Kỹ năng phân loại và phương pháp giải chi tiết bài tập trắc nghiệm Vật lý 12 (Trọng tâm): Phần 2

Nối tiếp nội dung phần 1 tài liệu Phân loại và phương pháp giải chi tiết bài tập trắc nghiệm Vật lý 12, phần 2 giới thiệu tới người đọc các nội dung: Dao động và sóng điện từ, dòng điện xoay chiều, sóng ánh sáng, lượng tử ánh sáng,... Mời các bạn cùng tham khảo nội dung chi tiết.

ChiTtfng IV DAO DQNG VA SONG DI|N Tlf

P H A N I : T O M T A T L I T H U Y E T

L

B

Van de 1: D A O D O N G D I E N T L /

1 Dao dOng di$n tC( trong mach LC

a) Mach dao dong

Dat Uo = ^ UAB = Uocos(o)t + (p)

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 = , chu ky

T = In

(0

= 2Tt^/LC yd tan so f =

271 V L C "

N h a n x e t : Cirdng do dong dien trong mach n h a n h pha - s o v d i dien

tich va hi§u dien the giijra h a i b a n cifc cua t u dien, h i f u di?n the cung

pha dien t i c h cua t u dien

3, Nang lUOng di^n tQ trong mgch dao d^ng LC

• N a n g lifpng tufc thcfi ciia t u dien:

Wc = - C u ' = - C cos2((ot + (p)

2 2 Dat w, oc = lcu^ = i c 4 = ^ 2C

N h a n x e t : Nang lifpng dien trUdng cua t u dien Wc va nang Itfcfng t i f

T - ,

trucmg cua cupn cam W L bien thien tuan hoan v6i chu k y — , tan so 2f, tan so

2

goc 2 CO Doi vdi nang liWng dien tii W la mot hkng 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 tieu hao nang lupng ( d o R c h ^ n g han) V i vay, dao dpng se dCrng l a i sau k h i tieu hao h e ' t nang lupng H i e n

tiipng nay gpi l a dao dpng dien tii tit dan Gia t r i R cang Idn t h i sU t ^ t

dan cang nhanh

5 Dao d6ng di$n tCf duy tri, h§ tU dao d6ng

Muon CO dao dpng dien tif khong t ^ t dan ta c 6 the duy t r i bhng each cung c a 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 LC dugrc duy t r i

o n din h t h i d a o dpng trong khung LC dupc duy t r i o n dinh vdi t a n so rieng Wo

ciia 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

a) Dao dong dien tii cUang bilc

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 dien the bien t h i e n theo t h d i gian u - Uo.cosw.t c h ^ n g h a n t h i dong dien trong mach LC se khong the d a o dpng theo t a n so g o c riehgco,, m a phai bie'ii t h i e n theo t a n so goc w Qua t r i n h nay dupc gpi l a d a o dpng

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 khung dat gid t r i C L T C d a i

Van de 2: D I E N TL/ T R U C l N G

1 Dien trUdng bi^n thiSn tCl trUdng bidn thiSn

a) TU tritang bien thien: PQii mot tiT trircfng bien t h i e n theo thcfi gian thi

no sinh ra mot dien trUdng xoay, tiJc l a dudng sufc ciia dien trUdng nay

khep k i n va bao boc xung quanh duomg siJc tir

b) Dien truang bien thien: K h i mot dien tri/dng bien t h i e n theo t h d i gian

t h i no sinh ra mot tij' triTcfng xoay Dudng sufc t i ^ ciia tCr trir6ng nay khep

k i n va bao boc xung quanh du&ng sure dien triTcfng

2 Dien tC/ trUdng

• M o i bien t h i e n theo t h d i gian ciia t\i trUdng deu sinh ra trong khong

gian xung quanh mot dien truorng xoay bien t h i e n theo t h d i gian, va

nguoc l a i

Ket luan: Dien trUdng bien t h i e n va tCr trUcfng bien t h i e n cung ton tai

trong k h o n g gian Chung c6 the chuyen hoa Ian nhau trong mot truang

thong nhat duoc goi l a dien tCf trUdng

• TCr trUdng bien t h i e n cang nhanh t h i cUcfng do dien trUcJng xoay cang Idn

va n g U O c l a i

V^ande 3; S O N G D I E N TL/

1 S6ng di§n tQ 1^ gi?

Song dien t\i la sU Ian truyen cua dien ti^ trifdng trong khong 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 dien tCr (Hz)

- Song dien tii la song ngang Trong qua t r i n h truyen song t a i mot diein

bat k y t r e n phuong truyen, vectd E , vector B luon vuong g6c nhau va

vuong goc vdi phuong truyen song

- Song dien tiT cung tuan theo cac d i n h luat phan x a , khiic x a va cung co

the giao thoa vdi nhau

- Qua t r i n h truyen song dien tir trong khong gian mang theo nSng luong

2 SU truyen s6ng dien tQ quanh Tr^i Dit

- K h i truyen song dien t\i trong thong t i n quanh T r a i DS't phu thuoc vao

cac yeu to nhu: Budc song, dieu k i e n moi trUcfng t r e n mat dat va t i n h chat ciia bau k h i quyen

* S o n g d a i v a s o n g t r u n g ;

- Song nay bi phan xa d t a n g dien l i va c6 kha nSng di vong quanh T r a i Dat qua nhieu Ian p h a n xa giUa t a n g dien l i va mat dat NgU6i ta diing song nay trong truyen t h a n h va truyen h i n h t r e n mat dat

- Song dai i t bi nUdc hap t h u nen diing de thong t i n dudi nUdc

* S o n g n g S n : phan xa d t a n g dien l i , phan xa t r e n mat dat nhieu I a n , do

do truyen dupc xa t r e n mat dat

* S o n g c i / c n g S n : Song nay cd nSng lupng Idn nhat, va khong bi t a n g dien l i phan xa va hap t h u nen truyen thang Song nay dUprc iJng dung

de thong t i n trong cU l i vai chuc k m hoac truyen thong qua ve t i n h (thong t i n vu t r u )

TRAC NGHIEM LI THUYET

'^^u 1 Su bien t h i e n ciia dong dien i trong mot mach dao dpng l$ch pha nhU

j | the nao so vdi sU bien t h i e n ciia dien t i c h q ciia mot ban tu dien

A i C l i n g pha vdi q B i lech pha n vdi q

C i sdm pha — so vdi q

2 D i tre pha — so vdi q 2

Cau 2. Mot m a c h dao d o n g LC c6 dion trcf t h u a n b h n g k h o n g Khi trong mac}^

CO d a o d o n g d i e n til tiX do v d i b i e u t h i J c d i e n t i c h t r e n hkn tu d i e n

• q = quCos(cut + cp) t h i g i a t r i c u c d a i c u a ciTdng do d o n g d i e n t r o n g m a c h l a

A (0 Qo D. CO qo

C a u 3. Chu k i dao dong rieng cua dao dong dien tii tii do trong mach dao don,,

LC (c6 dien trd thuan khong ddng ke) la

1 2n „ „ 1

A T = 2HVLC B. T =

V L C •

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 (Ill) D (IV)

QSiU 10. Mot mach dao dong dien til LC g6m cuon day thuan cam c6 dp tu cam

L khong doi va tu dien c6 dien dung C thay doi duoc Biet dien trd cua day

dSn la khong dang ke va trong mach c6 dao dong dien tit rieng Khi dien

1

^^'^'-'^ ^ ^ f c dung CO gia t r i Ci thi chu k i dao dong rieng ciia mach la Tj Khi dien dung

Cau 4. Tan so dao dong cua dien tH tu do ciia mach LC c6 di|n trd thuan WBr'c<3 gia tri C2 = I6C1 thi chu ki dao dong dien tU rieng trong mach la

khong dang ke la

A f ' B. f=27iVLC c f = — V L C D f = 2n •

V L C •

27:VLC ' 271

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 trd 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 bilu thiJc tinh nSng lupnf,r

dien tCr trong mach?

