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TRNG I HC NG CNG N THI
TRNG I HC NG
KHOA CNG NGH K THUT IN IN T****** CNG N THI H LIN THNG NM 2014MN: L THUYT MCH IN 1 V C S K THUT IN
PHN 1: MCH IN TUYN TNH CH XC LP NGUN HNH SIN1. Mch in l g? Cc bin s c bn ca mch in.
2. Kt cu hnh hc ca mch in l g?
3. Dng in l g? in p l gi? Ngun in p l g? Quan h gia in p v sc in ng trn ngun? Ngun dng in l g?
4. Pht biu nh lut m?
5. in tr R l g? Lut m cho phn t tr? Biu thc cng sut v nng lng tiu th trn in tr?
6. in cm L l g? Lut m cho phn t cm? Biu thc cng sut v nng lng t trng trn in cm?
7. in dung C l g? Lut m cho t in? Biu thc cng sut v nng lng in trng trn in dung?
8. Pht biu nh lut KF1? s phng trnh c lp vit theo lut KF1 cho mt mch l bao nhiu?
9. Pht biu lut KF2? Xc nh vng c lp bng cch no? S phng trnh c lp vit theo lut KF2 cho mt mch in l bao nhiu? 10. H phng trnh ca mch in tuyn tnh, xc lp, hnh sin di dng i s ca bin phc
11. Tng tr phc ca mt nhnh Tam gic tng tr
12. S phc ca mch in
13. H phng trnh ca mch di dng i s theo bin phc
14. Cc loi cng sut trong mch in
Cng sut tc dng
Cng sut phn khng
Cng sut biu kin - Tam gic cng sut
Cng thc chung tnh cng sut
H s cng sut
15. Phng php dng in nhnh
16. Phng php dng in vng
17. Phng php in th nh (nt)
Phng php in th nt
Phng php in th p dng cho mch c 2 nt
18. nh ngha mch ba pha, phn loi mch ba pha
19. c im mch ba pha i xng ni Y-Y (ngun i xng, ti i xng, cch ni ngun ba pha, cch ni ti ba pha). Cch gii mch ba pha i xng ni Y-Y.
20. c im ca mch ba pha i xng ni (-( (Quan h gia in p dy v in p pha, Quan h gia dng in dy vi dng in pha). Cch gii mch ba pha i xng ni (-(.
21. c im mch ba pha i xng ni Y-(. Cch gii mch ba pha i xng ni Y-(22. Tnh mch ba pha 4 dy khng i xng ni Y-Y , Y-( , (-( S dng phng php in th nt gii
Tnh dng in cc pha ca ti
Tnh dng in dy trung tnh 23. Tnh cng sut mch in ba pha
BI TP IN HNH Bi 1: Vit phng trnh Kirchhoff 1 v 2 cho hnh v sau:Cho bit:
; ;
;
; ;
Tm dng in trong cc nhnh bng cc phng php hc.
Tnh P, Q, S ca mch in
Bi 2: Tnh cc dng in v trn hnh v. T suy ra v
Tm dng in trong cc nhnh bng cc phng php hc.Tnh P, Q, S ca mch in
Bi 3: Cho mch in nh hnh v: Bit ; ;
; ; ; ; ;
Vit biu thc dng in chy trong cc nhnhTnh P, Q, S ca mch in
Bi 4: Cho mch in nh hnh v.
Bit:
Tm dng in trong cc nhnh bng phng php dng in nhnh
Tm dng in trong cc nhnh bng phng php dng in vng
Tm dng in trong cc nhnh bng phng php th nh
Bi 5: Cho mch in nh hnh v
Tm dng in trong cc nhnh bng phng php dng in nhnh
Tm dng in trong cc nhnh bng phng php dng in vng
Tm dng in trong cc nhnh bng phng php th nh
Bi 6: Cho mch in nh hnh 4.
