2/5/20151Chng 2PHNG TRNH VI PHN V NG DNGNi Dung1. Phng trnh vi phn cp 12. Phng trnh vi phn cp 23. ng dng phn gii mch in Mch DC (Direct Current) Mch bc nht RL, RC Mch bc hai RLC (ni tip, song song, tng qut) Mch AC (Alternating Current)2/5/20152MCH DC BC HAI KHNG CNGUN KCH THCH NGOIRLC Ni tipRLC Song SongRLC Ni Tip Khng C Ngun Ngoi (1) Gi s ti < 0, t & cun d tr 0 = 0, 0 = 0 Khi > 0, p dng KVL: +

+ 1

= 0 2

2 +

+

= 0 2/5/20153RLC Ni Tip Khng C Ngun Ngoi (2) = 0 0 + 0

+0 = 0 0

= 1

0 +0 Nhc li, p ng t nhin ca mch bc nht cdng hm m =

RLC Ni Tip Khng C Ngun Ngoi (3)

2 +

+ 1

ptt 2 +2 +02 = 0 = 2

0 =1

lun cho 2 nghim, nn nghim ca c dng

q = 1

1

+2

2

2/5/20154RLC Ni Tip Khng C Ngun Ngoi (4) = 1

1

+2

2

> 0: 1 2

q = 1

1

+2

2

= 0: 1 = 2 =

q = 1 +2 < 0: 1,2 = = ; = 02 2

q = 1cos +2sinRLC Song Song Khng C Ngun Ngoi (1) Gi s ti < 0, t & cun d tr 0 = 0, 0 = 0 p dng KCL:

+1

+

= 0 2

2 + 1

+

= 0 2/5/20155RLC Song Song Khng C Ngun Ngoi (2)Phn tch tng t, ta cptt 2 +2 +02 = 0 =12

0 =1

lun cho 2 nghim, nn nghim ca c dng

q = 1

1

+2

2

RLC Song Song Khng C Ngun Ngoi (3) = 1

1

+2

2

> 0: 1 2

q = 1

1

+2

2

= 0: 1 = 2 =

q = 1 +2 < 0: 1,2 = = ; = 02 2

q = 1cos +2sin2/5/20156MCH DC BC HAI C NGUNKCH THCH NGOIRLC Ni tipRLC Song SongRLC Ni Tip C Ngun Ngoi (1) Khi > 0, KVL:

+ + =

=

2

2 +

+

=

Nhc li RLC nt khng c ngun ngoi

2

2 +

+

= 0C cngdng ptrVi phn2/5/20157RLC Ni Tip C Ngun Ngoi (2) p ng ton phn = o +q =

+q p ng qu q xem li phn mch bc 2 khng c ngun ngoiRLC Ni Tip C Ngun Ngoi (3) V d: Cho kha K ng trong mt gian di sau chuyn trng thi ti thi im = 0. Tm () v() khi > 0 cho cc trng hp sau:a. = 5b. = 4c. = 12/5/20158RLC Ni Tip C Ngun Ngoi (4) Khi < 0, do kha K ng trong thi gian di, nnta xem t nh mch h v cuc nh ni tt. Do , gi tr ban u (0) chnh l dng mch chnh: 0 =245 +1 = 4 V gi tr ban u (0) chnh l HT qua = 1: 0 = 0 1 = 4 Khi > 0, mch tr thnh RLC ni tip = 2 = 2.5; =1

= 2; 1 = 1; 2 = 4;RLC Ni Tip C Ngun Ngoi (4) Bi v > , nn p ng ca mch v(t) nh sau: =

+ 1

+2

4

= 1

42

4 0 = 24 +1 +2 = 4 1 +2 = 20 Dng qua cun cng chnh l dng qua t, do : =

=0 0 = 0

0

= 0

=40.25=161 42 = 16 1 = 643 & 2 = 432/5/20159RLC Ni Tip C Ngun Ngoi (5) Phng trnh v(t) tr thnh = 24 +643 +43

4

Phng trnh i(t) tr thnh =

=0.25 643 163 4

= 163 43

4 ()RLC Ni Tip C Ngun Ngoi (6) = 4 = 24 19,2 1 + 2 = 4,8 +9,6 2 = 1 = 24 + 21,7sin 1,9 +12cos 1,9 0,5 = 3,1sin 1,9 +12cos 1,9 0,5 2/5/201510RLC Song Song C Ngun Ngoi (1) Khi > 0, KVL:

+ +

=

=

2

2 + 1

+

=

Nhc li RLC // khng c ngun ngoi

2

2 + 1

+

= 0C cngdng ptrVi phnRLC Song Song C Ngun Ngoi (2) p ng ton phn = o +q =

+q p ng qu q xem li phn mch bc 2 khng c ngun ngoi2/5/201511RLC Tng Qut Tm p ng ton phnca v khi > 0AC Tn Hiu Dng SINE =

sin + hoc =

cos +

: bin : tn s gc : pha2/5/201512AC Bin Pha (Phasors) Bin Pha l mt s phc c trng cho bin v pha ca mt tn hiu dng SINE. =

cos +min thi gian =

bin pha Theo cng thc Euler:

= cos sin =

=

cos + =

=

t =

=

Cng Thc Lng Gicsin = sincos cos sincos = cos cos sinsinsin = sincos = cos sin

2= cos cos

2= sin2/5/201513AC Bin Pha (Phasors)Min thi gian Bin Pha

cos +

sin +

2

cos +

sin +

2Tnh Cht Ca Bin Pha =

=

o hm

Tch phn

2/5/201514Chng Minh

=

sin + =

cos + +

2=

2

=

=

sin +

=

cos + +

2

=

2

=

V D Chuyn cc t/h sau thnh dng bin Pha = 6 cos 50

6 = 6

6 = 4 sin 30 +

5 = 4 710= 4 cos 30 +

5 +

2= 4 cos 30 +7102/5/201515V D Chuyn cc t/h sau thnh dng Sine = 3 + 4 = 5cos + 126= 5 1260 = 8

9 = 8 cos +718= 8

9

2= 8718Bi Tp Xc nh dng () v () bng pp bin Phc = 1 +2

1 = 4 cos( + 300)

2 = 5 sin 200 = 1 +2

1 = 10 sin( 300)

2 = 20cos +4504 + 8 3

= 50cos 2 + 7502

+ 5 + 10 = 50 cos 5 3002/5/201516Mi Quan H Gia Dng ThTr Khng Tng ng2/5/201517Mch AC Tm phng trnh 0 Ht Chng 2