Kỹ năng phân loại và phương pháp giải chi tiết bài tập trắc nghiệm Vật lý 12 (Trọng tâm): Phần 2
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• N a n g lufong t i J c t h d i c i i a c u p n c a m ;
ChiTtfng
IV. DAO DQNG VA SONG D I | N Tlf
W L = - hi' = -L
2
B,.
P H A N
2
If, sin'(cot + ip )
w«,.= i L I - l L ^ , ;
de
• Nang l u p n g d i e n tii cua m a c h dao dpng:
1: D A O D O N G D I E N T L /
nang lUgng tit trildng luon luon chuyen hoa cho nhau. nhitng, tdng nang
N h a n x e t : Nang lifpng d i e n trUdng cua t u dien Wc va n a n g Itfcfng t i f
M a c h d a o d o n g L C g o m t u d i e n C d a diTcfc
t i c h d i e n , n o i vdi c u p n d a y c6 d o t U c a m L
B
A
D i p n tror R c u a c u o n
L
day k h o n g dang ke.
b) Moi quail he giUa q, i, u trong mach LC
T
,
.
.
trucmg cua cupn cam W L bien t h i e n tuan hoan v6i chu k y — , t a n so 2f, t a n so
2
goc 2 CO . Doi vdi nang liWng dien tii W l a mot h k n g so khong thay doi.
4. Dao dOng di$n tC/ t^t d i n
Trong c a c mach dao dpng thuc luon cd t i e u hao n a n g lupng ( d o R c h ^ n g
* q = qocos(u)t + cp)
han). V i vay, dao dpng se dCrng l a i sau k h i t i e u hao h e ' t n a n g lupng. H i e n
* V d i lo = qo-w
tiipng n a y g p i l a dao dpng dien tii tit d a n . Gia t r i R cang I d n t h i sU t ^ t
dan cang n h a n h .
* UAB = , — = — cos((ot + q))
5. Dao d6ng di$n tCf duy tri, h§ tU dao d6ng
'c c
D a t Uo = ^
Muon CO dao dpng dien tif khong t ^ t dan t a c 6 the duy t r i bhng each cung c a p
UAB = Uocos(o)t + (p).
them nang lupng cho mach de bo vao phan nang lupng tieu hao trong moi chu
ky. K h i viec cung c a p nang lupng tii p i n cho khung dao dpng L C dugrc duy t r i
Vay: Dien tich, hieu di^n thi' trin hoi ban tu di^n va cUang do dong dien
trong mach bien thien dieu hba vai ciing tan so goc Im =
T =
In
= 2Tt^/LC yd tan so f =
(0
o n d i n h t h i d a o dpng trong k h u n g L C dupc duy t r i o n d i n h v d i t a n so rieng Wo
, chu ky
c i i a mach. Ta gpi day l a h e t U d a o dpng.
6. Dao dong di^n tCl caang bQc. Su cpng hudng
271
a) Dao dong dien tii cUang bilc
VLC"
K h i t a m a c mach d a o dpng L C n o i t i e p v d i nguon dien ngoai c d hieu
N h a n x e t : Cirdng do dong dien t r o n g mach n h a n h p h a - s o v d i dien
3,
= ^hll = hhng so-.
liigng dien tii la khong ddi.
a) Mach dao dong
kin.
=^CUl
Vay: Trong qua trinh dao dong cua mach, nang lugng dien triCang va
1. Dao dOng di$n tC( trong mach LC
thanh mot mach
'
W L = WoLsin^(cot + cp)
I: T O M T A T L I T H U Y E T
W = Wc + W , = ^
Van
= i
dien t h e bien t h i e n theo t h d i gian u - Uo.cosw.t
ch^ng
han
t h i dong
t i c h va hi§u dien t h e giijra h a i b a n cifc cua t u d i e n , h i f u di?n t h e cung
dien t r o n g mach L C se k h o n g t h e d a o dpng theo t a n so g o c riehgco,, m a
pha dien t i c h cua t u dien.
p h a i bie'ii t h i e n theo t a n so goc w. Qua t r i n h n a y dupc gpi l a d a o dpng
Nang lUOng di^n
tQ trong mgch dao d^ng LC
• N a n g lifpng tufc thcfi ciia t u d i e n :
dien tii cUdng buTc.
•
b) Sii cong hudng
H i e n tifpng bien dp c i i a dao dpng d i | n t r o n g k h u n g d a t g i d t r i
Wc = - C u ' = - C
2
Dat
w,oc
2
cos2((ot + (p)
= lcu^ = i c 4 = 2C
^
Wc = Woccos^((ot + (p)
khi
(I)
C L T C
dai
= cOp.
Litu y: Khi mach dao dong c6 R Ian thi dinh cong hiiang thdp (bien do
nhd) va ngiigc Igi.
t f n g dung: D o n g t r o n g c a c mach Ipc, mach chpn song, mach khuech dai...
Van de 2: D I E N TL/ T R U C l N G
1. Dien trUdng bi^n thiSn
a) TU tritang
tCl trUdng bidn thiSn
bien thien:
Vain de 4: T R U Y E N T H O N G B A N G S O N G D I E N TL/
1 Nguyen tSc truyen thOng bSng s6ng dign tQ
So( do k h o i c i i a
PQii mot tiT trircfng bien t h i e n theo thcfi gian thi
thong phat thanh v a thu thanh
no s i n h ra mot d i e n trUdng xoay, tiJc l a dudng sufc ciia d i e n trUdng n a y
khep k i n va bao boc xung q u a n h duomg siJc tir.
b) Dien
truang
bien thien:
^' 6
K h i mot dien t r i / d n g b i e n t h i e n theo t h d i gian
t h i no s i n h ra mot tij' triTcfng xoay. Dudng sufc t i ^ ciia tCr trir6ng nay khep
7
k i n va bao boc xung quanh du&ng sure dien triTcfng.
8
9
[
2. Dien tC/ trUdng
• M o i bien t h i e n theo t h d i gian ciia t\i trUdng deu s i n h r a tr ong khong
gian xung quanh m o t d i e n truorng xoay bien t h i e n theo t h d i gian, va
nguoc l a i .
.
1. May p h a t dao dong cao t a n
6. A n t e n t h u .
2. Bien dien.
7. Chon song.
Dao dpng cao t a n .
8. Tach song.
K e t l u a n : D i e n trUdng b i e n t h i e n va tCr trUcfng bien t h i e n cung t o n t a i
4. Khuyech dai cao t a n .
9. Khuyech d a i a m t a n
trong k h o n g gian. Chung c6 the chuyen hoa I a n nhau tr ong mot truang
5. A n t e n p h a t .
10. Loa
t h o n g n h a t duoc goi l a d i e n tCf trUdng.
• TCr trUdng bien t h i e n c a n g n h a n h t h i cUcfng do dien trUcJng xoay c a n g Idn
va n g U O c l a i
2. SU truyen s6ng dien tQ quanh Tr^i Dit
- K h i t r u y e n song d i e n t\i t r o n g t h o n g t i n quanh T r a i DS't phu thuoc vao
cac yeu to n h u : Budc song, dieu k i e n m o i trUcfng t r e n m a t dat va t i n h
chat ciia bau k h i quyen.
V^ande 3; S O N G D I E N TL/
* Song dai va song trung;
1. S6ng di§n tQ 1^ gi?
Song dien t\i l a sU Ian t r u y e n cua d i e n ti^ t r i f d n g t r o n g k h o n g gian
2. Oac d i l m , tinh ch^t cua s6ng di§n ta
- Truyen duoc trong moi trUdng vat chat va trong ca chan khong vci budc song:
^ = J (c = 3.10
m/s); f: T a n so ciia song d i e n tCr (Hz).
- Song dien tii l a song ngang. T r o n g qua t r i n h t r u y e n song t a i mot diein
- Song nay b i p h a n xa d t a n g d i e n l i va c6 k h a nSng di vong quanh T r a i
Dat qua nhieu I a n p h a n xa giUa t a n g dien l i va m a t dat. NgU6i t a d i i n g
song nay t r o n g t r u y e n t h a n h va t r u y e n h i n h t r e n m a t dat.
- Song d a i i t b i nUdc hap t h u n e n d i i n g de t h o n g t i n dudi nUdc.
* S o n g n g S n : p h a n xa d t a n g dien l i , p h a n xa t r e n m a t dat n h i e u I a n , do
do t r u y e n dupc xa t r e n m a t dat.
* S o n g c i / c n g S n : Song nay cd nSng lupng I d n n h a t , va k h o n g b i t a n g
bat k y t r e n phuong t r u y e n , vectd E , vector B luon vuong g6c nhau va
dien l i p h a n xa va hap t h u nen t r u y e n t h a n g . Song nay dUprc iJng dung
vuong goc v d i phuong t r u y e n song.
de t h o n g t i n t r o n g cU l i v a i chuc k m hoac t r u y e n t h o n g qua ve
- Song dien tiT cung t u a n theo c a c d i n h luat p h a n x a , khiic x a va cung co
the giao thoa v d i nhau.
tinh
(thong t i n vu t r u ) .
- Qua t r i n h t r u y e n song dien tir t r o n g k h o n g gian mang theo nSng luong.
TRAC NGHIEM LI THUYET
'^^u 1. Su bien t h i e n ciia dong d i e n i t r o n g mot mach dao dpng l$ch pha nhU
j | the nao so v d i sU b i e n t h i e n ciia dien t i c h q ciia m o t ban tu dien
A. i
Cling
pha v d i q
C. i sdm pha — so v d i q
2
B. i lech pha n v d i q
D. i t r e pha — so v d i q.
2
Cau 2. Mot
CO d a o
mach
dong
dao
dien
dong
til
LC c6 dion trcf
tiX d o
vdi
bieu
thuan bhng khong.
thiJc
dien
tich
tren
K h i trong mac}^
hkn
tu
dien
• q = quCos(cut + cp) t h i g i a t r i c u c d a i c u a c i T d n g d o d o n g d i e n t r o n g m a c h
B. ^
V2
A. (0 .Qo
C.
D.
^
Qo
la
CO. q o .
C a u 3. Chu k i dao dong rieng cua dao dong dien tii t i i do trong mach dao don,,
LC (c6 dien trd thuan khong ddng ke) la
1
2n
„ „
1
D. T =
A. T = 2 H V L C
C. T =
B. T =
V
L
C •
V'LC
•j2nhC
^^'^'-'^
C a u 4. Tan so dao dong cua dien tH tu do ciia mach LC c6 di|n t r d thuan
khong dang ke la
A.f.
'
27:VLC
B. f=27iVLC
'
c. f = — V L C
271
D. f =
2n
•
VLC •
C a u 5. Mot mach dao dong dien tCr L C gom tu dien c6 dien dung C va cuon day
thuan cam c6 do t i i cam L . Biet day dan c6 dien t r d thuan khong dang ke va
trong mach c6 dao dong dien tH rieng. Goi Qo, U,) Ian luot la dien tich cuc
dai va hieu dien the cue dai cua tu dien, lo la cUdng dp dong dien cUc dm
trong mach. Bieu thufc nao sau day khong phai la b i l u thiJc tinh nSng lupnf,r
dien tCr trong mach?
Cu^
.
D. W = - C U ^
c. w = | L I ^
B. W =
A. W =
2
2C
Cau 6. Trong mach dao dong LC gom mot tu dien c6 dien dung C v^ cupn day thuan
cam CO dp t U cam L dang c6 dao dong dien tU tiT do v6i hieu dien the cUc dai giiia
hai ban cUc cua tu dien la UQ. Dong dien trong mach c6 gia t r i cUc dai
D. Io = U o i l .
LC
C
C a u 7. Trong mach dao dong LC c6 dien t r d thuan bang khong t h i
A. NSng lifpng tit trudng tap trung d cupn cam va bien thien tuan hoan VO'l
chu k i bang nijfa chu k i rieng cua mach.
B. NSng lupng dien trUdng tap trung d tu dien v^ bien thien vdi chu k i bang
chu k i dao dpng ri§ng cua mach.
C. Nang li^ong tCr trUdng tap trung d tu dien va bien thien vdi chu k i bang
niia chu ki dao dpng rieng ciia mach.
D. Tai moi thcfi diem t t h i tong cua nSng lircfng dien trudng va nSng lupng
trudng luon tang.
C a u 8. K h i noi ve dien tif triTdng, phat bieu nao sau day khong diing?
A. Dien tich diem dao dpng theo thdi gian sinh ra dien til trifdng trong
khong gian xung quanh no.
B . TCr trudng bien thien theo thdi gian sinh ra dien tii trUdng xoay.
C. Dien tCr trudng Ian truyen trong chan khong vdi van toe bang van toe aiil^
sang trong chan khong.
D. Dien trudng bien thien theo thdi gian sinh ra tif tri/dng xoay.
9. Gpi (I): Giao thoa song di#n tCr.
): Cong hu'dng dao dong dien tit.
(III) : Phan xa song dien tir.
(IV) : Khiic xa song dien tii.
Mach chon song trong may thu song v6 tuyen dien hoat dong dira tren hien tuong
A. ( I )
B. (II)
C. ( I l l )
D. (IV).
QSiU 10. Mot mach dao dong dien til LC g6m cuon day thuan cam c6 dp t u cam
L khong doi va tu dien c6 dien dung C thay doi duoc. Biet dien t r d cua day
dSn la khong dang ke va trong mach c6 dao dong dien tit rieng. K h i dien
^ ^ f c dung CO gia t r i Ci thi chu k i dao dong rieng ciia mach la Tj. K h i dien dung
WBr'c<3 gia t r i C2 = I6C1 t h i chu k i dao dong dien tU rieng trong mach la
A. T2 = 16T,
B. T2 = 8T1
C. T2 = Ti/4
D. T2 - 4 T i .
Cau 11- Phat bieu nao sau day sai k h i noi ve dien til trudng?
A. Khi mot dien trUdng bien thien theo thdi gian, no sinh ra mot tU trUdng xoay.
B. Dien trudng xoay la dien trUdng giCra hai ban tu tich dien trai dau dat
song song each nhau mot doan d.
C. Khi mot til trucfng bien thien theo thdi gian, no sinh ra mot dien trudng xoay.
D. Dien trudng xoay la dien trudng ed dudng siie la nhCing dudng eong k i n .
Cau 12. Phat bieu nao sau day sai khi noi ve nang lupng cua mach dao dpng
dien tCr LC c6 dien trd thuan khong dang ke?
A. Nang lupng dien til cua mach dao dpng hkng n&ng lUPng dien trUdng cue
dai d tu dien.
B. N a n g lupng dien td cua mach dao dpng hkng n&ng iMng til trUdng cUc dai
d cupn cam.
C. Tai mpi thdi diem thi nSng lu'ptng dien til cua mach dao dpng hkng tong nang
luang dien trudng ciia tu dien va nang lu'png tCf trudng cua cupn cam.
D. Nang lupng dien trUdng va nang luprng tU trudng khong bien thien theo
thdi gian.
Cau 13. Mot mach dao dpng dien tU LC gom tu dien ed dien dung C va cuon
day thuan cam ed dp t u cam L. Biet dien t r d ciia day dan khong dang ke va
trong mach ed dao dpng dien tU rieng. Nang lupng dien tU trong mach
A. bien thien tuan hoan
B. bien thien nhung khong dieu hda
C. khong doi theo thdi gian
D. bien thien nhUng khong tuan hoan.
CSu 14. Mot mach dao dpng dien tCr LC, cd di$n t r d thuan khong dang ke.
Dien ap giufa hai ban tu dien bien thien dieu hda theo thdi gian vdi tan so f.
Phat bieu nao sau day sai?
A. N a n g lUPng dien tCr bien thien tuan hoan vdi chu k i T
B. N a n g lupng tU trudng bien thien tuan hoan vdi tan so 2f
!j C. N a n g lupng dien tU bkng nang luprng dien trUdng eUc dai
f D. Nang lupng dien tU b l n g nang lUPng tU trUdng cue dai.
15. Nang lupng dien trUdng trong tu dien ciia mot mach dao dpng bien
thien nhu the nao theo thdi gian?
