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7/26/2019 4- Dao ng in T
1/10
VT L[3K] 2014 - 2015 Thy Lm Phong
Gieo hnh vi gt thi quen. Gieo thi quen gt tnh cch. Gieo tnh cch gt s phn. ( Dick Lyles) 1
U
uO
M'2
M2
M'1
M1
-U U0
01
-U1Sng Sng
Tt
Tt
TNG HP KIN THC V DNG IN XOAY CHIU
Bi 1: I CNG V DNG IN XOAY CHIU
Nguyn tc to ra dng in xoaychiu:da trn hin tng cm ng in t (L hin tng c s bin thin ca t trngqua mt khung dy kn th trong khung xut
hin mt sut in ng cm ng sinh ramt dng in cm ng )
Cch to ra dng in xoay chiu:Cho khung dy dn din tch S, c N vngdy, quay u vi tn s gc trong in
trng u c cm ng t B (B trc
quay) th trong mch c dng in bin thin iu havi tn s gc gi l DXC.
T thngc phng trnh: = NBScos(t + ) Wb (V-be) = N ocos(t + ) Wb
(Trong o= BS l t thng cc i qua mi vng dy, S l din tch ca khung quay, N l s vngdy qun vo khung quay, l gc hp gia php tuyn ca khung v cm ng t B)
Sut in ng trong khung dy: e = - ' = NBSsin(t + ) = E ocos(t + -2
)
(Trong E o= NBSl sut in ng cc i quacc cun dy)
Khi nim v gi tr hiu dng, gi tr tc thi, gi tr cc i:
i: dng in tc thi, I: gi tr hiu dng, I o: gi tr cc i. Tng t ta c: I =I o
2, U =
U o
2, E =
E o
2
( E l sut in ng dng cho ni pht sinh dng in, U l hiu in th ni tiu th dng in )
Cc biu thc in p v dng in xoay chiu:
+ Biu thc in p tc thi v dng in tc thi:
u= U ocos(t + u)V v i= Iocos(t + i) A
Vi = u il lch pha ca uso vi i, c2 2
Cng thc tnh thi gian n hunh quang sng trong mt chu k:
Khi t in p u= U ocos(t + u) vo hai u bng n,bit n ch sng ln khi u U1.
t =
4
vicos =
U 1
U o, (0 < ZChay
1
LC
> 0 th unhanh pha hn i
7/26/2019 4- Dao ng in T
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VT L[3K] 2014 - 2015 Thy Lm Phong
Gieo hnh vi gt thi quen. Gieo thi quen gt tnh cch. Gieo tnh cch gt s phn. ( Dick Lyles) 3
+ Khi ZL< ZChay1
LC < 0 th uchm pha hn i
+ Khi ZL= ZChay1
LC = 0 th ucng pha vi i.
Lc Max
UI =
Rgi l hin tng cng hng dng in
Bi 3:CNG SUT V H S CNG SUT CA MCH R - L - C
Cng sut tiu th trn mt mch in l: P = UICos = UIRZ
= U2RZ2
= RI2 .
( ty d kin bi m ta tnh cng sut ph hp, c th tm c gc = u- i)
Mch in ch tiu th cng sut khi c in tr R hoc r.
+ Nu mch ch c L hoc C th cng sut P = 0
H s cng sut: cos =PUI
=RZ
=U RU
c bit, nu mch c R-r-L-C th cos
= R + rZ
= UR+ U
r
U
+ Cng sut ph thuc vo cos, s dng hiu qu in nng tiu th th ta phi mc thm vomch nhng t in c in dung ln. Qui nh trong cc c s s dng in th cos 0,85.
Ch : nu mch in u,i c 2 thnh phn nh u = U 1+ U2cos(t) hay i = I
1+ I
2cos(t)th:
+ thnh phn U 1( hay I1) c xem l phn khng i ( dng in 1 chiu ),
+ thnh phn U 2( hay I2) c xem l thnh phn xoay chiu
c bit:+ nu mch l R-C th I 1v U
1khng tn ti do C khng dng in i qua P = RI
22.
+ nu mch l R-L th I 1v U1tn ti nhngZ
L= 0 i vi D1CP = RI
12 + RI
22 .
