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III. DNG IN XOAY CHIU 1. CNG THC TNH NHANH V MT S DNG TON:
1.1)VIT BIU IN P V CNG DNG IN: Phng php:
* Tnh tng tr Z: 2 2. ( )
( ) ( )1 ( )
.
L
L C
C
Z L
Z R Z Z viZ
C
* Tnh bin I0 hoc U0 bng nh lut m: 0
0 0 0.U
I U I ZZ
* Tnh lch pha ca u so vi i: u u i
i
Vi:
2 2
L CZ Z
tg
R
* Vit biu thc: + Nu cho:
0. ( . ) ( )ii I c t A os 0. os( . ) ( )u u iu U c t V vi
+ Nu cho
0. os( . ) ( )
uu U c t V
0. os( . ) ( )
i i ui I c t A vi
Ch :
+ Nu cun dy khng thun cm
2 2( ) ( )
( 0)L
L C
L
Z R RL ZL ZC
R th Z Ztg
R R
+ Nu on mch thiu phn t no th cho tr khng ca phn t bng 0
on mch
Tng tr 2 2
CR Z 2 2
LR Z L CZ Z
tg
CZ
R L
Z
R 2
2
+ Nu cho: 0. ( . ) ( )ii I c t A os
in p tc thi hai u in tr thun R:
0 0 0. os( . ) ( ) .
R R i Ru U c t V vi U I R
in p tc thi hai u cun thun cm:
0 0 0. os( . ) ( ) .
L L i L L Lu U c t V vi U I Z
Welcome PCNew Stamp
Welcome PCStamp
-
in p tc thi hai u t in:
0 0 0. os( . ) ( ) .
C C i C C Cu U c t V vi U I Z
1.2) XC NH GI TR CC PHN T R, L, C C TRONG ON MCH KHNG PHN NHNH:
Phng php: * Da vo cc d kin cho tnh gi tri tng tr Z ca on mch ang xt
ri s
dng cng thc 2 2( ) ( ) L CZ R Z Z . T suy
ra: , ,L CZ Z R cn tm.
D kin cho S dng cng thc Ch
lch pha
L C
Z Ztg
Rhoc os
Rc
Z Thng tnh
os
RZ
c
Cng sut P hoc nhit lng Q
2. . os . .
RP U I c U I I R
Thng s dng tnh
I:P
I
R
ri mi p dng
nh lut m tnh tng
trU
Z
I
Cng hiu dng v in p hiu dng
CR L XY
L C XY
UU U UI
R Z Z Z
Nu cho n d kin th ta s tm c ( 1)n d kin
Ch : C th s dng cng thc trc tip tnh:
*Cng sut ca dng in xoay chiu: 2
2
2. . os . . .
R
UP U I c U I I R R
Z
2 2
2 2 2. ( ) .
L C
U UZ R R Z Z R
P P
* H s cng sut os oac c h :
.os
os
RP U R R
c ZU I U Z c
2
2 2( )
os
L C
RR Z Z
c
* in p hiu dng hai u mi phn t in:
.
. ; . ; . vi I = .
.
R
R L L C C L
L
C
C
UZ R
U
U UU I R U I Z U I Z Z Z
Z U
UZ Z
U
-
22 2 2
2
2 2 2
2
2 2 2
( ) .
( ) .
( ) .
L C
R
L C L
L
L C C
C
UR Z Z R
U
UR Z Z Z
U
UR Z Z Z
U
Ch : Tt c cc cng thc sau khi c bin i nh trn ta c th a v gii
phng trnh bc 2 hoc a v dng 2 2A B gii. 1.3) MCH IN THAY I DO NG
NGT KHA K: * Hin tng on mch: Xt mt on mch c tng tr l XZ v mt dy ni
AB c in tr khng
ng k theo hnh bn. V in tr ca dy ni khng ng k nn: + in th ti A (
)AV gn bng in th ti B ( )BV : A BV V
+ Ton b dng in khng i qua phn t XZ m i qua dy ni AB.Hin tng
trn
gi l hin tng on mch * Kt qu: + Khi c hin tng on mch phn t no ta
ci th xem nh khng c( khuyt) phn t trong mch. + Ni(chp) hai im A, B
hai u dy ni ri v mch li. 1.4) XC NH CU TO(HOC GI TR CC PHN T) CA
MCH IN: (Bi ton hp kn X) Phng php: * Tnh cht ca mch in:
: u nhanh pha hn i 2
: u nhanh pha hn i mt gc v ngc li hay
mch c tnh cm khng.
: u chm pha hn i mt gc v ngc li hay mch c
tnh dung khng.
* Da vo lch pha ca u so vi i, ca 1u so vi 2u ri v gin vec-t.