1 A. Bien thien dieu hda theo thdi gian vdi chu k i 2T
1
C. Song d i $ n til cQng b i p h d n xa va khuc xa k h i gSp m a t phSn cdch giCa h a i
m o i truomg
B. Big'n t h i e n dieu hoa theo t h 6 i gian v d i chu k i T
T
C. B i e n t h i e n t u a n hoan vdi chu k i —
D. Song dien t i f t r u y e n dupc t r o n g rhi,
long, k h i va ke ca chan k h o n g .
fiu 23. K h i n d i vi song d i e n tCf, p h a t bieu nao dudi day k h o n g diing?
D. K l i o n g b i e n t h i e n dieu hoa theo t h d i gian.
A. S6ng d i e n tCr cung b i p h a n xa v a khiic xa k h i gap m a t p h a n each giCa h a i
m o i trUdng
C a u 16. T r o n g mot mach dao dong LC gom cupn day t h u a n cam c6 dp t\J cam L
k h o n g doi va t u dien c6 d i e n dung C t h a y doi dupc. Chpn cau d i i n g
B. Song dien til c h i t r u y e n dUpe t r o n g m o i triTcfng v a t cha't d a n h o i m a
k h o n g t r u y e n dUpe t r o n g chan k h o n g
A. Chu k i dao dong r i e n g cua mach k h o n g doi k h i dien dung C cua t u dioi,
thay doi
B. Chu k i dao dong r i e n g cua mach g i a m k h i tSng d i $ n dung C cua t u dien
C. Song dien tCr la song ngang.
C. T a n so dao dong r i e n g cua mach tSng k h i tSng dien dung C cua t u dien
D. Song dien tii Ian t r u y e n t r o n g chan k h o n g v d i v a n toe e = 3. lO^m/s.
D. T a n so dao dong r i e n g ciia mach tSng gap doi k h i dien dung C cua tn
C a u 24. Song d i e n til va song eP hoc k h o n g cd ehung t i n h ehat nao sau day?
dien g i a m gia t r i dien dung cua t u dien d i 4 Ian
C a u 17. M o t cuon day t h u a n cam (cam t h u a n ) c6 dp tU cam L m ^ c n o i t i e p vdi
m o t t u dien c6 dien dung C t h a n h m ^ t mach dao dong (con gpi la mach dao
dong LC). T a n so dao dong d i e n til tiT do cua mach nay phu thupc vao
C. d i ^ n dung C va dp tU cam L ciia mach dao dong
C. K h o n g t r u y e n d U P c t r o n g chan k h o n g
C. Song t r u n g
D. Song dai.
A. Song eire n g ^ n
B. Song n g ^ n
C. Song t r u n g
D. Song d a i .
A. T r o n g song dien tCr t h i dao dong ciia dien t r i f d n g va ciia til trWdng t a i m o t
d i e m l u o n luon ddng pha v d i nhau.
ca h a i vectct E , B luon vuong goc v d i phudng t r u y e n song.
B. E va B luon vuong goc v d i nhau v a t r i i n g v d i phiTong t r u y e n sdng.
C a u 19. K h i p h a t bieu ve song dien tU, p h a t bieu nao diTdi day k h o n g dung?
A. Song cue n g ^ n t r u y e n dupe t r o n g chan k h o n g .
C. Cac sdng n g ^ n k h o n g the t r u y e n d i xa t r e n m a t da't.
B. Song ngin c6 t a n so' nho h o n t a n so song d a i
C. Song cue n g a n dupe dCmg t r o n g t h o n g t i n vu t r u
D. Song d a i dupc d i i n g de t h o n g t i n dudi nude
A. Song d i e n tii l a song dpc
B. Song d i e n tir Ian t r u y e n t r o n g chan k h o n g v d i v a n toe e = 3. lO^m/s.
B. Song n g ^ n
C a u 27. Chpn eau d u n g
D . La song ma t r o n g qua t r i n h t r u y e n song t h i E luon vuong goo v d i B v;i
C a u 22. K h i n d i ve song d i ^ n ti^, p h d t bieu nao sau day sai?
A. Song cue n g ^ n
dat la l o a i song v6 t u y e n nao?
C a u 18. Song dien til
A. Chi Ian t r u y e n t r o n g m o i triTdng r ^ n , long, k h i v d i v f i n toe 3. lOm/s.
B. Lk song doc
C a u 21. Song dien tii
A. L a song dpc
' B. vera c6 song ngang, vCra c6 song dpc
C. Chi phan xa ma k h o n g khiie xa k h i gap mat phan each giijfa h a i m 6 i triTdng
D. M a n g nSng lupng.
D. M a n g n a n g li/png.'
C a u 26. Song dien tU dCing t r o n g t h o n g t i n giu-a cac n h a du h a n h vu t r u v a m a t
D. hieu dien the cUc d a i giiJa h a i b a n t u dien ciia mach dao dong.
D. Song d i e n tiT k h o n g t r u y e n di/pe t r o n g chan k h o n g
C. Qiao thoa, n h i l u xa
tuyen nao?
B. dien t i c h cUc d a i cua b a n t u d i e n t r o n g mach dao dong
B. Song d i e n tCr ehi t r u y e n dupc t r o n g chat k h i
C. Song d i e n tii k h o n g t r u y e n d i f p c t r o n g chat r d n
B. T r u y e n dupe t r o n g chan k h o n g
C a u 25. Song d i e n tCr d i i n g t r o n g . t h o n g t i n giOa cac t a u n g a m l a l o a i song v6
A. dong dien cUc dai chay t r o n g cupn day ciia mach dao dong
C a u 20. Phat bieu n^o sau day dung k h i n d i ve song d i ^ n tCr?
A. Song dien tCf m a n g nftng lupng.
A. Phan xa
D. Cac song ngan v6 tuyen khong phan xa tot tren tang dien l i v a t r e n m a t da't.
C a u 28. Phat bieu nao sau day la dung k h i n o i v a cac l o a i sdng v6 tuyen?
A. Song d a i ehii yeu dupe d i i n g de t h o n g t i n dudi nuTdc.
,
I
B. Song t r u n g c6 the t r u y e n dupe ra't xa vao ban ngay.
C. Song n g a n cd n a n g lu'png nho h p n song t r u n g v a sdng d a i .
D. Sdng cue n g ^ n b i t a n g d i e n l i v a m a t da't p h a n xa.
C a u 29. Chpn p h a t bieu dung
A. Sdng d a i cd n a n g lirpng t h a p i t b i niTdc ha'p t h u .
B. Sdng t r u n g p h a n xa dupe t r e n t a n g dien l i v a o ban dem n e n ehung t r u y e n
dupe xa.
C. Sdng cue n g ^ n k h o a n g bo p h a n xa hoae ha'p t h u t r e n t a n g d i ? n l i .
D. T a t ca deu dung.
P H A N
II:B A I T A P T R A C
N G H I E M
BAI T A P M A U
+ BAI T A P L U Y E N T A P
B a i 1. Mach dao dong gom t u dien c6 dien dung 400 p F va m o t cuon cam c6
do tir cam 0,04 H . T i n h chu k y dao dong r i e n g v ^ t a n so r i e n g ciia mach
dao dong.
I . D a o d p n g d i ^ n txJf
* D i e n t i c h tuTc t h d i q - qocos(a)t + cp)
* Dong d i e n tufc t h d i i = Iocos((ot + 9+ ^ ) v
* Hieu dien t h e tufc thcfi u =
1
T =— =
CO " f
LiCu
•
•
•
•
,
T = 27tv^
• W , , , e n : Nang lufcJng d i e n (J)
• Wtc,: N a n g liTong tCf (J)
I m F r: l O - ' F
l u F = lO-'F
to =
f =
* Nang
y:
L : He so tir cam (H).
C: D i e n dung cua t u d i e n (F)
ZL: Cam k h a n g cuon day ( Q )
W: Nang lirong d i e n i\l (J)
X: Budc s(5ng (m)
W
= W j . e n + Wtc
= Wj,^„
max -
• Chu k y dao dong r i e n g cua mach dao dong
= 400.10"'- (F)
T = 27IX/LC
. L = 0,04 ( H )
Tim T = ?
f =?
T = 2,512.10 '(s)
• T a n so r i e n g ciia mach dao dong
f = i
T
=
1
—
2,512.10
Tom tat
• I„
Wti, m a x
f * 0.04.10'(Hz)
Hiidng
ddn gidi
D i e n t i c h cifc d a i cua t u dien
600 m A
T - 2n. S i
I,.
= 600.10-^A
^
q,, =
T i m q„ r: ?
w
-
B a i 2. Mach dao dong L C c6 chu k y dao dong r i e n g cua mach l a 0,004 (s),
cudng do dong dien ciTc d a i qua mach l a 600 mA. T i m d i e n t i c h cifc d a i ciia
tu dien.
• T = 0,004 (s)
mach dao dgng
Hit&ng ddn gidi
= 400 p F
. C
I n F = lO-'F
I p F = lO'^^F.
27rVLC
litgrng trong
Tom tat
T . 12.
L - 0,004.600.10 '
2n
2.3,14
q^ = 3 , 8 2 . 1 0 - ^ ( 0 .
B a i 3. Cucmg do dong dien trong mach dao dong c6 dang: i = Iocos2000t A. Tu
dien c6 dien dung la 4t.iF. T i m he so tif cam ciia cuon day thuan cam.
' di§n m a x —
I I . S o n g d i g n tH
A =
Tom tat
C.T
Hitdng
ddn gidi
V a n toe s6ng d i e n tii t r o n g chan khong: |c - 3.10" (m/s)|
i = Incos2000t A
TCr bieu thiJc cUdng do dong dien qua mach dao dong
Chii y:
• Mach dao dong c6 t a n so goc co, t a n so' f va chu k y T t h i nSng lucfng dien
C = 4MF = 4 . 1 0 " ^
Ta CO u) = 2000 rad/s
Ci nt C2 thi:
C, //
thi:
1
1
1
Tf ^ T| '
1 _ 1 ^ 1
Tim L
ma
1
CL) =
VLC
1
co^C
200014.10'^
L = 0,0625(H)
= Tf + T^' ,
B a i 4. Mach dao dong L C c6 L = 400 m H va C = 4 nF. Biet hieu dien thg' cUc d a i
.
giOra hai ban cUc ciia tu la 400 V . T i m circfng do dong dien circ dai qua mach.
Tom
HU&ng dan
tat
. L = 400 mH
= 400.10'^H
. C = nF = 4.10"'' F
. Vo = 400V
Tim lo = ?
Ta c6:
CU
= W,
^ I o = Uo -
=400
VL
IJai 8. Mach dao dong LC ciia mot mdy thu v6 tuyen gom ( u u n d ; i v i l u i a u
cam CO L = 1 mH, tu dien c6 dien dung C = 4 PF. Tim b U c J c song ma mach
thu duoc.
'
gidi
4.10"
400.10"
I„ = 0,04(A)
B a i 5. Mach dao dong LC c6 dien dung C = 200 pF. Hieu dien the cUc dai hai
ban cifc cua tu la 200 V. Tim nang lucfng dien tCf cua mach.
Tom tat
• C = 200 pF
- 200.10-^^ (F)
• Uo = 200 V
Tim W = ?
Hitcfng ddn gidi
Ta
CO
CU^
W = 4.10-''(J)
• - = 10"*' (s)
4
Tim C = ?
Hii&ng ddn gidi
Ta c6: Khoang t h d i g i a n giiJa h a i Ian l i e n t i e p ma
nSng lu'Ong t i i trUomg b a n g nftng lu'Ong dien
T
tru'6ng l a —.
nen: - = 10"" (s) -> T = 4.10"'' (s)
4
ma T = 2;: N/LC ^ C =
rj,2
(4.10^6^2
47i'L
4.3,1410,02
C«2,02.10-"(F)
B a i 7. Mach dao dong gom cuon day c6 di?n trd r, he so i\i cam 32 mH va
mot tu C = 0,2 |.iF. De duy t r i mot hieu dien the cu'c dai la 4 V tren tu
dien t h i ta phai cung cap cho mach mot cong suat trung binh la 50 ^iW.
Tim dien t r d cuon day?
Tom tdt
= 32 mH
= 32.10"^ H
• C = 0,2 laF
= 0,2.10-** F
. Uo = 4 V
. P = 50 p,W
= SO.IO'^W
Tim r = ?
. L
Hiidng ddn gidi
Cirdng do dong dien cifc dai
nen
I =
_ 0,01
V2 ~ V2
(1) (2)
K \
C
: c
—
os
L
C := 200 nF, L = 2 mH, 'g-.. 2 V
Tai thdi diem t = 0, khoa K chuyen tif v i t r i (1) sang (2). Lap bieu thiJc
bieu dien sir phu thupc cua:
a) dien tich tren tu dien C vao thdri gian t.
b) dong dien qua mach vao thdi gian t.
Tom tdt
• C = 200 nF
= 200.10"' F
• L = 2 mH
= 2.10"^ H
. . ^ = 2 V = Uo
a) Lap q (t)
b) Lap i (t)
HUdng ddn gidi
a) Bieu thii-c bieu dien sii phu thuoc cua di?n tich
tren tu dien C vao thdi gian t.
q = qo.cos(cot + (p)
1
1
= 5.10^ rad/s
CO =
VLC
^2.10^200.10"'
* qo = C.Uo = 200.10"^2 = 4.10"^ (C)
(2) nen
* (p = ? Do t = 0 khoa K chuyen tCr (1)
^ q = qo nen ta c6:
q - qocos(a)t + cp) ^ qp = qocos(5.10''.0 + (p) o coscp = 1 <-> (p = 0
Vay: q = 4.10"'.cos(5.10't)(C)
b) Bieu thiJc bieu dien sir phu thuoc ciia dong dien qua mach v^o thcfi gian t.
i = q'(t) - I4.10-'cos(5.10''t)l'
^ i ^ -4.10"^5.10'^sin(5.10''t) <^ i = +0,02sin(5.10' + TT)
(A)
P =r.I^-.r=^ =
T = 2%sIhC = 27rv'lO '.4.10
a. 3,97.lO"'' (s)
Budc song ma mach thu duoc la
>t = c.T = 3.10^3,97.10^^ - I » 119,l(m)
Bai 9. Cho mach dien nhuf hinh ve
B a i 6. Mach dao dong L C c6 L = 0,02 H va tu dien c6 dien dung C. Biet cu
sau khoang thdi gian la 10"® (s) t h i nang lu'Ong tii trirdng b^ng nSng luong
dien trifdng. Tim dien dung C ciia tu dien.
