Download - Tuyenchondien Xoay Chieu Kho Diem 10hocmaivncom.thuvienvatly.com.4f260.39098

BI TP V IN XOAY CHIU HAY V KH Cu 1. t mt in p xoay chiu vo hai u on mch L, R, C mc ni tip theo th t . in p hai u

cc on mch cha L,R v R,C ln lt c biu thc: uLR = 150sos(100t + /3) (V); uRC =

50 6 sos(100t - /12) (V). Cho R = 25 . Cng dng in trong mch c gi tr hiu dng bng: A. 3

(A). B. 3 2 (A) . C. 2

23 (A). D. 3,3 (A

Gii: V gin vc t nh hnh v ta c

MON = 12

5)

12(

3

MN = UL + UC

OM = URL = 75 2 (V)

ON = URC = 50 3 (V)

p dng L cosin cho tam gic OMN:

MN = UL + UC = 12

5cos.222

RCRLRCRL UUUU 118 (V)

UR2 = ULR

2 UL

2 = URC

2 UC

2 -----> UL

2 UC

2 = ULR

2 URC

2 = 3750

(UL + UC )(UL - UC ) = 3750-----> UL + UC = 3750/118 = 32 (V)

Ta c h phng trnh UL - UC =118 (V)

UL + UC = 32 (V)

Suy ra UL = 75 (V) -----> UR = 222 75 LRL UU = 75 (V)

Do I = UR/R = 3 (A). Chn p n A Cu 2. t mt n p xoay chiu vo hai u on mch gm in tr thun R, cun dy thun cm L v t in C c in dung thay i. Khi C = C1 in p hiu dng trn cc phn t UR = 40V, UL = 40V, UC =

70V.Khi C = C2 in p hiu dng hai u t l UC = 50 2 V. in p hiu dng gia hai u in tr l:

A. 25 2 (V). B. 25 (V). C. 25 3 (V). D. 50 (V).

Gii: Khi C = C1 UR = UL ----> ZL = R

in p t vo hai u mch; U = 22 )( CLR UUU = 50 (V)

Khi C = C2 ------> UR = UL

U = 2

2

2 )'(' CLR UUU = 50 (V)-----> UR = 25 2 (V). Chn p n A

Cu 3. Cho mch in xoay chiu gm 3 phn th ni tip: in tr R; cun cm L = 4

1(H) v t in C. Cho

bit in p tc thi hai u on mch u = 90cos(t + /6) (V). Khi = 1 th cng dng in chy qua

mch i = 2 cos(240t - /12) (A); t tnh bng giy. Cho tn s gc thay i n gi tr m trong mch c gi tr cng hng dng in, hiu in th gia hai bn t in lc l:

A. uC = 45 2 cos(100t - /3) (V); B. uC = 45 2 cos(120t - /3) (V);

C uC = 60cos(100t - /3) (V); D. uC = 60cos(120t - /3) (V);

Gii:

T biu thc ca i khi = 1 ta c 1 = 240

ZL1 = 2404

1= 60

Gc lch pha gia u v i : = u - i = 4

)12

(6

-----> tan = 1

O UR

N UCR

ULR

M

http://www.hocmaivn.com/




R = ZL1 ZC1; Z1 = 2451

245

I

U

Z12 = R

2 + (ZL ZC)

2 = 2R

2----> R = 45

R = ZL1 ZC1 ---> ZC1 = ZL1 R = 15

ZC1 = C1

1

----> C =

3600

1

15.240

11

11

CZ

(F)

Khi mch c cng hng

222 )120(

3600

1.

4

1

11

LC

----> 2 = 120

Do mch cng hng nn: ZC2 = ZL2 = 2 L = 30 ()

I2 = 245

245

R

U(A); uc chm pha hn i2 tc chm pha hn u gc /2

Pha ban u ca uC2 = 326

UC2 = I2,ZC2 = 30 2 (V)

Vy uC = 60cos(120t /3) (V). Chn p n D,

Cu 4 .Cho mt mch in gm bin tr Rx mc ni tip vi t in c 63,8C F v mt cun dy c in

tr thun r = 70, t cm 1

L H

. t vo hai u mt in p U=200V c tn s f = 50Hz. Gi tr ca Rx

cng sut ca mch cc i v gi tr cc i ln lt l

A. 0 ;378,4W B. 20 ;378,4W C. 10 ;78,4W D. 30 ;100W

Gii:

P = I2R=

R

ZZR

U

ZZR

RU

CLCL

2

2

22

2

)()(

Vi R = Rx + r = Rx + 70 70

ZL = 2fL = 100; ZC = 610.8,63.314

1

2

1

fC50

P = Pmax khi mu s y = R + R

3500 c gi tri nh nht vi R 70

Xt s ph thuc ca y vo R:

Ly o hm y theo R ta c y = 1 - 2

3500

R; y = 0 -----> R = 50

Khi R < 50 th nu R tng y gim. ( v y < 0)

Khi R > 50 th nu R tng th y tng

Do khi R 70 th mu s y c gi tr nh nht khi R = 70.

Cng sut ca mch c gi tr ln nht khi Rx = R r = 0

Pc = 4,378)( 22

2

CL ZZr

rUW

Chn p n A Rx = 0, Pc = 378 W





Cu 5. Cho mch in nh hnh v

t vo hai u AB mt in p xoay chiu c gi tr hiu dng v tn s khng i. lch pha ca uAN v

uAB bng lch pha ca uAM v dng in tc thi. Bit 3 120 3( )AB AN MNU U U V . Cng dng

in trong mch 2 2I A . Gi tr ca ZL l

A. 30 3 B. 15 6 C. 60 D. 30 2

V gin vc t nh hnh v:

AB = UAB UAB = 120 3 (V)

AM = UAM = Ur + UL

AN = UAN UAN = 120 3 (V)

AE = Ur

EF = MN = UMN = UR UMN = UR = 120 (V)

AF = Ur + UR ; EM = FN = UL ; NB = UC NAB = MAF suy ra MAN = FAB T UAB = UMN suy ra UL

2 = (UL UC)

2 -------> UC = 2UL suy ra NAF = FAB

V vy MAN = ANM ----> tam gic AMN cn MN = AM hay UAM = UR = 120(V) Ur

2 + UL

2 = UAM

2 = 120

2 (1)

(Ur + UR)2 + (UL UC)

2 = UAB

2

hay (Ur + 120)2 + UL

2 = 120

2 (2)

T (1) v (2) ta c Ur = 60 (V); UL = 60 3 (V)

Do o ZL = 61522

360

I

U L (), Chn p n B

Cu 6. Mt on mch AB gm hai on mch AM v BM mc ni tip. on mch AM gm in tr thun

R1 mc ni tip vi t in c in dung C, on mch MB gm in tr thun R2 mc ni tip vi cun cm

thun c t cm L. t in p xoay chiu u = U0cos t (U0 v khng i) vo hai u on mch AB th

cng sut tiu th ca on mch AB l 85 W. Khi LC

12 v lch pha gia uAM v uMB l 900. Nu

t in p trn vo hai u on mch MB th on mch ny tiu th cng sut bng:

A. 85 W B. 135 W. C. 110 W. D. 170 W.

Gii:

Khi LC

12 trong mch c cng hng ZL = ZC

v cng sut tiu th ca on mch c tnh theo cng thc

N R

A B

M C L,r

A Ur E UR F

I

UAN N

UAM

M

UAB B

L R1 C M R2 B A





P = 21

2

RR

U

(1). Ta c: tan1 =

1R

ZC ; tan2 = 1R

Z L

Mt khc: 2 - 1 = 900

------> tan1. tan2 = -1

Do 1R

ZC

1R

Z L = -1 -------> ZL = ZC = 21RR (2)

Khi t in p trn vo on mch MB th cng sut tiu th trn on mch

P2 = I22 R2 = 22

2

2

2

LZR

RU

=

212

2

2

2

RRR

RU

21

2

RR

U

= P = 85W. Chn p n A

Cu 7: Cho mch in nh hnh v. t vo hai u on

mch in p xoay chiu u=120 6 cos(100 t)(V) n nh,

th in p hiu dng hai u MB bng 120V, cng sut

tiu th ton mch bng 360W; lch pha gia uAN v uMB

l 900, uAN v uAB l 60

0 . Tm R v r

A. R=120 ; r=60 B. R=60 ; r=30 ;

C. R=60 ; r=120 D. R=30 ; r=60

Gii: V gin vc t nh hnh v

OO1 = Ur

UR = OO2 = O1O2 = EF

UMB = OE UMB = 120V (1)

UAN = OQ

UAB = OF UAB = 120 3 (V) (2)

EOQ = 900

FOQ = 600

Suy ra = EOF = 900 600 = 300.