Cu^

2 2C

Cau 6. Trong mach dao dong LC gom mot tu dien c6 dien dung C v^ cupn day thuan

cam CO dp tU 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 tri cUc dai

LC D Io = U o i l C

VO'l

Cau 7. Trong mach dao dong LC c6 dien trd thuan bang khong thi

A NSng lifpng tit trudng tap trung d cupn cam va bien thien tuan hoan

chu ki bang nijfa chu ki 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 ki bang

niia chu ki dao dpng rieng ciia mach

D Tai moi thcfi diem t thi tong cua nSng lircfng dien trudng va nSng lupng

trudng luon tang

C a u 8. Khi 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

A. T2 = 16T, B. T2 = 8T1 C. T2 = Ti/4 D. T2 - 4 T i

Cau 11- Phat bieu nao sau day sai khi 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 kin

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 tu cam L Biet dien trd 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 trd 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 ki 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 ki 2T

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 Bien t h i e n t u a n hoan vdi chu k i —

D K l i o n g bien t h i e n dieu hoa theo t h d i gian

C a u 16 Trong mot mach dao dong LC gom cupn day t h u a n cam c6 dp t\J cam L

khong doi va t u dien c6 dien dung C thay doi dupc Chpn cau diing

A Chu k i dao dong r i e n g cua mach khong 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 giam k h i tSng di$n dung C cua t u dien

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 T a n so dao dong r i e n g ciia mach tSng gap doi k h i dien dung C cua tn

dien g i a m gia t r i dien dung cua tu dien di 4 Ian

C a u 17 M o t cuon day t h u a n cam (cam thuan) c6 dp tU cam L m ^ c n o i tiep vdi

mot tu 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 dien til tiT do cua mach nay phu thupc vao

A dong dien cUc dai chay t r o n g cupn day ciia mach dao dong

B dien t i c h cUc dai cua ban t u d i e n t r o n g mach dao dong

C d i ^ n dung C va dp tU cam L ciia mach dao dong

D hieu dien the cUc dai giiJa h a i ban t u dien ciia mach dao dong

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 vfin toe 3 lOm/s

B Lk song doc

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

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

ca hai vectct E , B luon vuong goc vdi phudng t r u y e n song

C a u 19 K h i phat bieu ve song dien tU, phat bieu nao diTdi day khong dung?

A Song cue ng^n t r u y e n dupe t r o n g chan khong

B Song ngin c6 t a n so' nho hon t a n so song dai

C Song cue ngan dupe dCmg t r o n g t h o n g t i n vu t r u

D Song dai dupc diing de t h o n g t i n dudi nude

C a u 20 Phat bieu n^o sau day dung k h i ndi ve song d i ^ n tCr?

A Song dien tCf mang nftng lupng

B Song dien tCr ehi t r u y e n dupc t r o n g chat k h i

C Song dien tii k h o n g t r u y e n difpc t r o n g chat r d n

D Song dien 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 a u 21 Song dien tii

A La song dpc

' B vera c6 song ngang, vCra c6 song dpc

C Chi phan xa ma khong khiie xa k h i gap mat phan each giijfa hai m6i triTdng

D Mang nSng lupng

C a u 22 K h i ndi ve song d i ^ n ti^, phdt bieu nao sau day sai?

A Song dien tii la song dpc

B Song dien 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

C Song di$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

moi truomg

D Song dien tif t r u y e n dupc t r o n g rhi, long, k h i va ke ca chan khong

fiu 23 K h i n d i vi song dien tCf, phat bieu nao dudi day khong diing?

A S6ng dien tCr cung bi 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

moi trUdng

B Song dien til chi t r u y e n dUpe t r o n g m o i triTcfng vat cha't dan 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

C Song dien tCr la song ngang

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

C a u 24 Song dien til va song eP hoc k h o n g cd ehung t i n h ehat nao sau day?

A Phan xa B Truyen dupe t r o n g chan k h o n g

C Qiao thoa, n h i l u xa D M a n g n a n g li/png.'

C a u 25 Song dien tCr diing trong t h o n g t i n giOa cac tau ngam la loai song v6

tuyen nao?

A Song cue n g ^ n B Song n g ^ n

C Song t r u n g D Song dai

C a u 26 Song dien tU dCing t r o n g t h o n g t i n giu-a cac nha du h a n h vu t r u v a m a t

dat la loai song v6 tuyen nao?

A Song eire n g ^ n B Song n g ^ n

C Song t r u n g D Song dai

C a u 27 Chpn eau dung

A Trong song dien tCr t h i dao dong ciia dien trifdng va ciia til trWdng t a i mot

diem luon luon ddng pha v d i nhau

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 Cac sdng n g ^ n k h o n g the t r u y e n di xa t r e n m a t da't

D Cac song ngan v6 tuyen khong phan xa tot tren tang dien l i v a tren mat da't

C a u 28 Phat bieu nao sau day la dung k h i noi v a cac loai sdng v6 tuyen?

A Song d a i ehii yeu dupe diing de t h o n g t i n dudi nuTdc

, B Song t r u n g c6 the t r u y e n dupe ra't xa vao ban ngay

I C Song ngan cd n a n g lu'png nho hpn song t r u n g v a sdng dai

D Sdng cue n g ^ n bi t a n g dien l i v a m a t da't phan xa

C a u 29 Chpn phat bieu dung

A Sdng d a i cd nang lirpng thap i t b i niTdc ha'p t h u

B Sdng t r u n g phan xa dupe t r e n t a n g dien l i vao ban dem nen ehung t r u y e n

dupe xa

C Sdng cue n g ^ n khoang 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 I I : B A I T A P T R A C N G H I E M

+ B A I T A P L U Y E N T A P

I Dao dpng d i ^ n txJf

* Dien tich tuTc t h d i q - qocos(a)t + cp)

* Dong dien tufc t h d i i = Iocos((ot + 9+ ^ ) v<Ji lo = wQo

* Hieu dien the tufc thcfi u = = ^cos({ot + cp) = UoCos((ot + cp)

• C: Dien dung cua t u dien (F)

• ZL: Cam khang cuon day ( Q )

• W: Nang lirong dien i\l (J)

B a i 1 Mach dao dong gom tu dien c6 dien dung 400 pF va mot cuon cam c6

do tir cam 0,04 H T i n h chu k y dao dong rieng v^ t a n so rieng ciia mach dao dong

Tom tat

C = 400 pF

= 400.10"'- (F) L = 0,04 (H)

L = 0,0625(H)

B a i 4 Mach dao dong L C c6 L = 400 m H va C = 4 nF Biet hieu dien thg' cUc dai giOra hai ban cUc ciia tu la 400 V T i m circfng do dong dien circ dai qua mach

Bai 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

Bai 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

Ta c6: Khoang thdi g i a n giiJa h a i Ian l i e n tiep ma

nSng lu'Ong tii trUomg b a n g nftng lu'Ong dien

Bai 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 thi ta phai cung cap cho mach mot cong suat trung binh la 50 ^iW

Tim dien trd cuon day?

r = 1(Q)

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 '

Tom tdt

L = 1 mH

= I.IO"'' H C = 4 pF

^ 4.10"'^ F Tim 1 = ?