Tom tat
. L = 0,02 H
Hiidng ddn gidi
Chu ky dao dong rieng ciia mach
Tom tdt
. L = 1 mH
= I.IO"'' H
. C = 4 pF
^ 4.10"'^ F
Tim 1 = ?
hay
50.10"
0,01
V2 j
r = 1(Q)
i = 0,02.cos 5.10H + - (A)
2
Cau
BAI T A P T R A C N G H I E M
11. Mach dao dong c6 tu C dUgre tich dien q = 2.10"^sinl007tt(C). Bieu thiJc
cUdng do dong dien qua mach la
A. i = 10.10" sinlOOTtt ( A )
C a u 1. Mach dao dong gom L =
( m H ) , tu C =
200 rad/s
B. 300 rad/s
C. 444 rad/s
i = 20071.10 ".sini 1007rt + - (A)
D. i = 40071.10 •'sinl007it ( A )
V
2j
C a u 12. M a c h dao dong L C c6 dong dien i = 40071.10'".sin(1007it + 7 i ) ( A ) . D i e n
D . 500 rad/s
tich ciia t u bien ddi theo t h d i gian c6 dang
C a u 2. M a c h dao dong L C c6 C = 2 0 0 ( n F ) . T a n so dao dong ciia mach
A . q = 2.10"^sin lOTit -
500Hz. Do tu cam ciia cuon cam t r o n g m a c h dao dpng l a
A. 0,507 H
B. 0,607 H
C. 0,707 H
D. 0,807 H
C. SO.IO^'^ F
A. 1 2 , 5 6 . 1 0 ' s
D. 1,54.10'F
C a u 4. M a c h dao dong L C c6 L = 2 0 0 0 ( n H ) , C = 3000(pF). Chu k y dao
dono
C. 1 , 4 . 1 0 ' s
A. 3 , 6 . 1 0 " F
D . 1,53.10"'s
C a u 5. M a c h dao dong L C c6 L = 5 0 0 0 ( n H ) , chu k y dao dong r i e n g ciia mach hi
B. 0,03 F
D.4.10"' s
C. 0,04 F
A. 0,12 M J
D . 0,05 F
D . 8.10"" C
B . 0,24 M J
C . 0,36
D. 0,48 ^lJ
dai giuTa 2 b a n t u la 3 0 0 ( V ) . T i m eUdng dp dong dien eUc d a i qua m a c h
C. 0,05 H
A.*0,42A
D. 0,04 H
C a u 7. M a c h dao dong ciia m p t m a y t h u v6 tuyen c6 L = 5(f.iH) va C = l , 6 ( n F )
C. 168,5 m
B.»0,5A
C.aO,65A
D.«0,8A
C a u 17. Mach dao dong L C c6 L = 500(|.iH), hieu dien the cUc d a i giCfa 2 ban t u la
U„ = lOO(V), cirdng dp dong dien ciTc dai I„ = 2 ( A ) . Dien dung cua t u dien la
M a y CO the t h u du'oc song v6 t u y e n c6 bu'dc song la
B. 100 m
C. 4.10"'C
C a u 16. M a c h dao dpng L C c6 L = 2 0 0 { m H ) , C = 4 0 0 ( n F ) , hieu d i $ n the ciTc
C = 5|.iF . Do tiT cam cupn day 1^
B. 0,03 H
B. 2.10" F
ban t u la 2 0 0 ( V ) . N a n g lifong dien tir cua mach l a
C a u 6. Cirdng do dong d i e n t r o n g mach dao dong l a : i = I „ s i n 2 0 0 0 t ( A ) . T u dien
A. 150 m
B.2.10"^ s
C a u 15. M a c h dao dong L C c6 dien dung C = 6 ( p F ) , hieu dien t h e cure d a i d 2
2.10"^(s). D i e n dung ciia t u la
A. 0,02 H
B. 24,5.10^ s
cue d a i ciia t u dien la 1 2 ( n C ) . D i e n dung ciia t u la
B. 10"' s
A. 0,02 F
(C)
C a u 14. M a c h dao dong c6 n&ng li/prng dien tii ciia mach la 2 ( m J ) , dien t i c h
r i e n g ciia mach la
A. 1 0 ' s
B.q =4.10-'-'.sin lOOTtt + 2
ciia t u la 4000(nC). Chu k y dao dong cua mach la
dong la 60Hz. D i e n dung ciia t u la
B. 2 0 . 1 0 ' F
(C)
C. q = 47t.l0"^sin 1007lt + - (C)
D. q = 87i.lO"'sin(1007tt) (C)
2
C a u 13. Cho dong d i e n qua m a c h c6 dang i = 2.10 l c o s a ) t ( A ) . D i e n t i c h cifc d a i
C a u 3. M a c h dao dong L C c6 L = 0,04(H), t a n so dao dong r i e n g ciia m a c h dao
A . 1,76.10" F
B. i = 2007t.l0 "sinlOOTtt (A)
C
n
71
A.
( F ) . T a n so goc cua mach la
A. 0,2 m F
D . 190 m
B . 0,2.10'"' F
C. 4 nF
D . 8 nF
C a u 8. M a c h dao dong ciia m o t m a y t h u v6 tuyen g o m L = 8(|.iH) va t u dien
C a u 18. Mach dao dpng L C c6 dien tich cUc dai h a i ban tu la q,, = 3 ( n C ) , dien dung
bien t h i e n tCr 20(pF) den 4 5 0 ( p F ) . H o i m a y c6 t h e b a t duofc cac song vo
C = 2(|iF). T i m nSng lu'png dien tnrdng ciia t u dien va nang lUcfng tir trircJng cua
tuyen dien t r o n g day song nao?
A. 200 m <
>L
< 500 m
C. 753,6 m < ?i < 3574,6 m
cupn cam k h i nang lirpmg dien trUcfng bang 3 Ian nang lUcfng tiT trircfng.
B. 700 m <
< 900 m
D . 23,8 m < ^ .< 113,04 m .
,,
A. 1.6875.10"''J; 2,65.10"" J
6 . 1 , 5 . 1 0 " " J ; 5,625.10"'"J
C. 1,6875.10 " J ; 5,625.10 ' ' J
D . 1,5.10"''J; 1,6875.10""J
C a u 9. Song dien tii c6 biTdc song 100 (m) t h i t a n so la
A. 3 M H z
B. 4 M H z
C. 3,5 M H z
D. 20 M H z
BAI TAP LUYEN TAP
C a u 10. M a c h dao dong L C l i tucfng c6 L k h o n g d d i . K h i t u d i e n c6 d i e n dung
Ci t h i t a n so m a c h la i^ = 5 0 ( M H z ) . K h i thay t u C^ b ^ n g
m a c h la l O O ( M H z ) . N e u diing C, n d i t i e p
A. 100 M H z
B. 50 M H z
t h i t a n so
t h i t a n so' ciia mach la
C. 120 M H z
D. 111,8 M H z
i 1. M a c h dao dpng cua m o t m a y t h u v6 tuyen c6 L = 3 2 ( n H ) . H o i de b ^ t
dupe song 120(m) t h i t u c6 d i e n dung bao nhieu?
Bap so: 0,126 n F
B a i 2. T r o n g m a c h dao d p n g c u a m o t m d y t h u v 6 t u y e n c6 d i # n d u n g b i e n doj
tCr 5 6 ( p F )
den
667(pF).
t u c a m t r o n g gidri b a n
D d p so:
M u o n b a t diTtfc s o n g t i r 4 0 ( m )
den
2 6 0 0 ( m ) t h i dg
D a p so: 5 , 6 5 6 . 1 0 ' * C
n^o?
p a i 13- M a c h d a o d p n g L C co C = 2 ) . i F . N g i T d i t a t h a y r & n g cur s a u t h d i g i a n
8,04|iH < L < 2 , 8 5 m H
B a i 3. M a c h d a o d o n g c u a m o t m d y t h u v6 t u y e n d i e n c6 d p t i r c a m b i e n t h i e ^
tir 0,5(MH) d e n
t h e b a t dMc
D a p so:
l O ( n H ) , t u d i e n b i e n t h i e n tH 1 0 { p F ) d e n
song trong day song
4 , 2 m
B a i 4 . M a c h dao
n^o?
H o i p h a i cung c a p cho m a c h
mot
n h i e u d e d u y t r i d a o d p n g c u a n o k h i h i $ u d i ^ n t h e cUc d a i
m a c h l a T, = 3 ( s ) , k h i C =
t h i c h u k y dao d p n g cua m a c h l a T j = 4 ( s ) . K h i
d e m 2 t u C i v a C2 g h e p s o n g s o n g t h i c h u k y dao d p n g cua m a c h l a bao n h i e u ?
la
T, = 6 ( s ) , k h i L = L j
t h i c h u k y dao d p n g cua m a c h
la
N e u d i e u c h i n h L co g i a t r i L = L , + L j t h i c h u k y cua m a c h l a b a o
D a p so: 10 s
B a i 7. M a c h d a o d p n g L C co c h u k y l a 0 , 0 0 4 ( s ) , d o n g d i ? n cUc d a i
Tj =8(s).
nhieu?
a) B i e u thii'c d i e n t i c h t r e n cdc b a n t u d i e n t h e o t h d i g i a n
b ) B i e u thufc l i i e u d i e n t h e g i U a h a i b a n t u d i p n t h e o t h d i g i a n
200(mA).
- ) V
2
2
B a i 15. M a c h c h p n s o n g c u a m o t m a y t h u v 6 t u y e n g o m m o t c u p n d a y t h u a n
c a m CO d p t y c a m L = 4 m H v a m o t t u x o a y C^. T u x o a y co d i ^ n d u n g b i e n
t h i e n tCr 20 p F d e n 140 p F . K h i goc x o a y b i e n t h i e n tiT 0° d e n 6 0 ° . B i e t d i e n
B a i 8. M a c h dao d p n g L C co
a) B a d e s o n g , m a m a c h t h u 6MC k h i t u C b i e n t h i e n tCr 20 p F d e n 140 p F
b) BiTdc s o n g m a m a c h t h u dMc
D a p so: a ) 5 3 2 , 8 7 m
B a i 16. M a c h c h p n s o n g c i i a m o t m a y t h u v 6 t u y e n g o m c u p n d a y t h u a n c a m
goc
x o a y b i e n t h i e n t i ^ 0° d e n
L = 0,5 m H , c i r 6 n g d p d o n g d i ^ n cifc d a i l a
4(A).
P H A N III: D A P A N LI T H U Y E T
T R A C N G H I E M C H l / O N G IV
J
ciia t u d i e n v a n S n g l i i g n g tii t r u c r n g ciia cupn c a m k h i n S n g l u g n g d i ^ n bang
n a n g lirgng tir
C a u 1. C h p n C. i s d m p h a 2
so v d i q
C a u 2. C h p n D . In = co.qo.
8.10-'M.
B a i 1 0 . M a c h dao d p n g L C co L = 2 n H , C = 4 n F . N g i r d i t a t h a y r&ng t a i t h & i
C a u 3. C h p n A .
T = 27rN/LC
d i e m m a h i e u d i e n t h e c u a t u d i e n l a 4 V t h i ciicrng d p d p n g d i e n co g i d t r i
C a u 4. C h o n A . f =
2 n i A . T i m n a n g lifcrng d i e n tii c i i a m a c h dao
C a u 5. C h p n A .
dpng
^
27:VLC
D a p s o : 3,2004.10"'* J •
B a i 1 1 , M a c h d a o d p n g L C co L = 4 M H . T h d i d i e m k e tCr l u c d i e n t i c h cua t u
d i e n CO g i d t r i ciTc d a i c h o d e n k h i d i e n t i c h c u a t u d i e n b & n g 0 I d 4 ^is. Ti«^
D a p so: 1,62.10"*' F
I 8 O " . B i e t birdc s o n g m a m a c h b d t dirge k h i
D a p so: 4 , 2 4 . 1 0 " ^ H .
B a i 9. M a c h dao d p n g L C co L = 80 ( m H ) , I , , = 2 ((.lA). T i m n S n g lucfng d i e n t r U d n g
d i e n d u n g ciia t u d i e n
co
goc x o a y c i i a t u C^ I d 120° co g i d t r i 140 m . B i e t d i e n d u n g c i i a t u l a h a m b a t
T i m n S n g l u o n g d i e n cUc d a i c u a t u d i e n
D a p so: 8 . 1 0 " ' ' J ;
k h i t u x o a y C^ co goc x o a y l a 3 0 "
1409,8 m ; b) 1065,75 m
n h a ' t d o i v d i goc x o a y . T i m h e so tiT c a m L .
1,27.10 " C
D a p so: 4 . 1 0
- ) C; b. u = 2 0 c o s ( 2 5 . 1 0 ' ' t -
L v a t u x o a y C^. T u x o a y cd d i e n d u n g b i e n t h i e n t i i 10 p F d e n 190 p F . K h i
T i m d i ^ n t i c h cUc d a i c u a t u d i ^ n
D a p so:
hkng
20 n i A . T i m
d u n g c i i a t u l a h a m b a t n h a t d o i v d i goc x o a y . T i m :
5 s
B a i 6. M a c h d a o d p n g L C co L t h a y d o i k h i L = L , t h i c h u k y d a o d p n g cua
mach
H
D a p so: a. q = 8 . 1 0 " % s ( 2 5 . 1 0 ' ' t -
1,33.10"'^ W
B a i 5. M a c h d a o d p n g co L C co C t h a y d o i dugc k h i C = C i t h i c h u k y dao d p n g cua
D a p so:
D a p so: 5.07.lO"*'
G i a sOr or t h d i d i e m b a n d a u t h i c i r d n g d p d o n g d i e n d a t g i a t r i ciTc d a i
t r e n t u 1^ 5 (V)?
D a p so:
10"^ s t h i n a n g l i f d n g tCr t r U c d i g cua c u p n c a m l a i d a t g i a t r i cifc d a i . T i m h #
so t i r c a m c i i a c u p n d a y
B a i 14. M a c h dao d p n g g o m m o t t u d i e n co d i e n d u n g C = 4 n F v d L = 4 m H .
c u o n c a m c6 L = 28 (|.iH) v a m o t d i e n t r d t h u a n
1 ( Q ) , m o t t u d i e n co d i e n d u n g 3 0 0 0 ( p F ) .
cong suat bao
5 0 0 ( p F ) . M a c h c6
133,2m
dong gom
i 12. M a c h dao d o n g L C co L = 4 m H vk C = 1 nF. L u c circrng dp d 6 n g d i $ n q u a
m a c h l a 2 m A t h i d i e n t i c h ciia t u d i $ n l a 4 n C . T i m d i e n t i c h ciTc d a i ciia t u
C o n g thii'c s a i l a W =
.
2
^ S u 6. C h p n B .
U 23. Chpn B .
C a u 7. C h o n A .
T r o n g m a c h d a o d p n g L C c6 d i e n t r d t h u a n b ^ n g k h o n g t h i n&ng lUOng
S o n g d i e n tCr t r u y e n dupe t r o n g m o i t r u d n g v a t c h a t d a n h o i v a k e ca t r o n g
chan
trirdng t a p t r u n g d cupn c a m v a b i e n t h i e n t u a n h o a n v d i chu k i b k n g
T r u y e n dupe t r o n g c h a n
C a u 8. C h p n B .
D u n g se l a tijT t r i T d n g b i e n t h i e n t h e o t h 6 i g i a n s i n h r a d i e n t r U d n g x o a y
Q^u
khong
25. C h p n D .
Song dai
C a u 9. C h p n B .
Q^u
M a c h c h p n s o n g t r o n g m a y t h u s o n g v 6 t u y e n d i e n b o a t d p n g dufa t r e n h i e ,
26. C h p n A .
S o n g cue
t u p n g c p n g h u d n g dao d p n g d i e n tii
ngan
CaU 27. C h p n A .
C a u 10. C h p n D .
Do
khong
^ g u 24. C h p n B .
chu k i r i e n g ciia m a c h
- > T = 27T>/EC
T r o n g s o n g d i e n tiT t h i d a o d p n g e u a E v a q u a B t a i m o t d i e m l u o n d o n g
- > K h i C t a n g 16 I a n t h i T t S n g 4 I a n .
pha v d i n h a u .
C a u 11. C h p n B .
C a u 28. C h p n A .
D u n g se l a d i e n t r i r c r n g x o d y l a d i e n t r U d n g c6 d i f d n g sure 1^ nhCTng dudng
Song d a i dupe d u n g de t h o n g t i n d u d i nude,
cong k i n .
u 29. C h p n D .
C a u 12. C h p n D .
Nang l u p n g d i e n t r i T d n g v a n S n g l u p n g tis t r u d n g b i e n t h i e n t h e o t h d i g i a n ,
n S n g l u p n g d i e n tii l a m o t h a n g so k h o n g d d i .
DAP A N BAI TAP TRAC N G H I E M C H l / a N G IV
C a u 13. C h p n C.
Nang l u p n g d i e n tuT k h o n g d o i t h e o thcri g i a n .
C a u 1 . CD = -^L^
C a u 14. C h p n A . •
N a n g l u p n g d i e n tCr k h o n g d o i t h e o t h d i g i a n .
dao dpng bien thien
T
t u a n hoan v d i chu k i — .
2
C a u 16. C h p n D .
Do
f =
; =
27iVLC
~
27tvLC
=> L
0,507 ( H )
~
z=> C
1,76.10"^ ( F )
Chpn A .
C a u 3. f =
27tVLC
Chpn A .
C a u 17. C h p n C.
1
C S u 2. f =
- > T a n so d a o d o n g r i e n g c i i a m a c h t a n g g a p doi k h i die:;
d u n g C c u a t u d i e n g i a m g i a t r i d i e n d u n g c i i a t u d i f n di 4 I a n
Do f =
rad/s
C h p n C.
C a u I S . C h p n C.