Xt tam gic OEF: EF2 = OE

2 + OF

2 2.OE.OFcos300

Thay s ---------> EF = OE = 120 (V) Suy ra UR = 120(V) (3)

UAB2 = (UR + Ur)

2 + (UL UC)

2

Vi (UL UC)

2 = UMB

2 Ur

2 ( xt tam gic vung OO1E)

UAB2 = UR

2 +2UR.Ur + UMB

2 . T (1); (2), (3) ta c Ur = 60 (V) (4)

Gc lch pha gia u v i trong mch:

= FOO3 = 30

0 ( v theo trn tam gic OEF l tam gic cn c gc y bng 300)

T cng thc P = UIcos ----->

I = P / Ucos 360/(120 3 cos300) = 2 (A): I = 2A (5)

Do R = UR/I = 60; r = Ur /I = 30. Chn p n B

Cu 8. t in p xoay chiu u = 100 2 cost (c thay i c trn on [100 200; ] ) vo hai u

on mch c R, L, C mc ni tip. Cho bit R = 300 , L =

1(H); C =

410 (F).

UAN Q

O3

UL

UL + UC

O

UC

Ur O1 UR

O2

UAB F UMB E

UR + Ur

L,r

R A B

C N

M





in p hiu dng gia hai u L c gi tr ln nht v nh nht tng ng l

A.100 V; 50V. B.50 2 V; 50V. C.50V; 3

100v. D. .

3

100;

53

400VV

Gii:

Ta c UL = IZL; UL=

22

4

4

282

2

2

42

22 1110.71

101

)2(11

)1

(

U

LC

LR

C

UL

CLR

LU

Xt biu thc y = 2

4228 110.710

XX

Vi X = 2

1

> 0. Ly o hm y theo X ta thy y > 0:

gi tr ca y tng khi X tng, tc l lhi 2 hay gim. Vy khi tng th UL tng

Trong khong 100 200 UL = ULmax khi = 200. --->

ULmax =

22

4

4

28 1110.71

10

U

53

400

14

7

16

1

100

1

.4

110.7

10.16

110

22

4

48

28

U

(V)

UL = ULmin khi = 100. --->

ULmin =

22

4

4

28 1110.71

10

U

3

100

171

100

1110.7

10

110

22

4

48

28

U

(V)

Chn p n D.

Cu 9.. Cho mch in xoay chiu khng phn nhnh AD gm hai on AM v MD. on mch MD gm

cun dy in tr thun R = 40 3 v t cm L = 5

2 H. on MD l mt t in c in dung thay i

c, C c gi tr hu hn khc khng. t vo hai u mch in p xoay chiu uAD = 240cos100t (V).

iu chnh C tng in p (UAM + UMD) t gi tr cc i. Gi tr cc i l:

A. 240 (V). B. 240 2 (V). C. 120V. D. 120 2 (V)

Gii:

Ta c ZL = 100 .2/5 = 40-----> ZAM = 8022 LZR

t Y = (UAM + UMD)2

.

Tng (UAM + UMD) t gi tr cc i khi Y t gi tr cc i

Y = (UAM + UMD)2 = I

2( ZAM

2 +ZC

2 + 2ZAM.ZC) = 22

222

)(

)2(

CL

CAMCAM

ZZR

ZZZZU

Y = 640080

)6400160(

)40(40.3

)16080(2

22

22

222

CC

CC

C

CC

ZZ

ZZU

Z

ZZU





Y = Ymax khi biu thc X= 640080

)6400160(2

2

CC

CC

ZZ

ZZ= 1+

640080

2402 CC

C

ZZ

Z c gi tr cc i

------->X = 640080

2402 CC

C

ZZ

Z =

806400

240

C

CZ

Z

c gi tr cc i

X = Xmax khi mu s cc tiu, -----> ZC2 = 6400 -----> ZC = 80

tng in p (UAM + UMD) t gi tr cc i khi ZC = 80

(UAM + UMD)max = )( CAM ZZZ

U = 2240

80

160.2120

)8040(40.3

)8080(2120

22

(V)

Chn p n B: (UAM + UMD)max = 240 2 (V)

Cu 10. Mt cun dy khng thun cm ni tip vi t in C trong mch xoay chiu c in p u=U0cost(V) th dng in trong mch sm pha hn in p u l 1 v in p

hiu dng hai u cun dy l 30V. Nu thay C1=3C th dng in chm pha hn u gc 2 = 90

0 - 1 v in p hiu

dng hai u cun dy l 90V. Tm U0.

Gii: Cc ch s 1 ng vi trng hp t C; ch s 2 ng vi t 3C

V gin vc t nh hnh v: Ta c ZC2 = ZC1/3 = ZC/3

Do Ud = IZd = I22

LZR : Ud1 = 30V; Ud2 = 90V

Ud2 = 3Ud1 -----> I2 = 3I1

UC1 = I1ZC

UC2 = I2ZC2 = 3I1ZC/3 = I1ZC = UC1 =UC

Trn gin l cc on OUC; Ud1U1; Ud2U2 biu in UC U1 = U2 =U in p hiu dung t vo mch. Theo bi ra 2=90

0-1 .

Tam gic OU1U2 vung cn ti O

Theo hnh v ta c cc im UC; U1 v U2 thng hng. on thng UCU1 U2 song song v bng on OUd1Ud2

Suy ra U1U2 = Ud1Ud2 = 90 30 = 60V

Do OU1 = OU2 = U1U2/ 2

Suy ra U = 60/ 2 = 30 2 -----> U0 = 60V

2 UR2 I O 1 UR1

UC

U1

U2

Ud2 UL2

Ud1 UL1





Cu 11: Mch in xoay chiu R, L, C mc ni tip. in p hai u on mch l 0u U cos t . Ch

c thay i c. iu chnh thy khi gi tr ca n l 1 hoc 2 ( 2 < 1 ) th dng in hiu

dng u nh hn cng hiu dng cc i n ln (n > 1). Biu thc tnh R l

A. R = 1 22

( )

L n 1

B. R = 1 2

2

L( )

n 1

C. R = 1 2

2

L( )

n 1

D. R = 1 2

2

L

n 1

Gii: I1 = I2 =Imax/n ------> Z1 = Z2 -----> 1 L - C1

1

= - 2 L +

C2

1

-------> 2 L-=C1

1

m I1 = Imax/n

------>

)1

(1

1

2

CLR

U

= R

U

n

1--------->n

2R

2 = R

2 +( 1 L -

C1

1

)

2 = R

2 + ( 1 L -2 L )

2

------> (n2 1)R2 = ( 1 -2 )

2L

2 -------> R = 1 2

2

L( )

n 1

. Chn p n B

Cu 12. t mt in p u = U0 cos t ( U0 khng i, thay i c) vo 2 u on mch gm R, L, C mc ni tip tha mn iu kin CR2 < 2L. Gi V1,V2, V3 ln lt l cc vn k mc vo 2 u R, L, C. Khi tng dn tn s th thy trn mi vn k u c 1 gi tr cc i, th t ln lt cc vn k ch gi tr cc i khi tng dn tn s l

A. V1, V2, V3. B. V3, V2, V1. C. V3, V1, V2. D. V1, V3,V2.

Gii: Ta gi s ch ca cc vn k l U1,2,3

U1=IR =22 )

1(

CLR

UR

U1 = U1max khi trong mch c s cng hng in: ----->12 =

LC

1 (1)

U2 = IZL = 22

2

22

22222 21

)1

(y

U

C

L

CLR

UL

CLR

LU

U2 = U2max khi y2 = 2

2

2

42

211

LC

LR

C

c gi tr cc tiu y2min

t x = 2

1

, Ly o hm y2 theo x, cho y2 = 0 ----->x = 2

1

= )2(

2

2CRC

LC

)2(

2

22

2

2

RC

LC

=)2(

22CRLC

(2)