T
phn t ca mch. C th:
-
+ Nu 0 th mch thun tr(ch c R) + Nu 02
th mch c tnh cm
khng( Phi c R,L).
+ Nu 02
th mch c tnh dung khng( Phi c R,C). + Nu
2
th mch
c L hoc L v C vi (ZL> ZC).
+ Nu 2
th mch c C hoc L v C vi (ZL< ZC)
1.5) QUAN H GIA CC GI TRI HIU DNG CA CC IN P (S o ca Vn- k):
Phng php:
Cch 1: * S dng cng thc: 2 2 2( )R L C
U U U U v
; osL C R
R
U U Utg c
U U
* Hoc s dng cc cng thc cho tng loi on mch:
V d:
2 2 2
2 2 2
2 2
2 2 2
(1)
(2)
( ) (3)
( ) (4)
RL R L
RC R L
LC L C
R L C
U U U
U U U
U U U
U U U U
Gii cc phng trnh trn tm ra
, , ..............R L CU U U hoacso ch cuaVon Ke
Cch 2: S dng gin vec-t Fresnel * V gin vec-t Fresnel v nn v theo
quy tc 3 im( V cc vec- t lin tip nhau)
* p dng nh l hm s cos(hoc sin) tnh cos ( sin )hoac
* Da vo h thc lng trong tam gic tnh , , , ......R L CU U U U
1.6) BI TON LIN QUAN N LCH PHA GIA CC IN P 1 2va uu :
Phng php:
* S dng cng thc lch pha gia hai in p 1 2va uu : 1 1 22
u u uu i i
Trong : 1
2
:
:
1
2
olech pha cua u so vi i
olech pha cua u so vi i
ui
ui
Ch :
C th dng phng php gin vec-t Fresnel gii dng ton trn.
-
Nu 1 2va uu lch pha nhau 2
hay1 1 2
22u u u
u i i
. Ta lun c:
1 2( ).( ) 1u u
i i
tg tg
V d: Xt on mch theo hnh bn. Bit lch pha ca
2vi laAN MBu so u
. Tm h thc lin h gia , ,L CR Z Z .
Hng dn: Ta c ( ).( ) 1 1AN MB
CLu u
ii
ZZtg tg
R R
Kt qu::(CTTN) 2 2.L CL
R Z Z hay RC
1.7) BI TON CC TR (cc i hoc cc tiu): Phng php: Cch 1: * Bin i
biu thc C cn tm cc tr v dng phn s
( )( )
C: bieu thc can tm cc tr
vi D: la ai lng hang so trong mach(thng la U hai au oan
mach)
la ham so vi bien so la ai lng b thay oi cua mach ie
DC
f XY f X
L C
n( Thng la R, Z , Z ,f)
T max min
min m
( )
( )ax
C f X
C f X
* Kho st cc tr ca hm s ( )Y f X .
Ch : Xt cc tr ca hm s ( )Y f X bng cc cch sau;
Hin tng cng hng mI ax khi L CZ Z
Dng bt ng thc Csi cho 2 s , 0A B A . Vi
min
2 . 2 .A B AB A B AB A B
Dng o hm kho st hm s ( )Y f X
Nu ( )Y f X c dng phng trnh bc 2 2( ) . .Y f X a X b X c
min( ) 0.b
o: X= -
2.a
Y f X a Khi
* Tnh nhanh mt s trng hp c th: a) Tm gi tr cc i ca cng sut tiu
th ca mch:
S dng cng thc: 2 2
2
22 2. .
( )( )
Uvi I=
ZL CL C
U UP R I R
Z ZR Z ZR
R
+ Khi L, C hoc f thay i(R khng i):
-
Kt qu:(CTTN) Khi L hoc C thay i th:
2
max ( ).(machxay raconghng he qua hien tng cong hng)L CU
P Z Z XemR
+ Khi R thay i: ( p dng bt ng thc Csi cho hai s dng
2
L CZ -Z
vaA R BR
Kt qu:(CTTN) Khi R thay i th:
2
max
2.
2. 2o : cos = hay =
4L C
UP R Z Z Khi
R
b) Tm ;ax ax ax
hoacR L Cm m mU U U khi R, L, C thay i trong on mch RLC:
Tm axR m
U khi R thay i: Ta c
2 2 2
2
. .( ) ( )
1
R
L C L C
U UU R I R
R Z Z Z Z
R
Kt qu:(CTTN)Khi R thay i th: axR L Cm
U U Z Z
Tm axL m
U khi L thay i:
Ta c: 2 2 2 2 2 2
2 2
. .( ) ( ) 2.