Bit:
Tm dng in trong cc nhnh bng phng php dng in nhnh
Tm dng in trong cc nhnh bng phng php dng in vng
Tm dng in trong cc nhnh bng phng php th nhBi 7: My pht in ba pha u sao, c dy trung tnh, in p pha UP = 240V, mc vo ti l cc bng n c tr khng cc pha l ; ; . in tr ca dy ni c th b qua. Xc nh dng in trong cc dy pha v dy trung tnhBi 8: Ngun in ba pha u sao, c sc in ng pha i xng, , cung cp cho ti ba pha u sao, c tr khng ln lt l: ; ; . Tng tr dy trung tnh . Xc nh in p v dng in pha ca ti.
Bi 9: Cho mch 3 pha i xng Ud = 220V cung cp cho 2 ti:Ti 1: R1 = 4( , X1 = 3(Ti 2: l ng c c P2 = 7 kW , cos(2 = 0,6 , ( = 0,9 , ni tam gic
Tnh:
a) tnh dng in trn cc pha ca ti
b) dng in trn ng dy chnh
c) tnh cng sut tiu th, cng sut phn khng v cng sut biu kin ca ton mch.
Bi 10: Cho ngun ba pha i xng c . Tm dng trn cc dy v cng sut tc dng ca ti. Bit: R = 40( ; L = 0,1H ; ; ( = 200 rad/s
PHN 2: QU TRNH QU TRONG MCH IN TUYN TNH
1. nh ngha qu trnh qu
Nguyn nhn xy ra qu trnh qu
Thi im xy ra qu trnh qu - mc thi gian qu trnh qu
H phng trnh m t qu trnh qu - bi ton qu trnh qu
2. Lut ng m
Lut ng m 1: uC(0) = uC(-0)
Lut ng m 2: iL(0) = iL(-0)3. S kin
nh ngha s kin
Phn loi s kin: S kin c lp uC(0), iL(0) v S kin ph thuc
4. Cch tnh s kin c lp uC(0), iL(0)Vi xc lp c mt chiu:
a. lp phng trnh i s tnh UC c; IL c
b. ti t = 0 c uC(-0) = UC c; iL(-0) = IL c
c. dng lut ng m suy ra uC(0) = uC(-0) v iL(0) = iL(-0)Vi xc lp hnh sin:
a. dng s phc c lp h phng trnh i s vi nh phc b. tnh Cxl c suy ra uCxl c(t) thay ti t = 0 c uC(-0), c. tnh Lxl c suy ra iLxl c(t) thay ti t = 0 c iL(-0)b. Dng lut ng m suy ra s kin c lp uC(0) = uC(-0); iL(0) = iL(-0)5. Gii bi ton qu trnh qu cp 1 R-C; R-L tuyn tnh bng phng php tch phn kinh in
5.1.Tnh s kin c lp: uC(0), iL(0) t s xc lp c.
5.2. Tnh s m c trng p t s hin hnh i s ha theo p khng ngun (ch no c L thay bng PL; ch no c C thay bng ). Xc nh tng tr vo theo p t 1 ca mt nhnh bt k Zv(P) = 0 gii c P.
5.3. Chn bin qu (thng chn uCq trong mch qu c t in C. Chn iLq trong mch qu c cun cm L) v t di dng xp chng: uCq = uCxl + uCtd (nu chn bin qu l uCq) hoc iLq = iLxl + iLtd (nu chn bin qu l iLq) . Trong uCtd = Ae; iLtd = Be
5.4. Tnh nghim xc lp mi
a. Vi xc lp mt chiu: lp phng trnh i s tnh UCxl mi; ILxl mi; ti t = 0 c uCxl mi(0) = UCxl mi; iLxl mi(0) = ILxl mib. Vi xc lp mi hnh sin: Dng s phc lp h phng trnh i s vi nh phc tnh Cxl mi ( suy ra uCxl mi(t), thay ti t = 0 c uCxl mi(0), tnh Lxl mi ( suy ra iLxl mi(t), thay ti t = 0 c iLxl mi(0)
5.5. Tnh hng s tch phn
a. Biu thc qu trnh qu
uC q(t) = uCxl mi(t) + A (*) (nu chn bin qu uC q)
hoc iL q(t) = iLxl mi(t) + B (**) (nu chn bin qu iL q)
b. Phng trnh tnh hng s tch phn
ti t = 0 c phng trnh tnh hng s tch phn A
uC(0) = uCxl mi(0) + A
Gii A = uC(0) - uCxl mi(0) vi uC(0) l s kin c lp tnh v uCxl mi(0) tnh phn tnh nghim xc lp mi.