U3 = IZC = 23

22

222222 )21

()1

(y

U

C

L

CLRC

U

CLRC

U





U3 = U3max khi y3 = L24 +(R2 -2

C

L )2 +

2

1

C c gi tr cc tiu y3min

t y = 2 , Ly o hm ca y3 theo y, cho y3 = 0

y = 2 = 2

2

2

2

2

1

2

2

L

R

LCL

RC

L

32 =

2

2

2

1

L

R

LC (3)

So snh (1); (2), (3):

T (1) v (3) 32 =

2

2

2

1

L

R

LC < 1

2 =

LC

1

Xt hiu 22 - 1

2 =

)2(

22CRLC

-LC

1=

)2()2(

)2(22

2

2

2

CRLLC

CR

CRLLC

CRLL

>0

(V CR2

< 2L nn 2L CR2 > 0 )

Do 22 =

)2(

22CRLC

> 12 =

LC

1

Tm lai ta c 32 =

2

2

2

1

L

R

LC < 1

2 =

LC

1 < 2

2 =

)2(

22CRLC

Theo th t V3, V1 , V2 ch gi tr cc i Chn p n C

Cu 13 . on mch AB gm on AM ni tip vi MB. on AM goomg in tr R ni tip vi cuonj dy thun cm c t cm L thay i c. on MB ch c t in C. in p t vo hai u mch

uAB = 100 2 cos100t (V). iu chnh L = L1 th cng dng in qua mch I1 = 0,5A, UMB = 100(V),

dng in i tr pha so vi uAB mt gc 600. iu chnh L = L2 in p hiu dng UAM t cc i. Tnh

t cm L2:

A.

21(H). B.

31(H). C.

32 (H). D.

5,2(H).

Gii:

Ta c ZC =100/0,5 = 200, 360tantan0

R

ZZ CL -----> (ZL ZC) = R 3

Z = U/I = 100/0,5 = 200

Z = RZZR CL 2)(22 ------> R = 100

UAM = I.ZAM =

2212

22222

22

100

)100(4001

2)(

L

L

L

CLCLCL

L

Z

Z

U

ZR

ZZZZR

U

ZZR

ZRU

UAM =UAMmin khi y = 22100

100

L

L

Z

Z

= ymax c gi tr cc i

y = ymax khi o hm y = 0------> ZL2 200ZL -100 = 0

-------> ZL = 100(1 + 2 )

--------> L =

21(H) Chn p n A.





Cu 14. Cho mch in RLC mc ni tip theo th t R, L, C trong cun dy thun cm c t

cm L thay i c, in tr thun R=100 . t vo hai u on mch hiu in th xoay chiu c

tn s f=50Hz. Thay i L ngi ta thy khi 1L=L v khi

12

LL=L =

2th cng sut tiu th trn on

mch nh nhau nhng cng dng in tc thi vung pha nhau. Gi tr L1 v in dung C ln lt

l:

A. -4

1

4 3.10L = (H);C= (F)

2 B.

-4

1

4 10L = (H);C= (F)

3

C. -4

1

2 10L = (H);C= (F)

3 D.

-4

1

1 3.10L = (H);C= (F)

4

Gii: Do cng sut P1 = P2 -----> I1 = I2 ------> Z1 = Z2

Do (ZL1 ZC)2 = (ZL2 ZC)

2. Do ZL1 ZL2 nn ZL1 ZC = ZC ZL2 = ZC -

2

1LZ

----> 1,5ZL1 = 2ZC (1)

tan1 = R

ZZ CL 1 = R

Z L

4

1 v tan2 = R

ZZ

R

ZZ CL

CL

21

2 = R

Z L

4

1

1 + 2 = 2

------> tan1. tan1 = -1 -----> ZL1

2 = 16R

2 ----. ZL1 = 4R = 400

----> L1 =

41 LZ

(H)

ZC = 0,75ZL1 = 300 ----> C = 3

10

.

1 4

CZ(F)

Chn p n B

Cu 15: Cho 3 linh kin gm in tr thun R=60, cun cm thun L v t in C. Ln lt t in p

xoay chiu 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 i1= 2 cos 10012

t

(A) v i2= 7

2 cos 10012

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 cos(100t+

3 )(A) . B. 2 cos(100t+

3 )(A).

C. 2 2 cos(100t+

4 )(A) . D. 2cos(100t+

4 )(A).

Gii: Ta thy cng hiu dng trong on mch RL v RC bng nhau suy ra ZL = ZC lch pha 1 gia u v i1 v 2 gia u v i2 i nhau. tan1= - tan2

Gi s in p t vo cc on mch c dng: u = U 2 cos(100t + ) (V). Khi 1 = (- /12) = + /12 2 = 7/12

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

-- ZL = R 3

U = I1 2 2

12 120LR Z RI (V)





Mch RLC c ZL = ZC trong mch c s cng hng I = U/R = 120/60 = 2 (A) v i cng pha vi u =

U 2 cos(100t + /4) .

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

Cu 16. Cho mch RLC ni tip. Khi t in p xoay chiu c tn s gc ( mch ang c tnh cm khng). Cho thay i ta chn c 0 lm cho cng dng in hiu dng c gi tr ln nht

l Imax v 2 tr s 1 , 2 vi 1 2 = 200 th cng dng in hiu dng lc ny

l ax

2

mII .Cho 3

4L

(H). in tr c tr s no:

A.150. B.200. C.100. D.125.

Gii: I1 = I2 -----> Z1 = Z2 ------> (ZL1 ZC1)

2 = (ZL2 ZC2)

2 ----> ZL1 + ZL2 = ZC1 + ZC2

L(1 + 2) = 21

21

21

)11

(1

CC

-----> LC =

21

1

------> ZC1 = ZL2

Imax = R

U 2; I1 =

Z

U =

2

11

2 )( CL ZZR

U

=

R

U

2

2

-------> 4R2 = 2R

2 + 2(ZL1 ZC1)

2

R2 = (ZL1 ZL2)

2 = L

2 (1 - 2)

2 -----> R = L (1 - 2) =

200

4

3 = 150(). Chn p n A

Cu 17: Mt mch in xoay chiu gm cc linh kin l tng mc ni tip theo th t R, C v L. t vo

hai u on mch mt in p xoay chiu u = U0cos(t /6). Bit U0, C, l cc hng s. Ban u in p hiu dng hai u in tr R l UR = 220V v uL = U0Lcos(t + /3), sau tng R v L ln gp i, khi URC bng

A. 220V. B. 220 2 V. C. 110V. D. 110 2 .

Gii: Hiu pha ban u ca uL v i: UL - i = 2

---> i =

3

-

2

= -

6

Do ta c u, i cng pha, MCH C CNG HNG: nn: ZL = ZC v U = UR = 220 (V) Khi tng R v L ln gp i th R = 2R, ZL = 2ZL

URC = 22

22

)'('

'

CL

C

ZZR

ZRU

=

22

22

)2('

'

CC

C

ZZR

ZRU

= U = 220V. Chn p n A

Cu 18: t mt in p xoay chiu u = U0cos(100t+ ) vo hai u mt on mch gm R, L, C mc

ni tip (L l cun cm thun). Bit 410

C F

; R khng thay i, L thay i c. Khi 2

L H

th biu

thc ca dng in trong mch l 1 2 os(100 t /12)i I c A . Khi 4

L H

th biu thc ca dng in

trong mch l 2 2 os(100 t / 4)i I c A . in tr R c gi tr l

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

Gii:

Ta c ZC = 100; ZL1 = 200; ZL2 = 400

tan1 = R

ZZ CL 1 = R

100 ----.>1 = +

12





tan2 = R

ZZ CL 2 =R

300 = 3tan1 ----.>2 = +

4

-------> 2 - 1 = 4

-

12

=

6

tan(2 - 1) = tan6

=

3

1

tan(2 - 1) =3

1

tan31

tan2

tantan1

tantan

1

2

1

12

12

-----> tan1 =

3

1

-----> R

100=

3

1------> R = 100 3 () Chn p n A

Cu 19. Trong gi thc hnh mt hc sinh mc ni tip mt qut in xoay chiu vi in tr R, ri mc vo hai u mch in p xoay chiu c gi tr hiu dng 380V. Bit qut c cc gi tr nh mc 220V

88W. Khi hot ng ng cng sut nh mc th lch pha gia in p hai u qut v dng in qua

n l , vi cos = 0,8. qut hot ng ng cng sut th R =?

