cách viết biểu thức trong đoạn mạch xoay chiều vật lý 12 cực hay

7/31/2019 cch vit biu thc trong on mch xoay chiu vt l 12 cc hay

1/21

DNG : VIT BIU THC u HOC i trong mch in xoay chiuI.KIN THC CN NH:

a) on mch ch c in tr thun: uRcng pha vi i : I =R

UR

b) on mch ch c t in C: uC tr pha so vi i gc2

.

- L m: I =C

C

Z

U; vi ZC =

C

1l dung khng ca t in.

-t in p 2cosu U t= vo hai u mt t in th cng dng in qua n c gi tr hiu dngl I. Ti thi im t, in p hai u t in l u v cng dng in qua n l i. H thc lin h gia cci lng l :

Ta c: 122

12

2

2

2

20

2

20

2

=+=+CC U

u

I

i

U

u

I

i

2 2

2 2

u i2

U I+ =

-Cng dng in tc thi qua t: 2 cos( )2

i I t

= +

c) on mch ch c cun dy thun cm L: uL sm pha hn i gc2

.

- L m: I =L

L

Z

U; vi ZL = L l cm khng ca cun dy.

-t in p 2cosu U t= vo hai u mt cun cm thun th cng dng in qua n c gitr hiu dng l I. Ti thi im t, in p hai u cun cm thun l u v cng dng inqua n l i. H thc lin h gia cc i lng l :

Ta c:2 2 2 2

2 2 2 20 0L L

i u i u1 1

I U 2I 2U+ = + =

2 2

2 2

u i2

U I+ =

-Cng dng in tc thi qua cun dy: 2 cos( )2

i I t

=

d) on mch c R, L, C khng phn nhnh:

+t in p 2 cos( )uu U t = + vo hai u mch

+ lch pha gia u v i xc nh theo biu thc: tan =R

ZZ CL =1

LC

R

; Vi u i =

+ Cng hiu dng xc nh theo nh lut m: I =Z

U.

Vi Z =2

CL

2

)Z-(ZR + l tng tr ca on mch.Cng dng in tc thi qua mch: 2 cos( ) 2 cos( )i ui I t I t = + = +

+ Cng hng in trong on mch RLC: Khi ZL = ZC hay =LC

1th

Imax =R

U, Pmax =

R

U2, u cng pha vi i ( = 0).

Khi ZL > ZC th u nhanh pha hn i (on mch c tnh cm khng).Khi ZL < ZC th u tr pha hn i (on mch c tnh dung khng).R tiu th nng lng di dng to nhit, ZL v ZC khng tiu th nng lng in.

e) on mch c R, L,r, C khng phn nhnh:

+t in p 2 cos( )uu U t = + vo hai u mch

1

CA B

R L

NM

CBA

LA B

CA B

R L,r

NM


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+ lch pha gia uAB v i xc nh theo biu thc:

tan = L CZ Z

R r

+

=1

LC

R r

+. Vi u i =

+ Cng hiu dng xc nh theo nh lut m: I =Z

U.

Vi Z = 2 2L C(R+r) (Z - Z )+ l tng tr ca on mch.

Cng dng in tc thi qua mch: 2 cos( ) 2 cos( )i ui I t I t = + = + + Cch nhn bit cun dy c in tr thun r

-Xt ton mch, nu: Z 22 )(CL

ZZR + ;U 22 )( CLR UUU + hoc P I2R hoc cos

Z

R

th cun dy c in tr thun r 0.

-Xt cun dy, nu: Ud UL hoc Zd ZL hoc Pd 0 hoc cos d 0 hoc d 2

th cun dy c in tr thun r 0.

II. PHNG PHP 1: (PHNG PHP TRUYN THNG):a) Mch in ch cha mt phn t ( hoc R, hoc L, hoc C)- Mch in ch c in tr thun: u v i cng pha: = u - i = 0 Hay u = i

+ Ta c: 2 os( t+ )ii I c = th 2 os( t+ )R iu U c = ; viR

R

UI = .

+V d 1: in p gia hai u mt on mch in xoay chiu ch c in tr thun R= 100 c biu

thc u=200 2 cos(100 )( )4

t V

+ . Biu thc ca cng dng in trong mch l :

A. i= 2 2 cos(100 )( )4

t A

C.i= 2 2 cos(100 )( )4

t A

+

B. i= 2 2 cos(100 )( )2t A + D.i= 2cos(100 )( )2t A

+Gii :Tnh I0 hoc I= U /.R =200/100 =2A; i cng pha vi u hai u R, nn ta c: i = u = /4

Suy ra: i = 2 2 cos(100 )( )4

t A

+ => Chn C

-Mch in ch c t in:

uC tr pha so vi i gc2

. -> = u - i =-

2

Hay u = i -

2

; i = u +

2

+Nu cho 2 os( t)i I c = th vit: 2 os( t- )2

u U c

= v L m:C

C

UI

z= vi

1C

ZC

= .

+Nu cho 2 os( t)u U c = th vit: 2 os( t+ )2

i I c

=

+V d 2: in p gia hai u mt on mch in xoay chiu ch c t c in dung C=410

( )F

c biu

thc u= 200 2 cos(100 )( )t V . Biu thc ca cng dng in trong mch l :

A. i= )()6

5100cos(22 At

+ C.i= 2 2 cos(100 )( )

2t A

+

B. i= 2 2 cos(100 )( )2

t A

D.i= )()6

100cos(2 At

Gii : Tnh 1.

CZC

= =100, Tnh Io hoc I= U /.ZL =200/100 =2A;

i sm pha gc /2 so vi u hai u t in; Suy ra: i = 2 2 cos(100 )( )2

t A

+ => Chn C

2


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-Mch in ch c cun cm thun:

uL sm pha hn i gc2

-> = u - i =

2

Hay u = i +

2

; i = u -

2

+Nu cho 2 os( t)i I c = th vit: 2 os( t+ )2

u U c

= v L m: LL

UI

z= vi LZ =

Nu cho 2 os( t)u U c = th vit: 2 os( t- )2

i I c

=

V d 3: Hiu in th gia hai u mt on mch in xoay chiu ch c cun cm c t cm L=

)(1 H

c biu thc u= )()3

100cos(2200 Vt

+ . Biu thc cng dng in trong mch l :

A. i= )()6

5100cos(22 At

+ C.i= )()

6100cos(22 At

B. i= )()6

100cos(22 At

+ D.i= )()6

100cos(2 At

Gii : Tnh LZ L= = 100.1/ =100, Tnh I0 hoc I= U /.ZL =200/100 =2A;

i tr pha gc /2 so vi u hai u cun cm thun, nn ta c:3 2 = -

6

Suy ra: i = )()6

100cos(22 At => Chn C

Trc nghim vn dng:Cu 1: in p gia hai u mt on mch in xoay chiu ch c in tr thun R= 200 c biu thc u=

200 2 cos(100 )( )4

t V


A. i= 2 cos(100 ) ( )t A C.i= 2 2 cos(100 ) ( )t A

B. i= 2 cos(100 ) ( )4

t A

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

t A

Cu 2: in p gia hai u mt on mch in xoay chiu ch c in tr thun R= 100 c biu thc u=

200 2 cos(100 )( )4

t V


A. i= 2 2 cos(100 )( )4

t A

C.i= 2 2 cos(100 )( )4

t A

+

B. i= 2 2 cos(100 )( )2

t A

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

t A

Cu 3: in p gia hai u mt on mch in xoay chiu ch c t c in dung C=410

( )F

c biu thc u=

200 2 cos(100 )( )t V . Biu thc ca cng dng in trong mch l :A. i= )()

6

5100cos(22 At

+ C.i= 2 2 cos(100 )( )

2t A

+

B. i= 2 2 cos(100 )( )2

t A

D.i= )()6

100cos(2 At

Cu 4: Cho in p hai u t C l u = 100cos(100t- /2 )(V). Vit biu thc dng in qua mch, bit

)(10 4

FC

=

A. i = cos(100t) (A) B. i = 1cos(100t + )(A)C. i = cos(100t + /2)(A) D. i = 1cos(100t /2)(A)

Cu 5: t in p 200 2 os(100 t)u c = (V) vo hai u on mch ch c t n c C = 15,9F (Ly1

=

0,318) th cng dng in qua mch l:3


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A. 2 os(100 t+ )2

i c

= (A) B.