Hoc ti t = 0 c phng trnh hng s tch phn B
iL(0) = iLxl mi(0) + BGii B = iL(0) - iLxl mi(0) vi iL(0) l s kin c lp tnh v iLxl mi(0) tnh phn tnh nghim xc lp mi.5.6. Thay hng s tch phn tnh vo biu thc c nghim qu
uC q(t) = uCxl mi(t) +[ uC(0) - uCxl mi(0)] (nu chn bin qu uC q)
hoc iL q(t) = iLxl mi(t) +[ iL(0) iLxl mi(0)] (nu chn bin qu iL q)BI TP IN HNH Bi 1: T bi ton qu sau khi ng kha K hnh (1)
Bit E= 6V = const, R = 5, r = 1, L = 100mH. Tnh s kin c lp iL(0)?Tnh s kin ph thucBi 2: T bi ton qu sau khi m kha K hnh 2
Bit E = 10V = const; R1 = 6; R2 = 4; C = 100
Tnh s kin c lp uC(0).
Bi 3: Cho mch in hnh 3
Sau khi ng kha K c qtq xy ra. Trc khi ng kha K mch in ch xc lp.
Bit E=8V = const; R1 = 4; R2 = 2; C = 100; L = 100mH. Tnh s kin c lp uC(0); iL(0)?Tnh s kin ph thucBi 4: Cho mch in hnh 4
Sau khi ng kha K c qtq xy ra trong mch. Bit trc khi ng K mch in trng thi xc lp.
Bit ; R1 = 10; L = ; R2 = 5.
Tnh s kin c lp iL(0)?Tnh s kin ph thucBi 5: Cho mch in hnh 5
Trc khi ng kha K mch in ch xc lp. Bit ; R1 = 10; C = ; R2 = 5. Tnh s kin c lp uC(0)?Tnh s kin ph thuc
Tnh pBi 6: Xc nh p trong mch qtq hnh 6
Bi 7: Tnh s m p trong bi ton qtq hnh 7
Bi 8: Cho mch in nh hnh 8
Bit ngun mt chiu E = 20V, R1 = 1(, R2 = 0,019k(, C = 20(F.
Trc khi ng kha K mch in ch xc lp, sau khi ng kha K thi gian nh c qtq xy ra trong mch.
Hy gii qtq xc nh cc dng in, in p qu trong mch, xc nh thi gian xy ra qtq.
Bi 9: Cho mch in nh hnh 9
Bit trc khi m K mch in ch xc lp.
Ngun in p hnh sin e(t) = 10sin5t, C = 2000(F, R1 = 40(, R2 = 60(.
Sau khi m kha K thi gian nh c qtq xy ra trong mch.
Tm in p qu trn in tr R1.
Bi 10: ng ngun in e(t) = V vo cun dy c in tr R = 10(, in cm L = 0,1H. Ti thi im t = 0 nh s hnh 10.
Hnh 10
Hnh 11
Sau khi ng K c qtq xy ra trong mch. Tnh cc in p, dng in qu trong mch hin hnh hnh 11
TI LIU THAM KHO
1. Nguyn Ngn Gio trnh c s k thut in 1 v 2. i hc Nng - 20032. Nguyn Qun Bi tp l thuyt mch in 1&2- i hc bch khoa Tp H Ch Minh, nm 2006Nguyn Bnh Thnh C S k thut in 1,2 Nh xut bn i hc v trung hc chuyn nghip H Ni, nm 2006.
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