Gii:

Gi r l in tr ca qut: P = UqIcos = I2r.

Thay s vo ta c: I = cosqU

P =

8,0.220

88= 0,5 (A); r =

2I

P= 352

Zqut = I

U q= 22 LZr = 440

Khi mc vo U = 380V: I = Z

U=

22)( LZrR

U

=

222 2 LZrRrR

U

R2 + 2Rr +

2

quatZ = 2)(

I

U------> R

2 + 704R +440

2 = 760

2

-----> R2 + 704R 384000 = 0------> R = 360,7

Cu 20. Ni hai cc ca my pht in xoay chiu mt pha vo hai u on mch AB gm R ni tip vi L thun. B qua in tr cun dy ca my pht. Khi r to quay u vi tc n vng/pht th cng

hiu dng l 1A. Khi r to quay u vi tc 3n vng/pht th cng hiu dng l 3 A..Khi r to

quay u vi tc 2n vng/pht th cm khng ca on mch AB tnh theo R l?

Gii: I = Z

U=

Z

E

Vi E l sut in ng hiu dng gia hai cc my pht: E = 2 N0 = 2 2fN0 = U ( do r = 0) Vi f = np n tc quay ca roto, p s cp cc t

Z = 222 LR

Khi n1 = n th 1 = ; ZL1 = ZZ

Khi n3 = 3n th 3 = 3; ZL3 = 3ZZ ---->

3

1

I

I=

3

1

E

E

1

3

Z

Z=

3

1

1

3

Z

Z------->

3

1

22

22 9

L

L

ZR

ZR

=

3

1

I

I=

3

1------>R

2 + 9

2

LZ = 3R2 +3

2

LZ

62

LZ = 2R2 ------>

2

LZ = R2/3-----> ZL =

3

R





- Khi n2 = 2n th 2 = 2; ZL2 = 2ZZ = 3

2R

Cu 21: Mt cun dy khng thun cm ni tip vi t in C trong mch in xoay chiu c in p

0. osu U c t (V) th dng in trong mch sm pha hn in p l 1 , in p hiu dng hai u cun dy l

30V. Bit rng nu thay t C bng t 'C 3C th dng in trong mch chm pha hn in p l 2 1

2

v in p hiu dng hai u cun dy l 90V. Bin 0 ?U

A. 60V . B. 30 2V C. 60 2V . D. 30V

Gii: Ud1 = 30 (V)

Ud2 = 90 (V) ----> 1

2

d

d

U

U= 3 ----> I2 = 3I1 -----> Z1 = 3Z2 -------.Z1

2 = 9Z2

2

------> R2 + (ZL ZC1)

2 = 9R

2 + 9(ZL -

3

1CZ )2 ----->2(R2 +ZL2 ) = ZLZC1

------> ZC1 = L

L

Z

ZR )(2 22

1

1

d

d

Z

U=

1Z

U-------> U = Ud1

1

1

dZ

Z= Ud1

22

2

1

2 )(

L

cL

ZR

ZZR

= Ud1 22

1

2

1

22 2

L

CLCL

ZR

ZZZZR

=

Ud1 22

22

2

22222 )(22

)(4

L

L

LL

L

LL

ZR

Z

ZRZ

Z

ZRZR

= Ud1 3)(4

2

22

L

L

Z

ZR = Ud1 1

42

2

LZ

R

tan1 = R

ZZ CL 1 ; tan1 = R

ZZ CL 2 = R

ZZ CL

3

1

2 12

-----> 1 + 2 =

2

-----> tan1 tan2 = -1 ( v 1 < 0)

R

ZZ CL 1

R

ZZ CL

3

1

= -1------>(ZL ZC1)(ZL - 3

1CZ ) = - R2 ------->

R2 + ZL

2 4ZL

3

1CZ + 3

2

1CZ = 0 --------> (R2 + ZL2 ) 4ZL

L

L

Z

ZR

3

)(2 22 +

2

222

3

)(4

L

L

Z

ZR = 0

----->(R2 + ZL

2 )[1-

3

8+

2

22

3

)(4

L

L

Z

ZR ] = 0 ----->

2

22

3

)(4

L

L

Z

ZR -

3

5 = 0----->

2

2

3

4

LZ

R =

3

1

---->2

24

LZ

R = 1------> U = Ud1 1

42

2

LZ

R = Ud1 2

Do U0 = U 2 = 2Ud1 = 60V. Chn p bn A

Cu 22 Ni hai cc my pht in xoay chiu mt pha vo hai u mch ngoi RLC, b qua in tr dy

ni, coi t thng cc i gi qua cun dy l khng i Khi rto quay vi tc n0 vng/pht th cng sut mch ngoi cc i.Khi rto quay vi tc n1 vng/pht v n2 vng/pht th cng sut mch ngoi

c cng gi tr Mi lin h gia n1, n2 v n0 l





A. 20 1 2.n n n B.

2 2 2

0 1 2n n n C. 22

2

1

2

2

2

12

0nn

nnn

D.

2

2

2

1

2

2

2

12

0

2

nn

nnn

Gii: Sut in ng ca ngun in: E = 2 N0 = 2 2fN0 = U ( do r = 0) Vi f = np n tc quay ca roto, p s cp cc t Do P1 = P2 -----> I1

2 = I2

2 ta c:

2

1

1

2

2

1

)1

(C

LR

=2

2

2

2

2

2

)1

(C

LR

-------> ])1

([ 2

2

2

22

1C

LR

= ])1

([ 2

1

1

22

2C

LR

---> C

L

CLR 2122

2

2

122

2

2

1

22

1 2

=

C

L

CLR 2222

1

2

222

2

2

1

22

2 2

---> )2)(( 2222

1C

LR = )(

12

2

2

1

2

1

2

2

2

C =

2

2

2

1

2

1

2

2

2

1

2

2

2

))((1

C

-----> (2C

L- R

2 )C

2 =

2

2

2

1

11

(*)

Dng in hiu dng qua mch

I = Z

E

Z

U

P = Pmac khi E2 /Z

2 c gi tr ln nht hay khi y =

22

2

)1

(C

LR

c gi tr ln nht

y =

2

22

222 21

1

C

L

CLR

=

2

2

2

42

211

1

LC

LR

C

y = ymax th mu s b nht

t x = 2

1

---> y =

22

2

2

)2( LxC

LR

C

x

Ly o hm mu s, cho bng 0 ta c kt qu x0 = 20

1

=

2

1C

2(2 )

2RC

L (**)

T (*) v (**) ta suy ra 2

2

2

1

11

=

2

0

2

2

0

2

2

2

1

211

fff hay

2

0

2

2

2

1

211

nnn ------>

2

2

2

1

2

2

2

12

0

2

nn

nnn

Chn p n D

Cu 23 : t in p xoay chiu vo mch RLC ni tip c C thay i c. Khi C= C1 = 410

F v C=

C2 = 410

2

F th UC c cng gi tr. UC c gi tr cc i th C c gi tr:





A. C = 43.10

4

F . B. C = 410

3

F C. C = 43.10

2

F. D. C = 42.10

3

F

Gii:

UC1 = UC2 ------>2

1

2

1

)( CL

C

ZZR

UZ

=

2

2

2

2

)( CL

C

ZZR

UZ

---->

2

1

22

C

L

Z

ZR - 2

1C

L

Z

Z +1 =

2

2

22

C

L

Z

ZR - 2

2C

L

Z

Z +1 ------>

(R2 + 2LZ )( 2

1

1

CZ-

2

2

1

CZ) = 2ZL(

1

1

CZ -

1

1

CZ) ------>

1

1

CZ +

1

1

CZ =

22

2

L

L

ZR

Z

(1)

UC = 22 )( CL

C

ZZR

UZ

= UCmax khi y = 2

22

C

L

Z

ZR - 2

C

L

Z

Z +1 = ymin ------>

y = ymin khi ZC = L

L

Z

ZR 22 ------>

CZ

1=

22

L

L

ZR

Z

(2)

T (1) v (2)-----> 1

1

CZ +

1

1

CZ =

CZ

2------> C =

2

21 CC = 43.10

4

(F). Chn p n A

Cu 24: Mt on mch gm cun cm c t cm L v in tr thun r mc ni tip vi t in c in dung C thay i c. t vo hai u mch mt hiu in th xoay chiu c gi tr hiu dng U v

tn s f khng i. Khi iu chnh in dung ca t in c gi tr C = C1 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 1 2 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. 2

52 3 os 100 ( )

12i c t A

B.