Bi 4: Cch S Dng Gin Vect Trt Gii Cc Bi Ton R-L-C v Hp en.PHNG PHP GIN VECT TRT( Gii ton in bng hnh hc )Chn ngang l trc dng in.(Chuyn do thy Chu Vn Bin bin son )Chn im u mch (A) lm gc.V ln lt cc vc-t biu din cc in p, ln lt t A sang B ni ui
nhau theo nguyn tc:+ L - ln.+ Cxung.+ Rngang.
di cc vc-t t l vi cc gi tr hiu dng tng ng. *Ni cc im trn gin c lin quan n d kin ca biton.* Biu din cc s liu ln gin .* Da vo cc h thc lng trong tam gic tm cc in p hoc gc cha
bit.
Gii thiu mt s gin thng dng.+ Gin R-L-C:Cho mch in gm cun cm thun c t cm L,
in tr thun R, t in c in dung C mc ni tip.
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VT L[3K] 2014 - 2015 Thy Lm Phong
Gieo hnh vi gt thi quen. Gieo thi quen gt tnh cch. Gieo tnh cch gt s phn. ( Dick Lyles) 4
+ Gin R-Lr:Cho mch in gm in tr thun R, cun dy c t cm L v in tr thun r mcni tip.
+ Gin Lr-R-C:Cho mch in gm cun dy khng thun cm, in tr thun R v t in c indung C mc ni tip.
+ Gin R-C-L:Cho mch in gm cun in tr thun R, t in c in dung C v cun dy c t cm L mc ni tip.
+ Gin R-C-Lr:Cho mch in gm cun in tr thun R, t in c in dung C v cun dykhng thun cm c in tr r mc ni tip.
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VT L[3K] 2014 - 2015 Thy Lm Phong
Gieo hnh vi gt thi quen. Gieo thi quen gt tnh cch. Gieo tnh cch gt s phn. ( Dick Lyles) 5
+ Gin C-R-Lr:Cho mch in gm t in c in dung C, in tr thun R v cun dy khngthun cm c in tr r mc ni tip.
+ Gin R-Lr-C:Cho mch in gm intr thun R v cun dy khng thun cm c in tr r vt in c in dung C mc ni tip.
+ Kinh nghimcho thy khi trong bi ton c lin quan n lch phahoc qu nhiu s liuth nngii bng phng php gin vc t s c li gii ngn gn hn gii bng phng php i s.
Bi 5A: BI TON CNG HNG CA MCH R - L - CKhi trong mch c hin tng Cng Hng in l khi: = u-
i= 0
Khi = 0 cos= 1 cos ln nhtv khi R = Z U R= UKhi = 0 tan= 0 Z L- Z C= 0 Z L= Z CU L= U C
Khi = 0 P max= UI =U2R
Khi Z L= ZCZ
min= R m I =
UZ
I max=UR
hay LC2 = 1=1LC
Khi u, i cng pha nhau( V gin vect th chng trng ln nhau )Khi ulch pha so vi U
Cmt gc 90
o .
V rt nhiu trng hp khc a v cc trng hp trn u c xem l Cng hng.
Bi 5B: Mch R - L - C - f - C Cc Phn T Thay i.Mch R Thay i.
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VT L[3K] 2014 - 2015 Thy Lm Phong
Gieo hnh vi gt thi quen. Gieo thi quen gt tnh cch. Gieo tnh cch gt s phn. ( Dick Lyles) 6
+ Chnh R = R
1
R = R2th
P1= P
2
I 1= I2R1+ R
2=
U2P
v R1R2= (Z
L- Z
C)
2 ( Tha mn PT Vi-et).
Khi tng ng ta c 1+ 2= 90
o hay sin
1= cos
2
+ Chnh R = Roth P
ton mchCc i Ro= |Z
L- Z
C| = R
1R
2( bng in tr nhm cn li).
Khi P max=U22Ro
( Vi Rong vi gi tr trn )
V H s cng sut cos=R
Z=R
R2 + (ZL- Z
C)
2=
R
R2 + R2
=1
2
+ Nu mch R-rL-C, chnh R = Roth P
trn RCc i Ro= r
2 + (Z
L- Z
C)
2
+ Nu mch R-rL-C, chnh R = Roth P
ton mchCc i Ro+ r = |Z
L- Z
C|
Mch L Thay i.