Hiidng ddn gidi

Chu ky dao dong rieng ciia mach

T = 2%sIhC = 27rv'lO '.4.10 a 3,97.lO"'' (s)

Budc song ma mach thu duoc la

a) dien tich tren tu dien C vao thdri gian t

b) dong dien qua mach vao thdi 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) hay i = 0,02.cos 5.10H + -

2 (A)

BAI T A P T R A C N G H I E M

C a u 1 Mach dao dong gom L = ( m H ) , tu C = ( F ) Tan so goc cua mach la

Ạ 200 rad/s B 300 rad/s C 444 rad/s D 500 rad/s

C a u 2. Mach dao dong LC c6 C = 200(nF) T a n so dao dong ciia mach

500Hz Do tu cam ciia cuon cam trong mach dao dpng la

Ạ 0,507 H B 0,607 H C 0,707 H D 0,807 H

C a u 3 Mach dao dong LC c6 L = 0,04(H), t a n so dao dong rieng ciia mach dao

dong la 60Hz Dien dung ciia t u la

Ạ 1,76.10" F B 2 0 1 0 ' F C SỌIỐ^ F D 1,54.10'F

C a u 4. Mach dao dong LC c6 L = 2000(nH), C = 3000(pF) Chu k y dao d o n o

rieng ciia mach la

Ạ 1 0 ' s B 10"' s C 1,4.10' s D 1,53.10"'s

C a u 5. Mach dao dong LC c6 L = 5000(nH), chu k y dao dong rieng ciia mach hi

2.10"^(s) Dien dung ciia t u la

Ạ 0,02 F B 0,03 F C 0,04 F D 0,05 F

C a u 6. Cirdng do dong dien trong mach dao dong l a : i = I„sin2000t(A) Tu dien

C = 5|.iF Do tiT cam cupn day 1^

Ạ 0,02 H B 0,03 H C 0,05 H D 0,04 H

C a u 7. Mach dao dong ciia m p t may t h u v6 tuyen c6 L = 5(f.iH) va C = l,6(nF)

May CO the t h u dúoc song v6 tuyen c6 búdc song la

Ạ 150 m B 100 m C 168,5 m D 190 m

C a u 8. Mach dao dong ciia m o t may t h u v6 tuyen gom L = 8(|.iH) va t u dien

bien thien tCr 20(pF) den 450(pF) H o i may c6 the bat duofc cac song vo

tuyen dien trong day song naỏ

Cau 11 Mach dao dong c6 tu C dUgre tich dien q = 2.10"^sinl007tt(C) Bieu thiJc

cUdng do dong dien qua mach la

Ạ i = 10.10" sinlOOTtt ( A ) B i = 2007t.l0 "sinlOOTtt (A)

C i = 20071.10 ".sini 1007rt +

-V 2j

(A) D i = 40071.10 •'sinl007it (A)

C a u 12 Mach dao dong LC c6 dong dien i = 40071.10'".sin(1007it + 7i)(A) Dien

tich ciia tu bien đi theo t h d i gian c6 dang

A q = 2.10"^sin lOTit - (C) B.q =4.10''.sin lOOTtt +

-2 (C)

C a u 14. Mach dao dong c6 n&ng li/prng dien tii ciia mach la 2 ( m J ) , dien tich

cue dai ciia t u dien la 12(nC) Dien dung ciia tu la

Ạ 3 , 6 1 0 " F B 2 1 0 " F C 4.10"'C D 8.10"" C

C a u 15. Mach dao dong LC c6 dien dung C = 6 ( p F ) , hieu dien the cure d a i d 2

ban t u la 200(V) N a n g lifong dien tir cua mach la

C a u 18. Mach dao dpng LC c6 dien tich cUc dai hai ban tu la q,, = 3(nC), dien dung

C = 2(|iF) T i m nSng lúpng dien tnrdng ciia tu dien va nang lUcfng tir trircJng cua cupn cam k h i nang lirpmg dien trUcfng bang 3 Ian nang lUcfng tiT trircfng

Ạ 1.6875.10"''J; 2,65.10"" J 6.1,5.10""J; 5,625.10"'"J ,, C 1,6875.10 " J ; 5,625.10 ' ' J D 1,5.10"''J; 1,6875.10""J

BAI TAP LUYEN TAP

i 1. Mach dao dpng cua m o t m a y t h u v6 tuyen c6 L = 32(nH) H o i de b^t dupe song 120(m) t h i t u c6 dien dung bao nhieủ

Bap so: 0,126 n F

B a i 5. M a c h dao 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 chu k y dao d p n g cua

m a c h l a T, = 3 ( s ) , k h i C = t h i chu 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 song t h i chu k y dao d p n g cua m a c h l a bao nhieu?

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 trUdng

ciia t u d i e n v a n S n g l i i g n g tii trucrng ciia cupn c a m k h i n S n g l u g n g d i ^ n bang

i 12 M a c h dao d o n g L C co L = 4 m H vk C = 1 nF Luc circrng dp d 6 n g d i $ n qua

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

D a p so: a) 5 3 2 , 8 7 m<X< 1 4 0 9 , 8 m ; b ) 1065,75 m

B a i 16 M a c h c h p n s o n g ciia m o t m a y t h u v6 t u y e n g o m c u p n d a y t h u a n c a m co

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

goc x o a y b i e n t h i e n t i ^ 0° d e n 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

goc x o a y ciia 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 ciia t u l a h a m b a t

nha't d o i v d i goc x o a y T i m h e so tiT c a m L

i = q'(t) = 2.10~^1007i.cosl007it i = 2.10"Vos.l007tt (A)

hay i = 2.10-^7isin(1007it + ^ ) (A)

B a i 6. De cho L = L j + L2 ( * )

ma T = 271 V L C ^ = 47I2LC

(T^ t i le thuan v d i L ) (*) _> ^ Tf + T| = 100 T = 10 s

B a i 13 Sau t h d i gian lO-'^ s t h i nSng lifdng tCr trudng cua cuon cam l a i dat gia

tri cue dai ^ - = 10-' (s) ^ T = 2.10-' (s) 2

T i m L = ? T = 271 N / L C L = - 5,07.lO"** H

B a i 14 Nhan xet: t = 0 t h i i = lo = 20.10-^ (A)

Bieu thiJc cUdng do dong dien c6 dang:

(*) -> I„ = I„cos(25.10\ + (Pi) <-» coscp, =: 1 <^ cpi = 0

Vay i - 20.10-lcos(25.10't) (A)

a) Bieu thufc dien tich t r e n cac ban tu dien theo t h d i gian

q ^ H 0 : l e o s ( 2 5 1 0 ' t - ^ )

c 4.10'' 2 -> u - 20.cos(25.10''t - - ) (V)

2

ai 15

a) ^1 = c.T, - 3.10«.27I7L:C7 = 3.10«.2.3,14 V 4 1 0 •'.20.10'^ = 532,87 (m) ^2 = C.T2 - 3.10^27T7LC7 = 3.10^.2.3,14^4.10 ^ 1 4 0 1 0 " ' ' = 1 4 0 9 , 8 m

ChiTtfng V DONG DIEN XOAY CHIEU

P H A N I: T O M T A T L I T H U Y E T

Van de 1: DONG DIEN XOAY CHIEU

MACH DIEN XOAY CHIEU CHJ CO DIEN TRCJ THUAN,

TU DIEN, CUON CAM

1 S u i t dien d^ng xoay chi^u

- T h e o d i n h luat c a m iJng d i e n tCf suat

d i e n dong c a m ilng xuat h i e n trong

k h u n g d a y c6 bieu thuTc:

e = Eocos(a)t + cpo)

- S u a t d i e n dong n a y b i e n t h i e n dieu

hoa theo t h 6 i g i a n theo d i n h luat

d a n g s i n , thildng goi t^t l a suat d i e n

dong xoay chieu

- C h u k i v a t a n so l i e n he bori: w = — = 27if

T

- S u a t d i e n dong do cac m a y p h a t d i e n xoay chieu tao r a cung c6 bieu

thiirc tuorng t\i nhif t r e n

2 Dien xoay c h i l u Ddng diSn xoay chiSu

- D i e n ap b i e n doi dieu h o a theo th&i gian l a d i e n ap xoay chieu

- D o n g d i e n c6 cudng dp b i e n doi dieu h o a theo t h 6 i g i a n l a dong dien

xoay cl^ieu

V d i mpt d o a n m a c h xoay c h i e u t h i bieu thufc d i e n ap h a i d a u d o a n m a c h

v a CLfdng dp dong d i | n qua m a c h c6 bieu thiirc:

u = UflCosCcot + cpu)

i = Iocos((ot + (pi)