Nang l i r p n g d i e n t r u d n g t r o n g t u d i e n c i i a m o t m a c h
« 444
C a u 4. T = 27t ^/LC = 1,53.10"'' (s)
Chpn D .
=
T a n so d a o d o n g r i e n g c i i a m a c h p h u t h u o c v a o L , C.
Cftu 5. T = 271 X / L C
27rVLC
^ C = 0,02 ( F )
Chpn A.
C a u 18. C h p n D .
S o n g d i e n t i r l a s o n g m a t r o n g q u a t r i n h t r u y e n s o n g t h i E l u o n v u o n g gc^
v d i E v a ca h a i vectcr E , B l u o n v u o n g gdc v d i p h u o n g t r u y e n s o n g .
C a u 19. C h p n B .
D i i n g l a s o n g n g d n c6 t a n so I d n hern t a n so s o n g d a i d o Xti 1§ n g h i c h v d i 1
Q
t h o n g q u a c o n g thiJc ^ - j • N e n k h i X n h o - > f I d n .
C a u 20. C h p n A .
C h p n C.
C f i u 7. X = c . T = 3 . 1 0 ^ 2 7 t ^ / L C = 168,5 ( m )
C h p n C.
. X; = C . T 2 = c.2n T L C ^
S o n g d i e n tii m a n g n a n g l u p n g
C a u 21. C h p n D .
S o n g d i e n tCr m a n g n a n g l u p n g
C a u 22. C h p n A .
D u n g p h a i l a s o n g d i e n tii l a s o n g n g a n g .
c . T , = e.27iVLC7 = 23,8 m
Cfiu8.
I
= 113,04 m
Chpn D .
•fiu 9. X = ^ ^ f = ^ .
= 3.10" H z ^ f = 3 M H z
Chpn A.
C a u 10. Klai C, n t Ca t h i ma f =
^
^
+
(*) -^f^
= — +—
= —
^
D A P A N B A I T A P L U Y E N T A P C H l / a N G rV
(*)
Bai
t i le n g h i c h v
I3ai2.
= 50^ + 100' ^ f = 111,8 ( M H z )
•h
Chon D .
Cau 11.
X = 3.10^27r^/LC ^
1. X = c.T
= c . T , = 310**.27t 7 L , C ,
->L,
=
q = 2.10^'*sin(100Ttt) (C)
i = q'(t) = 2.10~^1007i.cosl007it
^1
(3.10*)^(27t)'C,
Vay 8,04 n H < L < 2,85 m H
Bai3.
(A)
•
= c . T i = 3.10^27t7^C7 = 4,2 m
• A., = C.T2 = 3.10^27t7^c7 = 133,2 m
Nhdn xet: i n h a n h p h a hcfn q m o t goc ^ n e n
Vay 4,2 m < A. 133,2 m
v , i q „ . i i = i22!Ll2:l = 4.io« (C)
B a i 4.
• Wc
Vay q = 4.10-^sin(1007tt + - ) (C)
= WL
L T2
^ <^ lo « 0 , 0 5 1 7 A
CU^
2
2
Chon B .
C a u 13. i = 2.10-''cosMt (A) -> lo = 2.10-'' (A)
V2
2
« 1,33.10-' W « 1,33 m W
B a i 5. K h i C i // C2 t h i C = C i + C2 ( * )
ma T - 2Tt N / L C
°- = 0,12.10-'^ J = 0,12
= 47rlLC
(T^ t i le t h u a n v d i C )
Chpn A.
PTT2
C a u 16. W™«, = W , . _
^
2
T
= ^
2
^lo
=
(*) ->
fn
\Jn^
^L
=
0,42 (A)
= T f + T| = 3 ' + 4 ' = 25 ^ T = 5 (s)
B a i 6. De cho L = L j + L2 ( * )
ma T = 271 V L C
Chon A .
= WL^^^
'-:^
2
=t^
2
^ C
= ^
Chon B .
C a u 18. W = Wc + W,, (*)
ma Wc = 3W,., W =
(*)
(A)
'0,0517?
P = r . I ^ = 1.
C = 3,6.10-* (F)
Chon A.
C a u 17.
V2
Cong suat can cung cap l a
Chon A .
C a u 15. W =
0,0517
• I =
• T = 2 7 1 . - ^ = 12,56.10-' (s)
2C
= 8,04 n H
- > L2 = 2,85.10-' H = 2,85 m H
Chon C.
C a u 14. W =
= 8,04.10 ' H
. X., = C.T2 = 3.10^2317L2C2
i = 2 . 1 0 " V o s . l 0 0 7 t t (A)
hay i = 2.10-^7isin(1007it + ^ ) (A)
C a u 12. i = 4 0 0 7 T . 1 0 " ^ s i n ( 1 0 0 7 T t +
C = 0,126.10-'F = 0,126.nF
^
= 0,2.10'^ F = 0,2 ^iF
^
= 47I2LC
(T^ t i le t h u a n v d i L )
(*) _>
B a i 7. T
^ T f + T| = 100
271.
T = 10 s
qo « 1,27.10-" ( C )
^0
B a i 8. W .
= W,
LI
-°- = 4 . 1 0 - ' ( J )
-52- = 3 W L + W L ^ W L = 5,625.10" J
2C
Suy r a Wc = 3 W L = 1,6875. IQ-^' J ^ Chon C.
B a i 9. W = Wc + W L -> W,.^_^ = W L + W L ^
^
2
= 2 W L - > W L = 8.10-^* (J) = Wc
Bai
10. W = Wc + W L
Cu'
Li'
4.10^4'
2.10-''(2.10-')'
_> W =
+
=
+•
2
2
2
2
^ W = 3,2.10-' + 4 . 1 0 - ' ' = 3,2004.10-** J
B a i 1 1 . T h d i gian ke tit luc dien t i c h cua t u dien c6 gia t r i cUc d a i cho den k h i
T
d i e n t i c h ciia t u dien b i n g 0 la 4|.is -> — = 4(.ts
T = 16 |.ts = 16.10-*^ (s)
T
=
27IN/LC
12. • w =
Bai
C =
-V-
= 1,62.10"^ (F)
= 5.10^ rad/s
^ qo = 5 , 6 5 6 . i o - ' ( O
w
13. Sau t h d i gian lO-'^ s t h i nSng lifdng tCr t r u d n g cua cuon cam l a i dat gia
Bai
t r i. cue
. .dai ^
Tim
-2 = 10-' (s) ^ T = 2.10-' (s)
L = ? T = 271 N / L C
a) . ^1 = c.T, - 3.10«.27I7L:C7 = 3 . 1 0 « . 2 . 3 , 1 4 V 4 . 1 0 •'.20.10'^ = 5 3 2 , 8 7 (m)
. ^2 = C.T2 - 3 . 1 0 ^ 2 7 T 7 L C 7 = 3 . 1 0 ^ . 2 . 3 , 1 4 ^ 4 . 1 0 ^ 1 4 0 . 1 0 " ' '
Vay 532,87 m < A < 1409,8 m
b) Do d^ cho C = Co + kcp
,
nen t a c6:
- 5,07.lO"** H
L =
Bieu thiJc cUdng do dong dien c6 dang:
* lo = 20.10-'*A
pF
do
B a i 16. Ta c6:
= 3 . 1 0 ' . 2 . 3 , 1 4 V 4 . 1 0 ' . 8 0 . 1 0 " " = 1 0 6 5 , 7 5 (m)
C , = 8 0 p F , cp, = 0 °
C2 = 1 9 0 p F , (p2 - 1 8 0 "
Ma C = Co + kcp
nen t a c6 he
C i = C„ + kcp-,
[10 = C„ + k.O
C2=C„+kcp2
[ l 9 0 = Co + k . l 8 0
C,, = 1 0 p F
<-> •.
k=
1
pF
dp
Gia t r i cua C^ k h i cp = 120° la
C = Co + kcp = 10 + 1.120 = 130 P F
Mat khdc:
- ? i = lo.cos(a)t + (pi) (*)
= c.T = 3 . 1 0 ^ 2 7 t ^ / L C
140'
IQ
(*) -> I„ = I„cos(25.10\ + (Pi) <-» coscp, =: 1 <^ cpi = 0
Vay i - 20.10-lcos(25.10't) (A)
a) Bieu thufc dien t i c h t r e n cac b a n t u dien theo t h d i gian.
q = qo.cos(cot + cp,,)
In
*qo = ? q o = ^
=8.10-«(C)
CO
* cpq = ? (p, = cp^i + -
(p,j
zz
- -
rad
Vay q = 8.10-^cos(25.10^t - - ) (C)
2
b) Bieu thiic hieu dien the gii^a h a i ban t u dien theo t h d i gian.
q ^ H0:l.eos(25.10't-^)
c
k = 2
20(pF)
C = 8 0 pF
= c.T = 3 . 1 0 ' . 2 7 t N / L C
= 25.10' rad/s
ma t = 0 t h i i =
C„=
20 = C „ + k . O 1
"
140 = Co + k.60j
Budc song ma mach bat duoc luc nay
i = Io.cos((ot + cpi)
cfip
fC, =C„+k(pj
"
^
C 2 = Cg + kcpj
C = 20 + 2.30 ^
B a i 14. Nhan xet: t = 0 t h i i = lo = 20.10-^ (A)
*
= 1409,8 m
K h i goc xoay ciia t u C^ l a 3 0 " t h i gia t r i cua t u C^ l a C = Co + kcp
^/LC
• q» = q ' + 4
a i 15.
4.10''
-> u - 20.cos(25.10''t - - )
2
2
(V)
(3.10'r.4;i'.C
(3.10'r.4n'.130.10-"7i
^
L = 4,24.10"' ( H )
Luu y: Dien dp ticc thai va cudng do dong dien tiJCc thai tren mot
doan
mach giong nhau a chd bien thien dieu hoa cung tdn so, khdc nhau a bien
dp va pha dao
dong.
3. D o a n m a c h xoay chidu c h l c6 diSn t r d t i i u l n
ChiTtfng V. DONG DIEN XOAY CHIEU
a) Quan
N e u u = Uo.cos co .t t h i i = I„.cos co .t
P H A N I: T O M T A T L I T H U Y E T
Ddng
dien
b) Gian
Van de 1: DONG DIEN XOAY CHIEU
MACH DIEN XOAY CHIEU CHJ CO DIEN TRCJ THUAN,
TU DIEN, CUON CAM
cam
ilng
xuat
hien
a) Quan
trong
Suat
dien dong
theo
th6i
tra
vecta
luat 6m:
u
I
^
h o a c 1=
•
lo = ^
cu<5n day
he giCta u va i
trd thuan
c6 ciing
tdn so va cung pha
thuin
•x
^
R
c^m
^
)
D i e n ap h a i d a u cuon d a y t h u a n c a m n h a n h p h a h o n cu6ng dp d o n g d i e n
nay bien thien
gian
theo
dieu
dinh
mot goc
luat
b) Gidn
d a n g s i n , t h i l d n g goi t^t l a s u a t d i e n
dd
-.
vecta
dong xoay c h i e u .
- C h u k i v a t a n so l i e n he bori: w = —
T
-
= 27if
O
S u a t d i e n d o n g do c a c m a y p h a t d i e n xoay c h i e u tao r a c u n g c6 bieu
c) Dinh
ludt 6m:
I,, = ^
I
X
hoac I =
thiirc tuorng t\i nhif t r e n .
2. Dien
d) Tdc dung
xoay c h i l u . D d n g diSn xoay chiSu
D o n g d i e n c6 cudng dp b i e n doi d i e u h o a theo t h 6 i g i a n l a d o n g
dien
xoay cl^ieu.
v a CLfdng dp d o n g d i | n qua m a c h c6 b i e u thiirc:
cpu)
do:
Uo, cpu lo, ^>i' I a n luat l a c a c b i e n dp v a p h a b a n d a u cua h i e u
- C a n trd dong di^n.
he giita u vd i
N e u u = Uo.cos( w .t -
- ) t h i i = I„.cos 03 .t
h) Gidn
do
vecta
O
d i e n t h e h a i d a u d o a n m a c h v a cu'dng dp d o n g d i e n qua m a c h .
''—I
• D a i lupfng 9 = cpu - 9i goi l a dp l^ch p h a giOfa u v a i trong mot doan m a c h .
J
* (p > 0: u s d m p h a horn i .
* 9 < 0: u tre p h a h o n i .
* (p :r 0: u c u n g p h a i .
cam
D i e n ap h a i d a u tu d i e n t r e p h a hern cifdng dp d o n g d i ? n m p t goc
i = Iocos((ot + ( p i )
Trong
cua cugn
5. D o a n m a c h chf chQa tu diSn
a) Quan
V d i m p t d o a n m a c h xoay c h i e u t h i b i e u thufc d i e n ap h a i d a u d o a n m a c h
u = UflCosCcot +
chinh
- L a m cho d i e n a p b i e n t h i e n n h a n h p h a so v d i ciTdng do d o n g d i e n .
- D i e n a p b i e n doi d i e u h o a theo th&i g i a n l a d i e n a p xoay c h i e u .
-
vdi
thuan.
2
e = Eocos(a)t + cpo).
hoa
qua dien
N e u i = I„.cos CO .t t h i u = UQ.COSC M .t + -
k h u n g d a y c6 b i e u thuTc:
-
do
. Oogn m a c h chf chQa
T h e o d i n h l u a t c a m iJng d i e n tCf s u a t
dong
xoay chieu
R
1. S u i t dien d^ng xoay chi^u
dien
dien
dp d hai dau dien
O*
c) Dinh
-
he giUa u va i
^
u,
*
i
X
-.
2
3. Cflng hadng d i § n
ludt 6m: In =
c) Dinh
d) Tdc dung chtnh
a) Dinh
cua tu dien
nghia:
H i e n tiTOng cgng hirdng di§n t r o n g d o a n m a c h R L C \k h i | n
t u g n g m a c h R L C c6 Z L = Zc h a y co =
- L a m cho di§n dp b i e n t h i e n c h a m p h a so v d i ciTdng do d o n g d i e n .
- C a n trd dong di^n.
*ZL = Z C C ^ C O L = —
toC
Van 6e2: MACH CO R, L, C MAC NOI TIEP. CONG HUdNG DIEN
:
Z„,n
=
0) =
1
- =
VLC
= VR' +
0 = R (do Z L = Zc)
cp = 0 => i qua doan mach R, L , C cung pha
UAB = U R + U L + Uc.
• G i a suf i = IQ.COS CO .t t h i t a c6:
wd'i u h a i dau doan mach.
UR = UoRCOscot, U L - UoLCOsCcot + 7i/2), U c = Uoccos((i)t - nl2).
* Cudng dp hieu dung qua doan mach R L C dat gid t r i cUc d a i :
UoR = IQR, UQL = IQZL, UQC = IQZC-
- > U = UoRCOSCOt + UoLCOS((Ot+-) + Uoccos(cot
H i e u dien t h e u h a i dau doan
0
L.C
7R^ + ( Z , - Z C ) '
X e t d o a n m a c h R , L, C m ^ c no'i tie'p t h i t a c6:
K e t lu^n:
1
2
CO =
* T o n g t r d ciia doan m a c h RLC co gia t r i cifc t i e u
1. C ^ c giS tri tL/c thdi
vdi:
.
vL.C
b) Cdc he qua khi co cong huang dien trong mach RLC
^max —
--^).
Z
~ R •
* Cac d i e n ap tuTC thdi giuTa hai giuTa h a i ban t u dien va giOfa h a i dau cuon
m a c h l a t d n g h o p ciia b a dao
dong d i e u h o a c u n g t a n s o , n e n t a t i m d a o d o n g t o n g h o p u n ^ y b ^ n g
cam CO bien do bang nhau n h i m g ngu'Oc pha nen t r i e t tieu iSn nhau.
phiTOng p h d p v e c to quay.
C h i n h v i vay d i e n ap giCra h a i dau mach b ^ n g dien dp h a i dau R.
2. Gi^n d6 Fre - nen. Quan h$ giOa cadng d6 ddng di^n
diSn
UL
a) Gidn do vecta
Van de 3: CONG SUAT CUA DONG DIEN XOAY CHIEU
HE SO CONG SUAT
_
• UR = UoRCOs(cot) -> U R .