+ Chnh L = L
1
L = L 2th
P1= P
2
I 1= I2Z C=
Z L1+ Z
L2
2 v u=
i1+
i2
2
+ Chnh
L = L1
L = L 2 th
P1= P
2
I 1= I 2 v chnh L = L
3th U
Lmax2
L 3=
1
L 1+
1
L 2
+ Chnh L U LmaxZL=
R2 + ZC
2
Z C, U Lmax=
U
R R2 + Z
C
2 =
U
U RU R
2 + U
C
2
Khi U Lmax2 = U
2 + U
RC
2 U
Lmax
2 = U
2 + U
R
2 + U
C
2
U RCU tan
RC. tan= -1(1)U L.U
C= U
R
2 + U
C
2 (2)U
L
2 = U
R
2 + U
C
2 + U
2
(3)U L= U. 1 +U CU R
2
(4)1
U2+
1
U R2 + U
C
2
=1
U R2
+ Chnh L U RLmax2 2
42
C C
L
Z R ZZ th ax2 2
2 R
4RLM
C C
UU
R Z Z
(R-L mc ni tip)
+ Chnh L P max, I
max, U
Cmax, U
Rmax, v.v... ( Nhng phn t khc MAX ngoi U
Lmax)
CNG HNG ( Xem bi 5)
+ Mch L-RC c R v L thay i, Chnh L U RCkhng ph thuc vo R
khi U RC= U v ZL= 2Z
CLC
2 = 2
Mch C Thay i. ( tng t L )
+ Chnh C = C 1C = C 2
th P 1= P 2I 1= I
2
Z L=Z C1+ Z C2
2 v u=
i1+ i22
+ Chnh C = C
1
C = C 2th
P1= P
2
I 1= I2
v chnh C = C 3th U
CmaxC3=
1
2(C 1+ C
2)
+ Chnh C U CmaxZC=
R2 + ZL
2
Z L, U Cmax=
U
RR2 + Z
L
2 =
U
U RU R
2 + U
L
2
Khi U Cmax2 = U
2 + U
RL
2 U
Cmax
2 = U
2 + U
R
2 + U
L
2
U RLU tan
RL. tan= -1
+ Chnh C U RCmax2 2
42
L LC
Z R ZZ
th ax
2 22 R
4RCM
L L
UU
R Z Z
( R v C mc ni tip)
+ ChnhC P max, I
max, U
Lmax, U
Rmax, v.v... ( Nhng phn t khc MAX ngoi U
Cmax)
7/26/2019 4- Dao ng in T
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VT L[3K] 2014 - 2015 Thy Lm Phong
Gieo hnh vi gt thi quen. Gieo thi quen gt tnh cch. Gieo tnh cch gt s phn. ( Dick Lyles) 7
CNG HNG ( Xem bi 5)
+ Mch C-RL c C v R thay i, Chnh C U RLkhng ph thuc vo R
khi U RL= U v ZC= 2Z
LLC
2 = 0,5
Mch thay i. ( tng t vi tn s f )
+ Chnh
= 1
=
2
th
P1= P
2
I
1= I
2
, nu chnh = 3th I
maxhoc P
max32 =
1.
2
+ Chnh = 1= 2
th I 1= I2=
I maxn
(n > 1) th khi R =L( 1-
2)
n2 - 1 vi ( 1>
2)
Hoc R =| 1-
2|
C 12 n
2 - 1
+ Chnh U Lmaxth 2 =
2
2LC - R2C2v ax
2 2
2 .
4LM
U LU
R LC R C
=U
1 -Z C
2
Z L2
+ Chnh U
Cmaxth
2
=
2LC - R2C2
2L2C2 v ax 2 2
2 .
4CM
U LU
R LC R C
=
U
1 -Z L2Z C
2
+ Chnh= 1th U
Lmax
= 2th U
Cmax
= 3th U
Rmax
th 32 =
1
2.
