2 = 6 ; R3 = 8 ; Rt = 5 ; Tnh it?

VD3


Thevenin (14)



(4 6)84 6 8

1 2 31 2 3

R R RR R R

4,44

1 2 3( ) //tdR R R R e = 16 V; j = 2 A; R1 = 4 ; R2 = 6 ; R3 = 8 ; Rt = 5 ; Tnh it?

VD3


Thevenin (15)

ttd

tdt RR

ei 10,67 1,13A

4,44 5ti

10,67 Vtde 4, 44tdR

ttd

tdt RR

ei

e = 16 V; j = 2 A; R1 = 4 ; R2 = 6 ; R3 = 8 ; Rt = 5 ; Tnh it?

VD3


Thevenin (16)

22

td

td

eiR R

VD4e = 16 V; j = 2 A; R1 = 4 ; R2 = 6 ; R3 = 8 ; Rt = 5 ; Tnh i2?

tdR

3 1( // )td ab tR R R R R 3

13

8.5 4 7,088 5

t

t

R R RR R


Thevenin (17)

22

td

td

eiR R

e = 16 V; j = 2 A; R1 = 4 ; R2 = 6 ; R3 = 8 ; Rt = 5 ; Tnh i2?

tde

td a be 0b c td a

e

1 16 4.2 24 Va aR j e 24 Vtde

VD4


Thevenin (18)

22

td

td

eiR R

e = 16 V; j = 2 A; R1 = 4 ; R2 = 6 ; R3 = 8 ; Rt = 5 ; Tnh i2?

22

td

td

eiR R

2

24 1,84 A7,08 6

i

24 Vtde 7,08tdR

VD4


Thevenin (19)

11

td

td

e eiR R

VD5e = 16 V; j = 2 A; R1 = 4 ; R2 = 6 ; R3 = 8 ; Rt = 5 ; Tnh i1?

3 2( // )td ac tR R R R R 3

23

8.5 6 9,088 5

t

t

R R RR R

tdR


Thevenin (20)e = 16 V; j = 2 A; R1 = 4 ; R2 = 6 ; R3 = 8 ; Rt = 5 ; Tnh i1?

tde

td cae u ca caR i ( )caR j 3 2[( // ) ]( )tR R R j

32

3

8.5( ) 6 ( 2) 18,15V8 5

t

t

R R R jR R

VD5

11

td

td

e eiR R


Thevenin (21)e = 16 V; j = 2 A; R1 = 4 ; R2 = 6 ; R3 = 8 ; Rt = 5 ; Tnh i1?

11

td

td

e eiR R

1

18,15 16 0,16A9,08 4

i

18,15 Vtde 9,08tdR

VD5

11

td

td

e eiR R


Norton (1) Tng t nh l Thevenin Pht biu: Mt mng tuyn tnh 2 cc c th c thay

th bng mt mch tng ng gm c ngun dng jtd& in tr Rtd, trong : jtd: ngun dng ngn mch gia 2 cc Rtd: in tr trn hai cc khi trit tiu cc ngun


Norton (2)

Mng tuyn tnh 2 cc


Rtd

Mng tuyn tnh 2 cc

jtd

Norton (3)


R1 = 10, R2 = 20, R3 = 30, e = 30V.Tnh dng in chy qua R3?

VD1

e2R

1R3R

e2R

1R tdj

jtd: ngun dng ngn mch trn 2 cc

33

tdtd

td

Ri jR R

1Ri1

eR

3010

3A


2R

1R tdR 1 21 2

R RR R

6,67

6,67330 6,67

0,55A

tdjtdR 3

R


Norton (4)

tdt td

t td

Ri jR R

VD2


e = 16 V; j = 2 A; R1 = 4 ; R2 = 6 ; R3 = 8 ; Rt = 5 ; Tnh it?


Norton (4)

2tdj i


1 2 1

1 1a

e jR R R

e = 16 V; j = 2 A; R1 = 4 ; R2 = 6 ; R3 = 8 ; Rt = 5 ; Tnh it?

Gi s c = 01 1 16 24 6 4a

14, 40 Va 22

14, 40 2, 4 A6

aiR 2,4 Atdj

VD2


Norton (5)

Rtd: in tr trn hai cc khi trit tiu cc ngun

(4 6)84 6 8

1 2 31 2 3

R R RR R R

4,44

1 2 3( ) //tdR R R R

e = 16 V; j = 2 A; R1 = 4 ; R2 = 6 ; R3 = 8 ; Rt = 5 ; Tnh it?