1
L L L
L C L C C C
L L L
U U UU Z I Z
R Z Z R Z Z R Z Z
Z Z Z
t: 2 2 2( ) ( ). 2 . 1C CY f X R Z X Z X . Vi: 1
C
XZ
Do C
onst ; R= const ; Z = constU c nn ta suy ra: min
( )axL m
U Y f X
Vi: 2 2 0; 2. ; 1C Ca R Z b Z c . Suy ra: min( )Y f X khi
2.b
Xa
2 22 2
1.C L C C
L C
ZZ Z R Z
Z R Z
. Khi : 2 2.
ax
U=
RL Cm
U R Z
Kt qu:(CTTN)Khi L thay i th:
2 2
2 2
.
.ax
U=
R
L C C
L Cm
Z Z R Z
U R Z
Tng t: (CTTN)Khi C thay i th:
2 2
2 2
.
.ax
U=
R
L C L
C Lm
Z Z R Z
U R Z
Cch 2: Dng gin vec-t quay
Xt on mch RLC theo hnh bn. nh C maxC
U . Tm
-
maxC
U
Hng dn:
Ta c: ; ; ; ;AB AN L CRL L CAB U AN U U MN U NB U MB U U
Vi: 2 2
sin onstR
AN L
U Rc
U R Z
.
p dng nh l hm s sin trong AMN :
2 2. .sinC C LU U U
U R Zsin sin R
( U = const)
Vy: 0max
sin 1 90 :2
lech pha viL RL ABU hay u so u
2 2max . .R
UL RC L
U UU U R Z
R
Khi : 1 2. 1 . 1
L CL Z ZZtg tgR R
Hay: 2 2.L C LZ Z R Z
BNG TM TT: i lng bin thin trong mch RLC
Gi tr cc tr cn tm
Mi lin h vi cc phn t cn li trong mch
Ch :
R axR m
U U L CZ Z Hin tng cng hng
R 2
max2.
UP
R L C
R Z Z 2
2cos = hay =
4
L hoc C
2
max ; 1axos mU
P cR
; L CZ Z Hin tng cng
hng
L 2 2.
ax
U=
RL Cm
U R Z 2 2.L C CZ Z R Z
2lech pha viRCu so u
C 2 2.
ax
U=
RC Lm
U R Z 2 2.L C LZ Z R Z
2lech pha viRLu so u
5.8) MT S DNG TON KHC:
1. Dng in xoay chiu i = I0cos(2f.t + i) * Mi giy i chiu 2f ln *
Nu pha ban u i = 0 hoc i = th ch giy u tin i chiu 2f-1 ln.
2. Cng thc tnh khong thi gian n hunh quang sng trong mt chu k
Khi t hiu in th u = U0sin(t + u) vo hai u bng n, bit n ch sng ln
khi u U1.
-
4t
Vi 1
0
osU
cU
, (0 < < /2)
3. MchRLC khng phn nhnh c C bin i. Khi C = C1 hoc C = C2 th UC c
cng gi tr th UCmax khi
1 2
1 21 1 1 1( )2 2C C C
C CC
Z Z Z
Khi C = C1 hoc C = C2 th cng sut P c cng gi tr th: 1 2
2.C C LZ Z Z
4. MchRLC khng phn nhnh c R bin i.. Khi R = R1 hoc R= R2 1 2( )R
R th
P c cng gi tr th: 2
1 2. ( )L CR R Z Z
5. Mch RLC c thay i:
Khi 1
LC th IMax URmax; PMax cn ULCMin Lu : L v C mc lin tip
nhau
Khi 2
1 1
2
C L R
C
th ax2 2
2 .
4LM
U LU
R LC R C
Khi 21
2
L R
L C th ax
2 2
2 .
4CM
U LU
R LC R C
Vi = 1 hoc = 2 th I hoc P hoc UR c cng mt gi tr th IMax hoc
PMax hoc URMax khi 1 2 tn s 1 2f f f
6. Hai on mch R1L1C1 v R2L2C2 cng u hoc cng i c pha lch nhau
Vi 1 111
L CZ Ztg
R
v 2 22
2
L CZ Ztg
R
(gi s 1 > 2)ffff
C 1 2 = 1 2
1 21
tg tgtg
tg tg
Trng hp c bit = /2 (vung
pha nhau) th tg1tg2 = -1. Ch 1 : I CNG V DAO NG XOAY CHIU
5.1.Khung dy dn quay u vi vn tc gc o quanh mt trc cc ng cm ng t.
