phương pháp Điện thế nút

+ U -

R1

R3 R4 R5

B A

C

D

R2

+ U -

R1

R3 R4 R5

B A

C

D

R2 I1

I5

I2

I4 I3

I

PHNG PHP IN TH NT GII CC BI TON IN MT CHIU B Vn Khi

I. M u Cc dng nh lut m gip ta gii quyt c nhiu bi ton mch in. Bng cc php thu gn mch in nh dng in tr tng ng, ngun in tng ng ta a c nhiu mch in v cc dng n gin c th p dng nh lut m cho on mch, cho mch kn gii quyt. Tuy nhin, khi s cc ngun in, cc in tr nhiu, mc thnh cc mch hn hp phc tp th phng php dng nh lut m khng gii quyt c. Phng php in th nt nhm ti vic thnh lp mt h nhiu phng trnh bc nht gip ta v nguyn tc c th gii quyt bi ton mch in phc tp bt k. Nt mng ( gi tt l nt) l ni c t 3 on mch tr ln ni vo. Ring trng hp c bit ca mch kn n gin, mch in ch c 1 mt, ta vn c th xem l c 2 nt vi mi nt ch c 2 on mch ni vo v vn p dng c phng php in th nt. Trong phng php in th nt ta s ly in th ca cc nt lm n s. Khi tm c in th cc nt ta d dng tnh c hiu in th gia cc cp nt v sau dng nh lut m suy ra dng in trong cc on mch. Trong mng in c n nt, ta chn 1 nt xem nh ni t c in th bng 0, cn ( n 1) nt v ta c ( n 1) n s Vi ( i = 1, 2,, n 1). Cc n Vi ny l cc i lng i s, c th dng hay m. Trong phng php ny, khi gii bi ton v mch in, h phng trnh c thit lp t cc phng trnh v nt v cc phng trnh cho cc on mch gia cc nt. Cc phng trnh cho on mch chnh l biu thc nh lut m tng qut vit cho on mch khng phn nhnh nm gia cc nt. V ta ch quan tm n hiu in th gia cc nt, nn ta c th chn gc tnh in th l in th 1 nt bt k no, ngha l t in th nt bng 0. Nh vy s n s v in th gim i 1. Gii h phng trnh ta tm c cc in th ti cc nt cn li, nh tm c li gii cho bi ton. II. Phng php in th nt 1. Biu din cng v chiu dng in trn cc on mch mt cch hon ton ty . 2. Cho in th mt nt no bng khng. Ch chn nt no c th t in th khng nt suy ra ngay c in th mt s nt khc. 3. Vit phng trnh cng dng in ti nhng nt cha bit in th theo nh lut nt mch. 4. Vit phng trnh cng dng in trn cc on mch theo cng thc nh lut m tng qut. 5. Thay cc cng ny vo cc phng trnh nt tm cc in th cha bit. 6. Thay cc in th tm c vo cc phng trnh bc 4 tnh cc cng dng in. II. Cc bi tp v d Bi tp 1. Cho mch in nh hnh v: R1 = R4 = R5 = 1; R2 = R3 = 2; U = 6(V). Tm cng dng in qua cc in tr v in tr tng ng ca mch.

Gii Gi s dng in c chiu v cng qua cc on mch nh hnh v. Chn mc in th ti B (VB = 0). Khi UAB = VA VB = U= 6(v) VA = 6(V) Ti nt C: I1 = I5 + I2 (1) Ti nt D: I4 = I5 + I3 (2)

p dng inh lut m ta c: A C1 A C A1

V VI V V 6 V

R-

= = - = -

C B C22

V V VI

R 2-

= =

A D D33

V V 6 VI

R 2- -

= =

D B4 D4

V VI V

R-

= =

I5 = C D C D5

V VV V

R-

= -

Thay vo phng trnh (1) v (2) ta c: 6 VC = VC VD + CV2

VD = VC VD + D6 V

2-

Gii h phng trnh trn ta c: VD = 187

(V); VC = 247

(V)

Thay vo cc phng trnh dng in ta c: I1 = 187

(A); I2 = 127

(A); I3 = 127

(A); I4 = 187

(A); I5 =67

(A)

Ti nt A ta c I = I1 + I3 = 307

(A)

in tr tng ng ca mch: R = U 6 1430I7

= = W

Bi tp 2. Cho mch in nh hnh v: e1 = 65V, e2 = 39V, r1 = r2 = 0, R1 = 20, R2 = R3 = R4 = R5 = 10. Tm tt c cc cng dng in.