+ Chnh = 1= 2
th U C1= U
C2.Nu chnh = 3th U
Cmax
32 =
12( 1
2 +
22)
+ Chnh = 1= 2
th U L1= U
L2. Nu chnh = 3th U
Lmax
1
32=
12(
1
1
2
+1
2
2)
+ Chnh = 1v = 2= n
1th mch tiu th cng h s cng sut ngha l cos
2= cos
1
vi L = CR2 . Khi : tan1=
1
2-
2
1=
f1f2
-f2f1
(cng thc ny ch p dng khi L = CR2)
Ch : nu c thm r = R hay L = CR2 = Cr2 th tan1=
12
1
2-
2
1
Khi tnh c tanta dng 1 + tan2 =1
cos2
+ t in p xoay chiu u = U 2cost (U khng i, thay i c) vo hai u mch c R, L ,C
mc ni tip. Khi = 1th cm khng v dung khng ca on mch ln lt l Z L1v Z C1. Khi = 2th
trong on mch xy rahin tng cng hng. H thc ng l: 1= 2
Z L1Z C1
Bi 6: DNG IN XOAY CHIU BA PHA Dng in xoay chiu ba phal h thng ba dng in xoay chiu, gy bi ba sut in ng xoay chiu
cng tn s, cng bin nhng lch pha tng i mt l2
3
( Da trn hin tng ng in t )
Cu to: Phn ngl ba cun dy ging nhau gn c nh trn mt ng trn tm O ti ba v tr tcch nhau mt gc 120o . Phn Cml mt nam chm c th quay quanh trc O vi tc gc khng i.
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VT L[3K] 2014 - 2015 Thy Lm Phong
Gieo hnh vi gt thi quen. Gieo thi quen gt tnh cch. Gieo tnh cch gt s phn. ( Dick Lyles) 8
1 0
2 0
3 0
os( )
2os( )
3
2os( )
3
e E c t
e E c t
e E c t
trong trng hp ti i xng th
1 0
2 0
3 0
os( )
2os( )
3
2os( )
3
i I c t
i I c t
i I c t
Bi 7: HIU SUT TRUYN TI IN NNG.
Cng sut tiu th: P = UIcosv P hao ph= RI2 P
hao ph = R
P2U2cos
2
U tng n ln th P hao ph gim n2Trong :P l cng sut truyn i ni cung cp
U l in p ni cung cpcosl h s cng sut ca dy ti in
lR
S
l in tr tng cng ca dy ti in (lu :dn in bng 2 dy)
+ gim in p trn ng dy ti in: U = IR
+ Hiu sut ti in: H =P - P
P.100 %
S l tit din trn ca dy.Do ta c S = r2 = d24
(d = 2r : l ng knh ca dy )
T cc mi quan h t l thun - nghch ta c:P 1
P 2=
U 2U 1
2
=S 2S 1
=
r 2r 1
2
trong P = 100 - H
Trong qu trnh truyn ti in i xa, gim in p hiu dngtrn ng dy ti in mt pha bng n
ln (n < 1)in p cui ng dy ny. Coi rng cng dng in lun cng pha vi in p. cngsut hao phtrn ng dy gim m ln (m > 1)nhng vn m bo cng sut tiu th nhn c khng
i.Cn phitng in p a vo truyn ti :n + mm(n +1)
Bi 8: MY BIN PHot ng: da trn hin tng Cm ng in T.Tc dng: bin i in p ( v cng dng in ) ca dng xoay chiu m vn gi nguyn tn s,khng c tc dng bin i nng lng.
Cng thcquan trng nht ca my bin p:1 1 2 1
2 2 1 2
U E I N
U E I N = k
+ Nu k > 1 N 1> N2U
1> U
2: My h p
+ Nu k < 1 N 1< N2U
1< U
2: My tng p
i vi bi ton ny khi thay i s vng dy cc cun s cp (N 1) hay thay i cun th cp (N2) u
nh hng n U 1v U2.
Hiu sut my bin p: H =P 2P 1
=U 2I
2cos
2
U 1I1cos
1
Bi 9 : NHNG LU KHI GII BI TP IN XOAY CHIU
Hai on mch AM gm R1L1C1ni tip v on mch MB gm R2L2C2ni tip mc ni tip vi nhau cUAB= UAM+ UMB uAB; uAMv uMBcng pha tanuAB= tanuAM= tanuMB
7/26/2019 4- Dao ng in T
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VT L[3K] 2014 - 2015 Thy Lm Phong
Gieo hnh vi gt thi quen. Gieo thi quen gt tnh cch. Gieo tnh cch gt s phn. ( Dick Lyles) 9
Hai on mch R1L1C1v R2L2C2cng uhoc cng ic pha lch nhau
+ Vi 1 11
1
tan L C
Z Z
R
v 2 22
2
tan L C
Z Z
R
(gi s 1> 2)
C 12= 1 21 2
tan tantan
1 tan tan
+ Trng hp c bit = /2 (vung pha nhau) th tan1tan2= -1.
