67687677 Phuong Phap Giai Mach Cau Cuc Hay

8/3/2019 67687677 Phuong Phap Giai Mach Cau Cuc Hay

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Chuyn Vt l ---- ---- Trng THCS An Ngha

Li ni uMch cu thng c ni n qua cc bi ton nng

cao trong chng trnh bi dng hc sinh gii lp 9, thnhng l thuyt v cch gii mch cu th t c ti liu tham kho. V vy nhm gip cho cc bn ng nghip c

iu kin tm hiu v mch cu, t c thm t liu chovic bi dng hc gii. Ti xin gii thiu chuyn : Cchgii mch cu.

Chuyn gm : 2 phnI. Gii thiu mch cu v phn loi mch cu.II. cch gii cc loi mch cu

+ Mch cu cn bng+ Mch cu khng cn bng- Mch cu tng qut- Mch cu khuyt

Mc d c rt nhiu c gng nhng khng trnh khi thiust, rt mong cc bn ng nghip tham gia ng gp kin, chuyn c hon thin hn.

Xin cm n !

GV Thc hin : ng Xun Cnh Trang 1


2/7

Chuyn Vt l -------- Trng THCS An Ngha

PHNG PHP GII MCH CU

I/ M CH CU.- Mch cu l loi mch c dng ph bin trong cc php o in nh

( Vn k, am pe k, m k)1. H nh dng.- Mch cu c v:Trong : Cc in tr R1, R2, R3, R4gi l in tr cnh. R5 gi l in tr gnh

2. Phn loi mch cu.Mch cu cn bng

- Mch cu Mch cu ( tng qut)Mach cu khng cn bng

Mch cu khuyt

3. D u hiu nhn bit cc loa mch cua/ Mch cu cn bng.- Khi t mt hiu in th UAB khc 0 th ta nhn thy I5 = 0.- c im ca mch cu cn bng.

+ V in tr.4

2

3

1

4

3

2

1

R

R

R

R

R

R

R

R==

+ V dng in: I1 = I2 ; I3 = I4 Hoc2

4

4

2

1

3

3

1 ;R

R

I

I

R

R

I

I==

+ V hiu in th : U1 = U3 ; U2 = U4 Hoc4

3

4

3

2

1

2

1 ;R

R

U

U

R

R

U

U==

b/ Mch cu khng cn bng.- Khi t mt hiu in th UAB khc 0 th ta nhn thy I5 khc 0.- Khi mch cu khng 5 in tr th gi l mch cu khuyt.

II/ CCH GII CC LOI MCH CU1. M ch cu cn bng.* Bi ton c bn.Cho mch in nh HV.

Vi R1=1, R2=2, R3=3, R4= 6, R5 = 5.UAB=6V. Tnh I qua cc in tr?

* Gii:Ta c :

2

1

4

3

2

1==

R

R

R

R=> Mch AB l mch cu cn bng.

=> I5 = 0. (B qua R5). Mch in tng ng: (R1 nt R2) // (R3 nt R4)- Cng dng in qua cc in tr

I1 = I2 = ARR

UAB 2

21

6

21

=

+

=

+; I3 = I4 = A

RR

UAB 67.0

63

6

43

+

=

+

2. M ch cu khng cn bng.a. Mach cu hay cn gi l mch cu tng qut.

* Bi ton c bn. Cho mch in nh HV.Vi R1=1, R2=2, R3=3, R4= 4, R5 = 5.UAB=6V. Tnh I qua cc in tr?


R1

R2

R3 R4

R5

A BM

N

R1

R2

R3

R4

R5

A BM

N

R1

R2

R3

R4

R5

A B

M

N


3/7


* Gii:Cch 1. Phng php in th nt.

-Phng php chung.+ Chn 2hiu in th bt k lm 2 n.+ Sau qui cc hiu in th cn li theo 2 n chn.+ Gii h phng trnh theo 2 n

VD ta chn 2 n l U1 v U3.-Ta c: UMN = UMA + UAN = -U1 + U3 = U3 U1 = U5- Xt ti nt M,N ta c

I1 + I5 = I2 2

1

5

13

1

1

R

UU

R

UU

R

UAB

=

+ (1)

I3 = I4 + I5 5

13

4

3

3

3

R

UU

R

UU

R

UAB

+

= (2)

-T (1) v (2) ta c h phng trnh

2

1

5

13

1

1

R

UU

R

UU

R

U AB =+

251

1131 UUUUU AB =

+

5

13

4

3

3

3

R

UU

R

UU

R

UAB

+

=

543

1333 UUUUU AB +

=

Gii ra ta c U1 , U3. Tnh U2 = UAB U1 , U4 = UAB U3. Ap dng nh lutm tnh c cc dng qua in tr.

C ch2. t n l dng-Phng php chung.

+ Chn 1 dng bt k lm n.+ Sau qui cc dng cn li theo n chn.+ Gii phng trnh theo n

- VD ta chn n l dng I1.Ta c: UAB = U1 + U2 = I1R1 + I2R2 = I1 + 2I2 = 6

I2 = 11 5.03

2

6I

I=

(1)

- T nt M. I5 = I2 I1 = 3 -0.5I1 - I1 = 3 1.5I1I5 = 3 1.5I1 (2)

- Mt khc: U5 = UMN = UMA + UAN = -U1 + U3 = U3 U1= I3R3 I1R1 = 3I3 I1=5I5

=> I3 = 35.615

3

5.715

3

511115IIIII

=

=

I3 =3

5.615 1I (3)

- T nt N. I4 = I3 I5 =3

5.6151I

- 3 1.5I1 =3

1161

I

I4 =3

1161

I(4)

-Mt khc. UANB = UAN + UNB = U3 + U4 = I3R3 + I4R4 = 3I3 + 4I4 = 6

3.3

5.615 1I + 4.3

1161

I= 6

Gii ra ta c I1 1.1 A. Th vo (1), (2), (3), (4) ta tnh c cc I cnli.