=

2.100cos4

ti (A)

C.

=

2.100cos22

ti (A) D.

+=

2.100cos2

ti (A)

Cu 6 Xc nh p n ng .Cng dng in qua t in i = 4cos100 t (A). in dung l 31,8 F.Hiu in th t hai u t in l:

A- . uc = 400cos(100 t ) (V) B. uc = 400 cos(100 t +2

). (V)

C. uc = 400 cos(100 t -2

). (V) D. uc = 400 cos(100 t - ). (V)

Cu 7: Cho in p gia hai u 1 on mch xoay chiu ch c cun thun cm )(1

HL

= l :

100 2 1003

cos( t )(V )

. Biu thc cng dng in trong mch l :

A. i=5

2 1006

cos( t )( A )

C.i= 2 1006

cos( t )( A )

B. i= 2 100

6

cos( t )( A )

+ D.i= )()

6

100cos(2 At

Cu 8: t in p 200 2 os(100 t+ )u c = (V) vo hai u on mch ch c cun thun cm )(1

HL

=

th cng dng in qua mch l:

A.

+=

2.100cos22

ti (A) B.

=

2.100cos4

ti (A)

C.

=

2.100cos22

ti (A) D.

+=

2.100cos2

ti (A)

Cu 9: t in p 200 2 os(100 t)u c = (V) vo hai u on mch ch c cun thun cm L= 0,318(H)

(Ly 1 = 0,318) th cng dng in qua mch l:

A.

+=

2.100cos22

ti (A) B.

=

2.100cos4

ti (A)

C.

=

2.100cos22

ti (A) D.

+=

2.100cos2

ti (A)

Cu 10: t mt hiu in th xoay chiu vo hai u cun dy ch c t cm L= H2

1th cng dng in

qua cun dy c biu thc i=3 2 cos(100t+

6

)(A). Biu thc no sau y l hiu in th hai u on mch:

A u=150cos(100t+3

2

)(V) B. u=150 2 cos(100t-3

2

)(V)

C.u=150 2 cos(100t+3

2

)(V) D. u=100cos(100t+3

2

)(V)

b) Mch in khng phn nh nh (R L C)PHNG PHP TRUYN THNG):

- Phng php gii: Tm Z, I, ( hoc I0 )v

4


5/21

Bc 1: Tnh tng tr Z: Tnh LZ = .;1 1

2CZ

C fC = = v 2 2( )

L CZ R Z Z= +

Bc 2: nh lut m : U v I lin h vi nhau biU

IZ

= ; Io =Z

Uo ;

Bc 3: Tnh lch pha gia u hai u mch v i: tan L CZ Z

R

= ;

Bc 4: Vit biu thc u hoc i

-Nu cho trc: 2 os( t)i I c = th biu thc ca u l 2 os( t+ )u U c = Hay i = Iocos t th u = Uocos( t + ).-Nu cho trc: 2 os( t)u U c = th biu thc ca i l: 2 os( t- )i I c =

Hay u = Uocos t th i = Iocos( t - )* Khi: (u 0; i 0 ) Ta c : = u - i => u = i + ; i = u -

-Nu cho trc 2 os( t+ )ii I c = th biu thc ca u l: 2 os( t+ + )iu U c =

Hay i = Iocos( t + i) th u = Uocos( t + i + ).

-Nu cho trc 2 os( t+ )uu U c = th biu thc ca i l: 2 os( t+ - )ui I c =

Hay u = Uocos( t + u) th i = Iocos( t + u - )

Lu : Vi Mch in khng phn nhnh c cun dy khng cm thun (R L,r C) th:

Tng tr : 2 2( ) ( )L C

Z R r Z Z= + + v tan L CZ Z

R r

=

+;

V d 1 : Mch in xoay chiu gm mt in tr thun R = 50, mt cun thun cm c h s t cm1

( )=L H

v mt t in c in dung42.10

( )

=C F

mc ni tip. Bit rng dng in qua mch c

dng ( )5cos100=i t A .Vit biu thc in p tc thi gia hai u mch in.

Gii :

Bc 1: Cm khng:1

100 . 100= = = LZ L

; Dung khng: 41 1

502.10

100 .CZ

C

= = =

Tng tr: ( ) ( )2 22 250 100 50 50 2= + = + = L CZ R Z Z

Bc 2: nh lut m : Vi Uo= IoZ = 5.50 2 = 250 2 V;

Bc 3: Tnh lch pha gia u hai u mch v i:100 50

tan 150

= = =L C

Z Z

R

4 =

(rad).

Bc 4: Biu thc in p tc thi gia hai u mch in: 250 2 cos 100 4 = + u t (V).

V d 2: Mt mch in xoay chiu RLC khng phn nhnh c R = 100 ; C= 41

10. F

; L=2

H. cng

dng in qua mch c dng: i = 2cos100 t (A). Vit biu thc tc thi in p ca hai u mch v hai umi phn t mch in.Hng d n :

-Cm khng :2

100 200LZ L.

= = = ; Dung khng : 41 1

10100

CZ.C

.

= =

= 100

-Tng tr: Z = 2 2 2 2100 200 100 100 2L CR ( Z Z ) ( )+ = + =

-HT cc i :U0 = I0.Z = 2. 2100 V =200 2 V

5


6/21

- lch pha:2 00 1 00

tan 1100 4

L CZ Z

ra dR

= = = ;Pha ban u ca HT: =+=+=

40

iu 4

=>Biu thc HT : u = )4

100cos(2200)cos(0

+=+ ttU u (V)

-HT hai u R:uR= U0Rcos )( Rut + ; Vi : U0R= I0.R = 2.100 = 200 V;

Trong on mch ch cha R : uRcng pha i: uR= U0Rcos )( Rut + = 200cos t100 V

-HT hai u L :uL = U0Lcos )( Lut + Vi : U0L = I0.ZL = 2.200 = 400 V;

Trong on mch ch cha L: uL nhanh pha hn cd2 :

220

2

=+=+= iuL rad

=> uL = U0Lcos )( Rut + = 400cos )2100(

+t V

-HT hai u C :uC = U0Ccos )( Cut + Vi : U0C = I0.ZC = 2.100 = 200V;

Trong on mch ch cha C : uC chm pha hn cd2

:

220

2

=== iuL rad

=> uC = U0Ccos )( Cut + = 200cos )2100(

t V

V d 3: Mch in xoay chiu gm mt in tr thun R = 40, mt cun thun cm c h s t cm0 8,

L ( H )

= v mt t in c in dung 42

10C . F

= mc ni tip. Bit rng dng in qua mch c dng

3cos(100 )( )i t A=a. Tnh cm khng ca cun cm, dung khng ca t in v tng tr ton mch.b. Vit biu thc in p tc thi gia hai u in tr, gia hai u cun cm, gia hai u t in, gia hai

u mch in.Hng dn:

a. Cm khng:

0,8

100 . 80LZ L = = =

; Dung khng:

4

1 150

2.10100 .