• U L = UoLCOs(tt)t + nl2) ^ U L .
1. Gang su^t cue doan m a c h : D i e n nSng doan m a c h t i e u t h u t r o n g 1 s.
• U c = UocCOS((Dt - 7T/2) ^ U C .
• u=
b) Dinh
UR
* Cong suat turc t h 6 i p = ui
+ U L + U c -> U = U R + U L + U C
* Cong suat t r u n g b i n h , cung l a cong
ludt 6m cho doan mach RLC mac not tie'p. Tdng trd mach
suat toa n h i e t t r e n R
U = VU^R+(U,-Ucf
P = UIcoscp = RI^
U„
I = - h o a c i = - ! ^ ; V d i 2=
z
R
vdi coscp = — goi l a h e so cong suat
^R^TczT^Z^
z
Li
c) Do lech pha cua u so vol i
U , - U(.
Z, - Zc
+ Neu: Z L > Zc hay caL >
1
(oC
t h i u n h a n h pha h o n i (mach c6 t i n h cam
khang).
+ Neu: Z L < Zc hay M L < J _ t h i u cham pha h o n i (mach c6 t i n h dung
coC
khang).
+ Neu: Z L = Zc hay coL = — t h i u ciing pha vdi i (mach cong hucfng dien)
wC
* D i e n nSng t i e u t h u : W = P.t
2.
sd c6ng su^t
Tir g i a n do F r e - n e n cua mach R, L , C t a chdng m i n h diTOc:
UR „ „
R
coscp = — - Hoac cosq)= —
U
Z
Y nghia h e so cong suat
* Trifofng hofp coscp = 1 <=> cp = 0
Day l a triiorng hop doan mach d i e n xoay chieu c h i chufa R, hoSc m a c h
RLC n h i m g xay r a cong huorng. Liic nay P = U L
* TrtfoTng hofp coscp = 0 <=> cp = ±—
2
Day la trUdng hop doan mach xoay chieu k h o n g chiJa d i e n trd t h u a n
(Day la t r u o n g hop m a c h chi c6 cuon cam L hay t u di§n C hoac ca L va
C). Luc nay P = 0
4. Ccich mSc di$n 3 pha
a) Cdch mac hinh sao
Day pha 1
A,
A',
* T r i i c f n g hcfp 0 < coscp < ! < = > - — < ( p < 0 hoac 0 < cp < —.
Day t r u n g hoa
Luc nay: P = UIcoscp < U L D a y la t r i r d n g hop hay gap n h a t .
Van de 4: MAY PHAT DIEN XOAY C H I E U
A,
1. Nguyfin t3c hoat dong m^y ph^t dien xoay chi^u
a) Nguyen
tdc hoat dgng cua cdc loqi may phdt dicn
xoay chieu:
i2
Day pha 2
Dua t r e n
b) Co hai each too ra sudt dien dgng xoay chieu thuong dung trong cdc may dien
~ Tit trirdrng c6' d i n h , eac vong day quay t r o n g tijf triTdng.
Day pha 3
i3
h i e n tiTOng cam ufng d i e n t\i.
C a c cong thufc: Ud = Vs Up ya Id =Ip
b) Cdch mdc hinh tam gidc
Day pha 1
- TiT truorng quay, eac vong day dat co d i n h .
2. MSyph^t di0n xoay chi^u m6t pha
A i 1 B,.,
Co h a i bo p h a n c h i n h l a p h a n cam va p h a n iJng. .
- P h a n cam la n a m c h a m d i e n hoac n a m cham v i n h cOru. Do Ik p h i n tao
ra tii t r u d n g .
-
P h a n l i n g la nhOfng cuon day, t r o n g do xuat h i e n suat d i e n dong cam
Day pha 2
ufng k h i may hoat dong.
M o t t r o n g h a i p h a n dat co' d i n h , p h a n con l a i quay quanh m o t true.
P h a n CO d i n h goi l a stato, p h a n quay goi l a roto.
3. MSy phSt diSn xoay chi^u ba pha
a) Dong dien xoay chieu ba pha
Dong d i e n xoay chieu ba pha la he t h o n g ba dong dien xoay chieu, gay
boi ba suat d i e n dong xoay chieu co cung t a n so, cung b i e n do n h u n g
lech nhau ve pha la
—
3
b) Cdu tqo vd hoat dgng cua may phdt dien xoay chieu ba pha
Day pha 3
Cac cong thufc: Ud = Up va Id = V3 I , ,
Van de 5: DONG C O KHONG DONG BO BA PHA
1. Nguyen tSc hoat d6ng cua dSng cd kh6ng d6ng b6
a) Tu: trUdng quay. Su quay dong
bo
K h i quay mot n a m c h a m . q u a n h m o t true, tCr trUcfng do n a m c h a m gay ra
CO eac dxibng sufc tCr quay trong k h o n g gian. Do l a m o t tCf trUdng quay.
• M a y c6 cau tao gom Stato c6 ba cuon day r i e n g r e , hoan t o a n gio'ng
Neu dat giufa h a i cUe ciia m o t n a m cham h i n h chiJ U m o t k i m n a m cham
nhau quan t r e n ba l o i sdt dat lech nhau 120° t r e n m o t vong t r o n . Roto
v a quay deu n a m c h a m chuf U t h i k i m n a m c h a m quay theo v d i cung toe
la m o t n a m cham di$n.
K h i roto quay deu, eac suat d i e n dong cam iJng xuat h i ? n t r o n g ba c u p n
day CO cung b i e n do, cung t a n so nhiftig lech nhau ve pha 1^ — . N e u
3
do goc. Ta n o i k i m n a m cham quay dong bp vdri tii trUdng.
b) Su quay khong dong
bg
Thay k i m n a m cham b k n g m o t k h u n g day d a n k i n . K h u n g n a y co the
quay quanh true xx' t r u n g v d i true quay ciia n a m cham. N e u quay deu
n a m c h a m t a t h a y k h u n g day quay theo c u n g c h i e u , d e n m 6 t liic nao do
noi cac dau day cua ba cuon v d i ba mach ngoai giong nhau t h i t a co he
k h u n g day c u n g quay d e u nhitog v d i toe do goc
ba dong d i e n cung b i e n do, cung t a n so' n h u n g lech nhau ve pha 1^ — .
Dong CO hoat dong difa theo nguyen t^c n 6 i t r e n goi la dong eo k h o n g
dong bo.
< co c u a n a m c h a m .
2. Tgo ra ta trudng quay bSng ddng di0n ba pha
TCr trtrdng quay c6 the dUgrc tao r a bkng dong dien ba pha nhu sau: M^c ba
cuon day giong nhau, bo tri lech nhau 1/3 vong tron vdi mang di^n ba pha.
Trong ba cuon day c6 ba dong dien cung bien do, cung tan so nhung lech
Neu bo qua moi hao phi nSng luong trong may bien ap, cong suat trong
mach so cap va thu" cap hhng nhau.
u, •
2TC
pha nhau — . Moi cuon day deu gay d vung xung quanh true O mot tCr
3
trUdng ma cam iJng tCr c6 phUcfng n ^ m doc theo true cuon day va bien doj
tuan hoan vdi cung tan so a nhuiig lech pha nhau — .
o
3. C^u tao
hoat d6ng cua dSng c6 kh6ng dfing b6 ba pha
Dong cd khong dong bo ba pha c6 hai bo phan chinh:
- Stato C O ba cuon day gio'ng nhau quan tren ba loi sMt bo tri lech nhau
1/3 vong tron.
- Roto la mot hinh tru tao bdi nhieu la thep mong ghep l a i goi la roto
long soc.
I,
nen
U„
N,
N,
I,
•
Do do: Dung may bien dp l a m dien dp tang bao nhieu Ian thi ciidng do
dong dien giam bay nhieu Ian va nguoc lai.
2. Truy^n t^i dien
Cong suat hao phi tren day
AP = R I ' ^ AP = R
(Ucoscp)
Trong do: R: dien tror duong day ( Q )
P: cong suat truyen di (W)
'
U : dien dp d noi truyen di (V)
costp: he so cong suat cua mach dien
• Pi Cong suat ccf hoc c6 ich
• P: Cong suat tieu thu ciia dong cot
• H : Hieu suat dong ccf
Van de 6: M A Y B I E N A P - T R U Y E N T A I D I E N
1. M^y bi^n
C a c e a c h g i a m AP
C a c h thiir n h a t : G i a m dien.trd R cua dudng day. Day la edeh ton k e m vi
phai tdng tiet dien cua day, do do ton nhieu k i m loai l a m day va phai
tang sufe chiu dUng ciia cdc cot dien.
C a c h thvC h a i : Tdng dien dp U or noi phdt dien va giam dien dp cf ncfi tieu
thu dien tdi gid tri can thiet. Cach nay ed the thifc hien don gian bang
mdy bien dp, do do duoc dp dung rong rai.
May bien dp 1^ thiet bi hoat dong dxia tren hi#n tUorng cam ufng di?n tCr,
dung de tSng hoac giam di^n ap xoay chieu ma khong l a m thay doi t a n so
TRAC N G H I E M LI THUYET
ciia no.
a) Cdu tao va nguycn tdc hoat dong
May bien ap gom hai cuon day c6 s6 vong khac nhau quan tren mot loi
s^t kin. L o i thudng l a m bhng cdc la sdt hoac thep pha silic, ghep each
dien vdi nhau de giam hao phi dien nSng do dong Phu-eo. C a c cuon
C a u 1. Cudng do hieu dung ciia dong dien xoay chieu i = IQ eos(a)t + cpi) dugc
tinh theo cong thiJc
B. I =
A. I - I n V 2
day thucfng lam bkng dong de cd dien trd nho va duoc each di^n vdi loi.
May bien ap hoat dong dua tren hien tifcfng cam ufng dien tif. Mot trong hai
cuon
CLia
may bien ap duoe noi vdi nguon dien xoay ehilu, duoe gpi la cuon so
cap. Cuon thuf hai difdc noi vori tai tieu thu, diTOc goi la cuon thuf cap.
b) Su bien doi dien dp va cUang dp ddng dien qua may bien dp
U
N
Neu bo qua dien trd 2 cuon day thi: —-= — U,
Nj
Vdy: T i so dien dp d h a i dau cupn thiJ cap vd cupn so cap b l n g ti so
vong day ciia hai cuon day.
• Neu N2 > N i -» U2 > U i : Mdy tdng dp.
• Neu N2 < N i -> U2 < U i : Mdy h a dp.
^
2
C.I =
^
V2
D. I = 2I0.
C S u 2. Dat mot dien dp u = U \/2 cos(cot + (pj vao hai dau doan mach gom: dien
trd thuan R , cuon day thuan cam c6 dp tU cam L va tu dien eo dien dung C
mdc noi tiep. Cudng dp dong dien qua doan mach c6 gid tri h i | u dung l a
A. I =
f
V
u
1
a)C + - —
U
C. I =
R +
coL -
B. I =
/
J R
V
D. I =
coC
U
1
„
1 >2
- (oL - 1
coC>
U
-2
1
R2 + coL —
coC
Cau 3 . Dat vao hai dau mot cuon day thuan cam c6 dp t u cam L mpt di^n ap
u = Uo cos(a)t + — ). Cudng do dong dipn chay qua cupn day c6 bieu thiJc la
2
A. i = UoCoLcos(cot + - )
2
0)
=
- 7 = = ^
B. CO =
C. CO =
, —
D. CO = —
VLC
^/LC
VL
,
C
Cau 5. Dong dien chay qua mot doan mach R, L, C m^c noi tiep c6 bieu thiJc
i = locos (cot + cpi). Nhiet liJOng toa ra tren di§n t r d R trong khoang thdi gian
t (t rat Idn so vcii chu k i ciia dong dien) la
A. Q = R I ^ t
B. Q = R-Iot
A.Z,=
—
=
^
B . Z L = R ^ ( Z L - Z C )
B. i = ^ coscot
tt)L
C. i = H o c o s ( ( o t + - )
D. i = ^ cosCcot- ^ )
o)L
2
L
2
Cau 4 . Dat vao hai dau doan mach R, L, C mSc noi tiep mot dien dp xoay chie u
CO bieu thiJc u - UoCos(ct)t + cpu) v6i U„, cp^ la hkng so con co thay doi duoc
Cu&ng dp dong dien hieu dung trong mach dat gia t r i Idn nhat k h i tan so
goc CD thoa man
A.
vdi dien ap giOra hai dau doan mach. Moi lien he giOfa dien trd thuan R vdi
cam khang Z L ciia cupn day va dung khang ZQ cua tu dien la
C.Q=-Rlgt
D. Q = - ^ R ' l o t
Cau 6. Cho biet bieu thuTc cua di^n ap xoay chieu la u = Uo.cos(cot + cpu). Di?n ap
hieu dung la
A. U o = - ^
B. U6 = 2U
C. UO = UN/2
D . U = 2UO. .
N/2
Cau 7. Dat vao hai dau doan mach RLC noi tiep mot di§n ap xoay chieu
u = Uocoscot t h i dp lech pha giufa u va i ciia mach difpc tinh theo cong thiJc
T
-wC—^
. ^
(oL - Ceo
„ ^
Leo
A. tan cp =
B. tan cp =
R
R
T
1
^
Ceo
4.
coL + Cco
;
C. tan cp =
^
D . tan cp =
R
R
Cau 8. Dat mot dien ap xoay chieu u = UoCoscot vao hai dau mot doan mach
dign RLC khong phan nhanh. Dong dien cung pha dien ap a hai dau doan
mach dien n^y thi
A. Leo = —
B. Leo < —
C. Leo > —
D. co = - i Cco
Ceo
Ceo
LC
Cau 9. Dong dien di qua mot doan mach R, L , C m^c noi tiep co bieu thu'''
i = locoseot. Di$n ap giuTa hai dau doan mach nhanh pha hon cifdng dp dong
dien k h i
A. CO L <
B. CO L >
C. CO L =
D. oj > - i eoC
coC
coC
LC
Cau 1 0 . Cho doan mach dipn xoay chieu gom cupn day co dien trd thuan R'
mdc noi tiep vdi tu di|n. Biet di^n dp giCra hai dau cupn day l^ch pha -
(Z,-Zc)
-R2
•
Cau 1 1 - Doan mach dien xoay chieu A B chi chura mot trong cac phan tuT: dien
trd thuan, cupn day hoac tu dien. K h i dat dien ap u = UoCos(eot + - ) len hai
2
diu A va B t h i dong dien trong mach co bieu thuTc i - I(,cos(eot + —). Doan
2
mach A B chiJa
A. Dien trd thuan
B . Tu dipn
C. Cupn day thuan cam (cam'thuan) D. Cupn day c6 dien t r d thuan
Cau 1 2 . Doan mach dien xoay chieu gom dien t r d thuan R, cupn day thuan cam
L va tu dien C m^c noi tiep. K i hieu U K , U ] , , U C tuong lirng la dien ap tiJc thdi
d hai dau cac phan tilr R, L va C. Quan he ve pha ciia cac dien ap tiJc thdi la
A.
UR
C.
UR
sdm pha
tre pha
2
so vdi uc
B.
U L
tre pha - so vdi Uc
2
so vdi uc
D. u^ sdm pha — so vdi uc
2
2
Cau 1 3 . Dong dien xoay chieu trong doan mach chi cd dien t r d thuan
A. Cling tan so' va ciing pha vdi di|n ap d hai dau doan mach
B. Cijng tan so' vdi dien ap d hai dau doan mach va cd pha ban dau luon
bang 0
C. Cd gia t r i hieu dung t i le thuan vdi dien t r d ciia mach
D. Luon lech pha7r/4 so vdi dien ap d hai dau doan mach.
—
Cau 1 4 . Dat vao hai dau cupn day cd dp ty cam L mot dien dp u = UoCos2Ttft
(V). Giam cam khang ciia cupn day bhng each
. Giam tan so f cua dien ap u.