2

52 2 os 100 ( )

12i c t A

C. 2 2 2 os 100 ( )3

i c t A

D. 2 2 3 os 100 ( )3

i c t A

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

Zd = Z1 -----> 2

1

2 )( CL ZZr = 22

LZr --------> ZL ZC1 = ZL

-----> ZL = 2

1CZ (1)

Zd = ZC1 -----> r2 +ZL

2 = ZC!

2 ----->r

2 =

4

3 21CZ -------> r = 2

3 21CZ (2)

tan1 = 3

1

2

3

2

1

1

1

1

C

CC

CL

Z

ZZ

r

ZZ----> 1 = -

6





Khi C = C2 UC = UCmax khi ZC2 = 11

2

1

22

2

2

C

C

C

L

L ZZ

Z

Z

Zr

Khi Z2 = 12

1

2

112

1

2

2

2 33)22

(4

3)( CCCCCL ZZZ

ZcZZZr

tan2 = 3

2

3

22

1

1

1

2

C

CC

CL

Z

ZZ

r

ZZ----> 2 = -

3

U = I1Z1 = I2Z2 -------> I2 = I1 23

32

3

1

2

1 I

Z

Z(A)

Cng dng in qua mch

i2 = I2 )364

100cos(2

t = 2 )12

5100cos(2

t (A). Chn p n B

Cu 25. t vo hai u mch in gm hai phn t R v C vi R = 100 mt ngun in tng hp c

biu thc u = 100 + 100cos(100t + /4) (V). Cng sut ta nhit trn in tr R c th l:

A. 50W. B. 200W. C. 25W, D, 150W

Gii: Ngun in tng hp gm ngun in mt chiu c U1chieu = 100V v ngun in xoay chiu c

in p hiu dng U = 50 2 (V). Do on mch cha t C nn dng in 1 chiu khng qua R. Do

cng sut ta nhit trn R < Pmax (do Z > R)

P = I2R <

R

U 2 =

100

)250( 2 = 50W. Chn p n C: P = 25W.

Cu 26: Mt mch tiu th in l cun dy c in tr thun r = 8 ,tiu th cng sut P=32W vi h s

cng sut cos = 0,8 .in nng c a t my pht in xoay chiu 1 pha nh dy dn c in tr R=

4.in p hiu dng 2 u ng dy ni my pht l

A.10 5 V B.28V C.12 5 V D.24V

Gii: Dng in qua cun dy I = r

P = 2A;

Ud = cosI

P = 20V , I =

d

d

Z

U=

dZ

20-----> Zd =

2

20 = 10

Zd = 22

LZr -----> ZL = 22 rZ L = 6

I = Z

U-----> U = IZ = I 22)( LZRr = 2

22 612 = 12 5 (V). Chn p n C

Cu 27 Cho on mch xoay chiu RLC mc ni tip.t vo 2 u mch 1 in p xoay chiu c tn s

thay i c.Khi tn s ca in p 2 u mch l f0 =60Hz th in p hiu dng 2 u cun cm thun t cc i .Khi tn s ca in p 2 u mch l f = 50Hz th in p 2 u cun cm l

uL=UL 2 cos(100t + 1 ) .Khi f = f th in p 2 u cun cm l uL =U0L cos(t+2 ) .Bit UL=U0L

/ 2 .Gi tr ca bng:

A.160(rad/s) B.130(rad/s) C.144(rad/s) D.20 30 (rad/s)

Gii: UL = IZL = 22 )

1(

CLR

LU





UL =ULmax khi y = 2

22 )1

(

C

LR

= ymin

-------> 2

0

1

=

2

2C(2

C

L-R

2) (1) Vi 0 = 120 rad/s

Khi f = f v f = f ta u c U0L = UL 2 Suy ra UL = UL ------>

22 )

1(

CLR

= 22 )

'

1'(

'

CLR

------>

2 [ 22 )'

1'(

CLR

] = 2 [ 22 )

1(

CLR

]

( 2 -2 )( 2C

L-R

2) =

2

1

C(

2

2

'

-

2

2'

) =

2

1

C( 2 -2 )(

2'

1

+

2

1

)

-----> C2 ( 2

C

L-R

2) =

2'

1

+

2

1

(2) Vi = 100 rad/s

T (1) v (2) ta c 2

0

2

=

2'

1

+

2

1

-------> 2 =

2

0

2

2

0

2

2

= 2

0

2

0

2

-------> =

2222 120100.2

120.100

= 160,36 rad/s. Chn p n A

Cu 28. t in p xoay chiu u = 100 6 cos(100t) (V); vo hai u on mch mc ni tip gm in

tr thun R, cun cm thun c t cm L v t in c in dung C thay i c. iu chnh C in p hiu dng hai u t t gi tr cc i th thy gi tr cc i bng 200 V. in p hiu dng

hai u cun cm l bao nhiu vn?

Gii:

UC = UCmax = 200 (V) khi ZC = L

L

Z

ZR 22 ----->

ULUC = UR2 + UL

2 ------.> UR

2 + UL

2 =200UL

U2 = UR

2 +(UL UC)

2 -------> (100 3 )2 = UR

2 + UL

2 +200

2 400UL

-----> 30000 = 200UL + 40000 400UL ----> UL = 50 (V)

Cu 29. Mt cun dy khng thun cm ni tip vi t in C trong mch in xoay chiu c in p

0. osu U c t (V) th dng in trong mch sm pha hn in p l 1 , in p hiu dng hai u cun dy l

30V. Bit rng nu thay t C bng t 'C 3C th dng in trong mch chm pha hn in p l 2 12

v in p hiu dng hai u cun dy l 90V. Bin 0 ?U

Gii: Ud1 = 30 (V)

Ud2 = 90 (V) ----> 1

2

d

d

U

U= 3 ----> I2 = 3I1 -----> Z1 = 3Z2 -------.Z1

2 = 9Z2

2

R2 + (ZL ZC1)

2 = 9R

2 + 9(ZL -

3

1CZ )2 ----->2(R2 +ZL2 ) = ZLZC1 -----> R2 + ZL

2 =

2

1CL ZZ





11

d

d

Z

U=

1Z

U-------> U = Ud1

1

1

dZ

Z= Ud1 22

1

2

1

22 2

L

CLCL

ZR

ZZZZR

= Ud1 3

2

?

1 Z

ZC (*)

tan1 = R

ZZ CL 1 ; tan1 = R

ZZ CL 2 = R

ZZ CL

3

1

2 12

-----> 1 + 2 =

2

-----> tan1 tan2 = -1 ( v 1 < 0)

R

ZZ CL 1

R

ZZ CL

3

1

= -1------>(ZL ZC1)(ZL - 3

1CZ ) = - R2 ------->

R2 + ZL

2 4ZL

3

1CZ + 3

2

1CZ = 0 --------> 2

1CL ZZ 4ZL3

1CZ + 3

2

1CZ = 0 ---> 3

2

1CZ - 6

5 1CL ZZ = 0

---->3

1CZ - 6

5 LZ = 0 ----> ZC1 = 2,5ZL (**) ------> U = Ud1 32

?

1 Z

ZC = Ud1 2

Do U0 = U 2 = 2Ud1 = 60V.