VD2


Norton (6)


e = 16 V; j = 2 A; R1 = 4 ; R2 = 6 ; R3 = 8 ; Rt = 5 ; Tnh it?

Rtd: in tr trn hai cc khi trit tiu cc ngunjtd = 2,4 A

Rtd = 4,44

4,442,4 1,13A5 4,44t

i

VD2

tdt td

t td

Ri jR R


Norton (7)VD3e = 16 V; j = 2 A; R1 = 4 ; R2 = 6 ; R3 = 8 ; Rt = 5 ; Tnh i2?

tdR

3 1( // )td ab tR R R R R 3

13

8.5 4 7,088 5

t

t

R R RR R


Norton (8)e = 16 V; j = 2 A; R1 = 4 ; R2 = 6 ; R3 = 8 ; Rt = 5 ; Tnh i2?

tdj

i1

1tdj i j 1 3 1

1 1 1a

t

e jR R R R

10,43Va

11

1,39 AaeiR 1,39 2 3,39 Atdj

VD3


Norton (9)e = 16 V; j = 2 A; R1 = 4 ; R2 = 6 ; R3 = 8 ; Rt = 5 ; Tnh i2?

22

tdtd

td

Ri jR R

jtd = 3,39 A

Rtd = 7,08

27,083,39 1,83A

6 7,08i

VD3


Thevenin & Norton (1)

Mng tuyn tnh 2 cc

etd = Rtd jtd



etd = Rtd jtd

tdtd

td

eRj

etd = uh mchjtd = ingn mch

tdu

Ri

h mch

ngn mch

(Cch th 3 tnh in tr tng ng ca s Thevenin)



tdef

td

u eRi j

h mchngn mch 10,67 4, 44

2, 4efR

e = 16 V; j = 2 A; R1 = 4 ; R2 = 6 ; R3 = 8 ; Rt = 5 ; Tnh Ref ca mng mt ca?

10,67 Vtde 2, 4 Atdj

VD1


Thevenin & Norton (4) Vic p dng nh l Thevenin hoc nh l Norton gi l phng

php mng mt ca/mng 2 cc Cc mch in c xy dng da trn nh l Thevenin hoc

nh l Norton gi l s (tng ng) Thevenin hoc s (tng ng) Norton

S Norton c th rt ra c t s Thevenin & ngc li Rtd = tng_tr_vo_sau_khi_trit_tiu_ngun, hoc

,td Thevenintdtd Norton

EuR

i j h mch

ngn mch

hoc

,tduRi

vovo


Truyn cng sut cc i (1) Mt s mch in c thit k truyn cng sut ti ti Vin thng: cn truyn mt cng sut ti a n ti Bi ton: tm thng s ca ti (gi tr ca in tr) cng

sut truyn n ti t cc i S dng s Thevenin


Truyn cng sut cc i (2)

ttt Rip2

ttd

tdt RR

ei t

ttd

tdt RRR

ep2

0

pt

Rt

4

22

)()(2)(

ttd

ttdtttdtd

t

t

RRRRRRRe

dRdp

0t

t

dRdp

0)()(

23

23

2

ttd

ttdtd

ttd

tttdtd RR

RReRR

RRRe

t tdR R


Truyn cng sut cc i (3) Cng sut cc i s c truyn n ti nu ti bng in tr tng ng Thevenin (nhn t pha ti)

Rt = Rtd : gi l ho hp ti hoc phi hp ti

t tdR R


Truyn cng sut cc i (4)

tdt RR

e = 16 V; R1 = 4 ; R2 = 6 ; R3 = 2 ; R4 = 10 ;Tnh Rt n nhn c cng sut ln nht?

43

43

21

21

RRRR

RRRRRtd

07,410210.2

646.4

07,4tR

VD


Mch mt chiu1. Cc nh lut c bn2. Cc phng php phn tch3. Cc nh l mch4. Phn tch mch in bng my tnh


Phn tch mch in bng my tnh Mc ch: tit kim thi gian tnh ton S tm hiu:

Gii cc php tnh phc tp (v d phng trnh ma trn) M phng mch in

Phn mm: Matlab, OrCAD PSpice


Phng trnh ma trn

41205

942075884902

6713

4

3

2

1

iiii


M phng mch in (1) Bng m lnh (Tutsim, Spice, ) Bng giao din ho (Pspice, Circuit maker, Matlab,

Workbench, )


M phng mch in (2)e1 = 16 V; e2 = 9 V; j = 2 A; R1 = 4 ; R2 = 6 ; R3 = 2 ; R4 = 10 ; Rt = 5 ;Tnh cc dng in trong mch?

VD1

mach mot chieu 2015b mk

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