. Tdng dp tiT cam L ciia cupn day
. Tang dien ap U
. Giam dien ap U.
u 1 5 . Dat vao hai dau doan mach RLC khong phan nhanh mot dipn ap xoay
chieu u = UoCoseot. K i hieu U R , U L , U C tUctng iJng la dien ap hieu dung d hai
dau dien trd thuan R, cupn day thuan cam L va tu dien C. Neu U R = U^ =
\ Uc t h i dong dien qua doan mach
J , A . Sdm pha — so vdi dien ap d hai dau doan mach
t u dien vh hai dau cupn d a y thi so chi cua von ke tUOng ufng \k U , U c vd U L .
B . T r e p h a — so v d i d i e n a p 6 h a i d a u d o a n m a c h
2
B i e t U R = U C = 2 U i , . H e so' c o n g suat c u a m a c h d i $ n l a
C. S d m p h a 4
A . COS
so v d i d i e n a p d h a i d a u d o a n m a c h
4
C . i = Iocos(a)t - 7 i / 2 )
D . i = I o c o s ( u t + n)
C a u 17. D i e n a p h a i d a u d o a n m a c h l a u = U > / 2
45
B . cos (p =
1
\
C. cos q) = —
2
D . cos (p = — - .
2
A . C u d n g d p d o n g d i e n chay qua d i e n t r d l u o n c h a m pha 7i/4 so v d i d i e n dp d
hai dau doan mach.
u = UoCosMt t h i ciTdng d o d d n g d i $ n c h a y q u a n o cd b i e u thuTc l a
B . i = lo cos(cot + 7 i / 2 )
2
—;=
t u d i e n C. N e u d u n g k h a n g Z c b a n g R t h i
16. D a t v a o h a i d a u c i i a m o t d i e n t r d t h u a n R m o t d i e n d p x o a y chio
A . i = locoswt
=
2 3 . M o t d o a n m a c h d i e n x o a y c h i e u g o m d i $ n t r d thuan R m S c n d i tiep v d i
D . T r e p h a — so v d i h i e u d i e n t h e d h a i d a u d o a n m a c h .
Cau
(0
B. C u d n g d p d o n g d i | n q u a m a c h n h a n h pha 7t/2 so v d i d i e n dp d hai d a u
doan mach.
C. D i e n ap h a i d a u m a c h c h a m pha 7i/4 so v d i cUdng dp d o n g d i e n q u a d o a n
coscot v a c u d n g d p d o n g d i e i
mach.
q u a d o a n m a c h l a i = I N/2 cos(cot + - ) . D o a n m a c h chiJa d u n g cu Ih
A. R
B. Cupn day t h u a n c a m
C.Tudien
D . R, C
D . C u d n g d p d o n g d i e n qua^ d i e n t r d se c h a m pha 7t/2 so v d i d i e n dp d h a i
dau t u d i e n .
C a u 2 4 . D a t d i e n dp u = U o c o s o t v a o h a i d a u m o t c u p n d a y cd d p t U c a m L v a
d i e n trd t h u a n r k h a c k h o n g t h i d i e n ap h a i d a u c u p n d a y se
C a u 18. D a t m o t d i e n a p x o a y c h i e u u = U Q costot v a o h a i d a u m o t d o a n m a c h dien
c h i CO t u d i e n c6 d i e n d u n g C. B i e u thufc ciTdng dp d o n g d i e n t r o n g m a c h l a
A . S d m pha gdc nl2 so v d i cUdng d p d o n g d i e n q u a c u p n d a y .
A . i = U„coC. cos(a)t + n/2)
B . i = U Q W C . coswt
B. S d m pha m o t gdc k h d c n/2 so v d i cUdng d p d d n g d i e n q u a c u p n d a y .
C. i = UocoC. cos{(ot -
D. i =
71/2)
UQCJC.
cos((ot -
C. T r e pha m o t gdc TC/2 SO v d i cUdng d p d o n g d i e n q u a c u p n d a y .
^)
D . T r e pha m o t gdc k h a c nl2 so v d i cUdng dp d o n g d i e n q u a c u p n d a y .
C a u 2 5 . D a t d i e n dp u = U V 2
C a u 19. T a c d u n g c u a c u p n c a m d o i v d i d o n g d i e n x o a y c h i e u l a
coscot ( v d i U v a co k h o n g d o i ) v a o h a i d a u m o t d o a n
A. N g S n can h o a n t o a n d o n g d i e n xoay chieu.
m a c h R L C k h o n g phan n h d n h , x a c d i n h . D o n g d i e n c h a y t r o n g m a c h cd
B . G a y c a m k h a n g I d n n e u t a n so d o n g d i e n n h o .
A . G i a t r i tOfc t h d i phu thupc vao t h d i g i a n theo quy l u a t cua h a m so' s i n hoac cosin.
C. G a y c a m k h a n g n h o n e u t a n so' d o n g d i e n n h o .
B. G i d tri t d c t h d i t h a y d o i c o n c h i e u k h o n g t h a y d o i t h e o t h d i g i a n .
;
C. C h i e u v a cUdng d p d o n g d i e n cUc d a i t h a y d o i t h e o t h d i g i a n .
D . C h i cho p h e p d o n g d i e n d i qua theo m o t chieu.
D. C u d n g dp d o n g d i e n h i e u d u n g t h a y d o i theo t h d i gian.
C a u 2 0 . D a t m o t d i e n a p x o a y c h i e u u = U Q coscot v a o h a i d a u m o t d o a n mach
d i e n c h i c6 t u d i e n . N e u d i e n d u n g c u a t u d i § n k h o n g d o i t h i d u n g k h a n g cua
C S u 26. D a t d i e n dp u = Uocoscot v a o h a i d a u m o t d o a n m a c h c h i cd t u d i e n C t h i
tu dien
CUdng dp d o n g d i e n t d c t h d i c h a y t r o n g m a c h l a i. P h a t b i e u nao sau d a y d u n g ?
A . L d n h o n k h i t a n so c u a d o n g d i e n n h o .
A . D o n g d i e n i l u o n cimg pha vdi d i e n dp u.
B. D o n g d i e n i l u o n n g u o c pha v d i d i e n dp u .
B . N h o h o n k h i t a n so c u a d o n g d i e n n h o .
C. L d n h o n k h i t a n so c i i a d o n g d i e n l d n .
•
D . O c u n g t h d i d i e m , d o n g d i e n i c h a m pha nl2 so v d i d i e n dp u .
D . K h o n g p h u t h u p c t a n so c u a d o n g d i e n .
C a u 2 1 . D a t d i e n a p u = U Q coscot v d o h a i d a u d o a n mach' c h i c6 t u d i e n C i '
C. O c u n g t h d i d i e m , d o n g d i e n i n h a n h pha nl2 so v d i d i ^ n dp u .
C S u 2 7 . T r o n g m o t d o a n m a c h x o a y c h i e u k h o n g phan n h d n h , d i e n dp h a i d a u
m a c h s d m pha cp ( v d i 0 < cp < 0,57i) so v d i cUdng d p d o n g q u a m a c h .
c u d n g dp d o n g ' d i e n t d c t h d i c h a y t r o n g m a c h l a i. P h a t b i e u n a o sau d a y d u n g '
A . CJ c u n g t h d i d i e m , d o n g d i e n i n h a n h p h a 7r/2 so v d i d i e n a p u .
;> m a c h d o g o m
B. D o n g d i e n i l u o n c u n g p h a v d i d i e n a p u .
I
A . C u p n thuan c a m v a t u d i e n .
B. D i e n t r d t h u a n va t u di^n.
C. d c i j n g t h d i d i e m , d i e n a p u n h a n h p h a 7T/2 SO v d i d d n g d i e n i .
C. C h i cd c u p n d a y t h u a n c a m .
D. D o n g d i e n i luon c h a m p h a v d i d i e n ap u.
D. G o m d i e n t r d t h u a n va cupn d a y t h u a n c a m .
C a u 2 2 . M o t m a c h d i e n x o a y c h i e u k h o n g p h a n n h a n h g 6 m : d i | n t r d t h u a n H'
c u p n d a y t h u a n c a m L v a t u d i e n C. D a t v a o h a i d d u d o a n m a c h d i $ n iiP
28.
Doan
M o t m a y phat d i e n x o a y c h i e u m o t pha ( k i e u c a m d n g ) cd p cap cUc
• q u a y d e u vdi t a n so gdc n (vong/phut),
v d i so cap cUc b k n g so c u p n d a y c u a
x o a y c h i e u c6 t ^ n so v d d i e n a p h i e u d u n g k h o n g d o i . D u n g v o n k e ( v o n k''
phan d n g t h i t a n so c u a d o n g d i e n do m a y t a o r a l a f ( H z ) . B i e u thufc h e n h e
n h i e t ) c6 d i p n t r d r a t l d n , I a n l u p t d o d i e n a p d h a i d a u d o a n m a c h , h a i d A "
giUa n, p v a f I d
A.np=60f
B . n p = ^
C. p = ^
f
D. f =
f
U 36. K l i i c6 cong hirdng trong doan mach dien xoay chieu RLC khong phan nhanh
A. D i e n ap tu^c t h d i giCfa h a i dau d i e n t r d t h u a n cung pha v d i d i e n ap tUc
t h d i giOra hai ban t u d i e n
B. D i e n ap tUc t h d i giufa h a i diu d i e n t r d t h u a n cung pha v d i d i e n ap tUc
t h d i giufa dau cuon cam
C. Cdng suat t i e u t h u t r e n doan mach dat gia t r i l d n nha't
^
n
C a u 29. Dong cO d i e n xoay c h i l u 1^ t h i e t b i d i e n b i e n doi
A. D i e n n a n g t h a n h co nSng
B. D i e n n a n g t h a n h hoa n a n g
C. Co n a n g t h a n h d i ? n nSng
D. D i $ n nftng t h a n h quang n&ng
C a u 30. K h i n d i ve dong ccf d i e n k h o n g dong bo, p h a t bieu nao sau day diing?
D. Cudng dp dong d i e n tUc t h d i t r o n g mach sdm pha -
A. B i e n ddi co n a n g t h a n h d i e n nSng ciia dong d i e n xoay chieu.
B. Hoat dong dua t r e n h i e n tugng cam ufng dien txi v a suf dung ti^ trUdng quay.
C . T a n so quay ciia roto Idn hcfn t a n so cija dong dien xoay chieu qua dong co.
D. Roto cua dong co quay d6hg bo v d i tit trUdng quay t r o n g dong ccf.
C a u 31. K h i dong cd k h o n g dong bp b a p h a hoat dpng o n d i n i i v d i toe dp quay
cua tii trufdng k h o n g d o i t h i toe dp quay cua roto se
A. N h o hon toe dp quay ciia tii trUdng.
dat vao h a i dau doan mach
Cau 37. Doan mach d i e n xoay chieu k h o n g p h a n n h a n h gom cuon day cd dp tU
cam L, d i e n t r d t h u a n R va tu d i e n cd d i e n dung C. KJii dong d i e n ed t a n so'
•
D. Co the l d n h o n , nho h o n hoac bang toe dp quay ciia tCf trUdng.
tif cam L v a t u d i e n cd d i e n dung C m^c n o i tiep. B i e t hieu d i e n the hieu
dung h a i dau doan mach l a U , cam k h a n g Z L , dung k h a n g Zc (vdi Zc
ZjJ va
t a n so dong d i e n t r o n g mach k h o n g doi. T h a y doi R de'n gia t r i Ro t h i cdng
suat t i e u t h u cua doan m a c h d a t gia t r i ciTc d a i
C.R„
= ZL.
B.
Zc
D.
Pmax,
k h i do
R„=|Z,-Zcl
R„ = Z L + ZC.
C a u 33. D a t hieu d i e n the u = Uosinwt (Uo v a 03 k h o n g ddi) vao h a i dau dom;
mach RLC k h o n g p h a n n h a n h . B i e t dp t y cam v a d i e n dung difpc giOr k h o i u
doi. Dieu c h i n h t r i so' d i e n t r d R de cong suat t i e u t h u cua doan mach d::
cue dai. K h i do h e so cong suat ciia doan mach b k n g
A. ^
B. ^
D. 1.
C. ^
C a u 34. D a t vao h a i dau doan mach RLC k h o n g p h a n n h a n h m o t d i e n dp xoay
chieu u = Uocoscot (V) t h i dong d i e n t r o n g mach la i =locoscot (A). Doan mach
d i e n nay luon ed
A.
ZL
> Zc
B. Zl, < Zc
7=
chay qua doan mach t h i h f so cong suat cua doan mach n a y
27rVLC
C.
ZL
= Zc
D.
ZL = R.
Caxi 35. D a t d i e n a p u = uV2coscot vao h a i dau doan mach RLC k h o n g phan
n h a n h (dien t r d t h u a n R =^ 0). Chpn dp tuf cam cua cuon day v a d i $ n dunr!
ciia t u d i e n sao eho cam k h a n g b^ng dung k h a n g t h i
A. T d n g t r d ciia doan mach nho hcfn d i e n t r d t h u a n R
B. Cudng dp dong d i e n t r o n g doan mach n h a n h p h a so v d i d i e n a p u.
C. He so cong suat ciia doan mach la 0 < eosqx 1
D. Cirdng dp dong d i ^ n h i p u dung qua mach cd g i d t r i cUc d a i .
"
.
B. Phu thupc dipn t r d t h u a n ciia doan mach
C. B a n g 1
D. Phu thupc t d n g t r d ciia doan mach
C . L d n hcfn toe dp quay eiia tii' trUdng.
A. R o = f -
f =
A. 0 < coscp < 1
B. B^ng toe dp quay ciia tiT trUdng.
C a u 32. Doan mach d i e n xoay chieu gom bien t r d R, cuon day t h u a n cam cd do
v d i d i | n dp tiJc t h d i
3
C a u 38. M o t mdy bien dp cd hieu suat xap x i bang 100%, cd so' vdng day cuon
scf cap l d n h o n 20 I a n so v d n g day cuon t h i i cap. M a y bien dp nay
A. L a m g i a m t a n so' dong d i e n d cuon thuT cap 20 I a n
B. La may t a n g dp
C. La mdy ha dp
'
D. L a m t a n g t a n so' dong d i ? n d cuon scf cap 20 I a n .
C a u 39. M o t m d y bien dp dupc suf dung l a m m d y t d n g dp. D a t vao h a i dau cuon
so cap m o t d i e n dp xoay chieu. Bp qua mpi hao p h i t r o n g mdy. K l i i mach t h d
cap k i n t h i
1 A. So v d n g day ciia cuon thu" cap nho hen so' vdng day ciia cuon so cap
B. D i e n dp h i e u dung d h a i dau cuon thu" cap nho h o n d i e n dp h i e u d u n g d
hai dau cuon so cap
C. Cudng dp hieu dung ciia dong d i e n t r o n g cuon thU ca'p b^ng eUdng dp
dong dien hieu d u n g cua dong d i e n t r o n g eupn so cap
D. Cudng dp dong d i e n hieu dung t r o n g cuon t h U cap nho horn cudng dp dong
dien hipu dung ciia dong d i e n t r o n g eupn so cap.
§
jCSu 40. M o t m a y bien dp dung l a m mdy ha dp gom eupn so cap cd N i vdng,
eupn t h u cap cd N2 vdng. D a t vao h a i dau eupn scf cap m o t d i e n dp xoay
'f chieu cd gid t r i hi$u d u n g U i t h i d i e n dp hieu dung U2 d h a i dau eupn thU
cap thda m a n
A. U2 = 2Ui
B. U2 < U i
C. U2> U i
D. N2 > N , .
p S u 41. Gpi: ( I ) : G i a m cong sua't t r u y e n t a i
( I I ) : T d n g chieu d a i dudng day
( I I I ) : T d n g d i e n the trUde k h i t r u y e n t a i
(IV) : G i a m t i e t d i e n day va g i a m dien the trude k h i t r u y e n t a i
T r o n g qud t r i n h t r u y e n t a i d i e n n d n g , bien phdp l a m g i a m hao p h i t r e n
dudng day t a i d i ^ n dupe sii dung chii yeu h i e n nay la
A. ( I )
B'. ( I I ) '
C. ( I l l )
D. ( I V )
P H A N II. B A I T A P T R A C
+ BAI TAP LUYEN
Tom tat
(j) = 2.10"^cos(1007rt) (Wb; s)
al T i m is/,, =
b) V i e t bieu thufc suat
dien dong h a i dau
khung.