Cu 30. Ni hai cc ca mt my pht in xoay chiu mt pha vo hai u on mch AB gm in tr

thun R = 30 , mc ni tip vi t in C. B qua in tr cc cun dy ca my pht. Khi r to quay

vi tc n vng /pht th cng hiu dng trong on mch l 1A. . Khi r to quay vi tc 2n

vng /pht th cng hiu dng trong on mch l 6 A. Nu r to quay vi tc 3n vng /pht th

dung khng ca t in l:

A. 4 5 () B. 2 5 () C. 16 5 () D. 6 5 ()

Gii: I = Z

U=

Z

E

Vi E l sut in ng hiu dng gia hai cc my pht: E = 2 N0 = 2 2fN0 = U ( do r = 0) Vi f = np n tc quay ca roto, p s cp cc t

Z = 22

2 1

CR

Khi n1 = n th 1 = ; I1 = 1Z

E; ZC1 = ZC =

C

1

Khi n2 = 2n th 2 = 2; ZC2 = ZC1 /2 = ZC /2 ----> I2 = 2Z

E

2

1

I

I=

3

1

E

E

1

2

Z

Z=

2

1

1

2

Z

Z------->

2

1

22

2

2

4

C

C

ZR

ZR

=2

1

I

I=

6

1------> 6R

2 + 1,5 2CZ = 4R

2 +4 2CZ

2,52

CZ = 2R2 ------>

2

CZ = 2R2/2,5 = -----> ZC =

5

2R = 12 5 ()

- Khi n3 = 3n th 3 = 3; ZC3 = ZC /3 = 4 5 (). Chn p n A

Cu 31: Mch in xoay chiu, gm in tr thun R, cun dy thun cm c t cm L v t in c in dung C mc ni tip. t vo 2 u on mch mt in p xoay chiu u tn s 1000Hz. Khi mc 1

ampe k A c in tr khng ng k song song vi t C th n ch 0,1A. Dng in qua n lch pha so





vi in p hai u on mch gc /6 rad. Thay ampe k A bng vn k V c in tr rt ln th vn k

ch 20 V, in p hai u vn k chm pha hn in p hai u on mch /6 rad. t cm L v in

tr thun R c gi tr:

A. 3 /(40)(H) v 150 B. 3 /(2)v 150

C. 3 /(40) (H) v 90 D. 3 /(2)v 90

Gii:

Khi mc ampe k mch RL: I1 = 22

LZR

U

= 0,1 (A). Lc ny u sm pha hn i;

tan1 = R

Z L = tan6

=

3

1---> ZL =

3

R (1) v U = I1

22

LZR = 3

2,0 R (V) (2)

Khi mc vn k mch RLC: UC = UV = 20V

2 = -2

- (-

6

) = -

3

tan2 =

R

ZZ CL = - tan3

= - 3 ----> ZC ZL = R 3

----> ZC = R 3 + 3

R =

3

4R; Z2 =

22 )( CL ZZR = 2R

UC = 2Z

UZC = 3

2U----->

3

2U = 20 ---> U =

3

2,0 R= 10 3 ----> R = 150 ()

ZL = 3

R = 50 3 ----->2fL = 50 3 -----> L =

1000.2

350

=

.40

3

(H)

Chn p n A: L =.40

3

(H) ; R = 150 ()

Cu 32. Cho mch in nh hnh v: uAB = Uocost; in p hiu dng UDH = 100V; hiu in th tc thi

uAD sm pha 150o so vi hiu in th uDH, sm pha 105

o so vi hiu in th uDB v sm pha 90

o so vi

hiu in th uAB. Tnh Uo?

A. Uo = 136,6V. B. Uo = 139,3V. C. oU 100 2V . D. Uo = 193,2V.

Gii:

V gin nh hnh v. t lin tip cc vect

UAD ; UDH ; UHB

UAB = UAD + UDH + UHB Tam gic DHB vung cn.

UHB = UDH = 100V

UDB = 100 2 (V)

Tam gic ADB vung ti A c gc D = 75

0 ----->

UAB = UDB sin750 = 100 2 sin75

0

U0 = UAB 2 = 200sin750 = 193,18V

Hay U0 = 193,2 V

Chn p n D

Cu 33: Dng in i = 24cos t (A) c gi tr hiu dng l bao nhiu?

A D H B

300 45

0

H B

D

A





Gii: Ta c i = 24cos t = 2cos2t + 2 (A) Dng in qua mch gm hai thnh phn

- Thnh phn xoay chiu i1 = 2cos2t, c gi tr hiu dng I1 = 2 (A)

- Thnh phn dng in khng i I2 = 2 (A) C hai kh nng : a. Nu trong on mch c t in th thnh phn I2 khng qua mch. Khi gi tr hiu dng ca dng

in qua mch I = I1 = 2 (A) b. Nu trong mch khng c t th cng su ta nhit trong mch

P = P1 + P2 = I12R + I2

2 R = I

2R --------> I = 622

2

1 II (A)

Cu 34.

on mch AB gm mt ng c in mc ni tip vi mt cun dy. Khi t vo hai u AB mt in

p xoay chiu th in p hai u ng c c gi tr hiu dng bng U v sm pha so vi dng in l 12

.

in p hai u cun dy c gi tr hiu dng bng 2U v sm pha so vi dng in l 12

5. in p hiu

dng gia hai u on mch AB ca mng in l :

A. U 5 . B. U 7 . C. U 2 . D. U 3 .

Gii: Gi u1,u2 l in p gia hai u ng c v cun dy

u1 = U 2 cos(t + 12

). ; u2 = 2U 2 cos(t +

12

5).

T gin ta tnh c 2

ABU = U2

+ 4U2 - 2.2U

2 cos 120

0 = 7U

2

UAB = U 7 . Chn p n B

Cu 35: Cho mch xoay chiu R,L,C, c cun cm thun, L thay i c.iu chnh L thy ULmax= 2URmax.

Hi ULmax gp bao nhiu ln UCmax?

A 2/ 3 . B. 3 /2. C. 1/ 3 . D. 1/2

Gii:

Ta c UR = URmax = U v UC = UCmax = R

UZC khi trong mch c cng hng ZL = ZC

UL = ULmax khi ZL = C

C

Z

ZR 22 : (*)

ULmax =

122

22

L

C

L

C

Z

Z

Z

ZR

U=

L

C

Z

Z

U

1

= 2URmax = 2U

-----> 1 -L

C

Z

Z =

4

1-----> ZL =

3

4ZC (**)

T (*) v (**) suy ra ZC = R 3 Do UCmax = R

UZC = U 3

1200





Vy max

max

C

L

U

U=

3

2

U

U =

3

2, Chn p n A

Cu 36: Cho mch in xoay chiu RLC mc ni tip. in p xoay chiu t vo hai u on mch c

biu thc u = U 2 cost tn s gc bin i. Khi = 1 = 40 rad/s v khi = 2 = 360 rad/s th cng dng in hiu dng qua mch in c gi tr bng nhau. cng dng in trong mch t

gi tr ln nht th tn s gc bng

A 100(rad/s). B 110(rad/s). C 200(rad/s). D 120(rad/s).

Gii: I1 = I1 ----> Z1 = Z1 ------> (ZL1 ZC1)

2 = (ZL2 ZC2)

2

Do 1 2 nn (ZL1 ZC1) = - (ZL2 ZC2) ----> ZL1 + ZL2 = ZC1 + ZC2

(1 + 2)L = C

1 (

1

1

+

2

1

) ------> LC =

21

1

(*)

Khi I = Imax; trong mch c cng hng LC = 21

(**)

T (*) v (**) ta c = 21 = 120(rad/s). Chn p n D

Cu 37: Cho on mch xoay chiu mc ni tip gm on dy khng thun cm (L,r) ni vi t C Cun dy l mt ng dy c qun u vi chiu di ng c th thay i c.t vo 2 u mch mt HDT

xoay chiu.Khi chiu di ca ng dy l L th HDT hai u cun dy lch pha /3 so vi dng in. HDT

hiu dng 2 u t bng HDT hiu dng 2 u cun dy v cng dng in hiu dng trong mch l I..Khi tng chiu di ng dy ln 2 ln th dng in hiu dng trong mch l:

A. 0,685I B. I C. 2I/ 7 D. I/ 7

Cc thy cho e hi Khi tng chiu di ng dy ln 2 ln th L tng 2 ln th R c tng ko

Gii: Khi tng chiu di ng dy ln 2 ln (L tng 2 ln); th s vng dy ca mt n v chiu di n gim i 2 ln, t cm ca ng dy L gim 2 ln nn cm khn ZL gim hai ln cn in tr R ca ng dy

khng i.