Trong do: Uo, cpu lo, ^>i' I a n luat l a cac b i e n dp va p h a b a n dau cua hieu

d i e n the h a i d a u doan m a c h v a cu'dng dp dong d i e n qua m a c h

• D a i lupfng 9 = cpu - 9i goi l a dp l^ch pha giOfa u v a i trong mot doan mach

dp va pha dao dong

3 Doan m a c h xoay chidu chl c6 diSn t r d t i i u l n

c) Dinh ludt 6m: In =

d) Tdc dung chtnh cua tu dien

- L a m cho di§n dp b i e n t h i e n c h a m p h a so vdi ciTdng do dong dien

- C a n t r d d o n g d i ^ n

Van 6e2: MACH CO R, L, C MAC NOI TIEP CONG HUdNG DIEN

1 C ^ c giS tri tL/c thdi

X e t doan m a c h R , L, C m ^ c no'i tie'p t h i t a c6:

UAB = U R + U L + Uc

• G i a suf i = IQ.COS CO t t h i t a c6:

UR = UoRCOscot, U L - UoLCOsCcot + 7i/2), U c = Uoccos((i)t - nl2)

vdi: UoR = IQR, UQL = IQZL, UQC =

IQZC > U = UoRCOSCOt + UoLCOS((Ot+IQZC ) + Uoccos(cot IQZC IQZC ^ )

K e t l u ^ n : H i e u d i e n t h e u h a i d a u doan m a c h l a tdng h o p ciia b a dao

dong dieu h o a cung t a n so, n e n t a t i m dao dong tong h o p u n ^ y b ^ n g

t h i u n h a n h pha hon i (mach c6 t i n h cam

t h i u cham pha hon i (mach c6 t i n h dung

+ Neu: Z L = Zc hay coL = — t h i u ciing pha vdi i (mach cong hucfng dien)

wd'i u h a i dau doan mach

* Cudng dp hieu dung qua doan mach RLC dat gid t r i cUc dai:

C h i n h v i vay dien ap giCra hai dau mach b^ng dien dp hai dau R

Van de 3: CONG SUAT CUA DONG DIEN XOAY CHIEU

HE SO CONG SUAT

1 Gang su^t cue doan mach: D i e n nSng doan mach tieu t h u t r o n g 1 s

* Cong suat turc t h 6 i p = ui

* Cong suat t r u n g b i n h , cung l a cong

Y nghia h e so cong suat

* Trifofng hofp coscp = 1 <=> cp = 0

Day l a triiorng hop doan mach dien xoay chieu chi chufa R, hoSc mach 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 khong chiJa dien trd thuan

(Day la t r u o n g hop mach chi c6 cuon cam L hay t u di§n C hoac ca L va

C) Luc nay P = 0

* T r i i c f n g hcfp 0 < coscp < ! < = > - — < ( p < 0 hoac 0 < cp < —

Luc nay: P = UIcoscp < U L Day la trirdng hop hay gap nhat

Van de 4: MAY PHAT DIEN XOAY C H I E U

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: Dua t r e n

hien tiTOng cam ufng dien t\i

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

- TiT truorng quay, eac vong day dat co dinh

2 MSyph^t di0n xoay chi^u m6t pha

Co hai bo phan c h i n h la phan cam va p h a n iJng

- Phan cam la n a m cham dien hoac nam cham v i n h cOru Do Ik p h i n tao

ra tii trudng

- Phan ling la nhOfng cuon day, t r o n g do xuat h i e n suat dien dong cam

ufng k h i may hoat dong

Mot t r o n g h a i phan dat co' d i n h , phan con l a i quay quanh mot true

Phan CO d i n h goi la stato, phan quay goi la roto

3 MSy phSt diSn xoay chi^u ba pha

a) Dong dien xoay chieu ba pha

Dong dien xoay chieu ba pha la he thong ba dong dien xoay chieu, gay

boi ba suat dien dong xoay chieu co cung t a n so, cung bien do nhung

lech nhau ve pha la —

3

b) Cdu tqo vd hoat dgng cua may phdt dien xoay chieu ba pha

• May c6 cau tao gom Stato c6 ba cuon day r i e n g re, hoan toan gio'ng

nhau quan t r e n ba l o i sdt dat lech nhau 120° t r e n mot vong t r o n Roto

la mot nam cham di$n

K h i roto quay deu, eac suat dien dong cam iJng xuat h i ? n t r o n g ba c u p n

day CO cung bien do, cung t a n so nhiftig lech nhau ve pha 1^ — Neu

3 noi cac dau day cua ba cuon vdi ba mach ngoai giong nhau t h i ta co he

ba dong dien cung bien do, cung t a n so' nhung lech nhau ve pha 1^ —

4 Ccich mSc di$n 3 pha

a) Cdch mac hinh sao

Day t r u n g hoa

A , i 2 Day pha 2

i3 D a y p h a 3

C a c cong thufc: Ud = Vs Up ya Id =Ip

b) Cdch mdc hinh tam gidc

Day pha 1

A i 1 B,.,

Day pha 2

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 nam c h a m q u a n h mot true, tCr trUcfng do n a m cham gay ra

CO eac dxibng sufc tCr quay trong k h o n g gian Do l a mot tCf trUdng quay Neu dat giufa hai cUe ciia mot nam cham h i n h chiJ U mot k i m nam cham

va quay deu nam c h a m chuf U t h i k i m nam c h a m quay theo vdi cung toe

do goc Ta noi k i m nam cham quay dong bp vdri tii trUdng

b) Su quay khong dong bg

Thay k i m nam cham b k n g mot khung day dan k i n K h u n g nay co the

quay quanh true xx' t r u n g v d i true quay ciia nam cham Neu quay deu

nam c h a m t a t h a y k h u n g day quay theo cung chieu, d e n m6t liic nao do

k h u n g day cung quay deu nhitog vdi toe do goc < co c u a nam c h a m

Dong CO hoat dong difa theo nguyen t^c n6i t r e n goi la dong eo k h o n g

dong bo

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

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 CO 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 lai goi la roto

long soc

• 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

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 lam thay doi tan so

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 Loi thudng lam bhng cdc la sdt hoac thep pha silic, ghep each

dien vdi nhau de giam hao phi dien nSng do dong Phu-eo Cac cuon

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 C L i a 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 hai dau cupn thiJ cap vd cupn so cap b l n g ti so

vong day ciia hai cuon day

• Neu N2 > Ni -» U2 > U i : Mdy tdng dp

• Neu N2 < Ni -> U2 < U i : Mdy ha dp

Neu bo qua moi hao phi nSng luong trong may bien ap, cong suat trong

mach so cap va thu" cap hhng nhau

u, •

Do do: Dung may bien dp lam 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

C a c e a c h g i a m AP

C a c h thiir n h a t : Giam dien.trd R cua dudng day Day la edeh ton kem vi phai tdng tiet dien cua day, do do ton nhieu kim loai lam 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

CSu 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 hi|u dung la

u

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

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

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 khi tan so

goc CD thoa man

A. 0 ) = - 7 = = ^ B CO = C CO = , — D CO = —

V L C ^/LC V L , 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 trd R trong khoang thdi gian

t (t rat Idn so vcii chu k i ciia dong dien) la

Cau 7 Dat vao hai dau doan mach RLC noi tiep mot di§n ap xoay chieu

u = Uocoscot thi dp lech pha giufa u va i ciia mach difpc tinh theo cong thiJc

T - w C — ^ ^ (oL - Ceo „ ^ Leo

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 khi

A CO L < B CO L > C CO L = D oj > i

-eoC coC coC LC Cau 10 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

-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

2 mach A B chiJa

A Dien trd thuan B Tu dipn

C Cupn day thuan cam (cam'thuan) D Cupn day c6 dien trd thuan Cau 12 Doan mach dien xoay chieu gom dien trd 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 U R sdm pha so vdi uc B U L tre pha - so vdi Uc

2 2

C. U R tre pha — so vdi uc D u^ sdm pha — so vdi uc

2 2 Cau 13 Dong dien xoay chieu trong doan mach chi cd dien trd 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 trd ciia mach

D Luon lech pha7r/4 so vdi dien ap d hai dau doan mach

Cau 14 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 15 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 thi dong dien qua doan mach

J , A Sdm pha — so vdi dien ap d hai dau doan mach

A i = U„coC cos(a)t + n/2) B i = U Q W C coswt

C i = UocoC cos{(ot - 71/2) D i = U Q C J C cos((ot - ^ )

C a u 21. D a t d i e n ap u = U Q coscot v d o h a i d a u doan mach' c h i c6 t u d i e n C i '

cudng dp d o n g ' d i e n tdc t h d i chay t r o n g m a c h l a i P h a t bieu n a o sau d a y dung'

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 i e t U R = U C = 2Ui, H e so' c o n g suat cua m a c h di$n l a

A S d m pha gdc nl2 so v d i cUdng dp d o n g d i e n qua cupn d a y

B S d m pha m o t gdc k h d c n/2 so v d i cUdng dp d d n g d i e n qua cupn d a y

C T r e pha m o t gdc TC/2 SO v d i cUdng dp d o n g d i e n q u a cupn 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 qua cupn d a y

C S u 26 D a t d i e n dp u = Uocoscot vao h a i d a u m o t doan m a c h c h i cd t u d i e n C t h i

CUdng dp d o n g d i e n tdc t h d i chay t r o n g m a c h l a i P h a t bieu nao sau d a y dung?

A D o n g d i e n i l u o n cimg pha vdi d i e n dp u

28 M o t m a y phat d i e n x o a y chieu m o t pha ( k i e u c a m d n g ) cd p cap cUc

• quay 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 cua

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 giUa n, p v a f I d

A n p = 6 0 f B n p = ^ C p = ^ D f = ^

f f n

C a u 29 Dong cO dien xoay c h i l u 1^ t h i e t bi dien bien doi

A D i e n nang t h a n h co nSng B D i e n nang t h a n h hoa nang

C Co nang 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 ndi ve dong ccf dien khong dong bo, p h a t bieu nao sau day diing?

A Bien ddi co nang t h a n h dien nSng ciia dong dien xoay chieu

B Hoat dong dua t r e n hien tugng cam ufng dien txi v a suf dung ti^ trUdng quay

C Tan so quay ciia roto Idn hcfn tan so cija dong dien xoay chieu qua dong co

D Roto cua dong co quay d6hg bo vdi tit trUdng quay trong dong ccf

C a u 31 K h i dong cd khong dong bp b a p h a hoat dpng on d i n i i v d i toe dp quay

cua tii trufdng khong doi t h i toe dp quay cua roto se

A Nho hon toe dp quay ciia tii trUdng

B B^ng toe dp quay ciia tiT trUdng

C L d n hcfn toe dp quay eiia tii' trUdng

D Co the l d n hon, nho hon hoac bang toe dp quay ciia tCf trUdng

C a u 32 Doan mach dien xoay chieu gom bien t r d R, cuon day thuan cam cd do

tif cam L v a tu dien cd dien dung C m^c noi tiep Biet hieu dien the hieu

dung hai 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

tan so dong dien t r o n g mach k h o n g doi Thay doi R de'n gia t r i Ro t h i cdng

suat tieu t h u cua doan mach dat gia t r i ciTc dai Pmax, k h i do

A R o = f - B R „ = | Z , - Z c l

C R „ = Z L Zc D. R „ = Z L + Z C

C a u 33 Dat hieu dien the u = Uosinwt (Uo v a 03 khong ddi) vao hai dau dom;

mach RLC k h o n g p h a n n h a n h Biet dp t y cam v a dien dung difpc giOr khoiu

doi Dieu c h i n h t r i so' dien t r d R de cong suat tieu 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

C a u 34 Dat vao hai dau doan mach RLC khong p h a n n h a n h mot dien dp xoay

chieu u = Uocoscot (V) t h i dong dien t r o n g mach la i =locoscot (A) Doan mach

dien nay luon ed

A. Z L > Zc B Zl, < Zc C Z L = Zc D. Z L = R

Caxi 35 Dat dien a p u = uV2coscot vao hai dau doan mach RLC khong phan

n h a n h (dien t r d thuan R =^ 0) Chpn dp tuf cam cua cuon day v a di$n dunr!

ciia t u dien sao eho cam k h a n g b^ng dung k h a n g t h i

A Tdng t r d ciia doan mach nho hcfn dien t r d thuan R

B Cudng dp dong dien t r o n g doan mach n h a n h p h a so vdi dien ap u

C He so cong suat ciia doan mach la 0 < eosqx 1

D Cirdng dp dong d i ^ n hipu dung qua mach cd gid t r i cUc dai

•

U 36 Klii 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 dien t r d thuan cung pha vdi dien ap tUc

t h d i giOra hai ban tu dien

B Dien ap tUc t h d i giufa h a i diu dien t r d thuan cung pha v d i dien ap tUc

t h d i giufa dau cuon cam

C Cdng suat tieu t h u t r e n doan mach dat gia t r i l d n nha't

D Cudng dp dong dien tUc t h d i t r o n g mach sdm pha - vdi d i | n dp tiJc t h d i

3 dat vao hai dau doan mach

Cau 37 Doan mach dien xoay chieu khong phan n h a n h gom cuon day cd dp tU

cam L, dien t r d thuan R va tu dien cd dien dung C KJii dong dien ed t a n so'

f = 7 = chay qua doan mach t h i h f so cong suat cua doan mach nay

2 7 r V L C

A 0 < coscp < 1 "

B Phu thupc dipn t r d thuan ciia doan mach

C Bang 1

D Phu thupc t d n g t r d ciia doan mach

C a u 38 M o t mdy bien dp cd hieu suat xap x i bang 100%, cd so' vdng day cuon

§A L a m giam t a n so' dong dien d cuon thuT cap 20 Ian scf cap l d n hon 20 I a n so vdng day cuon t h i i cap M a y bien dp nay

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 di?n d cuon scf cap 20 Ian

C a u 39 M o t mdy bien dp dupc suf dung l a m mdy t d n g dp D a t vao h a i dau cuon

so cap mot dien 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 vdng day ciia cuon thu" cap nho hen so' vdng day ciia cuon so cap

B Dien dp hieu dung d hai dau cuon thu" cap nho hon dien dp hieu dung d hai dau cuon so cap

C Cudng dp hieu dung ciia dong dien t r o n g cuon thU ca'p b^ng eUdng dp dong dien hieu dung cua dong dien trong eupn so cap

D Cudng dp dong dien hieu dung t r o n g cuon thU cap nho horn cudng dp dong dien hipu dung ciia dong dien t r o n g eupn so cap

jCSu 40 M o t may 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 Dat vao hai dau eupn scf cap mot dien dp xoay

'f chieu cd gid t r i hi$u dung U i t h i dien 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

(II) : Tdng chieu dai dudng day ( I I I ) : Tdng dien the trUde k h i truyen t a i (IV) : G i a m t i e t dien day va giam dien the trude k h i t r u y e n t a i Trong qud t r i n h t r u y e n t a i dien ndng, bien phdp l a m g i a m hao p h i t r e n dudng day t a i di^n dupe sii dung chii yeu h i e n nay la

A (I) B' ( I I ) ' C ( I l l ) D ( I V )

* ct)„: T'lr thong cifc dai (Wb)

* N ; So vong day ciia khung

* B: Tuf trirdng (T)

* S: Di$n tich khung (m^)

* e: Suat dien dong t a i t h d i diem t (V)

* E,,: Suat dien dong cUc dai (V)

* E: Suat dien dong hieu dung (V)

I I T a n so' c i i a m a y p h a t d i ^ n

f ^ np * f: Tan so dong dien (Hz)

60

f = n.p * p: So' cap ciTc

* n: Toe do quay (vong/ phiit) hoac (v6ng/s)

B a i 1 Mot khung day quay quanh true eo dinh trong tCr trudng deu B ma tii^

thong bien doi theo t c6 dang: (j) = 2.10"^cos(1007rt) (Wb; s)

a) T i m tCf thong cUe dai ciia khung

b) Viet bieu thu'c suat dien dong hai dau khung day ,

Hii&ng ddn gidi

a) TCr thong cifc dai

TCr bieu thu'c de cho (|)o = 2.10"^ (Wb)

b) Bieu thurc suat dien dong d hai dau k h u n g

e = -(t)'(t) = -[2.10-'cosl007ttJ

-> e - 0,27isinl007Tt (V)

hay e = 0,27rcos l O O n t

-2 (V)

B a i 2 Mot khung day dan cd 500 vong day quan no'i tiep, moi vong c6 dien

ti'ch la 200 em^ Khung day dugc dat trong ti^ triTcmg deu B = 0,2T Luc t = 0 thi vecto phap tuyen cua khung hop vdi B mot goc — Cho khung day

6 quay deu quanh true (A) vuong goc vdi B vdi tSn so 40 v6ng/s

a) Viet bieu thu'c suat dien dong cam iJng theo t

b) Viet bieu thufc tijr thong theo t

c) Xac d i n h gia t r i ciia suat dien dpng cam iJng va cua tCf thong t a i t h d i diem — (s)

Hitdng dan gidi

a) Bieu thiJc suat dien dong cam iJng theo t

HUdng dan gidi

To'c dp quay cua r o t o

= l A o I, = 1 ( A )

I12 TCr h i n h ve t a c6:

220 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 suat l a 0,9 T i n h c o n g s u a t t i e u t h u ciia d p n g co

Tom tat

U,, = 2 2 0 73 ( V ) I„ = 2 A

coscp = 0,9

P , =

V a y P r: 3.220.2.0,9

Hiidng dan gidi

Nhaii xet: C o n g s u a t t i e u t h u ciia d p n g co b ^ n g 3

1.2 Suat dien dong liieu dung cua kliung la

C a u 2. M o t khung day c6 suat dien dong cam iJng c6 dang: e = 27t.lO"^.sinl20itt

(V) Chon ket qua dung

2 1 Suat dien dong cUc d a i c6 gid t r i la >

C a u 3 Mot cuon day det h i n h chO nhat c6 dien tich 54 cm^ c6 500 vong day qua;

deu vdi van toe 50 v6ng/s quanh mot true nam trong mat phang ciia day, troiig

tii tru'dng deu vuong goc vdi true quay c6 B = 0,1 T Chon ket qua dung

3.1 Tii t h o n g cUe dai qua cuon day la

A 0,54 Wb B.0,81 Wb C 0,27 Wb D.1,08 Wb

3.2. Bien dp cua suat dien dong la

A 84,78 V B 80 V C.82 V D.86 V

C a u 4. Phan cam ciia mot may phat dien xoay chieu gom 4 cap cifc Cac cuon

day cua phan iJng no'i tiep va c6 so vong tong cong la 600 vong Tii thony

cue dai qua moi vong day va to'c do quay cua roto phai c6 gia t r i nao de suat

dien dong eo gia t r i hieu dung 1^ 220 V va t a n so la 50Hz

• A 750 vong/ phiit; 1,65 mWb B 12,5 v6ng/s; 2,65 mWb

C. 750 vong/phut; 0,1 mWb D 12,5 vong/s; 0,3 mWb

C a u 5 Roto cua may phat dien xoay chieu eo 6 cap eUc De tao ra t a n so 50H/

t h i roto p h a i quay vdi toe do bao nhieu?

A 500 v6ng/s B 500 vong/phut

C a u 6. Co 2 may phat dien xoay chieu roto ciia may thu" n h a t c6 4 cap ciTc qua>

3000 vong/phut H o i roto ciia may thu" hai c6 6 cap eUc t h i p h a i quay v d i tot

do bao nhieu de c6 the dau 2 may song song

A 1000 vong/phut B 2000 vong/phut

C 3000 vong/phut D 4000 vong/phiit

C a u 7. Roto ciia mot may phat dien xoay chieu eo 4 cap cUc De tao ra 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

u 8. Stato ciia mot dong cor khong dong bp ba pha gom 9 cupn day Cho d6ng

dien xoay chieu ba pha c6 t a n so' 50 Hz vao dong ca Tii trudng t a i tarn cua stato quay v6i toe dp la

C a u 9. Stato eua mot dong cor khong dong bp ba pha gom 6 cupn day Cho dong

dien xoay chieu ba pha c6 t a n so 50 Hz vao dong ca Roto long soc ciia dong

cO CO the quay vdi toe dp la

C a u 12. M o t may phat dien xoay chieu 3 pha h i n h sao c6 h i ^ u di?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 moi pha la R = 3Q, 7 ^ = 4 0

Chon ket qua dung

12.1. Hieu dien the day ciia m a n g dien la

C a u 13. M o t may phdt d i f n xoay chieu 3 pha c6

Up = 2 2 0 ( V ) Cudng dp hieu dung qua

C a u 14. M o t dong ca khong dong b p ba pha dau theo kieu h i n h sao vao m a n g

dien 3 pha c6 hieu dien the day la 220V3{V) m^c h i n h sao Dong cP c6 cong

suat lO(kW) va he so cong suat 0,8 Chon k e t qua ddng

14.1. Hieu dien 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 dung qua moi 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

c) Dien hieu dung giiifa h a i diem A va N ; M va B

-Tom tat

R = 20 ( Q ) ; 0,2

CO = 1007r(rad/s)

= 0 Cam k h d n g cupn day

Z L = L M = ^ I O O T I

Z, = 2 0 Q

Cw 10"' Zc = 40Q

47r 10071 Tong t r d mach

2 0 V 2 I = 5>y2(A)

0 T i m U A N ; U M B

-• U A N = ?

1 = ^ - > U A N = I Z 'AN (1)

a) Cong suat toa nhiet 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

HiCdng dan gidi Tii bieu thiJc i = 4cosl007it (A)

'In = 4(A)

^ • 0) = 1007t(rad/s) 9i = 0

• Cam k h a n g cuon day

Z,, = L M = - I O O T I = 100 Q

• Dung k h a n g t u dien

1

Zc = Co) 10" = 200 Q

B a i 3. M o t bong den g h i (220 V - 100 W ) dmc mdc vao hieu dien the 200 V

a) T i m dien trcj bc3ng den va cu'dng do dong dien d i n h milc cua den

b) H o i den cc3 sang b i n h thuorng khong? V i sao?

c) T i n h dien nang tieu t h u cua den trong 1 gid

Tom tat

• D (220 V - 100 W ) U = 200 V

a) T i m R = ?

I,i„, - ? b) Den c6 sang binh thUcfng kliong? V i sao?

c) W - ? Cho t = 1 gicJ = 3600 s

Hii&ng dan gidi

a) Dien t r d bong den ( 2 2 0 V - lOOW)

50 , U 200 Vc5fr I = — =

va cLfdng dp dong dien hieu dung k h i dat vao h a i dau cupn day dien dp xoay chi^u n h u t r e n

i Z,, = Lw = 0,2 lOOn = 20 Q

The vao (*) ^ I.c =! V20' + 20' 20N/2 I = 1(A)

Bai 5 Mot den neon mdc vdri mang 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?

bao nhieu Ian den

sang, bao nhieu

• Trong 1 chu ky dong dien doi chieu hai

Ian d^n sdng, 2 Ian den t^t)

• 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 , ,

a) 1 chu ky clia dong dien

b) ^ chu ky cua dong

2 IV2 „ IN/2

lo - - ( - 1 - 1 ) CO

2.V2, V2

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) Xac d i n h nhCfng thcfi d i e m m a t a i do cac cudng do d o n g d i e n t r e n d a y d a t :

* G i a t r i ci/c d a i hoac cUc t i e u

B a i 9 Mach dien xoay chieu chi chiJa cuon day thuan cam c6 he so tif cajj^

L = — (H) Dat vao hai dau mach dien ap c6 dang: u = 400cos

Viet bieu thtife ciforng do dong dien tufc thcri qua mach

dong dien tufc t h d i qua

Tom tat

271

• i = %/2 sindOOTit) (A)

Viet bieu thufc dien ap

tufc t h d i hai dau tu dien

Hitdng dan gidi N/ian xet:

- 0 day de cho bieu thufc cUdng do dong di';'

tiJc t h 6 i CO dang sin Ta c6 the viet bit

thufc dien dp cung c6 dang sin r o i sau d' chuyen bieu thufc sang dang cos

- Cdcli khdc: Ta se chuyen bieu thiJc cu6ng "

d5ng dien tufc t h d i sang dang cos va sau ('^

viet b i l u thiJc di^n dp cung c6 dang cos

2 ( * ) - t cp,,^ = 0 - ^ = rad

Vay: u^ = 200v2sin lOOTit

-2 (V) Hay u, = 200V2cos(l007it - 7r)(V)

Bai 11 Cho mach dien n h u h i n h

i = 2cos 1007lt +

-4 (A)

- - — 1 0 0 7 1 27t

C S u 16. M o t d6ng di$n xoay chieu t r o n g moi giay dao chieu 100 I a n Chu

cua dong dien la

A 2s B 0,2 s C 0,02 s D.0,002 s

C a u 17. Cho mot den ong dung dong dien c6 t a n so 50Hz se

A Phdt s&ng hoac t ^ t 25 Ian moi giay

B Phat sang hoSc t ^ t 50 Ian moi giay

C Phat sang hoac tfit 100 Ian moi giay

D Phat sang hoac t a t 75 I a n m o i giay

C a u 18. Cho dong dien xoay chieu c6 dang: i = 2^2cosl007it(A) B i e n dp ciia

cUcfng dp dong dien la

A 2 A B.2V2 A C 4 A D 6 A

C a u 19 Bien do cua dien ap xoay chieu dat vao 2 dau doan mach la 220 (Vi

doan mach chi chu'a dien t r d R = lOOQ Cudng do dong dien hieu dung la

A 2,2 A B 2 A C ^ A D 4 A

V2

C a u 20. Cho doan mach dien xoay chieu c6 R, L , C m^c noi tiep Biei

R = 100Q, L = - ( H ) , C = ( F ) , dien ap cifc dai d 2 dau mach la

2 0 0 V 2 ( V ) , f = 50Hz Chon cau sai

A R L C B

I 1 1 /2222S2SN / / 1

A Cuorng do dong dien hieu dung la V 2 ( A )

B U R = l O O ^ ( V ) ; U , = 100N/2(V) ; U^ = 2 0 o V 2 ( V )

C Dien ap hieu dung 2 dau mach la 200(V)

D Cudng do dong dien cham pha hcfn dien ap U A B la — •

4

u= 100V2.cosl007rt(V) Biet R = 50Q , C = ^ ^ ^ F Chon cau dung

5v37r

A Cudng do dong dien hieu dung la 0,1 A

B U R = 5 0 V; Up = 5 0 V

C Dien ap hai dau mach n h a n h pha hcfn i la ^

D Cudng do dong dien hipu dung la l A va n h a n h pha hon dien ap h a i d i n

Q§iU 23. Mot bong den c6 ghi (220 V - 75 W) dugc mic vao hieu dien the 210 (V)

23.1. Den sang n h u the nao?

A Den sang b i n h thudng

B Den md hon binh thudng

C Den sang hon b i n h thudng

D Luc dau md, sau sang hon miJc b i n h thudng

23.2. Dien nSng tieu t h u ciia den t r o n g 2 phut la

C a u 27. Cho mach dien R, L , C n o i tiep Biet R = l O Q , L =

4000,

C a u 28. Cho mach dien gom R, L , C n o i tiep Biet R = 200Q , L = ( m H )

C = — ( ^ F ) , u^B = 400V2cosl007Tt(V) Bieu thuTc dien ap h a i dau t u C la

271

A Uc = 400cos 1 0 0 7 l t

-2 (V) B Uc = 400cos lOOTlt

-37t (V)

C Up = 400N/2COS lOOTit +

-4 (V) D Up = 400^/^cos lOOnt + • (V)

V^an cfe 3; Dp LECH PHA

Lt^u y : Neu U , vuong pha U 2 t h i : tancpj.tancpj = - 1

BAI TAP MAU

B a i 1 Cho mach dien gom R, L , C m^c n o i tiep n h u h i n h ve:

471

• «5AH = - - ra d

4 a) T i m R = ? b) Viet UAM-

b) D o l e c h p h a giiJa UAN v a " N H

-0,25 (H); f = 50 H z

Tom tat

R = 25 Q;

0,25

L = ( H ) ; f = 5 0 H z UAN v u o n g p h a v d i UAB a) T i m C ?

C S u 3 1 Mach di^n gom hgp k i n X chi chtira mpt trong 3 phan tiif R', L hoSc C

va R = lOO Di§n ap 2 dau mach sdm pha han cirdng d p dong di§n 1 goc i

4

Cho / = 50Hz Chpn cau dung

A Hop k i n X c h i J a L = ^ ( H ) B Hop k i n X c h d a C = (F)

n n

C Hop k i n X chiJa R' = lOQ D Khong xac d i n h dufcfc

C a u 32 Mach di|n gom hop k i n X chi chiJa mot trong 3 phan tuT R', L hoac C

10

va C = (F) Dien dp hai d i u mach tre pha hon cUorng dp dong di^n 1 goc

- Cho f = 50 Hz Chon cau diing

3

A Hop k i n chiJa L = — H B Hop k i n chufa R = ^ Q

10

C Hop k i n chiJa C = —7=^7: F D Khong xac d i n h d U p c

C a u 33 Cho mach dien n h u h i n h Biet L = ^ ^ ( n i H ) , C = ^ ( F ) ,

R = lOO(Q) T i m f, biet Uc lech pha - so vdi UAU va UAB khong doi

A R M L N C B

I • : 3- » - / 2 2 2 2 2 S V _ ^ / 1 A.50HZ B 50V2HZ C 60Hz D.lOOHz

C a u 34 Cho mach dien n h u h i n h Biet UAN l $ c h pha ^ so vdi UNB- T i m moi lien

he R, r, L , C Biet U A „ khong doi

BAI TAP MAU

j3ai 1- Cho mach dien n h u h i n h ve:

-Ih B

Dien t r d cac von ke r a t Idn

V, chi lOV; V 2 chi lOV; V 3 chi 20V T i m : a) So chi V

b) Dp lech pha giOra dien ap hai dau mach vdri cUcfng dp dong dien qua mach

c) He so cong suat mach

d) Dien ap hieu dung giiifa A vdi N va giu:a M vdi B

Tom tat

U K = lOV U,, = lOV Uc = 20V a) UAB = ?

U„

10 10V2 coscp =

UAN = Mi + K - VlO' + 10' ^ U^.^ = 10V2(V)

• f khong doi -> co = 27if khong doi

• Z L = Lw: khong doi (do L w khong doi)

1 : kliong doi (do C, w khong doi)

it A B : khong doi (do U Q khong doi)

b) T i m so chi V2 va so' chi V,3

Nhan xet: K h i R thay doi t h i U R thay doi t h a n h U ' R ; U L thay doi t h a n h

U ' L ; U C thay doi t h a n h U'c- d day Z L , Z C vSn khong doi

Ta c6: U A B = V ^ R + (U^ - U;,)' <-> 50 = ^^O' + ( U L - Up)'

C a u 36. Cho mach dien U ^ ^ = 1 0 0 V 2 ( V ) ; = 1 0 0 V 2 ( V ) ; U ^ , B = lOO(V) Thr

U A M , U ^ u Chon cau dung

C a u 37. Cho mach dien R, L , C m^c no'i tiep vao 2 diem A B Che

= 50V2(V), 'U„ = 5 0 ( V ) , U,^ = 2Uc T i m U ^ U ^

Dieu chinh R de:

a) Cong suat tieu t h u ciia mach ciTc dai T i n h gia t r i cong sua't nay

b) Cong siiat t r e n bien t r d cifc dai T i n h gia t r i ciia cong suat luc nay

Tom tat

r = 10 Q 0,4

ciTc d a i b) Cong suat tieu t h u t r e n

mach cUc dai

R + r = i Zi - Zc I

^ R + 1 0 = I 4 0 - 2 0 I ^ R = I C Q

• Cong suat tieu thu ciia mach cufc dai

Pn,ach = (R + r ) l ' • '(1) Vdi

IlUdng dan gidi

Cong suat tieu thu mach

B a i 2 Cho mach dien gom dien trot R = 20 Q mac n o i tiep v d i cupn day

thuan cam c6 he so tir cam L Dien ap dat vao 2 dau doan mach c6 dang

u = 40 V2 coslOOTTt (V) T i m he so t i i cam ciia cupn day de cong suat tieu

thu ciia mach la 64 W

-Hiidng dan gidi

Cong suat tieu t h u ciia doan mach

dien dung C = (F) Dien dp h a i dau mach c6 dang u = 200 N/2 coscot

(V) T i m t a n so' cua dong dien t r o n g mach de cong suat tieu t h u ciia mach

la 200 W

C =

Tom tdt 10'

(F) R = 100 Q

-•AB _ = 200 V Cong suat tieu t h u ciia doan mach

2

mach = RI^ = R U AB

LZAB J RUAB

C a u 4 1 Cho mach dien gom R, L , C m^c n o i tiep D i ^ n ap dat v^o h a i dau

mach CO dang: u = 200 V2 coslOO 7:t (V) Biet L = - (H); C = — ^ i F

71 71

T i m R de cong suat tieu t h u ciia mach la 200W

A 5 0 Q B 200Q C 150Q D lOOQ

C a u 42 Cho mach dien gom R, L , C mdc n o i tiep Dien dp dat vao h a i dau

mach CO dang: u = 40072 cos 1007tt (V) Biet R = 200Q ; L = 1 ^ m H T i m

D i e u k i ^ n 1: L ho&c C hoac f hoSc wthay doi

D i e u k i ? n 2: Co 1 trong 5 dau hieu:

- CiXcfng do dong dien hieu dung circ dai

- Chi so (A) ciTc dai; (V) ciTc dai (thong tha&ng t h i trif U L ma^i k h i L thay doi;

Uc max k h i C thay doi)

- Cong suat circ dai

- He so' cong suat c\ic dai

- Dien ap cung pha ca6ng dp dong dien

* Lirtf y:

- Neu mach c6 mpt t u dien: = => C

- Neu mach c6 h a i tu die

BAI TAP MAU

B a i 1 Cho mach dien R, L , C m^c noi tiep n h u h i n h ve

Dieu chinh L de cong suat tieu thu cua mach circ dai

a) T i n h L va cong suat Itic nay

b) Viet hieu thu"c dien ap tufc thcfi hai dau tu dien

Tom tat

R = 10 Q 10"^'

C = (F)

271

UAB = 400cos lOOTtt +

-3 (V) Dieu chinh L de cong suat

mach ciic dai

27: -.10071

a) Do L thay doi de cong suat tieu thu cua mach cUc dai

Mach xay ra hien tMng cong hirdng: Z L = Zg Z L = 20 Q

L = M ( H )

CO 10071 Cong suat tieu thu cua mach ciTc dai luc nay la:

P m a c h = R I

VcJi

(do Z L = Z C Z L - Z C = 0)

400 (2) -) I = = 20 V2 (A)

V a y ( l ) - > P „ , e h = 10(20 V 2 ) 2

R

(1) (2)

P,„,,, = 8000W b) Bieu thiJc dien ap tiJc thcfi hai dau tu dien

d i e n a p bao n h i e u de d i e n a p h a i dau m a c h c u n g p h a cUcfng do d o n g d i e n

1 L thay ddi dl UL cac dai

2 C thay ddi dd Uc cUc dai

Tim he so tif cam ciia cupn day di:

a) Vi chi gia tri ciTc dai

b) V2 chi gia tri cUc dai

UAB = 8OV2 cosdOOTtt) (V)

a) Vi chi gia t r i cUc dai

b) V2 chi gia t r i cifc dai

Hitdng dan gidi

Ceo = 20 Q a) Do L thay doi de so chi Vj ciic dai nen trong mach xay ra hien tifgng cong hirorng Luc nay

Q^u 55 Cho mach dien gom R, L, C m^c noi tiep Dien dp ddt vdo hai dau

doan mach c6 dang: u = UoCoslOOTtt (V) C thay doi R, L khong doi Biet R =

C&u 56 Cho mach dien gom R, L, C m^c no'i tiep Dien dp dat vao hai dau

doan mach c6 dang: u - UoCosl20 7it (V) C thay doi R, L jchong doi Biet

L = — ( H ) Dieu chinh C den gid t r i C =^^— (F) thi di§n dp hieu dung hai

7t 727t

dau tu di^n dat gia t r i cUc dai Gid t r i ciia dien trd R trong mach Id

A 12Q B 24Q C 36Q D 48Q

CSu 57. Cho mach dien gom R, L, C mdc noi tiep c6 C thay doi, R, L khong

doi Biet dien dp dat vao hai dau mach c6 dang: u = 200cosl00 7it (V), R =

50( Q ), L = — (H) Dieu chinh C den gid t r i sao cho di?n dp hi^u dung hai

Cuon sa Cuon thiJ

II Cac cong thu-c cap cS'p

II Cac cong thu-c cap

1 • C6ng thQc m^y b\6n

* Chu y:

U , > U , (N, > N , ) : Mdy tdng dp

^ U , < U , (N, < N j ) : Mdy ha dp

* I: Tong chieu d a i day (m)

* p : Dien t r d suat ciia kirn loai (Q.m )

* P , U : cong suat va dien ap truyen d i

Ptt: cong suat tieu t h u

p 2

Thi^dng xet: coscp = 1 k h i do AP = j^R

Truyen t a i d i xa: H ' =

B A I T A P M A U

B a i 1 M o t mAy bien ap c6 cupn so cap gom 4000 vong, cuon thiJ cap gom 2000

vong Biet hai dau cuon scf cap dMc noi vdri mang dien xoay chieu ma gia tri

dien ap hieu dung la 200 V Bo qua hao phi ciia may bien ap T i m :

a) Dien ap hieu dung d hai dau cuon thuT cap

b) N o i 2 dau cupn thu" cap v d i t d i R = 20 Q T i m ciTcfng dp dong d i ^ n qua

cupn scf cap va thu' cap

B a i 3 M o t may phat dien truyen d i mot cong suat P = 200 k W tren day dan

CO dien t r d R = 10 Q Hieu dien the tit t r a m phdt dien chuyen d i la U = 2

kV T i n h : a) Cirdng dp dong dien t r e n day do may cung cap Biet dong di$n cung pha vdi dien ap

b) Cong suat toa nhiet tren day

c) Cong suat noi tieu thu

d) Hieu suat t a i dien

0 Ptt = ?

d) H = ? d) Hieu suat t a i dien