NGHIEM
TAP
Van de 1: T L / T H O N G - S U A T D I E N D O N G C A M LfNG
TAN S O DONG DIEN - MAC MACH D I E N - DONG C O DIEN
PHl/CfNG
Hii&ng ddn gidi
a) TCr t h o n g c i f c dai
TCr bieu thu'c de cho
(|)o = 2.10"^ (Wb)
b) Bieu thurc suat d i e n dong d h a i dau k h u n g
e = -(t)'(t) = -[2.10-'cosl007ttJ
-> e - 0,27isinl007Tt (V)
I . Tijf t h o n g - S u a t d i ^ n d p n g c a m \?ng
B a i 2. M o t k h u n g day dan cd 500 vong day quan no'i t i e p , m o i v o n g c6 d i e n
ti'ch la 200 em^. Khung day dugc dat trong ti^ triTcmg deu B = 0,2T. Luc t = 0
Neu tif t h o n g qua k h u n g day c6 dang
*
e = 0,27rcos l O O n t - - (V)
2
hay
PHAP
= cD^cos((ot + cp) V(Ji 0 „ = N.B.S ; cp = (B,n) k h i t = 0
t h i vecto phap tuyen cua k h u n g hop v d i B mot goc
—. Cho k h u n g day
6
quay deu quanh true (A) vuong goc v d i B v d i tSn so 40 v6ng/s.
T h i bieu thiJc suat dien dong la
e = E„.sin((ot + 9) Vdri E„ = O^.a);
=
E
a) V i e t bieu thu'c suat d i e n dong cam iJng theo t .
Trong do:
* CD : TCr t h o n g t a i t h d i d i e m t (Wb)
* ct)„: T'lr t h o n g cifc d a i (Wb)
b) V i e t bieu thufc tijr t h o n g theo t .
c) Xac d i n h gia t r i ciia suat dien dpng cam iJng va cua tCf t h o n g t a i t h d i
diem — (s).
40
* N ; So vong day ciia k h u n g .
* B: Tuf trirdng (T)
* e: Suat dien dong t a i t h d i d i e m t (V)
Tom tat
• N = 500 vong
* E,,: Suat d i e n dong cUc dai (V)
• S = 200 cm^
* S: D i $ n t i c h k h u n g (m^)
»
* E: Suat dien dong hieu dung (V)
* f: T a n so dong dien (Hz)
f = n.p
* p: So' cap ciTc.
e = Eosin(cot + cp)
= 200.10-" m '
• B = 0,2 T
h • Luc t = 0
I I . T a n so' c i i a m a y p h a t d i ^ n .
f ^ np
60
Hitdng dan gidi
a) Bieu thiJc suat dien dong cam iJng theo t .
• f
* n : Toe do quay (vong/ phiit) hoac (v6ng/s)
*
Eo = N.B.S.M = N.B.S.27Tf = IQOn (V)
* CO =
27if =
* (p
27r.40
= ? t = 0 ^
cp
1. T^i ddi XL/ng m S c hinh sao
U j = V3Up ; I , = I,.
2. T i i ddi xQng m 3 c hinh tam gi^c
t =
0 ^
i,B )' = 6-
(p =
U , = U , ; I, = S i ,
Vay (j) = 2cos
80Ttt +
V
BAI T A P
MAU
a) T i m tCf t h o n g cUe d a i ciia k h u n g .
b) V i e t bieu thu'c suat d i e n dong h a i dau k h u n g day.
7:
6>
c) T i m e va (j) luc t = —
40
B a i 1. M o t k h u n g day quay quanh true eo d i n h t r o n g tCr t r u d n g deu B ma tii^
t h o n g bien doi theo t c6 dang: (j) = 2.10"^cos(1007rt) (Wb; s).
(Wb)
(s)
• e = leOTisin
<->
I
,
= (n,B) = -
Vay e = IGOnsin 807it + - (V)
6
40 vong/s = 40 Hz
b) Bieu thu'c til t h o n g theo t : <}) = (|)ocos(ojt + cp!
*
(t)o - N.B.S = 2 (Wb);
*
( 0 = 2ni - 8071 (rad/s)
I I I . Mac mang di^n
= 80n (rad/s)
40
• (j) = 2eos 8 0 r r . ^ + ^
40
6
e = 807t(V)
6;
(|) = l,73(Wb)
rad
io = I i + 1 2 +
B a i 3. R o l o c u a m o t i i u i y p h a t d i e n x o a y c h i e u c6 2 cSp cuc. D e t a o r a tAn so
Vdri
5 0 H z t h i r o t o p h a i q u a y v 6 i to'c do b a o n h i e u ?
I3
= I12 + I 3
I i = I2 = I 3
N e n t a c6 h i n h
tat
HUdng
p = 2 c a p cUc
To'c d p q u a y c u a r o t o
Tom
f = 50 H z
np
f =
N = ?
gidi
Ma
f.60
n =
60
dan
T i f h i n h v e t a c6 I12 = I i = I2 = I 3 = 2 A
I12 T i is n e n :
Io = I12 -
50.60
I3
* TrurOng h o p 2:
Up
200
R>
100
B a i 4. S t a t o c u a m o t d o n g co k h o n g d o n g bo b a p h a g o m 6 c u o n d a y . C h o
Up
200
d o n g d i e n x o a y c h i e u b a p h a c6 t a n so f v a o d o n g ccf. TiS t r u d n g t a i t a r n
R2
100
cua s t a t o q u a y v d i to'c d o l a 1 5 0 0 v o n g / p h u t . T i m t a n so f.
Up
100
n = 1500v6ng/phut .
Tom
Hiidng
tat
dan
100
gidi
= lA
o I , = 1 (A)
. I12
1,2=1, = I 2 = 2A •
• T a n so c i i a d o n g d i e n x o a y c h i e u b a p h a .
Tim f
I2 = 2 ( A )
TCr h i n h v e t a c6:
c a p cUc.
n - 1500 v o n g / p h u t
I, = 2 (A)
i . + 12+13
• T r o n g s t a t o c6 6 c u o n d a y tacfng ufng v d i p = 2
6 cuon d a y
Io=0
= 0
f =
np
f =
1500.2
f = 50Hz
60
60
Ma
ii2 T i ia
I12 >
Nen
I3
Io = I12 - I 3 = 2 - 1 = l A
o Io=lA
B a i 5. M o t m a n g d i e n 3 p h a m ^ c h i n h sao c6 h i e u d i ^ n t h e g i f f a h a i d a y p h a
l a 2 0 0 V3 ( V ) .
B a i 6. M o t d p n g co d i e n b a p h a m a cac c u o n d a y c u a d p n g co d a u k i e u h i n h
a) T i m h i e u d i e n t h e g i C a d a y p h a v a d a y t r u n g h 6 a .
sao
b ) T i n h c u d n g d p d 6 n g d i e n cdc d a y p h a v a d a y t r u n g h o a k h i cac t a i t i e u
mac vao m a n g
0,9. T i n h c o n g s u a t t i e u t h u c i i a d p n g co.
* C a c t a i g i o n g n h a u , m o i t a i t i e u t h u c6 R = 6 0 Q v a Z,, = 8 0 Q.
* T a i t i e u t h u thur 1 c6 R i = 100 Q , t i e u t h u t a i thuT 2 c6 R2 = lOOQ v a t a i
Tom
Hu&ng
• Ud = 2 0 0 V3 ( V )
1,1 =
U,i = >/3 U p
b)
Io = ?
I„ = 2 A
gidi
U
(V)
=
200(V)
Hiidng
Nhaii
=
T '_ Up
R2 = 1 0 0 Q
R3 - 2 0 0 Q
xet:
ma
1,1
dan
gidi
C o n g s u a t t i e u t h u c i i a d p n g co b ^ n g 3
Pd, = 3 P p = 3.Up.Ip.cos(p
coscp = 0,9
P,
xet:
I a n c o n g suat t i e u t h u cua m S i p h a .
Vdi:
. Up = ^
= 220 (V)
V3
» I p = Id = 2 ( A )
Trircfng h o p 1:
= 80 n
* R i = 100 Q
Nhqn
Up =
tat
73
?
* R = 60 Q
ZL
ddn
a) H i e u d i e n t h e giOra d a y p h a v a d a y t r u n g h o a
a) T i m U p
b) T i m
Tom
U,, = 2 2 0
tat
sao co h i e u d i e n t h e d a y l a
2 2 0 N/S ( V ) . B i e t r S n g c u d n g dp d o n g d i e n d a y l a 2 A v a h e so c o n g s u a t l a
t h u m ^ c h i n h sao t r o n g h a i t r i r d n g h o p .
t i e u t h u thuf 3 c6 R.-, = 2 0 0 Q.
d i e n b a p h a mSc h i n h
U„
200
VR^7Z[
100
= 2 A
V a y P r: 3 . 2 2 0 . 2 . 0 , 9
= I p - > I„ = 2 ( A )
BAI TAPTRAC NGHIEM
• CiTdng d p d o n g d i e n t r o n g d a y t r u n g h o a .
D o cdc t a i t i e u t h u g i o n g n h a u n e n c u d n g d p d o n g d i $ n hie^'
d u n g q u a m o i p h a c6 d p I d n g i o n g n h a u v a cac d o n g d i e n n a y l e c h p h a
P = 1188(W)
Cfiu
1. M o t k h u n g d a y c6 tCf t h o n g d a n g :
n h a u m o t goc 120". ( d o cac t a i t i e u t h u do'i xufng n e n goc l e c h p h a giOfa i'
qua d u n g
v a i t r o n g m o i p h a d e u g i o n g n h a u - > cac d o n g d i e n l e c h p h a n h a u 1 2 0 " )
I . l . TCr t h o n g cifc d a i t r o n g k h u n g l a
A. 2 W b
B. 2 m W b
(j) = 2 . 1 0 ~ l c o s l 0 0 7 t t ( W b ) .
C. 0,2 W b
D . 0,02 W b
Chpn ket
u 8. Stato ciia mot dong cor k h o n g dong bp ba pha gom 9 cupn day. Cho d6ng
d i e n xoay chieu ba pha c6 t a n so' 50 Hz vao dong ca. Tii t r u d n g t a i tarn cua
stato quay v6i toe dp la
1.2. S u a t d i e n dong liieu d u n g cua k l i u n g la
A. 1007t 72
mV
B. 200 7 t . l O ~ ' V
C . x/2 71.10"' V
D. 2 7t.lO~'V.
1.3. Bieu thiifc suat d i e n dong cam ufng cua k h u n g theo t l a
A.
e = 0 , 2 7 i . s i n l 0 0 7 t t (V)
C. e = - 0 , 2 7 i . s i n l 0 0 7 T t (V)
B.
e = 2.sinl007Tt
D. e -
nsinlOOTit
A.
(V)
(V).
B.27t m V
C. 7T.10-'* V
D. 4 7r m V
C. — Wb
30
D. —
Wb.
300
2.2. Tir t h o n g cUc d a i c6 gia t r i la
A. —
Wb
600
B. — m W b
60
B. <\> = —coslOOrit
30
(mWb)
D.1,08
Wb
B. 80 V
2000 v6ng/s
dien xoay chieu ba pha c6 t a n so 50 Hz vao dong ca. Roto l o n g soc ciia dong
A. 20 v6ng/s
B. 25 v6ng/s
C. 30 v6ng/s
D. 40 v6ng/s
C a u 10. M o t m a n g d i e n 3 pha h i n h sao eo hieu d i e n the pha la
100V3(V).
Hieu d i e n the day l a
A. 200N/3 V
B.300 V
C. 300^3 V
D.400V3V
, t r u n g hoa la lOO(V). H i e u d i e n t h e giiJa h a i day pha l a
B.lOO V
D. 200V3
V
C a u 12. M o t m a y p h a t d i e n xoay chieu 3 pha h i n h sao c6 h i ^ u d i ? n the pha la
2 2 0 ( V ) . T a i mSc h i n h sao do'i xiifng va t a i cua m o i pha l a R = 3 Q ,
7^=40.
Chon k e t qua dung
12.1. H i e u d i e n the day ciia m a n g d i e n la
A. 110 V
B.127 V
C. 200 V
D. 381
12.2. Cucfng dp dong d i e n hieu dung qua cac t a i la
3.2. B i e n dp cua suat d i e n dong la
A. 84,78 V
n 50^ vong
C.
3
s
c O CO the quay vdi toe dp la
A.100N/3 V
C. (i) = —cosl207it (Wb)
D. (b = —cosl207:t (mWb).
^
30
60
C a u 3. M o t cuon day det h i n h chO n h a t c6 dien t i c h 54 cm^ c6 500 vong day qua;
deu v d i v a n toe 50 v6ng/s quanh mot true n a m t r o n g m a t p h a n g ciia day, troiig
tii tru'dng deu vuong goc vdi true quay c6 B = 0,1 T. Chon k e t qua dung.
3 . 1 . Tii t h o n g cUe d a i qua cuon day la
A. 0,54 Wb
B . 0 , 8 1 Wb
C. 0,27 Wb
, 1000 v6ng/s
B.
C a u 11. M a n g d i e n 3 pha h i n h sao c6 hieu d i e n the giijfa m o t day pha va day
2.3. Bieu thufc tCr t h o n g c i i a k h u n g theo t la
A. d) = —cosl007it (mWb)
^
60
vong
phut
C a u 9. Stato eua m o t dong cor k h o n g dong bp ba pha gom 6 cupn day. Cho dong
C a u 2. M o t k h u n g day c6 suat d i e n dong cam iJng c6 dang: e = 27t.lO"^.sinl20itt
(V). Chon k e t qua dung
2 . 1 . Suat d i e n dong cUc d a i c6 gid t r i la
>
A. 271 V
50
3
C.82 V
D . 8 6 V.
C a u 4. P h a n cam ciia m o t m a y p h a t d i e n xoay chieu gom 4 cap cifc. Cac c u o n
day cua p h a n iJng no'i t i e p va c6 so vong t o n g cong la 600 vong. Tii thony
c u e d a i qua m o i vong day va to'c do quay cua roto p h a i c6 gia t r i nao de suat
d i e n dong eo gia t r i hieu dung 1^ 220 V va t a n so la 50Hz
• A. 750 vong/ p h i i t ; 1,65 m W b
B. 12,5 v6ng/s; 2,65 m W b
C. 750 vong/phut; 0,1 m W b
D. 12,5 vong/s; 0,3 m W b .
C a u 5. Roto cua m a y p h a t d i e n xoay chieu eo 6 cap eUc. De tao ra t a n so 5 0 H /
t h i roto p h a i quay v d i toe do bao nhieu?
A . 500 v6ng/s
B. 500 vong/phut
C. 50 v6ng/s
D. 250 vong/phut.
C a u 6. Co 2 m a y p h a t d i e n xoay chieu roto ciia m a y thu" n h a t c6 4 cap ciTc qua>
3000 vong/phut. H o i roto ciia m a y thu" h a i c6 6 cap eUc t h i p h a i quay v d i tot
do bao n h i e u de c6 the dau 2 m a y song song
A. 1000 vong/phut
B. 2000 vong/phut
C. 3000 vong/phut
D. 4000 vong/phiit.
C a u 7. Roto ciia m o t m a y p h a t d i e n xoay chieu eo 4 cap cUc. De tao r a t a n so
60 Hz t h i so roto p h a i quay v d i to'c dp bao nhieu?
A. 900 v6ng/s
B. 15 v6ng/s
C. 9000 vong/phut
D. 30 vong/s.
A. 40 A
B. 4,4 A
C. 44 A
D. 24 A
C. 2808 W
D. 17424 W
12.3. Cong suat t r e n m o i t a i la
A. 5808 W
B. 2000 W
C a u 13. M o t m a y p h d t d i f n xoay chieu 3 pha c6
Up = 2 2 0 ( V ) .