Ta c : tand = R

Z L = tan3

= 3 -----> ZL = R 3 ----> Zd = 2R

Ud = UC -------> ZC = Zd = 2R. --------> Z = 2R 32

Do I = 322 R

U (*)

Sau khi tng chiu di ng dy ZL = 2

LZ = 2

3R

I= 22 )'( CL ZZR

U

=

22 )22

3( RR

R

U

= 3823

2

R

U (**)

I

I '=

3823

324

= 0,6847 --------> I = 0,685I. Chn p n A

Cu 38 : 1 on mch RLC . khi f1 =66 Hz hoc f2 =88 Hz th hiu in th 2 u cun cm khng i , f

= ? th ULmax





A 45,21 B 23,12 C 74,76 D 65,78

Gii: UL = IZL = 22 )

1(

CLR

LU

UL1 = UL2 -----> 2

1

1

2

1

)1

(C

LR

= 2

2

2

2

2

)1

(C

LR

2

1

1

+

2

2

1

= 42C2(2

C

L - R

2 ) (*)

UL = ULmax khi 22 )

1(

CLR

LU

=

2

22 )1

(

C

LR

UL

c gi tr max

hay y = 2

22 )1

(

C

LR = ymin ------> 2

2

= 42C2(2

C

L - R

2 ) (**)

T (*) v (**) ta c 2

2

=

2

1

1

+

2

2

1

hay

2

2

f =

2

1

1

f+

2

2

1

f

f = 2

2

2

1

21 2

ff

ff

= 74,67 (Hz). Chn p n C

Cu 39:

Cho mch in nh hnh v. in p t vo hai u on mch c gi tr hiu dng khng i nhng tn s thay i c. Khi tn s f = f1 th h s cng sut trn on AN l k1 = 0,6, H s cng sut trn ton mch l k = 0,8. Khi f = f2 = 100Hz th cng sut trn ton mch cc i. Tm f1 ?

A. 80Hz B. 50Hz C. 60Hz D. 70Hz

Gii: cos1 = 0,6 ------> tan1 = 3

4

tan1 = rR

Z L

=

3

4 -----> ZL =

3

4(R + r) (*)

cos = 0,8 ------> tan = 4

3

tan = rR

ZZ CL

=

4

3 ------> ZL ZC =

4

3(R +r) (**)

C

L

Z

Z =

2

1 LC v 2

2 LC = 1 ------> C

L

Z

Z=

2

2

2

1

=

2

2

2

1

f

f-----> f1 = f2

C

L

Z

Z

* Khi ZL ZC = 4

3(R +r) ------> ZC =

12

7(R +r) --->

C

L

Z

Z =

7

16

----> f1 = 7

4 2f = 151,2 Hz Bi ton v nghim

** Khi ZL ZC = - 4

3(R +r) ------> ZC =

12

25(R +r) -->

C

L

Z

Z =

25

16

C L; r R

N B A M





f1 = f2 C

L

Z

Z = f2.

5

4 = 80Hz. Chn p n A

Cu 40: t mt in p 2 osu U c t (U, khng i) vo on mch AB ni tip. Gia hai im AM l mt bin tr R, gia MN l cun dy c r v gia NB l t in C. Khi R = 75 th ng thi c bin

tr R tiu th cng sut cc i v thm bt k t in C no vo on NB d ni tip hay song song vi t in C vn thy UNB gim. Bit cc gi tr r, ZL, ZC, Z (tng tr) nguyn. Gi tr ca r v ZC l: A. 21 ; 120 . B. 128 ; 120 . C. 128 ; 200 . D. 21 ; 200 .

Gii: PR = I2R =

22

2

)()( CL ZZrR

RU

=

rR

ZZrR

U

CL 2)( 22

2

PR = PRmax khi R2 = r

2 + (ZL ZC)

2. (1)

Mt khc lc R = 75 th PR = PRmax ng thi UC = UCmax

Do ta c: ZC = L

L

Z

ZrR 22)( =

LZ

rR 2)( + ZL (2)

Theo bi ra cc gi tr r, ZL ZC v Z c gi tr nguyn ZC nguyn th (R+r)

2 = nZL (3) (vi n nguyn dng)

Khi ZC = n + ZL ------> ZC ZL = n (4)

Thay (4) vo (1) r2 + n

2 = R

2 = 75

2. (5)

Theo cc p n ca bi ra r c th bng 21 hoc 128. Nhng theo (5): r < 75

Do vy r c th r = 21 T (5) -----> n = 72.

Thay R, r, n vo (3) ---> ZL = 128 Thay vo (4) ----> ZC = 200

Chn p n D: r = 21 ; ZC = 200 .

Cu 41: Mt mch tiu th in l cun dy c in tr thun r= 8 m, tiu th cng sut P=32W vi h

s cng sut cos=0,8. in nng c a t my pht in xoay chiu 1 pha nh dy dn c in tr

R= 4. in p hiu dng 2 u ng dy ni my pht l

A.10 5 V B.28V C.12 5 V D.24V

Gii: cos =dZ

r=0,8 -----> Zd = 10 v ZL = 6,

Cng dng in qua mch I = r

P = 2 (A)

in p hiu dng 2 u ng dy ni my pht l

U = I 22)( LZrR = 2 22 612 = 12 5 (V) Chn p n C

Cu 42. Mch xoay chiu RLC gm cun dy c (R0, L) v hai t C1, C2. Nu mc C1//C2 ri ni tip vi cun dy th tn s cng hng l 1 = 48 (rad/s). Nu mc C1 ni tip C2 ri ni tip cun dy th tn s

cng hng l 2 = 100 (rad/s). Nu ch mc ring C1 ni tip cun dy th tn s cng hng l A = 70 rad/s B. = 50 rad/s C. = 74 rad/s D = 60 rad/s

Gii:

C// = C1 + C2; Cnt = 21

21

CC

CC

; =

LC

1-----> C =

L21

C// = L21

1

------> C1 + C2 =

L21

1

(*)





Cnt = L22

1

------->

21

21

CC

CC

=

L22

1

----> C1C2 =

L22

1

L21

1

=

22

2

2

1

1

L(**)

T (*) v (**)-------> C1 + 222

2

1

1

L 1

1

C =

L21

1

(***)

C1 = L2

1

(****)

Thay (****) vo (***) L2

1

+

22

2

2

1

2

L

L

=

L21

1

------>

2

1

+

2

2

2

1

2

=

2

1

1

-----> 22

2

1 + 4 = 22

2 -----> 4 - 222 + 22

2

1 = 0 (*****)

Phng trnh c hai nghim = 60 rad/s v = 80 rad/s Chn p n D

Cu 43 : Mch R, L, C ni tip . t vo 2 u mch in p xoay chiu u = U0cost (V), vi thay i c. Thay i UCmax. Gi tr UCmax l biu thc no sau y

A. UCmax = 2

C

2

L

U

Z1

Z

C. UCmax = 2

L

2

C

U.

Z1

Z

B. UCmax = 2 2

2U.L

4LC R C

D. UCmax = 2 2

2U

R 4LC R C

Gii:

UC = 22 )_( CL

C

ZZR

UZ

=

C

1

22 )1

(C

LR

U

= C

1

2

2242 1)2(CC

LRL

U

UC = UCmax khi 2 =

2

2

2

2

L

RC

L

v UCmax = C

1

2

42

4

4

L

RC

LR

U

= 224

2

CRLCR

LU

UCmax = 224

2CRLC

L

R

U

=

)4(4

22

2

2

CRLCL

R

U

=

)4 2

242

L

CR

L

CR

U

=

)4

1(12

242

L

CR

L

CR

U

=

22

)2

1(1L

CR

U

=

2

222

4

)2(

1L

CRC

L

U

=

22

4

22

4

)2(

1 CLL

RC

L

U

=

2241 CL

U

=

2

2

1C

L

Z

Z

U

Chn p n C.

Cu 44: Trong mt gi thc hnh mt hc sinh mun mt qut in loi 180 V - 120W hot ng bnh

thng di in p xoay chiu c gi tr hiu dng 220 V, nn mc ni tip vi qut mt bin tr.(coi





qut in tng ng vi mt on mch r-L-C ni tip) Ban u hc sinh bin tr c gi tr 70 th o thy cng dng in hiu dng trong mch l 0,75A v cng sut ca qut in t 92,8%.