Mch in xoay chiu mt pha cha cc phn t R-L-C+ = ocos(t + ) (vi
o= NBS : t thng cc i v l gc hp gia php tuyn n v cm ng t B)
+ c bit sut in ng to ra in p xoay chiu l e = - ' = NBS.cos(t + ) ( vi E o= NBS)
TH1: Mch ch c L: khi : I =E
Z L=
NBSL
=NBS
L= hng s
D c thay i tc quay n th no ? th I khng i.
TH2: Mch ch c C: khi I =E
Z C=
NBS1
C
= NBSC2
Nu tc quay tng n ln th I tng n2 ln
TH3: Mch gmL-R. v I =E
Z LR=
E o
2 R2 + ZL
2
=NBS
2 R2 + ZL
2
, t =NBS
2= const I =
R2 + Z
L
2
Mi lin h nm ch tc quay rto t l vi , t l vi ZLtheon 1n 2
=f1f2
=
1
2=
Z L1Z L2
=Z C2Z C1
( Cc trng hp cn li tng t)
Ni hai cc ca mt my pht in xoay chiu mt phavo hai u on mch ngoi RLC ni tip. Bqua in tr dy ni, coi t thng cc i gi qua cc cun dy ca my pht khng i. Khi rto ca my
pht quay vi tc n1 vng/phtv n2vng/pht th cng sut tiu th mch ngoi c cng mt gitr. Khi rto ca my pht quay vi tc n vng/pht th cng sut tiu th mch ngoi t cc i.
Khi :1
1
2+
1
22=
2
o
2
hay1
f12+
1f2
2
=2
fo2
hay1
n 12
+1
n 22
=2
n o2
Cho mch in xoay chiu gm cun cm (L,r) ni tip vi t in, c cm khng v dung khng ln lt lZLv ZC. Bit in p ga hai u cun dy vung pha vi hai in p hai u mch. H s cng sut mch
c tnh:Cos =Z LZ C
S Vung Phaca 2 thnh phn bt k trong mch R-L-C.
TH1:Nu mch ch c C. Khi u Ci i2I o
2+
u2U o
2
= 1
Chng minh: Gi s u = U ocost (1). Do i sm pha hn u 1 gc /2 i= /2
i = I ocos(t + /2) = - Iosint (2).
T (1) u2
U o2= cos2t ,
i2I o
2
= sin2t. Cng v ta ci2I o
2
+u2
U o2= 1.
TH2:Nu mch ch c L. Chng minh tng t ta ci2I o
2
+u2
U o2
= 1
TH3: Nu u RCu u
RC
2
U oRC
2
+ u
2
U o
2= 1
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VT L[3K] 2014 - 2015 Thy Lm Phong
Gieo hnh vi gt thi quen. Gieo thi quen gt tnh cch. Gieo tnh cch gt s phn. ( Dick Lyles)10
TH4: Nu u RLu u RL
2
U oRL2+
u2U o
2
= 1 ( C nh vy ta m rng ra cc trng hp c s vung gc )
Bi ton lin quan n lch pha vung gc:+ TH1: 1+
2= 90
o tan
1.tan
2= 1
+ TH2: 1- 2= 90
o tan
1.tan
2= -1
+ TH3: | 1| + |2| = 90
o tan
1.tan
2= 1
Cc xc nh phn ttrong bi ton hp enX.+ Mch in n gin:
a. Nu on NB cng pha vi i X cha Ro
b. Nu on NB sm pha vi i mt gc /2 X cha cun cm L.c. Nu on NB tr phavi i mt gc /2 X chat in C.
+ Mch in phc tp:a. Mch 1
NuU ABcng pha vii X chchaL
NuU
ANU
NBX chaR
Vy X cha R v L
b. Mch 2
NuU ABcng pha vii X chaCNuU ANU
NBX chaR
Vy X cha R v C
+ Mch inc cun dy cha bit thun cm hay cha thun cm ?Nu cun dy lch pha vi i 1 gc = /2 cun dy ch c L
Nu cun dy lch pha vi 1 gc vi 0 < < /2 Cun dy c L v r( C th dng gin vecto trt chng minh. Thng ta gi s L thun cm ri bin lun )
CHC CC EM N TP HIU QU V T KT QU CAO NHT TRONG KTHI TUYN SINH I HC 2014- 2015
[email protected] - [email protected]
R
C
XA N B
R
L C
XA N B
R
L
XA N B