CZ

C

= = =

Tng tr: ( ) ( )2 22 240 80 50 50L CZ R Z Z= + = + =

b. V uRcng pha vi i nn : cos100R oRu U t= ;Vi UoR= IoR = 3.40 = 120V Vy 120cos100u t= (V).

V uL nhanh pha hn i gc2

nn: cos 100

2L oLu U t

= +

Vi UoL = IoZL = 3.80 = 240V; Vy 240cos 1002L

u t

= +

(V).

V uC chm pha hn i gc2

nn: cos 100

2C oCu U t

=

Vi UoC = IoZC = 3.50 = 150V; Vy 150cos 1002C

u t

=

(V).

p dng cng thc:80 50 3

tan40 4

L CZ Z

R

= = = ; 37o

370,2

180

= (rad).

biu thc hiu in th tc thi gia hai u mch in: ( )cos 100ou U t = + ;Vi Uo= IoZ = 3.50 = 150V; Vy ( )150cos 100 0,2u t = + (V).

6


7/21

V d 4: Mt on mch in xoay chiu gm mt in tr thun R = 80, mt cun dy thun cm c tcm L = 64mH v mt t in c in dung 40C F= mc ni tip.a. Tnh tng tr ca on mch. Bit tn s ca dng in f = 50Hz.b. on mch c t vo in p xoay chiu c biu thc 282cos314u t= (V). Lp biu thc cng

tc thi ca dng in trong on mch.Hng dn:a. Tn s gc: 2 2 .50 100f = = = rad/s

Cm khng: 3100 .64.10 20LZ L = =

Dung khng:6

1 180

100 .40.10CZ

C = =

Tng tr: ( ) ( )2 22 280 20 80 100L CZ R Z Z= + = + =

b. Cng dng in cc i:282

2,82100

oo

UI

Z= = = A

lch pha ca hiu in th so vi cng dng in:

20 80 3

tan

80 4

L CZ Z

R

= = = 37o

37

37180

o

i u

= = = = rad; Vy

372,82cos 314

180i t

= +

(A)

V d 5: Cho mch in nh hnh v. Bit1

10L

= H,

310

4C

= F v

n ghi (40V- 40W). t vo 2 im A v N mt hiu in th

120 2 cos100ANu t= (V). Cc dng c o khng lm nh hngn mch in.

a. Tm s ch ca cc dng c o.b. Vit biu thc cng dng in v in p ton mch.

Hng dn:

a. Cm khng:1

100 . 1010L

Z L

= = = ; Dung khng: 31 1

4010

100 .4

CZC

= = =

in tr ca bng n:2 2

m

m

4040

40

UR

P= = =

Tng tr on mch AN:

2 2 2 2

40 40 40 2AN CZ R Z= + = + =

S ch ca vn k:120 2

1202 2

oANAN

UU = = = V

S ch ca ampe k:120 3

2,1240 2 2

ANA

AN

UI I

Z= = = = A

b. Biu thc cng dng in c dng: ( )cos 100o ii I t = + (A)

Ta c :40

tan 140

CAN

Z

R

= = =

4AN

= rad

4i uAN AN AN

= = = rad;3

2 . 2 32

oI I= = = A

7


8/21

Vy 3cos 1004

i t

= +

(A).

Biu thc hiu in th gia hai im A, B c dng: ( )cos 100AB o uu U t = + (V)

Tng tr ca on mch AB: ( ) ( )2 22 240 10 40 50

AB L CZ R Z Z= + = + =

3.50 150o o ABU I Z = = = V

Ta c:10 40 3

tan 40 4

L CAB

Z Z

R

= = =

37

180AB

= rad

37

4 180 20u i AB

= + = = rad; Vy 150cos 10020AB

u t

= +

(V)

V d 6: S mch in c dng nh hnh v, in tr R = 40, cun thun cm3

10L

= H, t in

310

7

C

= F. in p 120cos100AFu t= (V).

Hy lp biu thc ca:a. Cng dng in qua mch.b. in p hai u mch AB.

Hng dn:

a. Cm khng:3

100 . 3010

LZ L

= = = ; Dung khng: 31 1

7010

100 .7

CZC

= = =

Tng tr ca on AF: 2 2 2 240 30 50AF LZ R Z= + = + = 12 0

2, 450

oAFo

AF

UI

Z = = A

Gc lch pha AF :30 37

tan 0,7540 180

LAF AF

Z

R

= = = rad

Ta c:37

0180

i uAF AF AF AF

= = = = rad; Vy

372,4cos 100

180i t

=

(A)

b. Tng tr ca ton mch: ( )2240 30 70 40Z = + = 2,4.40 2 96 2o oU I Z = = = V

Ta c:30 70

tan 140 4

L CAB AB

Z Z

R

= = = = rad

37 414 180 9

u AB i

= + = =rad Vy 4196 2 cos 100

90u t =

(V)

V d 7: Cho mch in xoay chiu nh hnh v, R = 100 , L l t cm ca cun dy thun cm,410

3C

= F, RA 0. in p 50 2 cos100ABu t= (V). Khi K ng hay khi K m, s ch ca ampe k

khng i.a. Tnh t cm L ca cun dy v s ch khng i ca ampe k.b. Lp biu thc ca cng dng in tc thi trong mch khi K

ng v khi K m.Hng dn:a. Theo bi, in p v s ch ampe k khng i khi K ng hay khi K m nn tng tr Z khi K m v khiK ng bng nhau

8

CA B

R L

F


9/21

( )22 2 2

m d L C C Z Z R Z Z R Z= + = +

( )2 2

L C CZ Z Z =

2

0

L C C L C

L C C L

Z Z Z Z Z

Z Z Z Z

= = = =

Ta c: 41 1

17310

100 . 3

CZC

= = = ; 2 2.173 346L CZ Z = = = 34 6

1,1

10 0

LZ

L

= = H

S ch ampe k bng cng dng in hiu dng khi K ng:

2 2 2 250

0,25100 173

A d

d C

U UI I

Z R Z= = = =

+ +A

b. Biu thc cng dng in:

- Khi K ng: lch pha :173

tan 3100

Cd

Z

R

= = =

3d

= rad

Pha ban u ca dng in:3d

i u d d

= = =

Vy 0, 25 2 cos 1003d

i t

= +

(A).

- Khi K m: lch pha:346 173

tan 3100

L Cm

Z Z

R

= = =

3m

=

Pha ban u ca dng in:3mi u m m

= = =

Vy 0,25 2 cos 1003

mi t

=

(A).

V d 8: Cho mch in nh hnh v :UAN =150V ,UMB =200V. lch pha UAM v UMB l / 2Dng in tc thi trong mch l : i=I0 cos 100t (A) , cun dy thun cm.Hy vit biu thc UAB

Hng dn:

Ta c : VUUUUUU CANCAN 1502R

2R =+=+= (1)

VUUUUUU LMBLMB 2002R

2R =+=+= (2)

V UAN v UMB lch pha nhau / 2 nn 1..1.

RR21 ==

UUUUtgtg CL hay U2R= UL.UC (3)

T (1),(2),(3) ta c UL=160V , UC = 90V , VU 120R =

VUUUU CLAB 139)(22

R =+= ; sradUUU

tg CL /53,012

7

R

==

=

vy uAB = 1392 cos(100t +0,53) V

V d 9: Cho mch in khng phn nhnh gm R = 100 3 , cun dy thun cm L v t in C =10-4 /2(F).t vo 2 u mch in mt hiu in th u = 100 2 cos 100 t. Bit hiu in th ULC = 50V ,dng

in nhanh pha hn hiu in th.Hy tnh L v vit biu thc cng dng in i trong mchHng dn:

Ta c = 100 rad/s ,U = 100V, == 2001

CZC

9

(Loi)

R CL

N MAB


10/21

Hiu in th 2 u in tr thun l: VUUU LC 35022

R ==

cng dng in AU

I 5,0R

R == v == 100I

UZ LCLC

V dng in nhanh pha hn hiu in th,m trn gin Frexnen,dng in c bi din trn trc honhvy hiu in th c biu din di trc honh ngha l ZL< ZC. Do

ZC-ZL =100ZL =ZC -100 =100 suy ra HZ

L L 318,0==

lch pha gia u v i : 631 === R

ZZtg CL ; vy 0,5 2 os(100 )( )6i c t A= +

3. TRC NGHIM:

Cu 11: Cho mch in xoay chiu c R=30 , L=

1(H), C=

7.0

10 4(F); in p 2 u mch l u=120 2 cos100 t

(V), th cng dng in trong mch l

A. ( )4cos 1004

i t A

= +

B. 4cos(100 )( )4

i t A

=

C. 2cos(100 )( )4i t A

= D. 2cos(100 )( )4i t A

= +Cu 12:Cho on mch gm R, L, C mc ni tip; R = 10 3 ; L = 0,3/ (H); C = 310 / 2 (F). t vo hai u on

mch mt hiu in th ( )100 2 cos 100u t= (V).a) Vit biu thc cng dng in trong mch

A. ( )5 2 os 100 / 6i c t = (A) B. ( )5 2 os 100 / 6i c t = + (A)

C. ( )5 os 100 / 6i c t = (A) D. ( )5 os 100 / 6i c t = + (A)b) Vit biu thc hiu in th hai u mi phn t R; L; C

A. ( )86,5 2 cos 100 / 6Ru t = + ; ( )150 2 cos 100 / 3Lu t = + ; ( )100 2 cos 100 2 / 3Cu t =

B. A. ( )86,5 2 cos 100 / 6Ru t = ; ( )150cos 100 / 3Lu t = + ; ( )100cos 100 2 / 3Cu t = C. A. ( )86,5 2 cos 100 / 6Ru t = ; ( )150 2 cos 100 / 3Lu t = + ; ( )100 2 cos 100 2 / 3Cu t =

D. A. ( )86,5 2 cos 100 / 6Ru t = + ; ( )150 2 cos 100 / 3Lu t = + ; ( )100 2 cos 100 2 / 3Cu t = +

Cu 13: Cho mch xoay chiu c R, L, C mc ni tip c R=30 , C=

410(F) , L thay i c cho hiu in th

2 u mch l U=100 2 cos100 t (V) , u nhanh pha hn i gc6

rad th ZL v i khi l:

A.5 2

117,3( ), cos(100 )( )6

3

LZ i t A

= = B. 100( ), 2 2cos(100 )( )6L

Z i t A

= =

C.5 2

117,3( ), cos(100 )( )63

LZ i t A

= = + C. 100( ), 2 2cos(100 )( )6L

Z i t A

= = +

Cu 14: Mt mch gm cun dy thun cm c cm khng bng 10 mc ni tip vi t in c in dung42 .10C F

= . Dng in qua mch c biu thc 2 2 cos100 )3

= +i t A

. Biu thc hiu in th ca hai u

on mch l:

A. 80 2 s(100 )6

= u co t

(V) B. 80 2 cos(100 )6

= +u t

(V)

C. 120 2 s(100 )6

= u co t

(V) D.2

80 2 s(100 )3

= +u co t

(V)

Cu 15: Mch in xoay chiu gm in tr 40R = ghp ni tip vi cun cm L. Hiu in th tc thi hai uon mch 80 s100=u co t v in p hiu dng hai u cun cm LU =40V Biu thc i qua mch l:

10


11/21

A.2

s(100 )2 4

= i co t A

B.2

s(100 )2 4

= +i co t A

C. 2 s(100 )4

= i co t A

D. 2 s(100 )4

= +i co t A

Cu 16: Mt on mch in gm in tr R = 50 mc ni tip vi cun thun cm L = 0,5/ (H). t vo hai uon mch mt in p xoay chiu u = 100 2 cos(100t - /4) (V). Biu thc ca cng dng in qua onmch l:

A. i = 2cos(100t - /2) (A). B. i = 2 2 cos(100t - /4) (A).C. i = 2 2 cos100t (A). D. i = 2cos100t (A).

Cu 17: Khi t in p khng i 30V vo hai u on mch gm in tr thun mc ni tip vi cun cm thun

c t cm1

4(H) th dng in trong on mch l dng in mt chiu c cng 1 A. Nu t vo hai u

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

A. i 5 2 cos(120 t )4

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

4

= + (A).

C . i 5cos(120 t )4

= (A). D. i 5 2 cos(120 t )

4

= + (A).

Cu 18: Cho on mch xoay chiu LRC mc ni tip hai u AB, L mc vo AM, R mc vo MN, C mc vo NB. Biuthc dng in trong mch i = I0 cos 100 t (A). in p trn on AN c dng ( )100 2 os 100 / 3ANu c t = + (V) vlch pha 900 so vi in p ca on mch MB. Vit biu thc uMB ?

A.100 6

os 1003 6MB

u c t

=

B, ( )100 os 100MBu c t=

C.100 6

os 1003 6MB

u c t

= +

D. 100 os 1006MB

u c t

=

Cu 19: t in p xoay chiu u = Uocos(100t +3

) (V) vo hai u mt cun cm thun c t cm L=

1

2

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

A. i = 2 3 cos(100t +6

) (A). B. i = 2 2 cos(100t -

6

) (A).

C. i = 2 2 cos(100t +6

) (A). D. i = 2 3 cos(100t -

6

) (A).

Cu 20: Xt on mch gm mt in tr hot ng bng 100, mt t in c in dung50

C F

= v mt cun

cm thun c t cm

3

H mc ni tip. Nu t vo hai u mt in p 200cos100u t= (V) th in p giahai u in tr hot ng c biu thc

A. 200 cos(100 )4

Ru t

= (V). B. 100 2 cos(100 )Ru t= (V).

C. 200 cos(100 )4

Ru t

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

Ru t

= (V).

Cu 21: Cho on mch in AB khng phn nhnh gm cun cm thun, t in c in dung thay i c, mt

in tr hot ng 100. Gia A, B c mt in p xoay chiu n nh 110 cos(120 )3

u t

= (V). Cho C thay i.

11


12/21

Khi C =125

3F

th in p hiu dng gia hai u cun c gi tr ln nht. Biu thc ca in p gia hai u cun

cm l

A. 220 cos(120 )2

Lu t

= + (V). B. 110 2 cos(120 )

2Lu t

= + (V).

C. 220 cos(120 )

6

Lu t

= + (V). D. 110 2 cos(120 )

6

Lu t

= + (V).

Cu 22: Cho mch in nh hnh v:

150 , 200AN MB

U V U V = = . lch pha gia uAN v uMBl2

. Dng in tc thi trong mch l

0 sin(100 )( )i I t A= , cun dy thun cm. Biu thc ca uAB l

A. 139 2 sin(100 0,53)ABu t V= + B. 612 2 sin(100 0,53)ABu t V= +

C. 139 sin(100 0, 53)ABu t V= + D. 139 2 sin(100 0,12)ABu t V= +Cu 23: Cho ba linh kin gm in tr thun 60R = , cun cm thun L v t in C. Ln lt t in p xoaychiu c gi tr hiu dng U vo hai u on mch ni tip RL hoc RC th biu thc cng dng in trong

mch ln lt l 1 2 os(100 )( )12

i c t A

= v 17

2 os(100 )( )12

i c t A

= + . Nu t in p trn vo hai u

on mch RLC ni tip th dng in trong mch c biu thc:

A. 2 2 os(100 )( )3

i c t A

= + B. 2 os(100 )( )3

i c t A

= +

C. 2 2 os(100 )( )4

i c t A

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

i c t A

= +

Cu 24: Cho on mch in xoay chiu gm cun dy thun cm v t in mc ni tip nhau. in p hiudng hai u cun cm l 150V, gia hai u t in l 100V.Dng in trong mch c biu thc i =I0cos(t +/6)((A) . Biu thc in p hai u on mch l

A. )2/100cos(250 = tu V. B. Vtu )2/100cos(250 += .

C. )3/2100cos(250 = tu V. D. )3/2100cos(250 += tu V.

Cu 25: t in p u = 120cos(100t +3 ) (V) vo hai u mt on mch gm cun cm thun mc ni

tip in tr thun R= 30 th in p hiu dng hai u cun cm l 60 V. Dng in tc thi qua onmch l

A. )12

t100cos(22i

+= (A). B. )6

t100cos(32i

+= (A).

C. )4

t100cos(22i

= (A). D. )4

t100cos(22i

+= (A).

4.Trc nghim vit biu thc u hoc i nng caoCu 26. Cho 3 linh kin gm in tr thun R= 60, cun cm thun L v t in C. Ln lt t in pxoay chiu c gi tr hiu dng U vo hai u on mch ni tip RL hoc RC th biu thc cng dng

12

AAAMMM

NNN BBB

CCC RRR LLL


13/21

in trong nch ln lt l i1=cos(100-)(A) v i2=cos(100+)(A). nu t in p trn vo hai u on mchRLC ni tip th dng in trong mch c biu thc:A. 2cos(100t+)(A) B. 2 cos(100t+)(A)C. 2cos(100t+)(A) D. 2cos(100t+)(A)

HD: Theo ( )1 2

01 02

=- 1RL RC

L C

I I Z ZZ Z

= = =

Mt khc( ) ( )

( )1 1 2

2

1u 1

u

u 2

- = 2 3

2 4 - =

i i i

i

+ = =

T ( ) ( ) ( )1

2 , 3 3 60 3

3

LL

ZZ

R

= = =

( )0 01 120 2RLU I Z V = =

Khi RLC nt cng hng: i= 0U

Rcos(100t+ u )= 2cos(100t+)(A)

Cu 27: t vo hai u mch in xoay chiu gm mt cun dy v mt t in mc ni tip mt in p

xoay chiu n nh c biu thc u =100 6 cos(100 )( ).4

t V

+ Dng vn k c in tr rt ln ln lt o

in p gia hai u cun cm v hai bn t in th thy chng c gi tr ln lt l 100V v 200V. Biuthc in p gia hai u cun dy l:

A. 100 2 cos(100 )( )2d

u t V= + . B. 200cos(100 )( )4d

u t V= + .

C.3

200 2 cos(100 )( )4d

u t V

= + . D.3

100 2 cos(100 )( )4d

u t V

= + .

Cu 28: Cho on mch gm R, L, C mc theo th t trn vo on mch AB. M l im gia L v C; Biuthc hiu in th tc thi gia hai im A v M l uAM = uRL = 200 cos100 t(V). Vit biu thc uAB?A. ( )200cos 100ABu t= (V) B. ( )200 2 cos 100ABu t= (V)

C. ( )200cos 100 / 2ABu t = (V) D. ( )200cos 100 / 2ABu t = + (V)Cu 29: Cho on mch in AB gm R, L, C mc ni tip vi R l bin tr. Gia AB c mt in p

0 os( )u U c t = + lun n nh. Cho R thay i, khi R = 42,25 hoc khi R = 29,16 th cng sut tiu thca on mch nh nhau; khi R = R0 th cng sut tiu th ca on mch t gi tr ln nht, v cng

dng in qua mch 2cos(100 )12

i t

= + (A). in p u c th c biu thc

A.7

140,4 2 os(100 )( )12

u c t V

= + B.5

70,2 2 os(100 )( )12

u c t V

=

C. 140,4 2 os(100 )( )3

u c t V

= D. 70,2 2 os(100 )( )3

u c t V

= +

Gii: R0 = 21RR =35,1 khi th CL ZZR =0 , t tnh c U0 v tan bn s tm c uCu 30: Cho on mch in xoay chiu AB khng phn nhnh gm mt cun cm thun, mt t in c

in dung C thay i c, mt in tr hot ng 100. Gia AB c mt in p xoay chiu lun n nh

u=110cos(120t- )3

(V). Cho C thay i, khi C =125

F3

th in p gia hai u cun cm c gi tr ln

nht. Biu thc ca in p gia hai u cun cm l

A. L

u =110 2cos(120t+ )6

(V). B. L

u =220cos(120t+ )6

(V).

C. L

u =220cos(120t+ )2

(V). D. L

u =110 2cos(120t+ )2

(V).

Gii: khi thay i c ULmax th CL ZZ = ,t sua ra U0L=I0R=220V

M khi th u,i cng pha ,t suy ra23

+=uL = 6

13


14/21

Cu 31:t in p xoay chiu 0

u=U cos 120t+ V3

vo hai u mt cun cm thun c t cm1

L= H.6

Ti thi im in p gia hai u cun cm l 4 0 2 V th cng dng in qua cun cm l 1A . Biu thcca cng dng in qua cun cm l

A.

i=3 2cos 120t- A.6

B.

i=3cos 120t- A.6

C.

i=2 2cos 120t- A.

6

D.

i=2cos 120t+ A.

6

p dng cng thc c lp : 1I

i

U

u20

2

20

2

=+ 202

2L

2

IiZ

u=+ I0 = 3A i =

623

=

Chon p n B

Cu 32: khi t dng in p xoay chiu vo hai u mch gm in tr thun R mc ni tip mt t in Cth biu thc dng in c dang: i1=I0 cos(t+ )(A).mc ni tip thm vo mch iin cun dy thun cm Lri mc vo in p ni trn th biu thc dng in c dng i2=I0 cos(t- )(A). Biu thc hai u on mchc dng:

A:u=U0 cos(t +)(V) B: u=U0 cos(t +)(V) C: u=U0 cos(t -)(V) D: u=U0 cos(t -)(V)Gii: Gi s u = U0 cos(t + ). Gi 1; 2 gc lch pha gia u v i1; i2

Ta c: tan 1= CZR

= tan( - /6); tan 2= L CZ ZR = tan( + /3);

Mt khc cng dng in cc i trong hai trng hp nh nhau, nn Z1 = Z2 --

ZC2 = (ZL ZC)2 ; ZL = 2ZC . V vy: tan 2= L CZ Z

R

= C

Z

R= tan( + /3);

tan( - /6) = - tan( +/3) tan( - /6) + tan( +/3) = 0 =>sin( - /6 + +/3) = 0 => - /6 + +/3 = 0 => = - /12 =>u=U0 cos(t -)(V).Chn C

Cu 33: Mt on mch gm cun cm c t cm L v in tr thun r mc ni tip vi t in c indung C thay i c. t vo hai u mch mt hiu in th xoay chiu c gi tr hiu dng U v tn s fkhng i. Khi iu chnh in dung ca t in c gi tr C=C 1 th in p hiu dng gia hai u t in

v hai u cun cm c cng gi tr v bng U, cng dng in trong mch khi c biu thc

12 6 os 100 ( )

4i c t A

= +

. Khi iu chnh in dung ca t in c gi tr C=C2 th in p hiu dng

gia hai bn t in t gi tr cc i. Cng dng in tc thi trong mch khi c biu thc l

A. 25

2 3 os 100 ( )12

i c t A

= +

B. 25

2 2 os 100 ( )12

i c t A

= +

C. 2 2 2 os 100 ( )3i c t A

= +

D. 2 2 3 os 100 ( )3i c t A

= +

Gii: Khi C = C1 UD = UC = U => Zd = ZC1 = Z1

Zd = Z1 => 212 )( CL ZZr + = 22 LZr + => ZL ZC1 = ZL => ZL = 21CZ (1)

Zd = ZC1 => r2 +ZL2 = ZC!2 => r2 =4

3 21CZ => r =2

3 21CZ (2)

tan 1 =3

1

2

32

1

11

1 =

=

C

CC

CL

Z

ZZ

r

ZZ----> 1 = -

6

Khi C = C2 UC = UCmax khi ZC2 =

1

1

2

1

22

2

2

C

C

C

L

L Z

Z

Z

Z

Zr==

+

Khi Z2 = 12

12

112

12

22 33)2

2(

4

3)( CCCCCL ZZZ

ZcZZZr ==+=+

14


15/21

tan 2 = 3

2

3

22

1

11

2 =

=

C

CC

CL

Z

ZZ

r

ZZ=> 2 = -

3

U = I1Z1 = I2Z2 => I2 = I1 23

32

31

2

1 ===I

Z

Z(A)

Cng dng in qua mch: i2 = I2)

364

100cos(2

++t= 2

)

12

5100cos(2

+t

(A). Chn B

Cu 34( H -2009): t in p xoay chiu c gi tr hiu dng 60 V 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 t in C th cng

dng in qua on mch l 2 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).Gii: Gi biu thc ca u = Uocos(100t + )

Ta thy : I1 = I2 suy ra Z1 = Z2 hay L C LZ Z Z = ZL = ZC/2

Lc u: 1tanL C L

Z Z Z

R R

= = i1 = Io cos(100t + + 1 ) + 1 = /4

Lc sau: 2tanL

Z

R = i2 = Io cos(100t + - 2 ) - 2 = - /12;

M 1 2 = = /12 Vy u 60 2 cos(100 t )

12

= + (V).Chn C

Gii 2: Ta thy I1 = I2 ----> (ZL ZC)2 = ZL2 =>. ZC = 2ZL

tan 1 =R

ZZ CL = -R

ZL (*) tan 1 =R

ZL (**) => 1 + 2 = 0

1 = u -4

; 2 = u +

12

=> 2 u -

4

+

12

= 0 => u =

12

Do u 60 2 cos(100 t )12

= + , Chn C

Cu 35. Cho 3 linh kin gm in tr thun R= 60, cun cm thun L v t in C. Ln lt t in pxoay chiu c gi tr hiu dng U vo hai u on mch ni tip RL hoc RC th biu thc cng dng

in trong nch ln lt l i1=cos(100-)(A) v i2=cos(100+)(A). nu t in p trn vo hai u on mchRLC ni tip th dng in trong mch c biu thc:A. 2cos(100t+)(A) B. 2 cos(100t+)(A)C. 2cos(100t+)(A) D. 2cos(100t+)(A)

Gii 1: Theo ( )1 2

01 02

=- 1RL RC

L C

I I Z ZZ Z

= = =

Mt khc( ) ( )

( )1 1 2

2

1u 1

u

u 2

- = 2 3

2 4 - =

i i i

i

+ = =

T ( ) ( ) ( )1

2 , 3 3 60 3 3

LL

ZZ

R = = = ( )0 01 120 2RLU I Z V = =

Khi RLC nt cng hng: i= 0U

Rcos(100t+ u )= 2cos(100t+)(A) Chn C

Gii 2: Ta thy cng hiu dng trong on mch RL v RC bng nhau suy ra ZL = ZC lch pha 1 giau v i1 v 2 gia u v i2 i nhau. tan1= - tan2Gi s in p t vo cc on mch c dng: u = U 2 cos(100t + ) (V).

15


16/21

Khi 1 = (- /12) = + /12 2 = 7/12tan1 = tan( + /12) = - tan2 = - tan( 7/12)tan( + /12) + tan( 7/12) = 0 sin( + /12 + 7/12) = 0Suy ra = /4 - tan1 = tan( + /12) = tan(/4 + /12) = tan /3 = ZL/R

ZL = R 3 v U = I1 2 2 12 120LR Z RI+ = = (V)Mch RLC c ZL = ZC => c s cng hng I = U/R = 120/60 = 2 (A) v i cng pha vi u:

u = U 2 cos(100t + /4) . Vy i = 2 2 cos(100t + /4) (A). Chn C

Cu 36:Cho ba linh kin: in tr thun 60R = , cun cm thun L v t in C. Ln lt t in p xoaychiu c gi tr hiu dng U vo hai u on mch ni tip RL hoc RC th biu thc cng dng in trongmch ln lt l 1 2 cos(100 /12)( )i t A = v 2 2 cos(100 7 /12)( )i t A = + . Nu t in p trn vo hai uon mch RLC ni tip th dng in trong mch c biu thc:

A. 2 2 cos(100 /3)( )i t A = + B. 2 cos(100 / 3) ( )i t A = +

C. 2 2 cos(100 / 4)( )i t A = + D. 2 cos(100 / 4) ( )i t A = +

Gii: Pha ban u ca i:32

=

= LC =>

cos01

0

II = = 2 chn A

Ta c th m rng bi ton ny nh sau:

Mc mch RL vo hiu in th u th dng in l i = I cos(t + )Mc mch RC vo hiu in th u th dng in l i = I cos(t + )Mc mch RLC vo hiu in th u th dng in l i = '0I cos(t + )Ta lun c mi quan h:(v gin hoc s dng cng thc tan ta d dng chng minh c):

cos

tan2

0'0

II

RZZ CL

LC

=

==

=

Vy bi ton ny trong mch RLC ta c th tnh v vit c biu thc ca: R,L,C,u,i,P ...

Cu 37: t in p 0u U cos t2

=

vo hai u on mch cha mt in tr thun v mt t in mc

ni tip. Khi , dng in trong mch c biu thc 0i I cos t 4

=

. Mc ni tip vo mch t th hai c

cng in dung vi t cho. Khi , biu thc dng in qua mch l

A. ( )0i 0, 63I cos t 0,147 (A)= B. ( )0i 0, 63I cos t 0, 352 (A)=

C. ( )0i 1, 26I cos t 0,147 (A)= D. ( )0i 1, 26I cos t 0, 352 (A)=

0 0 Cu U cos t i I cos t R Z2 4

= = =

mc thm t na th 02 02

2

2 5

tan 2

C C

I IZ Z

==

=

p n A

Cu 38: Cho mch in xoay chiu RLC. Cun dy thun cm L = 0,3/ (H), C = 44.10 / (F); R l bin tr. t

mch vo hiu in th ( )200 2 cos 100u t= Va) Vit biu thc uRkhi cng sut ca mch t cc i

A.( )

200cos 100 / 4R

u t

= V B.

( )200cos 100 / 4

R

u t

= +V

C. ( )100cos 100 / 4Ru t = V D. ( )100cos 100 / 4Ru t = Vb) Cho R = 20 , Hi phi ghp vi C mt t C1 nh th no v bng bao nhiu cng sut tiu th ca mch t cc i;Vit biu thc hiu in th gia hai u cun cm khi .

16


17/21

A. mc song song C1 = 0,637 mF B. mc ni tip C1 = 0,637 mFC. mc song song C1 = 0,637 F D. mc ni tip C1 = 0,637 F

Cu 39: on mch AC c in tr thun, cun dy thun cm v t in mc ni tip. B l mt im trn AC viuAB = cos100t (V) v uBC = cos (100t - ) (V). Tm biu thc hiu in th uAC.

A. u 2 2cos(100 t) VAC = B. u 2cos 100 t VAC 3

=

C. u 2cos 100 t VAC 3

= +

D.u 2cos 100 t V

AC 3

=

Cu 40:t in p xoay chiu vo vo hai u mt cun cm thun c t cm H2

1L

= th cng dng

in qua cun cm c biu thc i = I0cos(100t -6

) (V). Ti thi im cng tc thi ca dng in qua

cun cm c gi tr 1,5 A th in p tc thi hai u cun cm l 100 V. in p hai u cun cm c biu thc

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

) V. B. u = 125cos(100t +

3

) V.

C. u = 75 2 cos(100t +3

) V. D. u = 150cos(100t +

3

) V.

Cu 41:t vo hai u AMNB ca on mch RLC gm ni tip. M l im ni gia t in v cun dythun cm, N l im ni gia cun dy v in tr thun. Khi biu thc in p ca hai u on mch

NB l uNB = 60 2 cos(100t -3

) V v in p gia hai u on mch AN sm pha hn in p hai u

on mch AB mt gc3

. Biu thc ca in p hai u on mch AB l

A. u = 60 6 cos(100t -6

) V. B. u = 40 6 cos(100t -

6

) V.

C. u = 40 6 cos(100t + 6

) V. D. u = 60 6 cos(100t + 6

) V.

I II.PHNG PHP 2: DNG S PHC TM BIU THC i HOC uVI MY FX-570ES; MY FX-570ES PLUS (NHANH V HIU QU CHO TRC NGHIM)1.Tm hiu cc i lng xoay chiu dng phc: Xem bng lin hI LNG IN CNG THC DNG S PHC TRONG MY TNH FX-570ES

Cm khng ZL ZL ZL i (Ch trc i c du cng l ZL )

Dung khng ZC ZC - ZC i (Ch trc i c du tr l Zc )Tng tr:

=LZ L. ;1

=CZ.C

;

( )22

L CZ R Z Z= +

( )= + L CZ R Z Z i = a + bi ( vi a=R; b = (ZL -ZC ) )-Nu ZL >ZC : onmch c tinh cm khng-Nu ZL =u

i u i

Z

=> =u

Z

iCh : ( )= + L CZ R Z Z i ( tng tr phcZ c gch trn u: R l phn thc, (ZL -ZC ) l phn o)

Cn phn bit ch i sau gi tr b = (ZL -ZC ) l phn o, khc vi chi l cng dng in

17


18/21

2.Chn ci dt my tnh: CASIO fx 570ES ; 570ES PlusChn ch Nt lnh ngha- Kt quCh nhdng nhp / xut ton Bm: SHIFT MODE 1 Mn hnh xut hin Math.Thc hin php tnh s phc Bm: MODE 2 Mn hnh xut hin ch CMPLXDng to cc: r Bm: SHIFT MODE 3 2 Hin th s phc dng: AHin th dng cc: a + ib. Bm: SHIFT MODE 3 1 Hin th s phc dng: a+biChn n v o gc l (D) Bm: SHIFTMODE3 Mn hnh hin th ch DChn n v o gc l Rad (R) Bm: SHIFTMODE4 Mn hnh hin th ch R

Nhp k hiu gc Bm SHIFT (-)

Mn hnh hin th Nhp k hiu phn o i Bm ENG Mn hnh hin th i

3 .Lu Ch hin th kt qu trn mn hnh:Sau khi nhp, n du = c th hin th kt qu di dng s v t,mun kt qu di dng thp phn ta n SHIFT =( hoc nhn phm SD ) chuyn i kt qu Hin th.

4. Cc V d 1 :V d 1 : Mch in xoay chiu gm mt in tr thun R = 50, mt cun thun cm c h s t cm1

( )=L H

v mt t in c in dung42.10

( )

=C F

mc ni tip. Bit rng dng in qua mch c

dng ( )5cos100=i t A .Vit biu thc in p tc thi gia hai u mch in.

Gii :1

100 . 100= = = LZ L

;1

.... 50= = = CZC

. V ZL-ZC =50

-Vi my FX570ES : Bm MODE 2 mn hnh xut hin: CMPLX.-Bm SHIFTMODE 3 2 : dng hin th to cc:( r )-Chn n v o gc l (D), bm: SHIFTMODE3 mn hnh hin th DTa c : iZZRXIZiu

CLi)((... 0 +== 5 0 50 50= +X ( i ) ( Php NHN hai s phc)

Nhp my: 5 SHIFT (-) 0 X ( 50 + 50 ENG i ) = Hin th: 353.55339 45 = 250 2 45Vy biu thc tc thi in p ca hai u mch: u = 250 2 cos( 100 t + /4) (V).

V d 2: Mt mch in xoay chiu RLC khng phn nhnh c R = 100 ; C= 41

10. F

; L=2

H. Cng

dng in qua mch c dng: i = 2 2 cos100 t(A). Vit biu thc in p tc thi ca hai u mch?

Gii:2

100 200LZ L.

= = = ;1

= =CZ.C

........= 100 . V ZL-ZC =100

-Vi my FX570ES : Bm MODE 2 mn hnh xut hin: CMPLX.-Bm SHIFTMODE 3 2 : Ci t dng to cc:( r )-Chn n v o gc l (D), bm: SHIFTMODE3 mn hnh hin th D

Ta c : iZZRXIZiuCLi

)((... 0 +== 2 2 0 100 100= +> X ( i ) ( Php NHN hai s phc)Nhp my: 2 2 SHIFT (-) 0 X ( 100 + 100 ENG i ) = Hin th: 400 45

Vy biu thc tc thi in p ca hai u mch: u = 400cos( 100 t + /4) (V).

V d 3: Cho on mch xoay chiu c R=40 , L=

1(H), C=

6.0

10 4(F), mc ni tip in p 2 u mch

u=100 2 cos100 t (V), Cng dng in qua mch l:A. i=2,5cos(100 t+ )( )

4A

B. i=2,5cos(100 t- )( )

4A

18

Phm ENG nhp phn o i


19/21

C. i=2cos(100 t- )( )4

A

C. i=2cos(100 t+ )( )4

A

Gii:1

100 100= = = LZ L.

; 41 1

10100

0 6

= =CZ.C

.,

= 60 . V ZL-ZC =40

-Vi my FX570ES : Bm MODE 2 mn hnh xut hin: CMPLX.-Bm SHIFTMODE 3 2 : Ci t dng to cc:( r )

-Chn n v o gc l (D), bm: SHIFTMODE3 mn hnh hin th D

Ta c : i0

( ( )= =

+ u

L C

Uu

R Z Z iZ

100 2 0

40 40

=

+.

( i ) ( Php CHIA hai s phc)

Nhp 100 2 SHIFT (-) 0 : ( 40 + 40 ENG i ) = Hin th: 2,5 -45Vy : Biu thc tc thi cng dng in qua mch l: i = 2,5cos(100 t - /4) (A). Chn B

V d 4: Mt on mch in gm in tr R = 50 mc ni tip vi cun thun cm L = 0,5/ (H). t vohai u on mch mt in p xoay chiu u = 100 2 cos(100t- /4) (V). Biu thc ca cng dngin qua on mch l:

A. i = 2cos(100t- /2)(A). B. i = 2 2 cos(100t- /4) (A).

C. i = 2 2 cos100t (A). D. i = 2cos100t (A). Gii:

0 5100 50= = = L

,Z L.

; . V ZL-ZC =50 - 0 = 50

-Vi my FX570ES : Bm MODE 2 mn hnh xut hin: CMPLX.-Bm SHIFTMODE 3 2 : Ci t dng to cc:( r )-Chn n v o gc l (D), bm: SHIFTMODE3 mn hnh hin th D

Ta c : i0

( )

= =

+u

L

Uu

R Z iZ

100 2 45

50 50

=

+.

( i ) ( Php CHIA hai s phc)

Nhp 100 2 SHIFT (-) - 45 : ( 50 + 50 ENG i ) = Hin th: 2 - 90

Vy : Biu thc tc thi cng dng in qua mch l: i = 2cos( 100 t - /2) (A). Chn A

V d 5(H 2009): Khi t hiu in th khng i 30V vo hai u on mch gm in tr thun mc nitip vi cun cm thun c t cm L = 1/4 (H) th cng dng in 1 chiu l 1A. Nu t vo hai uon mch ny in p u =150 2 cos120t(V) th biu thc cng dng in trong mch l:

A. 5 2 c o s(1 2 04

=i t

B. 5cos(120 )(4

= +i t

C. 5 2cos(120 )( )4

= +i t A

D. 5cos(120 )(4

= i t

Gii: Khi t hiu in th khng i (hiu in th 1 chiu) th on mch ch cn c R: R = U/I =30

1

120 304

= = = LZ L.

; i =u 150 2 0

(30 30i)Z

=

+ ( Php CHIA hai s phc)

a.Vi my FX570ES : -Bm MODE 2 mn hnh xut hin: CMPLX.-Bm SHIFTMODE 3 2 : Ci t dng to cc:( r )

-Chn n v gc l (D), bm: SHIFTMODE3 mn hnh hin th DNhp my: 150 2 : ( 30 + 30 ENG i ) = Hin th: 5 - 45Vy: Biu thc tc thi cng dng in qua mch l: i = 5cos( 120 t - /4) (A). Chn Db.Vi my FX570ES : -Bm MODE 2 mn hnh xut hin: CMPLX.

-Chn n v gc l (R), bm: SHIFTMODE4 mn hnh hin th RNhp my: 150 2 : ( 30 + 30 ENG i ) = Hin th dng phc: 3.535533..-3.535533i

Bm SHIFT 2 3 : Hin th: 5 -4

Vy: Biu thc tc thi cng dng in qua mch l: i = 5cos( 120 t - /4) (A). Chn D

5.TRC NGHIM:

19


20/21

Cu 1. cho on mch R, L, C mc ni tip vi R= 100 , L=

1H, C=

2

10 4F. t in p xoay chiu vo

gia hai u on mch u LR , = 200 )2

100cos(2

+t (V). biu thc u c dng

A. Vtu )100cos(200 = B. Vtu )100cos(2200 =

C. Vtu )3

100cos(200

+= D. Vtu )4

100cos(2200

+=

Cu 2. Cho on mch R, L, C mc ni tip vi R=59 , L= 1 H. t in p xoay chiu

VtUu )100cos(2 = vo gia hai u on mch th )4

100cos(100

+= tu L . Biu thc uc l:

A. uc = 50 )2

100cos(

t (V) B . uc= 50 )4

100cos(2

t (V)

C. uc= 50 )4

3100cos(

t D. uc = 50 )

4

3100cos(2

t

Cu 3: Mt on mch gm mt t in c dung khng ZC = 100 v cun dy c cm khng ZL = 200 mc

ni tip nhau. Hiu in th ti hai u cun cm c dng VtuL )6100cos(100

+= . Biu thc hiu in th hai u t in c dng nh thno?

A. VtuC )3

100cos(50

=

B. VtuC )6

5100cos(50

= C.

VtuC )6

100cos(100

+= D. VtuC )2

100cos(100

=

Cu 4. Cho mch R,L,C, u = 240 2 cos(100t) V, R = 40, ZC = 60 , ZL= 20 .Vit biu thc ca dngin trong mchA. i = 3 2 cos(100t) A B. i = 6cos(100t)AC. i = 3

2cos(100

t +

/4) A D. i = 6cos(100

t +

/4)A

Cu 5. Cho mch in R,L,C cho u = 240 2 cos(100t) V, R = 40 , ZL = 60 , ZC = 20, Vit biu thcca cng dng in trong mchA. i = 3 2 cos(100t)A. B. i = 6cos(100t) A.C. i = 3 2 cos(100t /4)A D. i = 6cos(100t - /4)ACu 6. Cho mch R,L,C, R = 40, ZL = ZC = 40 , u = 240 2 cos(100t). Vit biu thc iA. i = 6 2 cos(100t )A B. i = 3 2 cos(100t)AC. i = 6 2 cos(100t + /3)A D. 6 2 cos(100t + /2)ACu 7. Cho mch R,L,C, u = 120 2 cos(100t)V. R = 30 , ZL = 10 3 , ZC = 20 3 , xc nh biu

thc i.A. i = 2 3 cos(100t)A B. i = 2 6 cos(100t)AC. i = 2 3 cos(100t + /6)A D. i = 2 6 cos(100t + /6)A

Cu 8: Mch in xoay chiu gm t in C =

410 F, cun dy thun cm L =

10

1H mc ni tip. Bit

cng dng in l i = 4cos(100t) (A). Biu thc in p hai u mch y l nh th no?

A. u = 236 cos(100t -) (V) B. u = 360cos(100t +2

) (V)

C. u = 220sin(100t -2

) (V) D. u = 360cos(100t -

2

) (V)

Cu 9: in p gia hai u mt cun dy c r =4 ; L=0,4(H) c thc: ))(3

100cos(2200 Vtu

+= .

Biu thc ca cng dng xoay chiu trong mch l:

20


21/21

A. i = 50cos(100t +12

)(A) B. i = 50 2 cos(100t -

12

)(A)

C. i = 50cos(100t -12

)(A) D. i = 50 2 cos(100t +

12

)(A)

Cu 10: Cho on mch xoay chiu AB gm hai on mch AN v NB mc ni tip. t vo hai u onmch AB mt in p xoay chiu n nh )V()3/t100cos(2200u AB += , khi in p tc thi gia

hai u on mch NB l )V()6/5t100sin(250uNB += . Biu thc in p tc thi gia hai u on

mch AN lA. )V()3/t100sin(2150u AN += . B. )V()3/t120cos(2150u AN += .

C. )V()3/t100cos(2150u AN += . D. )V()3/t100cos(2250u AN += .

Nguyn tc thnh cng: Suy ngh tch cc; Cm nhn am m; Hnh ng kin tr !Chc cc em hc sinh THNH CNG trong hc tp!

Su tm v chnh l: GV: on Vn LngEmail: [email protected] ; [email protected];

T: 0915718188 0906848238
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