Cudng
dp
hieu
dung
qua
R = lOOQ k h i no m^c vao A va B l a
A. 2,36 A
B. 2,81 A
C. 2,2 A
D. 2,2^3
B
A
C a u 14. M o t dong ca k h o n g dong b p ba pha dau theo k i e u h i n h sao vao m a n g
dien 3 pha c6 h i e u d i e n the day la 220V3{V) m^c h i n h sao. Dong c P c6 cong
suat l O ( k W ) va he so cong suat 0,8. Chon k e t qua ddng
14.1. H i e u d i e n the dua vao m 6 i pha cua dong ca c6 gia t r i la
A. 220 V
B. 200 V
C. 210 V
D. 240 V
14. 2. Cudng d p hieu d u n g qua m o i cuon day ciia dong ca c6 gia t r i la
A. 18,9 A
B. 56,8 A
C. 60,5 A
D. 80 A
I
Van de 2: C A M K H A N G - D U N G K H A N G - T O N G TRCJ -
B I E U THUfC
BAI T A P M A U
DIEN A P VA C U d N G DO DONG DIEN
B a i 1 . Cho mach d i f n nhir h i n h ve:
P H l / d N G PHAP
I , B i e u thiifc cifcfng dp d o n g d i ^ n tutc thoTi, d i ^ n a p t i J c t h d i
A
R
-c
i = Io.cos(cot + cp,)(A)
B
M
N
R = 20 (Q); L = ^
u = Uo.cos(cot + (p„)(V)
I I . C o n g thii'c q u a n
L
71
"
(H); C = —
An
D i e n ap d a t vao h a i dau mach c6 dang: U A B = 200 42 coslOOitt (V).
cpu, (Pi, 9
9u = 9i +
Tim:
taiKp =
b) Cudng dp dong dien hieu dung qua mach.
a) C a m k h a n g cuon day, dung k h a n g t u d i e n
Trong
R +r
Tom tat
^•' = 2
L =
I I L M p t so c o n g thu:c k h a c
0,2
a) Tijr bieu thiJc d i e j i dp h a i dSu mach:
U A B = 200 V2 coslOOTit (V)
(H);
Uo^^ = 200V2(V)
10 ^
1. C^m kh^ng cuon day Z\_ (Q)'
CO = 1007r(rad/s)
47:
= Leo
= 0
U A B = 200 N/2 coslOOTit (V).
2. Dung khSng tu di§n Zc (Q)
Cam k h d n g cupn day
) a) Z,,; Zc; Z = ?
b) I = ?
Co)
c)
3. Djnh luStOm
1=
R
ZL = L.M =
?
U M B =
Cw
10"'
R
'
z,
Zc
T o n g t r d mach
Z/^B
Z A B =
Z = V(R + r f + ( Z , - Z c ) '
Q = (R + r ) l H
p = (R + r ) ! '
Zc = 40Q
.10071
47r
°"
^.IOOTI
Z, = 2 0 Q
?
• Dung k h a n g t u d i e n
z^
Zc
UAN =
-
Hudng ddn gidi
R = 20 ( Q ) ;
71
Tt
t o n g t r d cua doan mach.
c) D i e n hieu dung giiifa h a i d i e m A va N ; M va B.
do: cp do lech p h a giCa d i e n ap va cifdng dp dong d i e n (rad)
Nhd:
(F)
VR'' + ( Z , - Z J
^
|Z^B = 2 0 V 2 ( Q )
b) Cudng dp dong d i e n hieu dung qua mach: I =
'AD
W = P.t
Ma
UAB =
Vay
1=
^
V2
200V2
~
^/2
200
= 200
(V)
I = 5>y2(A)
20V2
T r o n g do:
* P: Cong suat (W); Q: N h i e t l i / a n g (J); W: D i e n nfing t i e u t h u (J)
I k W . h = 3600000 (J)
0 Tim
•
UAN;
UAN =
UMB?
1=
^
- > U A N
AM
= I . Z'AN
(1)
V20^ + 20' = 20 V2 Q
(1) - > UAN = 5V2 .20V2 -e^
= 200(V)
Vdi
ZAN =
+
=
(1)
coscp =
Vdi
ZMB = yJiZ^-Zc?
(2)
UMB = 5X/2.20 ^. U „ „
coscp = - 7 =
100%/2
V2
42
2
1
= 2 p h u t = 1 2 0 s.
U MB
Q = R l ' t = 100(2 sl2 )'.120 -> Q = 96000(J) = 96(KJ)
(2)
U M H = I . Z MB
1
100
B a i 3. M o t bong den g h i (220 V - 100 W ) dmc mdc vao hieu dien t h e 200 V .
= V(20-40)^- ^ 20 (Q)
a) T i m dien trcj bc3ng d e n va cu'dng do dong dien d i n h milc cua den.
=100V2(V)
b) H o i den cc3 sang b i n h thuorng khong? V i sao?
c) T i n h dien n a n g t i e u t h u cua den t r o n g 1 gid.
B a i 2. Cho mach dien gom R, L , C m^c n o i tiep n h u h i n h :
R
A
P-
H
L
C
nsm''
I
R = 100 (Q); L =
1 ^
(mH); C .
Hii&ng dan gidi
Tom tat
B
• D (220 V - 100 W )
M
2K
. U = 200 V
(,F)
a) D i e n t r d bong d e n
(220V Ta c6: D (220 V - 100 W )
Q
a) T i m R = ?
i = 4cosl007Tt (A)
Tinh:
a) Cong suat toa n h i e t cua doan mach va he so cong suat mach.
b) N h i e t liTong toa r a t r o n g 2 phut.
b)
U.„„ = 2 2 0 V
I,i„, - ?
Den c6 sang
P,„.=100W
binh
c) W - ?
HiCdng dan gidi
'In = 4 ( A )
^
• 0) = 1007t(rad/s)
9i = 0
= - (H)
• C a m k h a n g cuon day
2K
Z,, = L . M = - . I O O T I
2K
• i = 4cosl007it (A)
a) T i n h P = ? coscp = ?
b) Q = ? Cho t = 2 phut
= 120 s.
= 100 Q
• Dung khang t u dien
Zc =
1
Co)
10"
271
= 200 Q
-.IOOTT
Cudng do dong dien h i e u dung qua mach
I =
In
V2
S
^2^2
(A)
a) • Cong suat t o a n h i e t ciia doan mach
P = R I ' = 100(2
N/2
f
P = 800W
He so cong suat mach
coscp =
Vdi
I^
ZAB =
/ I \
(1)
+(ZL -ZJ^*
= yj\0(f + (100 - 200)' = 100 V2 (Q)
220'
U = 200 V
100
R = 484Q
Iilin
Tii bieu thiJc i = 4cosl007it (A)
(mH)
ui„
R =
thUcfng kliong? V i sao?
Cho t = 1 gicJ = 3600 s.
Tom tat
• R = 100 Q
^
,
1000
lOOW)
-
100
220
I , . = - I A )
b) De n sang m d hcfn muTc b i n h t h u d n g l a do U < U,in,
c) D i e n nSng t i e u t h u ciia den t r o n g 1 gicf
W = P.t
'
(1)
Cong suat t i e u t h u ciia d e n
P = RI'
(2)
50
,
U
200
Vc5fr I = — =
(A)
R
484 121
(2)
P = 484.
f 50
121/
82,64 W
Nhdn xet: O day P < P,i,n, I < I j , , , do den sang mor hcfn miJc b i n h thi^dng.
The P vao (1)
(1) ^ W - P.t - 82,64.3600
^ W
297504(J) « 0,08264(KWh)
0 2
B a i 4. M o t cuon day c6 d i e n trcJ R va he so tu' cam l a
( H ) . K h i d a t vao
71
hai dau cuon day dien ap m o t chieu 40 V t h i cadng dp dong d i e n t r o n g
cupn day l a 2A. K h i d a t vao h a i dau cupn day dien ap xoay chieu u =
40cosl00Trt (V) t h i cudng dp dong d i e n h i e u dung qua cupn day l a I . T i m R
va cLfdng dp dong d i e n hieu dung k h i d a t vao h a i dau cupn day d i e n dp
xoay chi^u n h u t r e n .
^ ^ Hu&ng ddn
Tom tat
.L=^(H)
R, L
Nhii vdy:
• IT = 0,02 (s) thi dong dien doi chieu 2 Ian (-> sdng 2 Ian, t^t 2 Ian)
• Is thi dong di$n doi chieu 100 Ian (den sang 100 Ian, tdt 100 Ian),
b) TiJ' hinh ve ta thay
T T T T 2T 2.0,02 0,04 (,s ),
+ t„„ + t B u = - + - +
giai
U , L<
n
• Uic = 40 (V)
• lie = lA
• = 40cosl007tt (V)
Tim R va I^^ = ?
t ^ = = t c A
'xc •
Khi
vao 2 dau
dat
lie
R
day dien ap
-.R=:^ = ^
CUQII
R = 20(Q)
mpt
chieu
Vdi:
R
i„
f
i
0,2
Z,, = Lw =
The vao (*) ^ I.c =!
V20' + 2 0 '
c)
(V)
55N/2 (V)
sang
ã
ô^
i-tAt
+ t o e + t c o + toD =
3
0,02
•-.St
=
-
^
=
—
—
Cudng d o d o n g
I = 2A, t a n s o
T
+
T
+
12
+
12
T
=
—
12
0,02 , ^
—
=
3
(S)
3
=2
dien qua m o t doan m a c h xoay
la f
= 50 Hz. Tinh til
a ) c h u k y c i i a d o n g d i e n k e tis
Hitcfng dan
o
-I—
giai
liic t =
luc t
- 0
chieu
thi i
c6 g i a t r i
= 0. Tinh
hieu
di?n
ililltllM
+55v/2
— I
B
+ 110N/2(VI
a)
q =
i-T
£^i.dt
=
J^'^IoSinwt.dt
.. . . . » - _ . . V . . . * . . , . . , ' < . J - J . . . - . . , J „ . L A ' . J
q = lo
l(
<-> q = lo —(coscot - cosO) ^ q = lo
T
T
Uo '
0.
Hitdng ddn giai
Bieu thiic cudng do dong dien qua doan mach
dien xoay chieu c6 dang: i = losincot
(do t = 0 thi i = 0 nen (pi = 0)
dq
-> d q = i.dt
ma
1 =1
dt
->
D
0.
n i j a c h u k y d o n g d i e n k e tif l u c t -
Tom tat
I = 2 A, f :r 50 Hz
t = 0 thi i = 0
Tim q trong:
a) 1 chu ky clia dong dien.
b) ^ chu k y cua dong
tHi
-110v/2(V)
a) Hoi trong 1 s c6
bao nhieu Ian den a) Nhan xet:
sang, bao nhieu
• Do hieu dien the hieu dung la 110 V
Ian den t^t.
U^/2 = 110 V2 (V)
b)
• Trong 1 chu k y dong dien doi chieu hai
trong 1 chu ky.
Ian d^n sdng, 2 Ian den t^t).
t.
ma T = - = — = 0,02 s
c)
f 50
I'sang
—
lUOng q u a t i e t d i e n c u a m a c h trong:
I.. = 1(A)
-5572
_
"lit
dung
Bai 5. Mot den neon mdc vdri m a n g dien xoay chieu c6 dien ap hieu dung la
110 V va tan so 50 Hz. Bi§'t rkng d^n sang khi dien ap giiJa hai ciTc khong
nho hmi 55 42 (V).
a) Hoi trong mot giay c6 bao nhieu Ian den sang, bao nhieu Ian den t^t?
b) Tim thdi gian den sang va thdi gian den t^t trong 1 chu ky?
c) Ti so giijfa thcfi gian den sang va thcri gian den tat trong mot chu ky bao nhieu?
-Uo
A I—
iDO
0,04
^saiig
B a i 6.
b)
Tom tat
. U = llOV
• f = 50 Hz
• Den sang khi u >
T
^
12
lOOn = 20 Q
20N/2
=
- t
twt -
^
.u -±^ = i^ = 2 0 V 2
tAc
0,04 (s)
thi:
• Khi dat vao 2 dau cuon day dien dp xoay chieu thi:
(*)
V
+
b) q = J^^i.dt = J ^ I „ s i n w t . d t
q =
It
•
T
If
^ q = lo
27:
IV2 „
IN/2
lo
cosM 2n cosO
0)
lo - - ( - 1 - 1 )
CO
2.V2,
—.coswt
CO
q =0
1,
T cosO)"1
= Io —(cosco
—.coscot
CO
— cosco
cosO
col^
CO.2
2
03
[ sincot.dt =
V2
2
B a i 7. C h o do t h i b i e u d i e n sir p h u t h u p c c i i a i v a o t n h u h i n h v e s a u :
t = — ^ + — ( k = 1,2,3...)
300
100
i(A)
* C u d n g d p d o n g d i $ n t r e n d a y d a t g i a t r i cue d a i
+1-
O
\600
T a c6: i - 2cos lOOnt + 3
\
/
\
m a i = + 2 A —> +2 - 2cos lOOnt + 3
\)
-2
<-> lOOTTt + -
3
cos lOOTlt + -
= 1
3;
= k27I <^ t = - — + — ( k = 1,2,3...)
300
50
* C u d n g d p d o n g d i e n t r e n d a y c6 g i a t r i i = 0
T a c6: i - 2cos lOOnt + 3
a) V i e t b i e u thiJc oxibwg do d o n g d i e n tiJc t h d i t h e o t .
b ) X a c d i n h nhCfng thcfi d i e m m a t a i do cac c u d n g do d o n g d i e n t r e n d a y d a t :
m a i = 0 - » 0 = 2cos lOOTit + 3
* G i a t r i ci/c d a i h o a c cUc t i e u .
* G i a t r i eye d a i .
* G i a t r i b a n g 0.
lliCdng
tNhqn
dan
xet:
• Luc t = 0 t h ii = +1A va i dang
giam
^
T
= -1^
12
600
•s - > — =
12
600
Tom
Viet
= —rad
+ q)|)
coscpi =
=
2
o
(D, = — r a d
3
D o l u c t = 0, i = l A v a d a n g g i a m n e n c h o n (pi = — ( r a d )
3
lOOTit + •
(A)
( k = 0,1,2,3...)
71
i = 2cos lOOTlt + •
3j
Nhdn
thu'c
hieu
dan
R
gidi
B
xct:
c h u k y n a y h i e u d i e n t h e tufc t h d i h a i d a u
mach
c u n g c h i n h l a h i e u d i e n t h e tiJc t h d i h a i
d a u d i e n t r d R.
UH =
.cos(a)t + (p„^ )
* U,,^ = lo.R = 2.100 = 200 ( V )
*
(1)
=
IOOTI
(rad/s)
—
6
(*)
(rad)
M a t k h a c cpu = 0
N
N ee n
n (*)
(*) ^
- > (p„
cp , =
= - + 0 = - (rad)
"
6
6
lOOTlt + 3]
( A ) . V i e t b i e u thiJc h i e u
B i e u thufc h i e u d i e n t h e tufc t h d i h a i d a u R c6 d a n g
De cho cpi =
* C u d n g d p d o n g d i e n t r e n d a y d a t g i a t r i cifc d a i h o a c ciTc t i e u
M a i = ± 2 A - > ± 2 = 2cos lOOnt + 3
bieu
dien the h a i dau mach.
'P„„ = 9, + q)u
(A)
b ) X a c d i n h cac t h d i d i e m m ^ t a i d o :
T a c6:
A
cos^
'
i = 2COS
100
HUdng
d
1 = loCOSCwt + (Pi) ( * )
Vay:
tat
i = 2cos lOOrrt + •
Do t = 0 t h i i = l A v a d a n g g i a m n e n
1 = 2COS(100K.O
k
dien the h a i dau mach.
R = 100 Q
B i e u thufc cifdng do d o n g d i e n tiJc t h d i c6 d a n g :
^
600
= IOOT: (rad/s)
600
(*)
600
+
d o n g d i e n q u a m a c h c6 d a n g i = 2cos l O O T l t H - -
• K h o a n g thcfi g i a n tii liic i , = l A d e n iz = 0 l a
CO
1
B a i 8. M a c h d i e n x o a y c h i e u c h i chuTa d i e n t r o R = 1 0 0 Q. B i e u thiJc c u d n g d o
a) TCr do t h i t a t h a y In = 2 A
ma
3y
<-> lOOTtt + - = ( 2 k + D - 4^. t =
3
2
gidi
=^0
<-> cos lOOTtt + -
V a y : u ^ = 2 0 0 c o s lOOnt + - ( V ) .
.
'