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+ Ch : Nu dng i qua MN theo chiu ngc li th s c kt qu khc.C ch 3 . Dng phng php chuyn mch:

-Phng php chung:+Chuyn mch sao thnh mch tam gic v ngc li.( )+V li mch in tng ng, ri dng nh lut Om, tnh in tr ton

mch, tnh cc dng qua cc in tra/ Phng php chuyn mch : => .

- Lng hai mch vo nhau, sau tnh x,y, z theo R1, R2, R3.

Ta c: RAB =( )

YXRRR

RRR+=

++

+

321

32.1

(1)

RBC =( )

ZYRRR

RRR+=

++

+

321

31.2 (2)

RAC =( )

ZXRRR

RRR+=

++

+

321

21.3

(3)

Cng 3 phng trnh theo v ri chia cho 2 ta c.

ZYX

RRR

RRRRRR++=

++

++

321

133221 (4)

Tr (4) cho (1), (2), (3) ta c:

Z =321

32 .

RRR

RR

++; X =

321

31.

RRR

RR

++; Y =

321

21.

RRR

RR

++(5)

=> Tng qut: Tch 2 in tr kX, Y, X =

Tng 3 in tr

b/ Phng php chuyn mch : =>

- T (5) ta chia cc ng thc theo v.

12

2

1 .RX

ZR

R

R

Z

X== ; 13

3

1 .RY

ZR

R

R

Z

Y==


R1

R2

R3

x

y z

A

B C

R1

R2

R3

A

B C

y

xz

A

B C

X Y

Z

R1

R3

C

R2

A

B

A

B C

YX

Z

A

B C

R3

R2


5/7


Kh R2, R3 trong (5) suy ra:

3

133221

R

RRRRRRX

++= ;

2

133221

R

RRRRRRY

++= ;

1

133221

R

RRRRRRZ

++=

=>Tng qut: Tng cc tch lun phinX,Y,Z =

in tr vung gc

c/ Ap dng gii bi ton trn.* Theo cch chuyn tam gic thnh sao

- Mch in tng ng lc ny l: [(R1nt X) // (R3 nt Y)] nt Y- Tnh c in tr ton mch- Tnh c I qua R1, R3.- Tnh c U1, U3

+Tr v s gc- Tnh c U2, U4.- Tnh c I2, I4- Xt nt M hoc N s tnh c I5* Theo cch chuyn sao thnh tam gic.

Ta c mch tng ng: Gm {(Y// R3) nt (Z // R4)}// X.- Ta tnh c in tr tng ng ca mch AB.- Tnh c IAB.- Tnh c UAN = U3 , UNB = U4- Tnh c I3 , I4- Tr v s gc tnh c I1 = IAB I3 ; I2 = IAB I4- Xt nt M hoc N, p dng nh l nt mch tnh c I53. M ch cu khuyt:

Thng dng rn luyn tnh ton v dng in khng i.a. Khuyt 1 in tr( C 1 in tr bng khng vd R1= 0)


A B

M

N

R1

R3

x

z

yR1 R2

R3

R4

R5

A BM

N

R2

R3

R4

R5

A BM

N

A B

N

R3

R5

R4

R2

R1

R2

R3 R4

R5

A BM

N

A B

X

Y Z

R3 R4N


6/7


+ Phng php chung.- Chp cc im c cng in th, ri v li mch tng ng. Ap dng nh lutm gii nh cc bi ton thng thng tnh Iqua cc R. Tr v s gc xt ntmch tnh I qua R khuyt.

- Khuyt R1: Chp A vi M ta c mch tng nggm: {(R3 // R5) nt R4 } // R2

- Khuyt R2: Chp M vi B ta c mch tng nggm: {(R4 // R5) nt R3 } // R1

- Khuyt R3: Chp A vi N ta c mch tng nggm: {(R1 // R5) nt R2 } // R4

- Khuyt R4: Chp N vi B ta c mch tng nggm: {(R2 // R5) nt R1 } // R3

- Khuyt R5: Chp M vi N ta c mch tng nggm: {(R4 // R3) // (R2 //R4)

b. Khuyt 2 in tr. (c 2 in tr bng 0)

- Khuyt R1 v R3: chp AMN ta c mch tng ng gm : R2 // R4

V I5 = 0 nn ta tnh c I2 =2

R

UAB

, I4 =4

R

UAB

, I1 = I2 , I3 = I4

- Khuyt R2 v R4 tng t nh trn- Khuyt R1 v R5 : chp AM lc ny R3 b ni tt (I3 = 0), ta c mch tng nggm : R2 // R4. Ap dng tnh c I2, I4, tr v s gc tnh c I1, I5- Khuyt R2 v R5 ; R3 v R5 ; R4 v R5 tng t nh khuyt R1 v R5

c. Khuyt 3 in tr. (c 3 in tr bng 0)

- Khuyt R1, R2, R3 ta chp AMN. Ta c mch tng ng gm R2 // R4. Th cchgii vn nh khuyt 2 in tr- Khuyt R1, R5, R4 ta chp A vi M v N vi B. Ta thy R2, R3 b ni tt.

---------- Ht ----------

TI LIU THAM KHO- Vt l nng cao Trung hc c s


R2

R4

R5

A BM

N

A B

R2

R4

R2

R3

A BM

N

R2

R3


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( Nguyn Cnh Ho L Thanh Hoch)- Bi tp vt l dng cho hc sinh chuyn l THCS

( Nguyn Thanh Hi)


67687677 Phuong Phap Giai Mach Cau Cuc Hay

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