T thng qua khung bin thin vi:
tn s gc > o tn s f > fo tn s gc = o tn s gc < o 5.2.T
thng gi qua mt khung dy c din tch S gm N vng dy quay u vi vn tc gc
quanh trc trong mt t trng u c vect cm ng t B , c biu thc = 0cos(t +
).Trong :
0 = NBS l gc hp bi php tuyn ca mt phng khung dy thi im t = 0 vi
vct cm ng t
n v ca l Wb (v-be) C A,B,C u ng
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5.3.Mt khung dy din tch 1cm2, gm 50 vng dy quay u vi vn tc 120
vng/pht quanh trc t trng u B = 0,4T. Khi t = 0, mt phng khung dy c
v tr vung gc cc ng cm ng t. Biu thc ca t thng gi qua khung:
= 0,02cos(4t + /2)(Wb) = 0,002cos(4t)(Wb) = 0,2cos(4t)(Wb) =
2cos(4t + /2)(Wb)
5.4.Khung dy dn quay u vi vn tc gc o quanh mt trc cc ng cm ng t.
S cm ng bin thin vi:
tn s gc > o tn s gc = o tn s gc < o Khng c c s kt lun
5.5.Khung dy dn quay u vi vn tc gc quanh mt trc cc ng cm ng t ca mt
t trng u. T thng cc i gi qua khung v sut in ng cc i trong khung lin
h nhau bi cng thc :
Eo = o/ 2 Eo = o/ Eo = o/ 2 Eo = o
5.6.Khung dy dn c din tch S gm N vng dy, quay u vi vn tc gc
quanh mt trc cc ng cm ng t ca mt t trng u. Sut in ng cm ng sinh ra
trong khung dy c biu thc e = E0cos(t + ). Trong :
E0 = NBS l gc hp bi vct php tuyn ca mt phng khung dy vi vct cm
ng t khi t = 0
n v ca e l vn (V) C A,B,C u ng 5.7.Dng in cm ng
xut hin trong cun dy dn kn trong thi gian c s bin thin ca cc ng
cm ng t qua tit din cun dy
xut hin trong cun dy dn kn khi c cc ng cm ng t gi qua tit din S
ca cun dy
cng ln khi din tch S ca cun dy cng nh tng khi t thng gi qua tit
din S ca cun dy tng v gim khi cc t thng
gi qua tit din S ca cun gim 5.8.Dng in cm ng s KHNG xut hin khi
mt khung dy kn chuyn ng trong mt t trng u sao cho mt phng khung
dy:
Song song vi cc ng cm ng t Vung gc vi cc ng cm ng t
To vi cc ng cm ng t 1gc 0 < < 90o C 3 cu u to c dng in cm
ng 5.9.Trong cun dy dn kn xut hin dng in xoay chiu khi s ng sc t
xuyn qua tit din S ca cun dy
Lun lun tng Lun lun gim Lun phin tng, gim Lun khng i 5.10 i vi
dng in xoay chiu cch pht biu no sau y l ng?
A. Trong cng nghip, c th dng dng in xoay chiu m in.
B. in lng chuyn qua mt tit din thng dy dn trong mt chu k bng
khng.
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C. in lng chuyn qua mt tit din thng dy dn trong khong thi gian
bt k
u bng khng.
D. Cng sut to nhit tc thi c gi tr cc i bng 2 ln cng sut ta nhit
trung
bnh.
5.11 Cng dng in trong mch khng phn nhnh c dng i=2 2
cos100t(A).
Cng dng in hiu dng trong mch l : A. I=4A B. I=2,83A
C. I=2A D. I=1,41A
5.12 Hiu in th gia hai u on mch c dng u=141cos100t(V). Hiu in
th
hiu dng gia hai u on mch l :
A. U=141V B. U=50Hz C. U=100V D. U=200V
5.13 Trong cc i lng c trng cho dng in xoay chiu sau y, i lng no
c
dng gi tr hiu dng :
A. Hiu in th B. Chu k C. Tn s D. Cng sut
5.14Trong cc i lng c trng cho dng in xoay chiu sau y, i lng
no
khng dng gi tr hiu dng :
A. Hiu in th B. Cng dng in C. Tn s
D. Cng sut
5.15 Pht biu no sau y l ng?
A. Khi nim cng dng in hiu dng c xy dng da vo tc dng ha hc
ca dng in.
B. Khi nim cng dng in hiu dng c xy dng da vo tc dng nhit ca
dng in.
C. Khi nim cng dng in hiu dng c xy dng da vo tc dng t ca
dng in.
D. Khi nim cng dng in hiu dng c xy dng da vo tc dng pht
quang ca dng in.
5.16.Dng in xoay chiu l dng in: i chiu lin tc theo thi gian m
cng bin thin iu ho theo thi
gian
m cng bin thin tun hon theo thi gian C A,B,C u ng 5.17 .Nguyn tc
to dng in xoay chiu da trn:
Hin tng quang in. Hin tng t cm. Hin tng cm ng in t. T trng quay
5.18 .Cc n ng dng dng in xoay chiu c tn s 50Hz s pht sng hoc tt
50 ln mi giy 25 ln mi giy 100 ln mi giy Sng u khng tt
5.19 .Chn cu tr li sai. Dng in xoay chiu:
