Dong Dien Xoay Chieu Day Du

8/8/2019 Dong Dien Xoay Chieu Day Du

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V t L 12 Dngin Xoay Chiu

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CHNG VIN XOAY CHIU

CH IDNG IN XOAY CHIU MCH IN XOAY CHIU

A. TM TT KIN THC CBN

I. DNG IN XOAY CHIU1. Hiu in th dao ng iu ha. Cng dng in xoay chiu. Cc gi tr hiu dng. Dng in xoay chiu l dng in m cng bin thin iu ha theo thi gian theo phng trnh:

0 cos( )i I t = + Hiu in thhai u mch in xoay chiu cng bin thin iu ha cng tn s v khc pha so vi

dng in.

a. Chu k, tn s khung quay:2

2 fT

= =

Trong : f (Hz hay s dao ng/giy) : tn s, s dao ng lp li trong mt n v thi gian.T (s) : chu k, thi gian ngn nht m dao ng lp li nh c.

b. Tthng qua khung dy: cos BS t = Nu khung c N vng dy : 0cos cos NBS t t = = vi 0 NBS = Trong : 0 : gi tr cc i ca t thng.

( ), ;t n B n = : vectphp tuyn ca khungB (T); S (m2); 0 ( )Wb

c. Sut in ng cm ng+ Sut in ng cm ng trung bnh trong thi gian t c gi tr

bng tc bin thin t thng nhng tri du: E t

= v c ln : E t

=

+ Sut in ng cm ng tc thi bng o hm bc nht ca t thng theo thi gian nhng tri du:

0 0' sin sin ;e NBS t E t E NBS = = = =

d. Hiu in th tc thi: =0cos( t + ) = 2cos( t + )u U U

e. Cng dng in tc thi : =0cos( t + ) = I 2cos( t + )i I

Vi = u i l lch pha ca u so vi i, c2 2

2. Dng in xoay chiu i = I0cos(2ft + i). S ln dng in ichiu sau khong thi gian t.

* Mi giy i chiu 2f ln.* S ln i chiu sau khong thi gian t: 2tf ln.

* Nu pha ban u i =2

hoc i =

2

th ch giy u tin

i chiu (2f 1) ln.

3. t in p u = U0cos(2ft + u) vo hai u bng n hunhquang, bit n ch sng ln khi hiu in th tc thi t vo n l

1u U . Thi gian n hunh quang sng (ti) trong mt chu k.

Vi 1

0

osU

c

U

= , (0 < ZC hay1

LC > > 0 th u nhanh pha hn i.

+ Khi ZL < ZC hay1

LC < < 0 th u chm pha hn i.

+ Khi ZL = ZC hay1

LC = = 0 th u cng pha vi i. Lc Max

UI =

Rgi l hin tng cng

hng dng in.

6. Gin vct: Ta c:0 0 0 0

R L C

R L C

u u u u

U U U U

= + +

= + +uur uuur uuur uuur

R

L C

0U R

0U L

0U C

0U LC 0U AB

0I

O

i

0U R

0U L

0U C

0U LC

0U AB

0I

O

i

0U R

0U L

0U C

0U AB 0I

O

i

A B
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7. Cng sut ta nhit trn on mch RLC:* Cng sut tc thi: 0cos cos(2 )u iP UI U t = + + + * Cng sut trung bnh: 2cosUI I R= +P

8. in p 1 0 cos( )u U U t = + + c coi nh gm mt in p khng i U1 v mt in p xoay chiu

0 cos( )u U t = + ng thi t vo on mch.

II. BI TON CC TR CNG SUT CA MCH RLC

1. on mch RLC c R thay i:a. Nu U, R = const. Thay i L hoc C, hoc . iu kin axMP

T :2 2

2 2( ) Max L C L C

U U R Z Z

R Z Z R= = =

+ P P

(Mch xy ra hin tng cng hng in v h s cng sut cos 1 = )b. Nu L, C, , U = const. Thay iR. iu kin axMP

T :2

2 2( )L C

UR

R Z Z =

+ P . p dng bt dng thc C-si ta c

2 2

ax 2 2ML C

U U

Z Z R= =

P khi R = ZL- ZC

22 cos

2Z R = =

c. Mch RrLC c R thay i (hnh v)

Khi2 2

ax 2 2( ) AB M L C L C

U UR r Z Z

Z Z R r = = + =

+P

Khi2

2 2ax ( )2( ) R M L C

U R r Z Z

R r= = +

+P

d. Mch RrLC khi R bin i cho hai gi tr 1 2R R u cho cng sut 0 axM


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Vn dng phng php i s hay phng php gin vectta c :2 2

axL

C M

U R ZU

R

+= khi

2 2

2 2 2L

C

L

R Z LZ C

Z R L

+= =

+, khi RL ABU U v UAB chm pha hn i.

c. Khi RC RC MaxU U= ta c:2 2

2 2

2 2( )

C

RC C

L C

U R ZU I R Z

R Z Z

+= + =

+ .

Vn dng phng php o hm kho st RCU ta thu c: 2 2 0C L CRC MaxU Z Z Z R =

Khi2 24

2L L

C

Z R Z Z

+ += th ax 2 2

2 R

4RC M

L L

UU

R Z Z =

+ Lu : R v C mc lin tip nhau

d. Khi2 2

2 2

2 2( )

L

RL L

L C

U R ZU I R Z

R Z Z

+= + =

+ lun khng i vi mi gi tr ca R (R gia L v C), bin

i i s biu thc RLU ta c : ( 2 ) 0 2C C L C L Z Z Z Z Z = =

e. Khi RL RC U U (C R gia L v C): Dng gin vecthay 21 2tan .tan 1 L CZ Z R = =

f. Khi RL RC U U v , RL RC U a U b= = . Tm , , R L C U U U ?

+ Ta c:

2

22 2 2

2 2 2

( )

( )

L C R

L R L L C L

C

R C C L C

U U UU a

U U U U U aU b

U U U U U b

= + = + = =

+ = + =

v R C La b

U U Ub a

= =

+ Hoc dng gin vects cho kt qu nhanh hn.

3. on mch RLC c L thay i. Tm L :a. min, , , , , , cosMax R Max C Max RC Max AB Max Z I U U U P cc i,

Cu tr pha so

2

vi

ABu ? Tt c cc trng hp trn u lin

quan n cng hng inL C

Z Z =

b. RL RC U U (C R gia L v C): Dng gin vecthay 21 2tan .tan 1 L C Z Z R = = c. Khi

L MaxU ta c:

2 2 2 2 2 2 2

2

( ) 2 ( ) 21

L L L L L

L C L L C C C C

L L

UZ UZ U U IZ U

R Z Z R Z Z Z Z R Z Z

Z Z

= = = =+ + + +

+

Vn dng phng php o hm ta c :2 2

ax

C

L M

U R ZU

R

+= khi

2 22

2

1CL

C

R Z Z L CR

Z C

+= = + , khi RC ABU U v UAB nhanh pha hn i.

Lu : R v L mc lin tip nhau.

d. 2 2 RL L

U I R Z = + cc i (C R gia L v C). Dng phng php o hm 2 2 0 L C LZ Z Z R =

4. Mch RLC c thay i. Tm:a. min, , , , cosMax R Max AB Max Z I U P cc i, ...? Tt c cc

trng hp trn u lin quan n cng hng in.

2 1 1

2L C

Z Z f LC LC

= = =

b. Khi axC MU ta c : 2 22

4C Max

ULU

R LC R C = khi

2

2 2 2

1(2 ) 2

Rf

LC L = =

c. KhiaxL M

U ta c :2 2

2

4L Max

ULU

R LC R C =

khi 2 2

2 2

2(2 )

2f

LC R C = =

R L CA B

R L CA B


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d. Thay i f c hai gi tr 1 2f f bit 1 2 f f a+ = th 1 2 ?I I=

Ta c :1 1 2 2

2 21 2 ( ) ( ) L C L C Z Z Z Z Z Z = = = = h

21 2

1 2

1

2

chLC

a

= = + =

hay 1 2 1 21

LC

= = tn s 1 2 f f f =

5. Khi kha K mc song song vi L hoc C, khi ng hay mth Ing = Im

a. Kha / / :K C Zm= Zng2 2 2 2

0( )

2C

L C L

C L

Z R Z Z R Z

Z Z

= + = + =

b. Kha / / :K L Zm= Zng2 2 2 2 0( )

2L

L C C

L C

Z R Z Z R Z

Z Z

= + = + =

III. BI TON V PHA CA DAO NG

1. Mch RLC c C bin i cho hai gi tr C1 v C2a. C hai gi tr C1 v C2 cho lch pha gia dngin v hiu in th trong hai trng hp l nh nhau.

T1 2

2 2 2 21 2 1 2cos cos ( ) ( ) L C L C Z Z R Z Z R Z Z = = + = +

1 2( )

L C L C Z Z Z Z =

b. Ngoi ra, khi gp bi ton C bin thin C1, C2 lm cho hoc I1 = I2 hoc P1 = P2 th cm khng cng c

tnh trong trng hp 1 2 = tc l :1 2

2C C

L

Z ZZ

+= .

c. Khi 1C C= v 2C C= (gi s 2C C> ) th 1i v 2i lch pha nhau . Gi 1 v 2 l lch pha ca

ABu so vi 1i v 2i th ta c 1 2 1 2 > = .

+ Nu 1 2I I= th 1 2 2

= =

+ Nu 1 2I I th tnh1 2

1 21 2

tan tantan( ) tan

1 tan . tan

= =

+

d. Nu C bin thin, c hai gi tr C1, C2 lm cho hoc I1 = I2 hoc P1 = P2 hoc 1 2 = . Tm C c cnghng in. Ta c :

1 2

1 2

1 2 1 2

21 1 1 1 1( ) ( )

2 2C C CC C

Z Z Z C C C C C C

= + = + =+

e. Nu C bin thin, c hai gi tr C1, C2 lm cho hiu in th trn t bng nhau trong hai trng hp. Tm C hiu in th trn tt gi tr cc i th :

1 2

1 21 21 1 1 1 1( ) ( )2 2 2

C C C

C CC C C C Z Z Z

+= + = + =

3. Mch RLC vi L bin i, c hai gi tr L1 v L2a. Nu L bin thin, c hai gi tr L1, L2 cho hoc I1 = I2 hoc P1 = P2 hay cho cng ln ca s lch pha ca

u v i th dung khng CZ tnh c bao gicng bng trung bnh cng ca cm khng LZ theo biu thc :

1 2

2L L

C

Z ZZ

+=

b. Nu L bin thin, c hai gi tr L1, L2 cho hoc I1 = I2 hoc P1 = P2 hay cho cng ln ca s lch pha cau v i. Tm L c cng hng in max max max( , , 0, (cos ) 1, ,...)u i u i I I P P = = = = = = = th

bao gita cng thu c : 1 22

L LL += .

c. Nu cun dy thun cm vi L bin thin, c hai gi tr L1, L2 cho cng mt hiu in th trn cun dy. hiu in th trn cun dy t cc i th L c gi tr l :


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1 2

1 1 1 1

2 L L L= +

hay

1 2

1 2

2L LL

L L=

+

4. Mch ch cha t C hay cun dy thun cm L

S dng cng thc :

2 2

0 0

1 ( )i u

I U

+ =

cho hai dng ton thng gp sau :

a. Nu bi ton cho hai cp gi tr tc thi u v i, nu thay vo (*) ta s thu c h 2 phng trnh 2 n

cha U0, I0. Gii h => U0, I0, t tnh c CZ theo0

0C

UZ C

I=

b. Nu bi ton cho hai cp gi tr tc thi u v i, cho thm CZ cn tm U0, I0 th s dng thm h thc

0 0 CU I Z= ri thay vo (*) ta s c phng trnh mt n cha I0 (hoc U0 ) t tm c I0 (hoc U0 ).Ch : Cc bi ton i vi cun dy thun cm L cng lm tng t nh hai bi ton v t C ni trn.

5. Bi ton f bin thin c yu t cng hngLc u c tn s f, khi xy ra cng hng c tn s f.Nu : +

L C

Z Z

>=> khi cng hng ' ' '

L C L

Z Z Z

= gim => f > f

+ L CZ Z< => khi cng hng ' ' ' L C L Z Z Z = tng => f < f6. Bi ton nu c 2 cun dy hoc 2 tin

+ 1 2L nt L 1 cun dy c 1 21 2 1 2 L L LL L L Z Z Z L L L= + = + = +

+ 1 2

1 2 1 2

1 21 2

1 2 1 2

1 1 1 1 1 1/ / : L L

L

L L L L L

Z Z L L L L Z L

Z Z Z Z Z L L L L L= + = = + =

+ +

+1 2

1 21 2

1 2 1 2

1 1 1: C C C

C CC nt C Z Z Z C

C C C C C = + = + =

+

+ 1 21 2 1 2

1 2 1 2

1 1 1/ / :

C C

C

C C C C C

Z ZC C Z C C C

Z Z Z Z Z = + = = ++

7. Hai on mch AM gm R1L1C1 ni tip v on mch MB gm R2L2C2 ni tip mc ni tip vi nhauc UAB = UAM + UMB uAB; uAM v uMB cng pha tanuAB = tanuAM = tanuMB

8. Hai on mch R1L1C1 v R2L2C2 cng u hoc cng i c pha lch nhau

Vi 1 111

tan L CZ Z

R

= v 2 22

2

tan L CZ Z

R

= (gi s1 > 2)

C 1 2 = 1 21 2

tan tantan

1 tan . tan

=

+

Trng hp c bit = 2

(vung pha nhau) th 1 2tan . tan 1 = VD: * Mch in hnh 1 c uAB v uAM lch pha nhau

y 2 on mch AB v AM c cng i v uAB chm pha hn uAM AM AB =

AM ABtan tan

tan( ) tan1 tan .tan

AM AB

AM AB

= =

+

Nu uAB vung pha vi uAM th tan . tan = - 1 AM AB

1L CLZ ZZ

R R

=

* Mch in hnh 2: Khi C = C1 v C = C2 (gi s C1 > C2)th i1 v i2 lch pha nhau y hai on mch RLC1 v RLC2 c cng uAB

Gi 1 v 2 l lch pha ca uAB so vi i1 v i2 th c 1 > 21 - 2 =

R L CMA B

Hnh 2

N

R L CMA B

Hnh 1

N


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X

X

X

X

X

X

X

X

X

Nu I1 = I2 th 1 = - 2 =2

Nu I1 I2 th tnh 1 21 2

tan tantan

1 tan . tan

=

+

Ch : Cc dng mch: RL ni tip, RC ni tip, RLC ni tip m cun dy c in trtrong vcng thctng tr, nh lut Ohm, lch pha, hscng sut, lin hgia cc hiu in thhiu dng,

IV. BI TON HP KN (BI TON HP EN)

1. Mch in n gin:a. Nu

NBU cng pha vi i suy ra ch cha 0R

b. Nu NBU sm pha vi i gc2

suy ra ch cha 0L

c. Nu NBU tr pha vi i gc2

suy ra ch cha 0C

2. Mch in phc tp:a. Mch 1

NuAB

U cng pha vi i suy ra ch cha 0L

Nu ANU v NBU to vi nhau gc2

suy ra ch cha 0R

Vy cha ( 0 0, LR )

b. Mch 2Nu ABU cng pha vi i suy ra ch cha 0C

NuAN

U vNB

U to vi nhau gc2

suy ra ch cha 0R

Vy cha ( 0 0, CR )

B. MT S KIN THC TON HC CN VN DNG KHI GP CC DNG BI TM CC TR

1. Phng php 1: Dng bt ng thc C-si

+ p dng bt ng thc C-si cho 2 s dng a, b: 2a b ab+

( )

( )

min

max 2

a b ab

a bab

+ = + =

du = xy ra khi a = b

+ p dng cho n s hng: 1 2 1 2...

...n na a a

a a an

+ + + du = xy ra khi 1 2 ... na a a= = =

Lu : p dng: + Tch khng i khi tng nh nht.+ Tng khong i khi tch ln nht.

2. Phng php 2:

+ nh l hm s sin trong tam gic:

sin sin sin

a b c

A B C

= =

+ nh l hm s cosin trong tam gic: 2 2 2 2 cosa b c bc A= +

max max(cos ) 1 0; (sin ) 1 2

= = = =

R

L C

XA N B

R

C

XA N B

A

CB

c b

a


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3. Phng php 3: Da vo hm s bc 2: 2( ) ( 0) y f x ax bx c a= = + +

+ a > 0 th nh Parabol2

ax

b= c

2

min

4

4 4

ac by

a a

= =

+ a < 0 th nh Parabol2

ax

b= c

2

max

4

4 4

ac by

a a

= =

+ th:

4. Phng php 4: Dng o hmNi dung:+ Hm s y = f(x) c cc tr khi f(x) = 0+ Gii phng trnh f(x) = 0+ Lp bng bin thin tm cc tr+ V th nu bi ton yu cu kho st s bin thin

Ngoi cc phng php trn cn c mt s phng php khc kho st Max, min ca mt i lng vt l. Ty theo biu thcca i lng vt l c dng hm no m p dng bi ton gii.

C nhng hm s khng c cc tr, ch c tnh ng bin haynghch bin ta tm c Max, min trong min no .

Trong on [a,b]: f(b)Max khi x = b

f(a)min khi x = a

Di y l mt sbi ton tlun m t cho cc phng php trn.

Bi ton 1: Cho mch in xoay chiu nh hnh v.

1. Cho R = const. Thay i L hoc C hoc cng sut tiu th trn on mch AB l cc i.Phng php:

Cng sut tiu th trn mch:2

22 2

( )( ).

( ) ( )L C

U R rR r I

R r Z Z

+= + =

+ + P

Cc i lng bin thin u nm trong s hng 2( )L CZ Z

Nhn thy2

Max

U

R r= =

+P P khi hiu 0L CZ Z = , tc mch xy ra cng hng in.

=> Tnh c L hoc C hoc .

2. GiL, C v khng i. Thay i R, tm R :a. Cng sut tiu th trn mch AB cc i.

b. Cng sut trn R cc i.c. Cng sut tiu th trn cun dy cc i.

a > 0

ymin

2

b

a

y

x

O

a < 0

ymax

2

b

a

y

x

O

b

f(b)

f(a)

O a

x

y

A B

CR L

A B

CR L, r

A B

CR L


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Phng php:

a. Tm R MaxP ?

Ta c :

2 22

22 2

( )( )

( )( ) ( )( )

( )L CL C

U R r U R r I

Z Z R r Z Z R r

R r

+= + = =

+ + + +

+

P P

Dng bt ng thc C-si cho mu s ta c:

2

2( ) Max L C L C U R r Z Z R Z Z r

R r= + = = +P

b. Tm R R MaxP ?

Ta c :

2 22

2 2 2 2.

( ) ( ) ( )2

R R

L C L C

U R UR I

R r Z Z r Z ZR r

R

= = =+ + +

+ +

P P

Vn dng bt ng thc C-si cho s hng:

2 2( )L C

r Z ZR

R

+ +

22 2

0( ) ( )2( ) R Max L C U

R r Z Z RR r = = + =+P

Dng th:

c. Tm R r MaxP

?Ta c:

22

2 2( ) ( )r L C

rUrI

R r Z Z = =

+ + P suy ra

2

2 20

( )r Max L C

rUR

r Z Z= =

+ P

Bi ton 2: Cho mch in xoay chiu nh hnh v.a. Tm R RU cc i.

b. Tm L LU cc i.

c. Tm C CU cc i.

d. Tm ln lt RU cc i, LU cc i, CU cc iPhng php:

a. Tm R RU cc i.

Ta c: 2 2 2

2

( ) ( )1

R

L C L C

UR UU IR

R Z Z Z Z

R

= = =+

+

Suy ra :R Max

U U R= =

b. Tm L LU cc i.Cch 1: Dng phng php i s - Ly cc tr l ta nh.

Ta c:2 2

( )

L

L L

L C

UZU IZ

R Z Z

= =

+

2 2 22L

L L C C

UZ

R Z Z Z Z =

+ +

P

R0O

maxRP

R

A B

CR L

( )R O

( )R

U V

U


10/42



Chia c t v mu choL

Z v rt gn ta c: 2 2

2

21

L

C C

L L

U UU

y R Z Z

Z Z

= =+

+

min L MaxZ y . t1

L

xZ

= , ta c hm 2 1 y ax bx= + + vi2 2

2C

C

a R Z

b Z

= +

= (*)

V a > 0 nn2

min

4

4 4

ac by

a a

= = khi2

bx

a= (**)

Thay a, b (*) vo (**) ta c:

2 2

2 2

1 C CL

L C C

Z R Z Z L

Z R Z Z

+= =

+

v2 22 2

min 2 2

4

4C

L MaxC

U R Zac b Ry U

a R Z R

+= = =

+

Cch 2: Dng phng php o hm, kho st LU theo LZ .

2 2 2 2 2( ) 2

L L

L L L C L L C C

UZ UZ U IZ

R Z Z R Z Z Z Z = = =

+ + + Ly o hm, lp bng bin thin ta s thu c cc tr v dng ca th:

Cch 3: Dng gin vectri da vo php tnh hnh hc kho stTa c: AB AM MN NBu u u u= + +

Hay dng vect: AB AM MN NBU U U U = + +

Theo cch v cc vectni tip nhau, theo gin ny ta c:

AB

R

L

C

AB U U

AM U

MN AK U

NB U

= === =

=

p dng nh l hm s sin trong ABK ta c:sin

.sin sin sin sin sin

LL

U AB AK U U U

= = =

Trong KBN vung ti N ta c:

2 2,

sin R

R C C

UKN R

KB U R Z = = =

+

Nn2 2sin

. .sin

sin

C

L

U R ZU U

R

+= =

Lc ny ta thy LU ch ph thuc vo sin . Vy nn khi sin 1 = th:2 2

C

L L Max

U R ZU U

R

+= =

2 2c

c

R Z

Z

+

2 2cU R Z

R

+

U

0

0

ZL

UL

( )LZ O

U

ULmax

UL(V)

RU

K

AM

N

B

LU

I

ABU

CU


11/42



v khi sin 12

= = =

2 2

tan tan L C CLC C

Z Z R Z RZ

Z R Z

+ = = =

Ch : Khi L L MaxU U= , theo phng php gin vec tnu trn, in p gia cc phn t c mi lin h:2 2 2 2

L R C U U U U = + + c. Tm C CU cc i.

2 2 2 2 2( ) 2

C C

C C

L C L L C C

UZ UZ U IZ

R Z Z R Z Z Z Z = = =

+ + +

Chng minh tng t cu b ta c:2 2 2 2

L LCC Max

L

U R Z R ZU U C

R Z

+ += =

Ch : Biu thc tnh , L Max C Max

U U v ,L CU U ca hai bi ton trn c dng tng t, chi vai tr

ca LU v CU cho nhau.

d. Tm ln lt RU cc i, LU cc i, CU cc i

v RU cc i

2 22 21( ) ( )

C

R

L C

UZ URU IR

R Z Z R L

C

= = =+ +

1 10

RR MaxU U L

C LC

= = = (mch cng hng in)

Dng th:

vLU cc i

Ta c:2 2 2 2 2( ) 2

L LL L

L C L L C C

UZ UZ U IZ

R Z Z R Z Z Z Z = = =

+ + +

2 2 2 2 22 2 2 4 2

1 2 1 2 1( )

L

UL UL ULU

L L y R L R L

C C C C

= = =+ + + +

t 221

x y ax bx d = = + + vi

2

2

2

1

2

aC

L

b R C

d L

=

= =

(*)

UR

R O

maxRU


12/42



D thy minL MaxU y . V v a > 0 nn2

min

4

4 4

ac by

a a

= = khi

2

bx

a= (**)

Thay a, b, d (*) vo (**) ta c:

2 2 2

2 1 224

LL Max

ULU

LCR LC R C

RC

= =

vi iu kin 22L

RC

>

Dng th:

v CU cc i

Ta c: 2 22 2 2

2 2

1 2( )

C

C C

L C

UZ UU IZ

L R Z Z C R L

C C

= = =+ + +

2 4 2 22

2 1( )

C

U UU

L C yC L R

C C

= =+ +

t 2 2 x y ax bx d = = + + vi

2

2

2

2

1

a LL

b RC

dC

= =

=

(*)

D thy minC MaxU y . V v a > 0 nn2

min

4

4 4

ac by

a a

= = khi

2

bx

a= (**)

Thay a, b, d (*) vo (**) ta c:

2

2 2

2

2 1 24CC Max

LR

UL CUL R LC R C

= =vi iu kin 22L R

C>

Ch : Tn s gc trong 3 bi ton trn c mi lin h :2

R L C =

Bi ton 3: Cho mch in xoay nh hnh vHiu in th hai u on mch AB l:

85 2 cos100 ( ), 70 , 80AB

u t V R r = = = ,cun dy c L thay i c, tin c C bin thin.

a. iu chnh3

2L H

= ri thay i in dung C.

Tm C UMB cc tiu.b. iu chnh

310

7C F

= ri thay i in dung L.

Tm L UAN cc i.

( / )rad s O

U

ULmax

UL(V)

L

rK


13/42



Phng php:

a. Tm C UMB cc tiu.

Ta c:

2 2

2 2 2 2

2 2

( )

( ) ( ) ( ) ( )

( )

L C

MB MB

L C L C

L C

U r Z Z UU IZ

R r Z Z R r Z Z

r Z Z

+ = = =

+ + + +

+

2

2 2

21

( )

MB

L C

UU

R Rr

r Z Z

=+

++

d thy rng2

min( ) 0

L CMBU Z Z =

310150

15L C Z Z C F

= = =

b. Tm L UAN cc i.

Ta c:

2 2 2 2

2 22 2

( ) ( )( ) ( )

L L AN AN

L CL C

U R Z R ZU IZ U U y

R r Z Z R r Z Z

+ += = = =

+ + + +

minAN MaxU y

Trong :

2 2 2 2

2 2 2 2

70

( ) ( ) 150 ( 150)L

L C

R Z xy

R r Z Z x

+ += =

+ + + vi( 0)L x Z x= >

Ly o hm y theo x v rt gn ta thu c:

2 2

22 2

3000 80200 70 .300

150 ( 150)

x xy

x

+ +=

+

Cho2 2 17,22' 0 3000 80200 70 .300 0

284,55

x y x x

x

= = + + =

=

Bng bin thin:

Theo bng bin thin ta thy 2,11Maxy = khi 284,55x = tc l khi 284,55LZ = 0,906L

ZL H

= = th ( ) 85 2,11 123, 47 AN Max MaxU U y V = = =

Dng th:

2,11

1

-

x

y

y0

-17,22 0 284,55

+ + 0 -

0,1088

O ( )L

Z

85

123,47

UAN(V)

27,9

284,55


14/42



C. CC DNG TON V PHNG PHP GII

Dng 1 : Vit biu thc i hay u+ Nu i = tI cos0 th dng ca u l u = )cos(0 +tU .

+ Hoc u = tU cos0 th dng ca i l l i = )cos(0 tI

Vi22

00

0 )()( CL ZZrR

U

Z

U

I ++==v tan

rR

ZZCL

+

= ( Khi on mch khng c phn t no th in tr

ca phn t bng khng)

+ C th dng gin vector tm (

RU v trng trc

I ,

LU v vung gc trc

Iv hng ln,

CU v

vung gc trc

Iv hng xung , sau dng quy tc a gic ).+Lu : Khi 1 i lng bin thin theo thi gian thi im t0 tng th o hm bc nht ca n theo t s

dng v ngc li.

Dng 2 : Tnh ton cc i lng ca mch in

+ I = 2

0I

, U = 2

0U

, P = UIcos ,nu mch ch c phn t tiu thin nng bin thnh nhit th P = R

2

I

+ H s cng sut cos22 )()( CL ZZrR

rR

Z

rR

++

+=

+=

+ Ch ni n cng hng khi mch c R + r = const v lc :

rRZ +=min , 0= ,rR

UI

+=max ,

rR

UP

+=

2

max

+ Dng cng thc hiu in th : 222 )( CLR UUUU += , lun c UR U+ Dng cng thc tan xc nh cu to on mch 2 phn t :

Nu 2

= mch c L v C

Nu 0> v khc2

mch c R,C

Nu 0


15/42



+ C th dng th xc nh cc tr ( th hm bc 2)+ Mch RLC c thay i , tm :

1. Hiu in th hai u R cc i : =LC

1

2. Hiu in th hai u C cc i : =2

2

2

1

L

R

LC

3. Hiu in th hai u L cc i : =222

2

CRLC

Dng 4 : iu kin 2 i lng in c mi lin hvpha+ Hai hiu in th trn cng on mch cng pha : 21 = 21 tantan =

+ Hai hiu in th trn cng on mch vung pha :221

= 2

1 tan

1tan

=

+ Hai hiu in th trn cng on mch lch pha nhau gc :

1 2 = 212

tan tantan

1 tan .tan

=

m

B. CU HI V BI TP TRC NGHIM

Cu 1: Trong mt mch in xoay chiu th cun cmA. c tc dng cn trhon ton dng in xoay chiuB. c tc dng cn trdng in xoay chiu i qua v tn s dng in xoay chiu cng ln th n cn tr

cng mnh.C. c tc dng cn trdng in xoay chiu i qua v tn s dng in xoay chiu cng nh th n cn tr

cng mnh.D. khng nh hng g n dng in xoay chiu.

Cu 2:i vi on mch c R, L, C mc ni tip, bit in trthun R 0, cm khng ZL 0, dung khng ZC

0 th :A. in p hiu dng gia hai u on mch lun ln hn in p hiu dng trn mi phn t.B. in p hiu dng gia hai u on mch lun bng tng in p hiu dng trn tng phn t.C. in p tc thi gia hai u on mch lun bng tng in p tc thi trn tng phn t.D. in p hiu dng gia hai u on mch c th nh hn in p hiu dng trn in trthun R.

Cu 3: Dng in xoay chiu l dng in c tnh cht no sau y?A. Chiu dng in thay i tun hon theo thi gian.B. Cng bin i tun hon theo thi gian.C. Chiu thay i tun hon v cng bin thin iu ho theo thi gian.D. Chiu v cng thay i u n theo thi gian.

Cu 4: Tc dng ca cun cm i vi dng in xoay chiuA. Cn trdng in, dng in c tn s cng ln cng b cn trB. Cn trdng in, dng in c tn s cng nh b cn trcng nhiuC. Cn trdng in, cun cm c t cm cng b th cn trdng in cng nhiuD. Cn trdng in, dng in c tn s cng ln th t b cn tr

Cu 5: Khi xy ra hin tng cng hng dng in trong mch R, L, C mc ni tip th pht biu no sau ykhng ng?

A. in p hai u tin vung pha vi cng dng in.B. in p hai u cun dy thun cm vung pha vi cng dng in.C. in p hai u in trthun vung pha vi cng dng in.D. in p hai u on mch in cng pha vi cng dng in.

Cu 6: Pht biu no sau yng vi cun cm?A. Cun cm c tc dng cn trdng in xoay chiu, khng c tc dng cn trdng in mt chiu.

B. Cm khng ca cun cm thun t l nghch vi chu k dng in xoay chiu.C. Hiu in th gia hai u cun cm thun cng pha vi cng dng in.D. Cng dng in qua cun cm t l vi tn s dng in.


16/42



Cu 7: Mt on mch gm ba thnh phn R, L, C c dng in xoay chiu 0 cosi I t= chy qua, nhng phnt no khng tiu thin nng?

A. R v C B. L v C C. L v R D. Ch c L.Cu 8: Mt on mch gm R, L, C mc ni tip trong c L CZ Z> . So vi dng in hiu in th hai umch s:

A. Cng pha B. Chm pha C. Nhanh pha D. Lch pha2

rad

Cu 9: Hiu in th v cng dng in trong on mch ch c cun dy thun cm c dng

0 cos( )u U t = + v 0 0cos( ).4i I t I

= + v c gi tr no sau y?

A. 0 0 ; 4 I U L rad

= = B. 00 ; 4

U I rad

L

= =

C. 00 ; 2

U I rad

L

= = D. 0 0 ; 2

I U L rad

= =

Cu 10: Mt cun dy c in tr thun R, h s t cm L mc vo hiu in th xoay chiu 0 cosu U t= .Cng hiu dng ca dng in qua cun dy c xc nh bng h thc no?

A. 02 2 2UI

R L=

+B. UI

R L= +

C.2 2 2

UI

R L=

+D. 2 2. I U R L= +

Cu 11: t mt hiu in th 0 2 cos( )u U t = + vo hai u on mch gm: in trthun R, cun dythun cm (cm thun) c t cm L v tin c in dung C mc ni tip. Cng dng in qua onmch c gi tr hiu dng l:

A.2

2 1

UI

R CL

=

+

B.1

UI

R L

C

= +

C.2

1

UI

R LC

= +

D.2

2 1

UI

R LC

= +

Cu 12: Cho on mch xoay chiu nh hnh v.

Tng trca on mch l:

A.2

2 2

1

C Z R L

C

= +

B.

22

1 2

1 1 Z R L

C C

= + +

C.2

2

1 2

1 1 Z R L

C C

= +

D.

2

2

1 2

1

( ) Z R L

C C

= + +

Cu 13: Hai cun thun cm L1 v L2 mc ni tip trong mt on mch xoay chiu c cm khng l:

A. 1 2( )L Z L L = B. 1 2( )L Z L L = + C. 1 2

( )L

L LZ

= D. 1 2

( )L

L LZ

+=

C2C1

L2R L1


17/42



Cu 14: Tng trca on mch xoay chiu c tnh bng cng thc no sau y?

A. ( )22

L C Z R Z Z = + B.

2

2 L

C

ZZ R

Z

= +

C. ( )22

C L Z R Z Z = + D. ( )

22L C

Z R Z Z = + +

Cu 15: Chn cu sai trong cc cu sau: Mt on mch c ba thnh phn R, L, C mc ni tip nhau, mc vo

hiu in th xoay chiu 0 cosu U t= khi c cng hng th:

A. 2 1LC = B. 2 21( ) Z R LC

= +

C. 0 cosi I t= v0

0

UI

R= D. R CU U=

Cu 16: Hiu in th v cng dng in trong on mch ch c tin c dng0 cos( )4

u U t

= + v

0 cos( )i I t = + . I0 v c gi tr no sau y:

A. 003

;

4

U I rad

C

= = B. 0 0 ;

2

I U C rad

= =

C. 0 03

;4

I U C rad

= = D. 00 ; 2U

I rad C

= =

Cu 17: Hiu in th gia hai u mch in xoay chiu ch c in trthun:0 cos( )2

u U t V

= + . Biu thc

cng dng in qua on mch trn l nhng biu thc no sau y?

A. 0 cos( )2i I t

= + (A) B. 0 cos( )2

i I t

= (A)

C. 0 cosi I t= (A) D. 0 cos( )4i I t

= + (A)

Cu 18: Dng in xoay chiu 0 cos( )4i I t = + qua cun dy thun cm L. Hiu in th gia hai u cun dy

l 0 cos( )u U t = + . 0U v c cc gi tr no sau y?

A.0

0

;2

LU rad

I

= = B. 0 0

3;

4U L I rad

= =

C. 003

;4

IU rad

L

= = D. 0 0; 4

U L I rad

= =

Cu 19: Hiu in th v cng dng in trong on mch ch c cun dy thun cm c dng

0 cos( )6u U t

= + v 0 cos( )i I t = + . I0 v c gi tr no sau y?

A. 0 0 ; 3 I U L rad = = B. 00 2; 3

U I rad L

= =

C. 00 ; 3

U I rad

L

= = D. 0

0

;6

L I rad

U

= =

Cu 20: Trong mch in gm r, R, L, C mc ni tip. Gi Z l tng trca mch. lch pha gia hiu inth hai u mch v cng dng in trong mch c tnh bi cng thc:

A. tan L CZ Z

R r

=

B. tan L C

Z Z

R

=

C. tan L CZ Z

R r

=

+D. tan

R r

Z

+=

Cu 21: Trong mch in gm r, R, L, C mc ni tip. Gi Z l tng trca mch. lch pha gia hiu inth hai u mch v cng dng in trong mch c tnh bi cng thc:

A. sin L CZ Z

R r

=

B. sin

R r

Z

+=


18/42



C. sin L CZ Z

R r

=

+D. sin L C

Z Z

Z

=

Cu 22: Mt khung dy quay iu quanh trc trong mt t trng u B vung gc vi trc quay vi tc gc . T thng cc i gi qua khung v sut in ng cc i trong khung lin h vi nhau bi cng thc:

A. 002

E

= B. 002

E

= C. 00E

= D. 0 0E =

Cu 23: Mt vng dy phng c ng knh 10 cm t trong t trng u 1 .B T

= T thng gi qua vng dy

khi vctcm ng tB hp vi mt phng vng dy mt gc 030 = bng:A. 31,25.10 Wb B. 35.10 Wb C. 12,5 Wb D. 50 Wb

Cu 24: Mt khung dy t trong t trng u B c trc quay ca khung vung gc vi cc ng cm ngt. Cho khung quay u quanh trc , th sut in ng cm ng xut hin trong khung c phng trnh l:

200 2 cos(100 ) .6

e t V

= Sut in ng cm ng xut hin trong khung ti thi im1

.100

t s=

A. 100 2 V B. 100 2 V C. 100 6 V D. 100 6 V Cu 25: Mt khung dy t trong t trng u B c trc quay ca khung vung gc vi cc ng cm ng

t. Cho khung quay u quanh trc , th t thng gi qua khung c biu thc 1 cos(100 ) ( ).2 3

t Wb

= +

Biu thc sut in ng cm ng xut hin trong khung l:

A.5

50cos(100 ) .6

e t V

= + B. 50cos(100 ) .6

e t V

= +

C. 50cos(100 ) .6

e t V

= D.5

50cos(100 ) .6

e t V

=

Cu 26: t vo hai u on mch in ch c cun dy thun cm1

L H

= mt hiu in th xoay chiu

100 2 cos(100 ) .

6

u t V

= Pha ban u ca cng dng in trong mch l:

A.2

3i

= B. 0i = C. 3i

= D.3i

=

Cu 27: Cho on mch in RLC ni tip. K hiu ,R L

u u vC

u tng ng l hiu in th tc thi hai u

cc phn t R, L v C. Quan h v pha ca cc hiu in th ny l:

A.R

u tr pha2

so vi

Cu B.

Cu tr pha so vi

Lu

C.L

u sm pha2

so vi

Cu D.

Ru sm pha

2

so vi

Lu

Cu 28: Cho on mch in RLC ni tip. t vo hai u mt hiu in th xoay chiu n nh u th hiu in

th gia hai u cc phn t 3 ; 2 . R C L C U U U U = = lch pha gia hiu in th hai u mch v cng dng in trong mch l

A.6

= B.

6

= C.

3

= D.

3

=

Cu 29: Mt tin c dung khng 30. Chn cch ghp tin ny ni tip vi cc linh kin in t khc di

y c mt on mch m dng in qua n tr pha so vi hiu th hai u mch mt gc4

.

A. mt cun thun cm c cm khng bng 60B. mt in trthun c ln 30C. mt in trthun 15 v mt cun thun cm c cm khng 15D. mt in trthun 30 v mt cun thun cm c cm khng 60

Cu 30: Mt on mch in xoay chiu gm in trthun, cun cm thun v tin mc ni tip. Bit cmkhng gp i dung khng. Dng vn k xoay chiu (in trrt ln) o in p gia hai u tin v in pgia hai u in trth s ch ca vn k l nh nhau. lch pha ca in p gia hai u on mch so vicng dng in trong on mch l


19/42



A.4

. B.

6

. C.

3

. D.

3

.

Cu 31: Cho mch in xoay chiu RLC nh hnh v.( )VftUuAB 2cos2= . Cun dy thun cm c t cm

1,L H

= t din c

310C F

= , 40R = . Hiu in th uAM

v uAB

lch pha nhau2 . Tn s f ca dng in xoay chiu c gi tr l

A. 120Hz B. 60Hz C. 100Hz D. 50HzCu 32: Cho mch in xoay chiu nh hnh v, in p t vohai u mch l: ( )AB 0u U cos100 t V= . Cun dy thun cm

c t cm1

L H=

. Tin c in dung40,5.10

C F

=

.

in p tc thi uAM v uAB lch pha nhau2

. in trthun ca

on mch l:

A. 100 B. 200 C. 50 D. 75Cu 33 : Xt mch in xoay chiu RLC, hiu in th 2 u mch lch pha so vi c-ng dng in qua

mch 1 gc4

. Kt qu no sau y l ng?

A. ZC = 2 ZL B. RZZ CL = C. ZL = ZC D. ZL = 2ZCCu 34: Mt on mch in xoay chiu c dng nh hnh v.

Bit hiu in th uAE

v uEB

lch pha nhau2

.

Tm mi lin h gia R, r, L, C.A. R = LCr B. r = CRLC. L = CRr D. C = LRr

Cu 35:t in p u = U0cost vo hai u on mch mc ni tip gm in trthun R, tin v cun cmthun c t cm L thay i c. Bit dung khng ca tin bng R 3 . iu chnh L in p hiu dnggia hai u cun cm t cc i, khi

A. in p gia hai u in trlch pha6

so vi in p gia hai u on mch.

B. in p gia hai u tin lch pha6


C. trong mch c cng hng in.

D. in p gia hai u cun cm lch pha6


Cu 36:t mt in p xoay chiu c gi tr hiu dng U vo hai u on mch AB gm cun cm thun c t cm L, in trthun R v tin c in dung C mc ni tip theo th t trn. Gi UL, UR v UC_ln lt l

cc in p hiu dng gia hai u mi phn t. Bit in p gia hai u on mch AB lch pha2

so vi in

p gia hai u on mch NB (on mch NB gm R v C ). H thc no di y l ng?

A. 2 2 2 2R C LU U U U= + + . B.2 2 2 2C R LU U U U= + + .

C. 2 2 2 2L R CU U U U= + + D.2 2 2 2R C LU U U U= + +

Cu 37: Mt on mch xoay chiu gm R v cun dy thun cm L mc ni tip, 100R = , tn s dng in f =

50Hz. Hiu in th hiu dng 2 u mch U = 120V. L c gi tr bao nhiu nu umch v i lch nhau 1 gc3

,

cho bit gi tr cng sut ca mch lc .

A. 3L H

= B.1

3L H

= C. 1L H

= D.

1

2L H

=


20/42



Cu 38: Mt on mch gm tin c in dung310

12 3C F

= mc ni tip vi in tr 100R = , mc on

mch vo mng in xoay chiu c tn s f. Tn s f bng bao nhiu th i lch pha4

so vi u hai u mch.

A. f = 60 3 Hz B. f = 25Hz C. f = 50Hz D. f = 60HzDng dkin sau tr li cu 39, 40, 41

Mt on mch xoay chiu gm in trthun 100R = , mt cun dy thun cm c t cm 2L H

=

v mt tin c in dung410

C F

= mc ni tip gia hai im c hiu in th 200 2 cos100 ( )u t V=

Cu 39: Biu thc tc thi cng dng in qua mch l:

A. 2 2 cos(100 )( )4

i t A

= B. 2 cos(100 )( )4

i t A

=

C. 2 cos(100 )( )4

i t A

= + D. 2 cos(100 )( )4

i t A

= +

Cu 40: Hiu in th hai u cun cm l:

A. 400 2 cos(100 )( )4L

u t V= + B. 3200 2 cos(100 )( )4L

u t V= +

C. 400cos(100 )( )4L

u t V

= + D. 400cos(100 )( )2L

u t V

= +

Cu 41: Hiu in th hai u t l:

A.3

200 2 cos(100 )( )4C

u t V

= B. 200 2 cos(100 )( )4C

u t V

= +

C. 200cos(100 )( )2C

u t V

= D. 3200cos(100 )( )4C

u t V

=

Cu 42: Mt dng in xoay chiu c th nh hnh v.Phng trnh dng in trong mch l:

A. 4cos50 ( )i t A= B. 4cos100 ( )i t A=

C. 2 2 sin100 ( )i t A=

D. 2 2 sin(100 ) ( )i t A = +

Cu 43: Cho on mch xoay chiu gm R, L mc ni tip.0,4

40 , R L H

= = . on mch c mc vo

hiu in th 40 2 cos100 ( )u t V= . Biu thc cng dng in qua mch l:

A. cos(100 )( )

4

i t A

= B. cos(100 )( )

4

i t A

= +

C. 2 cos(100 )( )2

i t A

= D. 2 cos(100 )( )2

i t A

= +

Cu 44: Cho on mach xoay chiu gm R, L mc ni tip.0,2

20 , R L H

= = . on mch c mc vo

hiu in th 40 2 cos100 ( )u t V= . Biu thc cng dng in qua mch l:

A. 2cos(100 )( )4

i t A

= B. 2cos(100 )( )4

i t A

= +

C. 2 cos(100 )( )2

i t A

= D. 2 cos(100 )( )2

i t A

= +

Cu 45: Cho mch R, L , C mc ni tip R = 20 3 , L =0,6

H , C =

310

4 F

. t vo hai u mch in mt

in p u = 200 2 cos(100 t) V. Biu thc c-ng dng in trong mch

210

i (A)

2 2

2 2 t (s)

O


21/42



A. 5 2 cos 1003

i t

= + (A) B. 5 2 cos 100 6

i t

=

(A)

C. 5 2 cos 1006

i t

= +

(A) D. 5 2 cos 1003

i t

= (A)

Cu 46:t in p xoay chiu vo hai u on mch c R, L, C mc ni tip. Bit R = 10, cun cm thun c

L =

1

10 (H), tin c C =

310

2

(F) v in p gia hai u cun cm thun l Lu 20 2 cos(100 t )2

= + (V).Biu thc in p gia hai u on mch l

A. u 40cos(100 t )4

= + (V). B.u 40cos(100 t )

4

= (V)

C.u 40 2 cos(100 t )4

= + (V). D. u 40 2 cos(100 t )

4

= (V).

Cu 47: t in p xoay chiu c gi tr hiu dng 60V vo hai u on mch R, L, C mc ni tip th cng

dng in qua on mch l i1 = 0I cos(100 t )4

+ (A). Nu ngt b tin C th cng dng in qua on

mch l2 0i I cos(100 t )

12

= (A). in p hai u on mch l

A. u 60 2 cos(100 t )12

= (V). B. u 60 2 cos(100 t )6

= (V)

C. u 60 2 cos(100 t )12

= + (V). D. u 60 2 cos(100 t )

6

= + (V).

Cu 48: Khi t hiu in th khng i 30V vo hai u on mch gm in trthun mc ni tip vi cun

cm thun c t cm1

4H th dng in trong on mch l dng in mt chiu c cng 1A. Nu t

vo hai u on mch ny in p u 150 2 cos120 t= (V) th biu thc ca cng dng in trong onmch l

A. i 5 2 cos(120 t )4

= (A). B. i 5cos(120 t )4

= + (A).

C. i 5 2 cos(120 t )4

= + (A). D. i 5cos(120 t )

4

= (A).

Cu 49: t in p0 cos 100 3

u U t

=

(V) vo hai u mt tin c in dung 2.10

F. thi im in

p gia hai u tin l 150 V th cng dng in trong mch l 4 A. Biu thc ca cng dng introng mch l

A. 4 2 cos 1006

i t

= + (A) B. 5cos 100 6i t

= +

(A)

C.5cos 100 6i t

= (A) D.

4 2 cos 100 6i t

= (A)

Cu 50: t in p xoay chiu0 cos 100 ( )3

u U t V

= +

vo hai u mt cun cm thun c t cm

1

2L

= (H). thi im in p gia hai u cun cm l 100 2 V th cng dng in qua cun cm l 2A.

Biu thc ca cng dng in qua cun cm l

A. 2 3 cos 100 ( )6

i t A

=

B. 2 3 cos 100 ( )6

i t A

= +

C. 2 2 cos 100 ( )6

i t A

= +

D. 2 2 cos 100 ( )6

i t A

=

Cu 51:on mch xoay chiu nh hnh v, 2L H

= ;

C = 31,8 F, R c gi tr xc nh, i 2cos 100 t3

=

(A).

RB

CLA

M


22/42



Biu thc uMB c dng:

A.MBu 200cos 100 t 3

= (V) B.

MBu 600 cos 100 t 6

= +

(V)

C.MBu 200cos 100 t 6

= + (V) D.

MBu 600 cos 100 t 2

=

(V)

Cu 52: Hiu in th hai u on mch xoay chiu ch c t410

C F

=c biu thc

100 2 cos(100 )3

u t

= + V, biu thc cng dng in qua mch trn l nhng dng no sau y?

A. 2 cos(100 )2

i t A

= B. 2 cos(100 )6

i t A

=

C.5

2 cos(100 )6

i t A

= + D. 2 cos(100 )6

i t A

=

Cu 53: Mch in xoay chiu gm in tr 40R = ghp ni tip vi cun cm L. Hiu in th tc thi hai

u on mch 80cos100u t= v in p hiu dng hai u cun cm LU = 40V. Biu thc i qua mch l:

A.2

cos(100 )2 4i t A

= B.2

cos(100 )2 4i t A

= +

C. 2 cos(100 )4

i t A

= D. 2 cos(100 )4

i t A

= +

Cu 54: Mt on mch gm t410

C F

= v cun dy thun cm c t cm L =2

H mc ni tip. Hiu

in th gia 2 u cun dy l 100 2 cos(100 )3L

u t

= + V. Hiu in th tc thi hai u t c biu thc nh

th no?

A. 50 2 cos(100 )6C

u t

= V B.2

50 2 cos(100 )3C

u t

= V

C. 50 2 cos(100 )6C

u t

= + V D. 100 2 cos(100 )3C

u t= + V

Cu 55: Mt on mch gm t410

C F

= v cun dy thun cm c t cm L =2

H mc ni tip. Hiu

in th gia 2 u cun dy l 100 2 cos(100 )3L

u t

= + V. Hiu in th tc thi hai u t c biu thc nh

th no?

A. 50 2 cos(100 )6C

u t

= V B. 250 2 cos(100 )3C

u t

= V

C. 50 2 cos(100 )6

Cu t

= + V D. 100 2 cos(100 )

3Cu t

= + V

Cu 56: Mch xoay chiu gm R, L, C mc ni tip (cun dy thun cm), 100R = , 31,8C F= , h s cng

sut mch2

o s =2

c , hiu in th hai u mch 200cos100u t= (V) t cm L v cng dng in

chy trong mch l

A.2

, 2 cos(100 )4

L H i t

= = (A) B.2

, 2 cos(100 )4

L H i t

= = + (A)

C.2,73

, 2 3 cos(100 )3

L H i t

= = + (A) D.2,73

, 2 3 cos(100 )3

L H i t

= = (A)

Cu 57: Mt bn l 200V 1000W c mc vo hiu in th xoay chiu 100 2 cos100u t= (V). Bn l c

t cm nh khng ng k. Dng in chy qua bn l c biu thc no?A. 2,5 2 cos100i t= (A) B. 2,5 2 cos(100 )

2i t

= + (A)


23/42



C. 2,5cos100i t= (A) D. 2,5 2 cos(100 )2

i t

= (A)

Cu 58: Mt mch gm cun dy thun cm c cm khng bng 10 mc ni tip vi tin c in dung42.10

C F

= . Dng in qua mch c biu thc 2 2 cos(100 )3

i t A

= + . Biu thc hiu in th ca hai u

on mch l:

A. 80 2 cos(100 )6u t = (V) B. 80 2 cos(100 )6

u t = + (V)

C. 120 2 cos(100 )6

u t

= (V) D.2

80 2 cos(100 )3

u t

= + (V)

Cu 59: Cho on mch xoay chiu mc ni tip gm in trc R = 100 , tin c dung khong 200 ,

cun dy c cm khng 100 . in p hai u mch cho bi biu thc u = 200cos(120 t +4

)V. Biu thc

in p hai u tin

A. uc = 200 2 cos (100 t +4

)V. B. uc = 200 2 cos (120 t -2

)V.

C. uc = 200 2 cos (120 t )V. D. uc = 200cos (120 t - 4

)V.

Cu 60: on mch R , L , C mc ni tip c R = 40 ; L =5

1H; C =

6

10 3F. t vo hai u mch in p

u = 120 2 cos100 t (V). C-ng dng in tc thi trong mch l

A. 1,5cos(100 )4

i t A

= +

B. 1,5cos(100 )4

i t A

=

C. 3cos(100 )4

i t A

= +

D. 3cos(100 )4

i t A

=

Cu 61: Nu t vo hai u mt mch in cha mt in tr thun v mt tin mc ni t ip mt in p

xoay chiu c biu thc u = U0cos( t - 2 ) (V), khi dng in trong mch c biu thc i = I0cos( t - 4 ) (A).

Biu thc in p gia hai bn t s l:

A. uC = I0 R cos( t -3

4

)(V). B. uC = 0

U

Rcos( t +

4

)(V).

C. uC = I0ZC cos( t +4

)(V). D. uC = I0 R cos( t -

2

)(V).

Cu 62: Mt on mch xoay chiu gm R v C ghp ni tip. t gia hai u on mch in p xoay chiu c

biu thc tc thi 220 2 cos 100 ( )2

u t V

=

th cng dng in qua on mch c biu thc tc thi

4,4cos 100 ( )4

i t A =

. Hiu in th gia hai u tin c biu thc tc thi l:

A. ( )220cos 100 ( )Cu t V = B.3

220cos 100 ( )4C

u t V

=

C. 220 2 cos 100 ( )2C

u t V

= +

D. ( )220 2 cos 100 ( )Cu t V =

Cu 63: Mt on mch gm cun dy thun cm c t cm L =1

H5

mc ni tip vi tin c in dung

C =310

6

F. Dng in chy qua on mch c biu thc i = 2 2 cos(100t +

3

) (A). Biu thc in p hai u

on mch s l:

A. u = 80 2 cos(100t +6

) (V) B. u = 80 2 cos(100t -

3

) (V)


24/42



C. u = 80 2 cos(100t -6

) (V) D. u = 80 2 sin(100t -

6

) (V)

Cu 64 :Dng in chy qua on mch xoay chiu c dng i = 2cos100t (A), in p gia hai u on mch

c gi tr hiu dng l 12V, v sm pha3

so vi dng in. Biu thc ca in p gia hai u on mch l:

A. u = 12cos100t (V). B. u = 12 2 cos100t (V).

C. u = 12 2 cos(100t3 ) (V). D. u = 12 2 cos(100t

3+ ) (V).

Cu 65: Dng in xoay chiu c tn s 50 Hz. Trong 1s n i chiu bao nhiu ln?A. 25 ln B. 50 ln 100 ln D. 200 ln

Cu 66: Mt n ng hunh quang c di mt hiu in th xoay chiu c gi tr cc i 127V v tn s 50

Hz. Bit n ch sng ln khi hiu in th tc thi t vo n l u 90V. Tnh trung bnh thi gian n sngtrong mi pht l:

A. 30 s B. 40 s 20 s D. 10 sCu 67: Mt n ng hunh quang c di mt hiu in th xoay chiu c gi tr cc i 220V v tn s 50

Hz. Bit n ch sng ln khi hiu in th tc thi t vo n l u 110 2 V. Tnh trung bnh thi gian n

sng trong mi pht l:A. 30 s B. 40 s 20 s D. 10 sCu 68:Mt chic n nn t d-i mt in p xoay chiu 119V 50Hz. N ch sng ln khi in p tc thigia hai u bng n ln hn 84V. Thi gian bng n sng trong mt chu k l

A. t = 0,0100s. B. t = 0,0133s. C. t = 0,0200s. D. t = 0,0233s.Cu 69: Mt n neon c t di hiu in th xoay chiu c dng 100cos100 ( )u t V= . n s tt nu hiuin th tc thi t vo n c gi tr nh hn hoc bng 50V. Khong thi gian n tt trong mi na chu kca dng in xoay chiu l bao nhiu?

A.600

tt s= B.

300

tt s= C.

50

tt s= D.

150

tt s=

Cu 70:Ngi ta t gia hai bn tin mt in p xoay chiu 0 cos 100 .

3

u U t

=

in p t gi tr

cc i ti thi im:

A. 1 ,300 100

kt s k Z

= +

B. 1 ,300 100

kt s k Z

= +

C. ,100

kt s k Z = D.

1,

3 100

kt s k Z

= +

Cu 71: Mch R, L, C mc ni tip, t vo hai u on mch 0 cosu U t= , iu kin c cng h-ng

A. LC 2 = R2 B. R =L

CC.

1

LC = D. LC 2 = 1

Cu 72: Cho mch in xoay chiu nh hnh v. Cun dy c in tr

R v t cm 0,2L H

= , tin c in dung5

10 .C F

= t vo hai

u AB mt hiu in th xoay chiu c biu thc: 0 cos 2 ( )u U ft A= c tn c f thay i c. Xc nh f 2 ?L CZ Z=

A. 50 Hz B.50 2 Hz C. 100 2 Hz D. 500 Hz

Cu 73: Cho mch in xoay chiu gm in trthun R, cun dy thun cm L v tin C =1

(mF) mc ni

tip. Biu thc ca hiu in th gia hai bn tin l u = 50 32 os(100 )4

c t

(V). Cng dng in trong

mch khi t = 0,01s l:

A. - 5 2 A. B. 5 2 A C. 5 A D. 5 ACu 74: Biu thc cng dng in trong mt on mch xoay chiu AB l 4cos(100 )i t A = + . Ti thiim t = 0,04s cng dng in trong mch c gi tr.

A. i = 4 A B. i = 2 2 A C. i = 2 A D. i = 2 A

A B

CL, R


25/42



Cu 75: Cho on mch xoay chiu AB gm in trR = 100, tin410

C F

= v cun cm L =2

H mc

ni tip. t vo hai u on mch AB mt hiu in th xoay chiu c dng u = 200cos100t (V). Cng dng in hiu dng trong mch l :

A. I = 2A B. I = 1,4A C. I = 1A D. I = 0,5A

Cu 76: Cho on mch xoay chiu AB gm in trR = 60, tin C =

-410

F v cun cm L =0,2 H mc

ni tip. t vo hai u on mch AB mt hiu in th xoay chiu c dng u = 50 2 cos100t (V). Cng dng in hiu dng trong mch l :

A. I = 0,25A B. I = 0,50A C. I = 0,71A D. I = 1,00A

Cu 77:t vo hai u t in410

C F

= mt in p xoay chiu u = 141cos(100t)V. C-ng dng in

qua t in lA. I = 1,41A. B. I = 1,00A. C. I = 2,00A. D. I = 100A.

Cu 78:t vo hai u cun cm 1L H

= mt in p xoay chiu u = 141cos(100t)V. C-ng dng in

hiu dng qua cun cm lA. I = 1,41A. B. I = 1,00A. C. I = 2,00A. D. I = 100A.

Cu 79:t vo hai u cun cm 1L H

= mt in p xoay chiu 220V 50Hz. C-ng dng in hiu

dng qua cun cm lA. I = 2,2A. B. I = 2,0A. C. I = 1,6A. D. I = 1,1A.

Cu 80:C-ng dng in trong mch khng phn nhnh c dng i = 2 2 cos100t(A). C-ng dng inhiu dng trong mch l

A. I = 4A. B. I = 2,83A. C. I = 2A. D. I = 1,41A.Cu 81:t in p xoay nhiu c gi tr hiu dng 200V v tn s khng i vo hai u A v B ca on mchmc ni tip theo th t gm bin trR, cun cm thun c t cm L v tin c in dung C thay i. Gi

N l im ni gia cun cm thun v tin. Cc gi tr R, L, C hu hn v khc khng. Vi C = C1 th in phiu dng gia hai u bin trR c gi tr khng i v khc khng khi thay i gi tr R ca bin tr. Vi C =

21C th in p hiu dng gia A v N bng

A. 200 2 V B. 100 V C. 200 V D. 100 2 VCu 82: Biu thc hiu in th hai u mt on mch: u = 200 cos t (V). Ti thi im t, hiu in th u =

100(V) v ang tng. Hi vo thi im (4

Tt+ ), hiu in th u bng bao nhiu?

A. 100 V. B. 100 2 V. C. 100 3 V. D. - 100 V.

Cu 83: Ti thi im t, in p u = )2100cos(2200

t (trong u tnh bng V, t tnh bng s) c gi tr

100 2 V v ang gim. Sau thi im s300

1, in p ny c gi tr l

A. -100 2 V B. -100 V C. 100 3 V D. 200 V

Cu 84: Cho mch in nh hnh v vi UAB = 300V,UNB = 140V, dng in i tr pha so vi uAB mt gc (cos = 0,8), cun dy thun cm. Vn k V ch gi tr:

A. 100 V B. 200 VC. 300 V D. 400 V

Cu 85: Mt mch dao ng in t, cun dy thun cm c h s t cm 0,5mH, tin c in dung 0,5nF.Trong mch c dao ng in tiu ha. Khi cng dng in trong mch l 1 mA th in p hai u tinl 1V. Khi cng dng in trong mch l 0 A th in p hai u t l:

A. 2 V B. 2 V C. B. 22 V D. 4 V

RB

CLA

N

V


26/42



Cu 86: on mch gm mt in trni tip vi cun dy thun cm, khi vn k mc gia hai u in trsch vn k l 80V, mc gia hai u cun dy s ch l 60V. S ch vn k l bao nhiu khi mc gia hai u onmch trn?

A. 140V B.20V C. 100V D. 80VCu 87: Cho mch in xoay chiu nh hnh v; cundy thun cm. Hiu in th hiu dng gia A v B l

200V, UL

=3

8U

R

= 2UC

. Hiu in th hiu dng gia

hai u in trR l:A. 180V. B. 120V . C. 145V. D. 100V.

Cu 88:in p tc thi gia hai u on mch c dng u = 141cos(100t)V. in p hiu dng gia hai uon mch l

A. U = 141V. B. U = 50V. C. U = 100V. D. U = 200V.

Cu 89:t mt in p 120 2 os(100 )( )6

u c t V

= vo hai u mch in gm tin c dung khng 70

v cun dy c in trthun R, t cm L. Bit dng in chy trong mch 4 os(100 )( )12

i c t A

= + . Tng tr

ca cun dy l

A. 100. B. 40. C. 50. D. 70.Cu 90: Mch in xoay chiu gm RLC mc ni tip, c R = 30, ZC = 20, ZL = 60. Tng trca mch l

A. Z = 50 B. Z = 70 C. Z = 110 D. Z = 2500Cu 91: Cho on mch xoay chiu R, C mc ni tip. 100R = , 1,5

C RU U= , tn s ca dng in xoay chiu f

= 50Hz. Tng trca mch v in dung ca t c gi tr no sau y?

A.210

; 10115

C F Z

= = B.310

; 18015

C F Z

= =

C.310

; 1125

C F Z

= = D.410

; 141C F Z

= =

Cu 92:t vo hai u t in

410C F

= mt in p xoay chiu tn s 100Hz, dung khng ca t in lA. ZC = 200. B. ZC = 100. C. ZC = 50. D. ZC = 25.

Cu 93:t vo hai u t in410

C F

= mt in p xoay chiu u = 141cos(100t)V. Dung khng ca t in

lA. ZC = 50. B. ZC = 0,01. C. ZC = 1. D. ZC = 100.

Cu 94:t vo hai u cun cm1

L H

= mt in p xoay chiu u = 141cos(100t)V. Cm khng ca cun

cm lA. ZL = 200. B. ZL = 100. C. ZL = 50. D. ZL = 25.

Cu 95: Cho mt on mch xoay chiu gm cun dy c in trr, t cm L mc ni tip vi in trthun5R = . Bit hiu in th gia hai u on mch v cng dng in qua mch c biu thc:

100 2 cos(100 ) ( ), 2 2 cos(100 ) ( )6 2

u t V i t A

= + = + . Gi tr ca r bng:

A. 20,6 B. 36,6 C. 15,7 D. 25,6 Cu 96: Cho mt on mch xoay chiu nh hnh v. Hiu in thhai u AB c gi tr hiu dng U 240 2= V. Bit C LZ 2Z= . B quain trca cc dy ni v kha K.

Khi kha K ngt, dng in qua mch l:1i 4 2 cos(100 t ) (A).3

= +

Khi kha K ngt, dng in qua mch l: 2i 4 2 cos(100 t ) (A).6

= Gi tr ca R l:

A. 30 2 B. 60 C. 60 2 D. mt gi tr khc

A B

CR L

BAR, L

K

C


27/42



Cu 97:t in p xoay chiu c gi tr hiu dng khng i vo hai u on mch gm bin trR mc nitip vi tin. Dung khng ca tin l 100 . Khi iu chnh R th ti hai gi tr R1 v R2 cng sut tiu thca on mch nh nhau. Bit in p hiu dng gia hai u tin khi R = R1 bng hai ln in p hiu dnggia hai u tin khi R = R2. Cc gi tr R1 v R2 l:

A. R1 = 50, R2 = 100 . B. R1 = 40, R2 = 250 .C. R1 = 50, R2 = 200 . D. R1 = 25, R2 = 100 .

Cu 98: Cho on mach xoay chiu gm R, L mc ni tip, hiu in th hai u on mch c dng

100 2 sin100 ( )u t V= v cng dng in qua mch c dng 2cos(100 )( )4

i t A= . R, L c nhng gi

tr no sau y:

A.1

50 , R L H

= = B. 250 2 ,2

R L H

= =

C. 150 ,2

R L H

= = D.1

100 , R L H

= =

Cu 99: Cho mch in xoay chiu nh hnh v bn. Cun dy c

r= 10 ,L= H10

1

. t vo hai u on mch mt hiu in th

dao ng iu ho c gi tr hiu dng l U= 50V v tn sf= 50Hz.Khi in dung ca tin c gi tr l C1 th s ch ca ampe k l cci v bng 1A. Gi tr caR v C1 l

A. R = 40 v FC

3

1

10.2 = . B. R = 50 v FC

3

1

10= .

C. R = 40 v F103

1

=C . D. R = 50 v FC

3

1

10.2 = .

Cu 100: Cho mch in xoay chiu gm R, C ghp ni tip, hiu in th hai u mch c dng

50 2 cos100AB

u t= (V) v cng dng in qua mch 2 cos(1003

i t

= + ) (A). R, C c nhng gi tr no

sau y?

A.310

50 ;5

R C F

= = B.23.10

25 ;25

R C F

= =

C.210

25 ;25 3

R C F

= = D.35.10

50 ; R C F

= =

Cu 101: Mt on mch AB gm hai on mch AM v MB mc ni tip. on mch AM c in tr thun

50 mc ni tip vi cun cm thun c t cm

1H, on mch MB ch c tin vi in dung thay i

c. t in p u = U0 t100cos (V) vo hai u on mch AB. iu chnh in dung ca tin n gi tr

C1 sao cho in p hai u on mch AB lch pha

2

so vi in p hai u on mch AM. Gi tr ca C1 bng

A. F

510.8 B. F

510C. F

510.4 D. F

510.2

Cu 102: Cho mch in nh hnh v. Bit cun dy thun cm.t vo hai u mt in p xoay chiu 0 cos(100 ).u U t = +

iu chnh gi tr ca t cm L ta thy khi 13

L L H

= = v

2

1 L L H

= = thi dng in tc thi 1 2,i i tng ng u lch pha mt mt gc 4

so vi hiu in th hai

mch in. Tnh C?

A. 50C F

= B. 100C F

= C. 150C F

= D. 200C F

=

CRr, L

NM

A

R L CA B


28/42



Dng dkin sau tr li cu 103, 104t vo hai u mch RLC mt hiu in th xoay chiu: 120 2 cos100u t= (V). Bit 20 3R = ,

60CZ = v t cm L thay i (cun dy thun cm).

Cu 103: Xc nh L LU cc i v gi tr cc i ca LU bng bao nhiu?

A.ax

0,8; 120

Lm L H U V

= = B. ax

0,6; 240

Lm L H U V

= =

C. ax0,6 ; 120

Lm L H U V

= = D. ax

0,8 ; 240Lm

L H U V

= =

Cu 104: 120 3L

U V= th L phi c cc gi tr no sau y?

A. 0,6L H

= hoc1, 2

L H

= B.0,8

L H

= hoc1,2

L H

=

C. 0,4L H

= hoc0,8

L H

= D.0,6

L H

= hoc0,8

L H

=

Dng dkin sau tr li cu 105, 106Mt mch in R, L, C ni tip (cun dy thun cm). Hiu in th hai u mch 100 6 cos100u t= (V),

100 2R = ,2

L H= .Cu 105: Tin C c gi tr bng bao nhiu th UC max gi tr UC max bng bao nhiu?

A.510

3C F

= , UC max = 30V B.410

C F

= , UC max = 100V

C.510

3C F

= , UC max = 300V D.410

C F

= , UC max = 30V

Cu 106: Tin C c gi tr bng bao nhiu 200 2C

U = V?

A.410

3C F

= B.410

2,4C F

= hoc410

4C F

=

C. 4102,4

C F

= hoc 5103

C F

= D. 410

3C F

= hoc 4104

C F

=

Cu 107: Cho on mch R, L, C ni tip vi L c th thay i c. Trong R v C xc nh. Mch in c

t di hiu in th 2 cos .u U t= Vi U khng i v cho trc. Khi hiu in th hiu dng gia hai u

cun cm cc i. Gi tr ca L xc nh bng biu thc no sau y?

A. L = R2 +2 2

1

CB. L = 2CR2 +

2

1

C

C. L = CR2 +2

1

2C

D. L = CR2 +2

1

C

Cu 108:t in p xoay chiu c gi tr hiu dng 120V, tn s 50 Hz vo hai u on mch mc ni tip

gm in trthun 30 , cun cm thun c t cm 0,4 H

v tin c in dung thay i c. iu chnh

in dung ca tin th in p hiu dng gia hai u tin t gi tr cc i bngA. 150 V. B. 160 V. C. 100 V. D. 200 V.

Cu 109: Nu t vo hai u cun dy mt hiu in th mt chiu 9V th cng dng in trong cun dy l0,5A. Nu t vo hai u cun dy mt hiu in th xoay chiu c tn s 50Hz v c gi tr hiu dng l 9V thcng hiu dng ca dng in qua cun dy l 0,3A. in trthun v cm khng ca cun dy c gi tr l:

A. 18 ; 30L

R Z= = B. 18 ; 24LR Z= = C. 18 ; 12

LR Z= = D. 30 ; 18LR Z= =

Cu 110: Mch in xoay chiu gm R, cun dy thun cm L, t C mc ni tip. Hiu in thhai u mch50 2 cos100u t= (V), 30LU V= , 60CU V= . Cng sut tiu th trong mch l P = 20W. R, L, C c nhng gi

tr no sau y?


29/42



A.30,8 10

60 , ;12

R L H C F

= = = B.30,6 10

80 , ;12

R L H C F

= = =

C.30,6 10

120 , ;8

R L H C F

= = = D.31, 2 10

60 , ;8

R L H C F

= = =

Cu 111: Mt cun dy c t cm l1

4H mc ni tip vi tin C1 =

310

3

F ri mc vo mt in p xoay

chiu tn s 50Hz. Khi thay i t C1 bng mt t C2 th thy cng dng in qua mch khng thay i. indung ca t C2 bng

A.310

4

F B. 102

F C.310

2

F D.32.10

3

F

Cu 112: Mt mch in xoay chiu mc ni tip gm mt in tr, mt t in v mt cun dy thun cm c hs t cm L c th thay i, vi u l hiu in th hai u on mch v uRC l hiu in th hai u on mchcha RC, thay i L hiu in th hai u cun dy t gi tr cc i khi kt lun no sau y l sai ?

A. u v uRC vung pha. B.2 2 2

L Max RC U U U= +

C.2 2C

L

C

Z RZ

Z

+= D.

2 2C

LMax

C

U R ZU

Z

+=

Cu 113: t vo hai u mt on mch RLC khng phn nhnh mt hiu in th u = U 0cos t (V) th c-ng

dng in trong mch c biu thc i = I0 cos(3

t

)A. Quan h gia cc tr khng trong on mch ny tho

mn:

A. 3L CZ Z

R

= B. 3C L

Z Z

R

=

C.1

3C LZ Z

R

= D.

1

3L CZ Z

R

=

Cu 114: Mt on mch RLC khng phn nhnh gm in tr = 50R , tin c dung khng bng in trv

cun thun cm c t cm L thay i c. Mc on mch vo in p xoay chiu c in p hiu dngkhng i v tn s 50Hz. iu chnh L in p gia hai u cun dy cc i, gi tr ca L l

A. HL2

1= B. HL

2= C. HL

2

1= D. HL

1=

Cu 115:t in p xoay chiu c gi tr hiu dng khng i, tn s 50Hz vo hai u on mch mc ni tipgm in trthun R, cun cm thun c t cm L v tin c in dung C thay i c. iu chnh C n

gi tr F4

10 4hoc F

2

10 4th cng sut tiu th trn on mch u c gi tr bng nhau. Gi tr ca L bng

A.3

1H B.

2

1H C.

3

H D.

2H

Cu 116: Cho mch in nh hnh v,Bit

0,6L H

= ,

410C F

= , r = 30,

uAB = 100 2 cos100t(V). Cng sut trn Rln nht khi R c gi tr:

A. 40 B. 50 C. 30 D. 20Cu 117: Mt on mch xoay chiu gm R v cun dy thun cm L mc ni tip, 100R = , tn s dng in f

= 50Hz. Hiu in th hiu dng 2 u mch U = 120V. L c gi tr bao nhiu nu umch v i lch nhau 1 gc3

,

cho bit gi tr cng sut ca mch lc .A. P = 36W B. P = 75W C. P = 72W D. P = 115,2W

Cu 118: Trong mch in xoay chiu khng phn nhnh, hiu in th gia hai u on mch v cng

dng in trong mch ln lt l: u = 100cos100t (V) v i = 100cos(100t +3

) (mA). Cng sut tiu thu trong

mch l

RB

Cr, LA


30/42



A. 5000W B. 2500W C. 50W D. 2,5W

Cu 119: Dt mt in p u U 2 cos t= mt on mch RLC mc ni tip. Trong U, , R v C khng i.iu chnh L in p trn L t cc i. Chn biu thc sai

A. 2 2 2 2R L CU U U U= + + B.2 2L L CU U U U 0 =

C. 2 2L C CZ Z R Z= + D.2 2

CL max

U R ZU

R

+=

Cu 120: Cho mch in RLC mc ni tip. t mt in p xoay chiu u U 2 cos t (V)= vo hai u mchv thay i tn s fin p hai u tin t gi tr cc i. Khi , tn s f bng

A.1 1 R

f2 LC 2L

=

B.2

2

1 1 Rf

2 LC 2L=

C.2

2

1 Rf LC

2 2L=

D. 2 2

1 1f R L

2 LC=

Cu 121: Cho mch in RLC mc ni tip. t mt in p xoay chiu u U 2 cos t (V)= vo hai u mchv thay i tn s fin p hai u cun cm t gi tr cc i. Khi , tn s f bng

A.1 1 R

f 2 LC 2L= B. 2 21 2

f 2 2LC R C=

C.2 2

1 2f

2 2LC R C=

+D. 2 2

1 1f R L

2 LC=

Cu 122: Cho mch in RLC mc ni tip. t mt in p xoay chiu u U 2 cos t (V)= . Thay i lnlt R L CU , U , U t gi tr cc i vi 0 1 2, , . Mi lin h gia 0 1, v 2 l

A. 20 1 2 = B. 102

=

C. 2 1

02

=

D. 0 1 2 =

Cu 123:t in p xoay chiu u = U0cost c U0 khng i v thay i c vo hai u on mch c R,L, C mc ni tip. Thay i th cng dng in hiu dng trong mch khi = 1 bng cng dng inhiu dng trong mch khi = 2. H thc ng l :

A. 1 22

LC + = . B. 1 2

1

LC = .

C. 1 22

LC + = . D. 1 2

1

LC = .

Cu 124:t in p u = U 2 cost vo hai u on mch AB gm hai on mch AN v NB mc ni tip.on AN gm bin trR mc ni tip vi cun cm thun c t cm L, on NB ch c tin vi in dung

C. tLC2

11 = . in p hiu dng gia hai u on mch AN khng ph thuc vo R th tn s gc

bng

A.21 B.

221 C. 21 D. 21

Cu 125: on mch xoay chiu gm in trthun R, cun dy thun cm L v tin C mc ni tip. Dt vohiu in th xoay chiu mt hiu in th hiu dng c tn s khng i c tn s gc thay i c. khi

1 200 / rad s = = hoc 2 500 / rad s = = th cng sut tiu th trn on mch l nh nhau. cngsut tiu th trong on mch l cc i th tn s gc phi bng

A. 125 / rad s B. 40 /rad s C. 100 / rad s D. 200 / rad s

Cu 126:t in p u 100 2 cos t= (V), c thay i c vo hai u on mch gm in trthun 200

, cun cm thun c t cm 25

36

H v tin c in dung10

F mc ni tip. Cng sut tiu th ca on

mch l 50 W. Gi tr ca lA. 150 rad/s. B. 50 rad/s. C. 100 rad/s. D. 120 rad/s.


31/42



Cu 127: Mt mch in R, L, C mc ni tip. Bit h s cng sut ca mch ny l os = 1c . Nhn xt no sauy l sai.

A. Cng dng in qua mch t cc i.B. Mch tiu th cng sut ln nhtC. Hiu in th hiu dng hai u mch bng hiu in th hiu dng hai u cun dy.D. Hiu in thhai u mch cng pha vi cng dng in

Cu 128:Mt tin c in dung C = 5,3F mc ni tip vi in trR = 300 thnh mt on. Mc onmch ny vo mng in xoay chiu 220V - 50Hz. H s cng sut ca mch l :

A. 0,3331 B. 0,4472 C. 0,4995 D. 0,6662

Cu 129:Mt tin c in dung C = 5,3F mc ni tip vi in trR = 300 thnh mt on. Mc onmch ny vo mng in xoay chiu 220V - 50Hz. in nng m on mch tiu th trong mt pht l :

A. 4329J B. 1047J C . 1933J D. 2148J

Cu 130: Cho mch xoay chiu R, L, C khng phn nhnh, = 250R , 100 2RLU U V= = , VUC 200= . Cngsut tiu th ca mch l

A. 100 2 W B. 200 2 W C. 200 W D. 100 WCu 131: Cho mch xoay chiu khng phn nhnh RLC c R thay i -c. Hiu in th hai u mch l

( )tUu 100cos0= (V),2

104

=C F,8,0=L H . cng sut tiu th ca mch cc i th gi tr ca R bng

A. 120 B. 50 C. 100 D. 200

Cu 132:t in p u 100cos( t )6

= + (V) vo hai u on mch c in tr thun, cun cm thun v t

in mc ni tip th dng in qua mch l i 2cos( t )3

= + (A). Cng sut tiu th ca on mch l

A. 100 3 W. B. 50 W. C. 50 3 W. D. 100 W.Cu 133: Mch nh hnh v, C = 318 F, R bin i. Cun dythun cm, hiu in th hai u mch 0 cos100u U t= (V),

cng sut tiu th ca mch t gi tr cc i khi R = R0 = 50.Cm khng ca cun dy bng:

A. 40 B. 100 C. 60 D. 80

Cu 134: Cho mch in gm R, L, C mc ni tip. Bit 0,2L H

= , 31,8C F= , f = 50Hz, hiu in th hiu

dng hai u on mch l 200 2( )U V= . Nu cng sut tiu th ca mch l 400W th R c nhng gi tr nosau y:

A. 160 40 R hay R= = B. 80 120 R hay R= = C. 60R = D. 30 90 R hay R= =

Cu 135: Cho mch in gm R, L, C mc ni tip. Bit 1

L H=,

310

4C F

=, 120 2 cos100 ( )u t V

=, in

trphi c gi tr bng bao nhiu cng sut ca mch t gi tr cc i? Gi tri cc i ca cng sut l baonhiu?

A. ax120 , 60wmR P= = B. ax60 , 120wmR P= =

C. ax40 , 180wmR P= = D. ax120 , 60wmR P= = Dng dkin sau tr li cu 136, 137

Cho mch in gm R, L, C mc ni tip. hiu in th gia hai u A v B c biu thc

100 2 cos100 ( )u t V= . Cun cm c t cm2,5

L H

= , in trthun r = R = 100 . Tin c in dung

C. Ngi ta o c h s cng sut ca mch l os = 0,8c

Cu 136: Bit hiu in th gia hai u mchsm pha hn cng dng in qua mch. Gi tr ca C l baonhiu?

A.310

3C F

= B.410

C F

= C.410

2C F

= D.310

C F

=

RCL


32/42



Cu 137: cng sut tiu th cc i, ngi ta mc thm mt t c in dung C1 vi t C c mt b tinc in dung thch hp. Xc nh cch mc v gi tr C1

A. Mc song song,4

1

10

2C F

= B. Mc song song,4

1

3.10

2C F

=

C. Mc ni tip,4

1

3.10

2C F

= D. Mc ni tip,4

1

2.103

C F

=

Cu 138: Cho mch R, L, C ni tip, R l bin tr. Hiu in th hai u mch c dng: 200 2 cos100u t= (V);1,4L H

= ;

410

2C F

= . R c gi tr bao nhiu cng sut tiu th ca mch l 320W.

A. 25R = hoc 80R = B. 80R = hoc 45R = C. 25R = hoc 45R = D. 45R = hoc 80R =

Cu 139: Cho mch in xoay chiu RLC ni tip. Bit1

,2

L H

=

410,C F

= R l bin tr. Bit hiu in th hai u mch in l:

200 2 cos10 ( ).u t V= Thay i R cng sut trong mch t

cc i. Khi :A. max 100P W= B. max 200P W= C. max 400P W= D. max 500P W=

Cu 143: Mt on mch in xoay chiu gm mt cun dy c in trthun 5 v t cm235.10

L H

=

mc ni tip vi in trthun 30R = . Hiu in thhai u on mch l: 70 2 cos100u t= (V). Cngsut tiu th ca on mch l:

A. 35 2P = W B. 70P = W C. 60P = W D. 30 2P = WCu 144: Cho mt on mch in xoay chiu RLC mc ni tip. Bit 80R = . Hiu din th hai u mch inc biu thc: 200 2 cos 2 ( ),

ABu ft V = vi tn s f thay i c. Thay i f cng sut trong mch t cc

i. Gi tr cc i bng:

A. 125P W= B. 250 2P W= C. 250P W= D. 500P W= Cu 145: Cho on mch RLC nh hnh v,uAB = 100 2 cos100t(V). Thay i R n R0 thPmax = 200W. Gi tr R0 bng:

A. 75 B. 50C. 25 D. 100

Cu 146:t mt hiu in th xoay chiu c gi tr hiu dng U = 100V vo hai u on mch RLC ni tip,cun dy thun cm khng, R c gi tr thay i c. iu chnh R hai gi tr R1 v R2 sao cho R1 + R2 = 100th thy cng sut tiu th ca on mch ng vi hai trng hp ny nh nhau. Cng sut ny c gi tr l

A. 200W. B. 400W. C. 50W. D. 100W.

Cu 147: Mch in R, L, C mc ni tip. L = 0,6

H , C =-410 F

, f = 50Hz. Hiu in th hiu dng hai u

on mch U = 80V. Nu cng sut tiu th ca mch l 80W th gi trin trR lA. 30. B. 80. C. 20. D. 40.

Cu 148:t in p xoay chiu c gi tr hiu dng v tn s khng i vo hai u on mch gm bin trRmc ni tip vi tin c in dung C. Gi in p hiu dng gia hai u tin, gia hai u bin trv h scng sut ca on mch khi bin trc gi tr R1 ln lt l UC1, UR1 v cos1; khi bin trc gi tr R2 th ccgi tr tng ng ni trn l UC2, UR2 v cos2. Bit UC1 = 2UC2, UR2 = 2UR1. Gi tr ca cos1 v cos2 l:

A. cos1 =5

1; cos2 =

3

1. B. cos1 =

3

1; cos2 =

5

2.

C. cos1 = 51 ; cos2 = 52 . D. cos

1 = 22 1 ; cos2 = 21 .

RB

CLA

A B

CR L


33/42



Cu 149:t vo hai u mch in xoay chiu RLC ni tip c R thay i c mt in p xoay chiu lun nnh v c biu thc u = U0cos t (V). Mch tiu th mt cng sut P v c h s cng sut cos . Thay i R vgi nguyn C v L cng sut trong mch t cc i khi :

A. P =2

L C

U

2 Z Z, cos = 1. B. P =

2U

2R, cos = 2

2.

C. P =

2

L C

U

Z Z , cos =2

2 . D. P =

2U

R , cos = 1.

Cu 150: Trong mch in xoay chiu c R, L, C mc ni tip. Cho L, C khng i, R thay i. Thay i R chon khi R = Ro th cng sut Pmax. Khi :

A. Ro = |ZL ZC| . B. Ro = (ZL ZC)2. C. Ro = ZC ZL. D. Ro = ZL ZC.

Cu 151: Cho mt on mch in xoay chiu gm R ni tip cun dy (L, r) ni tip t C. Bit hiu in thhiu dng gia hai u on mch l U = 200V, tn s f = 50 Hz, in trR = 50, UR = 100V, Ur = 20V. Cngsut tiu th ca mch

A. 60 W B. 480W. C. 120W. D. 240W.

Cu 152: t hiu in th xoay chiu u = 160 2 cos100 t (v) vo hai u mt on mch xoay chiu thy biu

thc dng in l i = 2 cos(1002

t

+)A. Mch ny c nhng linh kin g ghp ni tip vi nhau?

A. C ni tip L B. R ni tip LC. R, L, C ni tip D. R ni tip C

Cu 153:t vo hai u n mch ch c mt phn t mt in p xoay chiu u = U0cos( t )6

(V) th dng

in trong mch l i = I0cos(2

t )3

(A). Phn t l :

A. cun dy thun cm. B. cun dy c in trthun.C. in trthun. D. tin.

Cu 154: Mt hp en cha mt phn tv mt linh kin no . Nu ta mc dng in mt chiu AI 2= quahp th thy cng sut l P, khi ta thay dng in trn bng dng in xoay chiu c c-ng hiu dng ng

bng 2 A th thy cng sut ch cn l2P . Phn tv linh kin trong hp X l

A. T in v iot B. Cun dy khng thun cmC. Cun dy thun cm v iot D. in tr thun v iot

Cu 155: Cho mt hp en X bn trong cha 2 trong 3 phn tR, L,C. t mt hiu in th khng i U = 100V vo hai u on mch th thy I = 1A. Xc nh cc phn ttrong mch v gi tr ca cc phn t.

A. Cun dy khng thun cm =100R B. Cun dy thun cm, = 100LZ C. Cun dy khng thun cm == 100LZR D. in tr thun v t in, == 100CZR

Cu 156:Cho on mch in xoay chiu gm hai phn t mc ni tip. Hiu in th gia hai u on mch

l: 100 2 os(100 )2

u c t V

= v cng dng in trong mch c biu thc: 10 2 os(100 )4

i c t A

= . Hai

phn t l?A. Hai phn t l RL. B. Hai phn t l RC.

C. Hai phn t l LC. D. Tng trca mch l 10 2

Cu 157:Mt on mch in t di hiu in th0 os( )4

u U c t

= V th cng dng in qua mch c

biu thc0 os( )4

i I c t A

= + . Cc phn t mc trong on mch ny l:

A. Ch c L thun cm B. Ch c CC. L v C ni tip vi LC2 < 1 D. B v C ng

Cu 158:Cho on mch AB gm bin tr ni tip vi hp kn X. Hp X ch cha cun thun cm L hoc t

C UAB = 200V khng i ; f = 50 Hz Khi bin trc gi tr: sao cho PAB cc i th I =2

A v sm pha hnuB. Khng nh no l ng ?

A. Hp X cha C =50

F B. Hp X cha L =1

H


34/42



C. Hp X cha C =200

F D. Hp X cha L =1

2H

Cu 159:on mch AM gm cun thun cm L, in tr thun R ni tip vi on mch MB gm hp kn X

uAB = 200 2 cos100tV R = 20 ; L =3

5H, I = 3A, uAM vung pha vi uMB . on mch X cha 2 trong 3

phn t Ro, L hoc Co mc ni tip. Khng nh no l ng ?

A. X cha Ro = 93,8 v ZC = 54,2 B. X cha Ro = 93,8 v ZL = 120 C. X cha ZC = 54,2 v ZL = 120 D. X cha =

3

3800R v ZC =

3

80 .

Cu 160: Mt on mch xoay chiu c 2 phn t mc ni tip R, C hoc cun dy thun cm. Hiu in th gia

hai u mch in v cng dng in qua mch c biu thc: 100 2 cos(100 )u t= V, 2cos(100 )4

i t A

= .

Mch gm nhng phn t no? in trhoc trkhng tng ng l bao nhiu?A. R, L; 40 , 30LR Z= = B. R, C; 50 , 50CR Z= = C. L, C; 30 , 30

L CZ Z= = D. R, L; 50 , 50LR Z= =

Cu 161: t mt in p xoay chiu0 cos 100 ( )3u U t V

= + vo hai u mt cun dy thun cm c

t cm1

.2

L H

= thi im in p tc thi hai u cun cm l 100 2 V th cng dng in tc thi

qua cun cm l 2 A. Biu thc ca dng in qua cun cm l:

A. 2 3 cos 100 ( )6

i t A

=

B. 2 3 cos 100 ( )6

i t A

= +

C. 2 2 cos 100 ( )6

i t A

= +

D. 2 2 cos 100 ( )6

i t A

=

Cu 162: Cho mch in xoay chiu nh hnh v. Bit cng dng in trong mch c biu thc

0 cos100 ( )i I t A= . in p trn on mch AN c dng 100 2 cos 100 ( )3u t V = + v lch pha 2 so

vi in p ca on mch MB. Biu thc in p ABu v h s cng sut ca mch l:

A.2 5

100 cos 100 ( ); cos3 6 7MB

u t V

= =

B. 2100 3 cos 100 ( ); cos3 7MB

u t V = + =

C.2 3

100 cos 100 ( ); cos3 6 7MB

u t V

= =

D.6

100 2 cos 100 ( ); cos3 7MB

u t V

= + =


35/42



CH IITRUYN TI IN NNG I XA. MY BIN P

MY PHT IN XOAY CHIUNG CKHNG NG B BA PHA

A. TM TT KIN THC CBN

I. TRUYN TI IN NNG I XA. MY BIN P

1. Bi ton truyn ti in nng i xa :+ Cng sut my pht : Ppht = UphtI.cos

+ Cng sut hao ph :2

2 2os

PP R

U c =

Trong : P l cng sut truyn i ni cung cpU l in p ni cung cp cos l h s cng sut ca dy ti in

lR

S= l in trtng cng ca dy ti in (lu :dn in bng 2 dy)

+ gim in p trn ng dy ti in: U = IR+ Gim hao ph c 2 cch :Gim R : cch ny rt tn km chi phTng U : Bng cch dng my bin th, cch ny c hiu qu

+ Hiu sut truyn ti .100%tt

tt

P PH

P

=

2. My bin p :a. nh ngha : Thit b c kh nng bin iin p xoay chiu.b. Cu to : Gm 1 khung st non c pha silc

( Li bin p) v 2 cun dy dn qun trn 2

cnh ca khung .Cun dy ni vi ngun ingi l cun scp. Cun dy ni vi ti tiu thgi l cun th cp.

c. Nguyn tc hot ng : Da trn hin tngcm ng in t.

Dng in xoay chiu trong cun scp gy ra bin thin t thng trong cun th cp lm pht sinh dngin xoay chiu.d. Cng thc :

N1, U1, I1 l s vng dy, hiu in th, cng dng in cun scpN2, U2, I2 l s vng dy, hiu in th, cng dng in cun scp

1 1 2 1

2 2 1 2

U E I N

U E I N = = =

U2 > U1 ( N2 > N1): My tng pU2 < U1 ( N2 < N1) : My h p

e. ng dng : Truyn ti in nng, nu chy kim loi, hn in

II. MY PHT IN XOAY CHIU

1. My pht in xoay chiu 1 pha :- Phn cm : L nam chm to ra t thng bin thin bng cch quay quanh 1 trc Gi l rto- Phn ng : Gm cc cun dy ging nhau cnh trn 1 vng trn.

Tn s dao ng:; n (vong/s)

; n (vong/phut)60

f np

np

f

=

=

; p: s cp cc t

Ch : M t my pht i n c 1 cp cc t mu n pht ra vi tn s 50Hz th phi quay vi tc 50 vong/sn = ; c 10 cp cc tmun pht ra vi tn s 50Hz th phi quay vi tc 5 vong/sn = . S

cp cc tng ln bao nhiu ln th tc quay gim i by nhiu ln.

U1N1

U2N2

U2N2

U1N1


36/42



2. My pht in xoay chiu 3 pha :a. Cu to v nguyn tc hot ng :

- My pht in xoay chiu ba pha l my to ra 3 sut in ng xoay chiu hnh sin cng tn s, cng bin

v lch pha nhau2

3

.

Cu to :

Gm 3 cun dy hnh tr ging nhau gn cnh trn mt vng trn lch nhau2

3

.

Mt nam chm quay quanh tm O ca ng trn vi tc gc khng i

Nguyn tc : Khi nam chm quay t thng qua 3 cun dy bin thin lch pha2

3

lm xut hin 3 sut in

ng xoay chiu cng tn s, cng bin , lch pha2

3

.

T thng gi qua khung dy ca my pht in = NBScos(t +) = 0cos(t + )Vi 0 = NBS l t thng cc i, N l s vng dy, B l cm ng t ca t trng, S l din tch ca vng dy, = 2f

Sut in ng trong khung dy: e = NSBcos(t + -2

) = E0cos(t + -2

)

Vi E0 = NSB l sut in ng cc i.Dng in xoay chiu ba pha l h thng ba dng in xoay chiu, gy bi ba sut in ng xoay chiu

cng tn s, cng bin nhng lch pha tng i mt l2

3

.

1 0

2 0

3 0

os( )

2os( )

32

os( )3

e E c t

e E c t

e E c t

= =

= +

trong trng hp ti i xng th

1 0

2 0

3 0

os( )

2os( )

32

os( )3

i I c t

i I c t

i I c t

= =

= +

My pht mc hnh sao: Ud = 3 UpMy pht mc hnh tam gic: Ud = UpTi tiu th mc hnh sao: Id = Ip

Ti tiu th mc hnh tam gic: Id = 3 Ip

Lu : my pht v ti tiu th thngchn cch mc tng ng vi nhau.

c. u im :- Tit kim c dy dn- Cung cp in cho cc ng c3 pha

Lu : my pht v ti tiu th thngchn cch mc tng ng vi nhau.

~

~ ~

1

23

0

K hiu My pht in ba pha

A3A1

B1 A2

B2B3A3 A1

B1

A2

Mc sao Mc tam ic

1B

2B3B

(1)

(2)3


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III. NG CKHNG NG B BA PHA

1. Nguyn tc hot ng :Khung dy dn t trong t trng quay s quay theo t trng

vi tc nh hn.Nguyn tc hot ng da vo hin tng cm ng in t v s dng

t trng quay.

2. ng ckhng ng b ba pha :- Cu to: Gm c 2 b phn chnh:

+ Stato : (phn ng) gm 3 cun dy ging nhau t lch2

3

trn 1

vng trn.+ Rto : (phn cm) Khung dy dn quay di tc dng ca t trng.

- Khi cho dng in xoay chiu 3 pha vo 3 cun dy y th t trng do 3 cun dy to ra ti tm O l t

trng quay: 03

2B B= vi B l t trng tng hp ti O, B0 lt trng do 1 cun dy to ra. T trng quay

ny s tc dng vo khung dy lm khung dy quay vi tc nh hn tc quay ca t trng. Chuyn ng

quay ca rto (khung dy) c s dng lm quay cc my khc.

B. CU HI V BI TP TRC NGHIM

Cu 1: Trong my pht in xoay chiu mt phaA. gim tc quay ca r to ngi ta gim s cun dy v tng s cp cc.B. gim tc quay ca r to ngi ta tng s cun dy v tng s cp cc.C. gim tc quay ca r to ngi ta gim s cun dy v gim s cp cc.D. gim tc quay ca r to ngi ta tng s cun dy v gim s cp cc.

Cu 2: Tm pht biu sai khi ni v my bin p:A. Khi tng s vng dy cun th cp, hiu in th gia hai u cun th cp tng.B. Khi gim s vng dy cun th cp, cng dng in trong cun th cp gim.

C. Mun gim hao ph trn ng dy ti in, phi dng my tng th tng hiu in th.D. Khi mch th cp h, my bin th xem nh khng tiu thin nng.

Cu 3: Thit b no sau y khng c kh nng bin i dng in xoay chiu thnh dng mt chiu?A. Hai vnh bn khuyn cng hai chi qut trong my pht in.B. Bn it mc thnh mch cu.C. Hai vnh khuyn cng hai chi qut trong my pht in.D. Mt it.

Cu 4: Trong cc cch mc dng n xoay chiu ba pha i xng theo hnh sao, pht biu no sau y l khngng:

A. Dng in trong dy trung ho bng khng.B. Hiu in th pha bng 3 ln hiu in th gia hai dy pha.

C. Dng in trong mi pha bng dng in trong mi dy pha.D. Truyn ti in nng bng bn dy dn, dy trung ho c tit din nh nht.

Cu 5: Chn cu ng trong cc cu sau: My bin th l mt thit bA. C tc dng lm tng hoc gim hiu in th ca dng in xoay chiu.B. C tc dng lm tng hoc gim cng ca dng in xoay chiuC. S dng in nng vi hiu sut cao.D. C A, B, C u ng.

Cu 6: Trong my pht in xoay chiu c p cp cc quay vi tn s gc n vng/giy th tn s dng in pht ra

A.60

npf = B. f np= C. 60pf

n= D.

60nf

p=

Cu 7: Chn cu sai trong cc cu sau:

A. Cng sut ca dng in xoay chiu c tnh bi cng thc 0 0 os2

U I cP = .

B. i vi nhng ng cin, ngi ta c th mc song song mt tin vo mch lm tng osc .


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C. Trong thc t, ngi ta thng dng nhng thit b s dng in xoay chiu c osc < 0,85.D. Khi on mch ch c cun thun cm, hoc tin hoc cun thun cm v tin th on mch

ny khng tiu thin nng.Cu 8: Pht biu no sau y l ng khi ni v my pht in xoay chiu mt pha.

A. My pht in xoay chiu mt pha bin cnng thnh nhit nng.B. My pht in xoay chiu mt pha hot ng nhs dng t trng quay.C. My pht in xoay chiu mt pha c th to ra dng in khng i.

D. B gp ca my pht in xoay chiu mt pha gm hai vnh bn khuyn v hai chi qut.Cu 9: iu no sau y l ng khi ni v my pht in xoay chiu?

A. Rto c th l phn cm hoc phn ngB. Phn quay gi l rto, phn ng yn gi l stato.C. Phn cm to ra t trng, phn ng to ra sut in ng.D. Tt c A, B, C u ng.

Cu 10: Chn cu ngA. Dng in xoay chiu mt pha ch c th do my phat in xoay chiu mt pha to ra.B. Ch c dng in xoay chiu ba pha mi to ra c t trng quayC. Dng in do my pht in xoay chiu to ra lun c tn s bng s vng quay trong mt giy ca

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Dong Dien Xoay Chieu Day Du