Chuyn Vt l ---- ---- Trng THCS Yn Lng Ngi thc hin: inh Vit c

Chuyn gm : 2 phn I. 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

PHNG PHP GII MCH CUI/ MCH CU. - Mch cu l loi mch c dng ph bin trong cc php o in nh ( Vn k, am pe k, m k) 1. Hnh dng. R1 R M 2 - Mch cu c v: B A Trong : Cc in tr R1, R2, R3, R4 R5 gi l in tr cnh. R5 gi l in tr gnh 2. Phn loi mch cu. R3 N R4 Mch cu cn bng - Mch cu Mch cu ( tng qut) Mach cu khng cn bng Mch cu khuyt 3. Du hiu nhn bit cc loa mch cu a/ 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. + V dng in:R1 R3 = R2 R4 R1 R2 = R3 R 4

I 1 R3 I 2 R4 = ; = I 3 R1 I 4 R2 U 3 R3 U1 R = 1 ; = + V hiu in th : U1 = U3 ; U2 = U4 Hoc U 2 R2 U 4 R4

I1 = I2 ; I3 = I4

Hoc

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 CU 1. Mch 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 : R1 A R2 B

M

R5 R3 N R4

R R1 1 = 3 = => Mch AB l mch cu cn bng. R 2 R4 2

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

Chuyn Vt l ---- ---- Trng THCS Yn Lng Ngi thc hin: inh Vit c - Cng dng in qua cc in tr I1 = I2 =U AB 6 = = 2A R1 + R2 1 + 2

;

I3 = I4 =

U AB 6 = 0.67 A R3 + R4 3 + 6

2. Mch cu khng cn bng. R1 R a. Mach cu hay cn gi l mch cu tng qut. M 2 * Bi ton c bn. Cho mch in nh HV. A R5 Vi R1=1, R2=2, R3=3, R4= 4, R5 = 5. UAB=6V. Tnh I qua cc in tr? R3 N R4 * 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 cU 1 U 3 U 1 U AB U 1 + = R1 R5 R2 U 3 U AB U 3 U 3 U 1 = + I3 = I4 + I5 R3 R4 R5

B

I1 + I5 = I2

(1) (2)

-T (1) v (2) ta c h phng trnhU 1 U 3 U 1 U AB U 1 + = R1 R5 R2 U 3 U AB U 3 U 3 U 1 = + R3 R4 R5

U 1 U 3 U 1 U AB U 1 + = 1 5 2 U 3 U AB U 3 U 3 U 1 = + 3 4 5

Gii ra ta c U1 , U3. Tnh U2 = UAB U1 , U4 = UAB U3. Ap dng nh lut m tnh c cc dng qua in tr. Cch2. 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 = - T nt M.6 I1 = 3 0.5 I 1 2

(1) (2)

I5 = I2 I1 = 3 -0.5I1 - I1 = 3 1.5I1 I5 = 3 1.5I1 - Mt khc: U5 = UMN = UMA + UAN = -U1 + U3 = U3 U1 = I3R3 I1R1 = 3I3 I1=5I5 => I3 =5 I 5 I 1 15 7.5 I 1 I 1 15 6.5 I 1 = = 3 3 3 15 6.5 I 1 I3 = 3

(3)

Chuyn Vt l ---- ---- Trng THCS Yn Lng Ngi thc hin: inh Vit c - T nt N. I4 = I3 I5 = I4 = -Mt khc. 3. li.6 11 I 1 3 15 6.5 I 1 6 11 I 1 - 3 1.5I1 = 3 3

(4)

UANB = UAN + UNB = U3 + U4 = I3R3 + I4R4 = 3I3 + 4I4 = 615 6.5 I 1 6 11 I 1 + 4. =6 3 3

Gii ra ta c I1

1.1 A. Th vo (1), (2), (3), (4) ta tnh c cc I cn

+ Ch : Nu dng i qua MN theo chiu ngc li th s c kt qu khc. Cch 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 tr a/ Phng php chuyn mch : => . - Lng hai mch vo nhau, sau tnh x,y, z theo R1, R2, R3. A R1 B R2 R3 y C B A x z C B R1 y R2 A x z R3 C

Ta c: RAB =

R1. ( R2 + R3 ) = X + Y (1) R1 + R2 + R3 R2. ( R1 + R3 ) =Y +Z RBC = R1 + R2 + R3 R3. ( R1 + R2 ) = X +Z RAC = R1 + R2 + R3R1 R2 + R2 R3 + R3 R1 = X +Y + Z R1 + R2 + R3

(2) (3)

Cng 3 phng trnh theo v ri chia cho 2 ta c. (4)R1 .R2 R1 + R2 + R3

Tr (4) cho (1), (2), (3) ta c: Z=R2 .R3 ; R1 + R2 + R3

X=

R1 .R3 ; R1 + R2 + R3

Y=

(5)

=> Tng qut: X, Y, X =

Tch 2 in tr k Tng 3 in tr

A b/ Phng php chuyn mch : A X R2 B R3 C B

Y

=> X C R2

A R1 R3 Z Y C

Z

Chuyn Vt l ---- ---- Trng THCS Yn Lng Ngi thc hin: inh Vit c

B - T (5) ta chia cc ng thc theo v.R X Z = 1 R2 = .R1 ; Z R2 X R Y Z = 1 R3 = .R1 Z R3 Y

Kh R2, R3 trong (5) suy ra:X = R1 R2 + R2 R3 + R3 R1 ; R3 Y = R1 R2 + R2 R3 + R3 R1 ; R2 Z= R1 R2 + R2 R3 + R3 R1 R1

=>Tng qut: X,Y,Z =

Tng cc tch lun phin in tr vung gc

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

R1 A

M

R2 B A

R1

M

x z

R5 R3 N R4

y

B

R3

N

- 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. R1 A M R2 B A X Y Z R3 N R4 {(Y// R3) nt (Z // R4)}// X. B

R5 R3 N R4

Ta c mch tng ng: Gm

Chuyn Vt l -------- Trng THCS Yn Lng Ngi thc hin: inh Vit c - 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 I5 3. Mch 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) R R2 M R 2 A B B A 3 R5 N R R R3 N R4 4 + Phng php chung. 5 - Chp cc im c cng in th, ri v li mch tng ng. Ap dng nh lut m gii nh cc bi ton thng thng tnh I qua cc R. Tr v s gc xt nt mch tnh I qua R khuyt. - Khuyt R1: Chp A vi M ta c mch tng ng gm: {(R3 // R5) nt R4 } // R2 - Khuyt R2: Chp M vi B ta c mch tng ng gm: {(R4 // R5) nt R3 } // R1 - Khuyt R3: Chp A vi N ta c mch tng ng gm: {(R1 // R5) nt R2 } // R4 - Khuyt R4: Chp N vi B ta c mch tng ng gm: {(R2 // R5) nt R1 } // R3 - Khuyt R5: Chp M vi N ta c mch tng ng gm: {(R4 // R3) // (R2 //R4) b. Khuyt 2 in tr. (c 2 in tr bng 0) R R2 M 2 B A A B R5 R4 N R4 - Khuyt R1 v R3: chp AMN ta c mch tng ng gm : R2 // R4 V I5 = 0 nn ta tnh c I2 =U AB , R2

I4 =

U AB , R4

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 ng gm : 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) M A R3 N R2 B R2

R3

Chuyn Vt l ---- ---- Trng THCS Yn Lng Ngi thc hin: inh Vit c - Khuyt R1, R2, R3 ta chp AMN. Ta c mch tng ng gm R2 // R4. Th cch gii 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 ----------