Gii Chiu v cng dng in trn cc on mch c biu din nh hnh v. Chn in th nt C bng 0 (VC = 0) ta c ngay: VB VC = e2 I2r2 = e2 = 39V VB = 39V Ti nt A: I1 = I3 + I2 Ti nt D: I2 = I4 + I5 (1) p dng nh lut m cho cc on mch: B A A1

1

V V 39 VI

R 20- -

= =

A D A D22

V V V VI

R 10- -

= =

A C A33

V V VI

R 10-

= = (2)

D C D44

V V VI

R 10-

= =

R2

R4 R1

E2

C

A

B

E1

D

R5

R6

R3

D B 1 D55

(V V ) e V 26I

R 10- + +

= =

I6 = I3 + I4 Thay vo (2) vo (1) ta c h phng trnh: 5VA 2VD = 39 3VD VA = - 26 Gii h ta c: VA = 5V; VD = - 7V Thay cc in th tnh c vo cc phng trnh (2) ta thu c: I1 = 1,7A; I2 = 1,2A; I3 = 0,5A; I4 = - 0,7A; I5 = 1,9A; I6 = - 0,2A. Du I4 v I6 chng t chiu dng in biu din trn mch khng ph hp vi chiu thc t. Bi tp 3. Cho mch in hnh v: E1 = 5(V); E2=3(V), r1 = r2 = 0(), R1 = R2 = R3 = R4 = R5 = R6 = 1(). Tnh cng dng in qua cc nhnh v hiu in th UAB.

Gii Gi s dng in c chiu v cng nh hnh v Chn mc in th ti D( VD = 0) Khi UAD = VA VD = E2 - I r2 = E2 = 3(V) VA = 3(V) Ti nt C: I2 = I4 + I6 (1) Ti nt B: I6 = I5 + I1 (2) A C C D C B

2 3 4 6

V V V V V VR R R R

- - - = +

+

C B B D B A 16 5 1

V V V V V V ER R R- - - +

= +

5VC -2VB = 3 VC 3VB = 2

Gii h phng trnh ta c: VC = 513

(V); VB = -7

13(V)

Vy : I1 = 1913

(A); I2 = 1713

(A); I3 = 1713

(A); I4 = 5

13(A); I5 = -

713

(A); I6 = 1213

(A); I = I2 I1 = - 0,15(A).

I v I5 chy theo chiu ngc li ta quy c. Bi tp 4. Cho mch in hnh v: E = 29(V); r = 1(); R1 = R5 =5(); R2 =2(); R3 =10(); R4 =30(); R6 =3(). Tnh cng dng in qua cc in tr v in tr tng ng ca mch AB.

Gii Gi s dng in c chiu v cng qua cc on mch nh hnh v. Chn mc in th ti B (VB = 0). Ti cc nt C, D v A, ta c:

R2

R4 R1

E2

C

A

B

E1

D R5

R6

R3

I2

I4

I6

I

I1 I5

R3 R4

R5 B A

C

D

R2 I1

I5

I2

I4 I3

I6 R6 E, r

A C C B C D

1 2 51 2 5

C DD B A D4 3 5

4 3 56 1 3

A CB A A D

6 1 3

V V V V V VR R R

I I IV VV V V V

I I IR R R

I I IV VV V E V V

R r R R

- - -= +

= + -- - = + = +

= + -- + -= +

+

A C C C D

A C DC DD A D

A C D

A C DA CA A D

V V V V V5 2 5 2V 9V 2V 0

V VV V V3V 6V 10V 0

30 10 511V 4V 2V 130V V26 V V V

4 5 10

- - = +- + =

-- = + + - = - - =-- -

= +

Gii h phng trnh ta c: VA = 15(V), VC =5(V), VD = 7,5(V) Thay vo cc phng trnh I ta c: I1 = 2(A); I2 = 2,5(A); I3 = 0,75(A); I4 = 0,25(A); I5 = 0,5(A); I6 = 2,75(A).

in tr tng ng ca mch AB l: ABABU 15

R 5, 4545I 2,75

= = W .

Bi tp 5. Cho mt mch t nh hnh v: C1 = C2 = 6F; C3 = 2F; C4 = C5 = 4F; U = 20V. Tm in tch tng t in v in dung ca mch.

Gii Gi s in tch cc bn t in c du nh hnh. Chn in th nt B bng 0 (VB = 0). Ta c ngay: VA VB = U = 20V hay VA = 20V. Ti nt C: q1 = q2 + q5 Ti nt D: q4 = q3 + q5 (1) in tch ca cc t in c tnh:

q1 = C1(VA VC) = 120.10-6 6.10-6VC

q2 = C2(VC VB) = 6.10-6VC

q3 = C3(VA VD) = 40.10-6 2.10-6VD (2)

q4 = C4(VD VB) = 4.10-6VD

q5 = C5(VC VD) = 4.10-6VC 4.10

-6VD Thay (2) vo (1) v rt gn ta c: 4VC VD = 30 4VC 10VD = - 40

Gii h phng trnh ta thu c: VC = 859

V; VD = 709

V.

Thay cc in th tnh c vo (2) ta c: q1 = 5

5710

9- C; q2 = 5

5110

9- C; q3 = 5

2210

9- C; q4 = 5

2810

9- C; q5 = 5

210

3- C.

in dung ca mch: C = qU

= 13q q

U+ =

79180

F.