Mun qut hot ng bnh thng th phi iu chnh bin tr nh th no? A. gim i 20 B. tng thm 12 C. gim i 12 D. tng thm 20

Gii : Gi R0 , ZL , ZC l in tr thun, cm khng v dung khng ca qut in. Cng su nh mc ca qut P = 120W ; dng in nh mc ca qut I. Gi R2 l gi tr ca bin tr khi

qut hot ng bnh thng khi in p U = 220V

Khi bin tr c gi tri R1 = 70 th I1 = 0,75A, P1 = 0,928P = 111,36W

P1 = I12R0 (1) ------> R0 = P1/I1

2 198 (2)

I1 = 2222

101 )(268

220

)()( CLCL ZZZZRR

U

Z

U

Suy ra

(ZL ZC )2 = (220/0,75)

2 2682 ------> ZL ZC 119 (3)

Ta c P = I2R0 (4)

Vi I = 22

20 )()( CL ZZRR

U

Z

U

(5)

P = 22

20

0

2

)()( CL ZZRR

RU

--------> R0 + R2 256 ------> R2 58

R2 < R1 ----> R = R2 R1 = - 12

Phi gim 12. Chn p n C Cu 45: t mt in p xoay chiu u vo hai u ca mt on mch gm in tr R mc ni tip vi mt t in c in dung C. in p t thi hai u in tr R c biu thc

uR = 50 2 cos(2ft + fi) (V). Vo mt thi im t no in p tc thi gia hai u on mch v hai

u in tr c gi tr u = 50 2 V v uR = -25 2 V. Xc nh in p hiu dng gia hai bn t in.

60 3 V. B. 100 V. C. 50 2 V. D. 50 3 V

Gii:

uR = 50 2 cos(2ft + ) (V). -----> UR = 50 (V)

Ti thi im t: u = 50 2 ;(V) uR = -25 2 (V) u = 2uR-----> Z = 2R

Z2 = R

2 + ZC

2 ------> ZC

2 = 3R

2 -----> ZC = R 3 ----->

UC = UR 3 = 50 3 (V) Chn p n D





Cu 46 : t mt in p u = 80cos(t) (V) vo hai u on mch ni tip gm in tr R, t in C v cun dy khng thun cm th thy cng sut tiu th ca mch l 40W, in p hiu dng UR = ULr = 25V; UC = 60V. in tr thun r ca cun dy bng bao nhiu? A. 15 B. 25 C. 20 D. 40

Gii: Ta c Ur

2 + UL

2 = ULr

2

(UR + Ur)2 + (UL UC)

2 = U

2

Vi U = 40 2 (V) Ur

2 + UL

2 = 25

2 (*)

(25+ Ur)2 + (UL 60)

2 = U

2 = 3200

625 + 50Ur + Ur2 + UL

2 -120UL + 3600 = 3200

12UL 5Ur = 165 (**) Gii h phng trnh (*) v (**) ta c

* UL1 = 3,43 (V) ----> Ur1 = 24,76 (V)

nghim ny loi v lc ny U > 40 2 * UL = 20 (V) ----> Ur = 15 (V)

Lc ny cos = U

UU rR = 2

1

P = UIcos -----> I = 1 (A)

Do r = 15 . Chn p n A

Cu 46: Mng in 3 pha c hiu in th pha l 120 V c ti tiu th mc hnh sao, cc ti c in tr l

R1 = R2 = 20 ; R3 = 40 . Tnh cng dng in trong dy trung ho.

A. 6 A B. 3 A C. 0 A D. 2 3 A

Gii: Dng in qua cc ti l I = R

U P I1 = I2 = 6A; I3 = 3 A

Dng in qua dy trung tnh i = i1 + i2 + i3 Dng phng php cng vc t ta c

I = I1 + I2 + I3

Gc gia i1, i2., i3 l 2 /3

t lin tip cc vc t cng dng in nh hnh v,

ta c tam gic u Theo hnh v ta c I = I3 = 3 A

Chn p n B: 3A

Cu 47: Cho mch in RLC, t in c in dung C thay i.

iu chnh in dung sao cho in p hiu dng ca t t gi tr cc i, khi in p hiu dng trn R l 75 V. Khi in

p tc thi hai u mch l 75 6 V th in p tc thi ca

on mch RL l 25 6 V in p hiu dng ca on mch l

A. 75 10 V. B. 75 3 V C. 150 V. D. 150 2 V

Gii: V gin vect nh hnh v. Ta thy

UC = UCmax khi = 900 tc khi uRL vung pha vi u

2

maxCU = U2 +

2

RLU

ULr

U

UC

UL

Ur UR

I3

I2 I1

I3

I

I1

I2

I

I3

I2

I1

O

C

UR

UR

L





Khi u = 75 6 V th uRL = 25 6 V ----> Z = 3ZRL hay U = 3URL

---> 2maxCU = U

2 + 2

RLU = 102

RLU .

Trong tam gic vung hai cnh gc vung U; URL; cnh huyn UC ng cao thuc cnh huyn UR ta c: U.URL = URUC

3 2RLU = 10 URLUR ----> 3URL = 10 UR = 75 10

----> URL = 25 10 (V). Do U = 75 10 (V). p n A

Cu 48: Mt mch tiu th in l cun dy c in tr thun r= 8, tiu th cng sut P=32W vi h s

cng sut cos=0,8. in nng c a t my pht in xoay chiu 1 pha nh dy dn c in tr R=

4. in p hiu dng 2 u ng dy ni my pht l

A.10 5 V B.28V C.12 5 V D.24V

Gii: cos =dZ

r=0,8 -----> Zd = 10 v ZL = 6,

Cng dng in qua mch I = r

P = 2 (A)

in p hiu dng 2 u ng dy ni my pht l

U = I 22)( LZrR = 2 22 612 = 12 5 (V) Chn p n C

Cu 49: Mt cun dy khng thun cm ni tip vi t in C thay i c trong mch in xoay chiu

c in p u = U0 cost (V). Ban u dung khng ZC, tng tr cun dy Zd v tng tr Z ton mch bng

nhau v u bng 100. Tng in dung thm mt lng C =

310.125,0 (F) th tn s dao ng ring

ca mch ny khi l 80 rad/s. Tn s ca ngun in xoay chiu bng:

A. 80 rad/s. B. 100 rad/s. C. 40 rad/s. . D.50 rad/s.

Gii: Do ZC = Zd = Z.--------> UC = Ud = U. = 100I

V gin vc t nh hnh v. ta suy bra UL = Ud/2 = 50I

---> 2ZL = -----ZL = 50

Vi I l cng dng in qua mch

ZL = L; ZC = C

1----->

C

L= CLZZ = 5000 (*)

= )(

1

CCL = 80-------> L(C+ C) =

2)80(

1

(**)

5000C(C+C) = 2)80(

1

---->

C2

+(C)C - 5000.)80(

12

= 0----> C2

+

310.125,0 C -

5000.)80(

12

= 0--->

C2

+ 8

10. 3C -

4.8

102

6

= 0 ------> C = 8

10. 3F

ZC = C

1 = 100 -----> =

CZC

1 = 80 rad/s. Chn p n A

UC

Ud

U

UL





Cu 50 Mt cun dy khng thun cm ni tip vi t in c in dung C thay i c trong mch

in xoay chiu c in p u = U0cost (V). Ban u dung khng ZC v tng tr ZLr ca cun dy v Z ca

ton mch u bng 100. Tng in dung thm mt lng C = 0,125.10-3/ (F) th tn s dao ng ring

ca mch ny khi l 80 rad/s. Tn s ca ngun in xoay chiu bng

A. 80rad/s B. 100rad/s C. 40rad/s D. 50rad/s

Gii: V gin vect

ZC = ZLr = Z = 100-----. ZL = 2

CZ = 50

ZL.ZC =C

L = 5000 (2) -----> L = 5.103C (*)

2

0 = )(

1

CCL -----> 5.10

3C

2 + 5.10

3C.C - 2

0

1

= 0

---> 5.103C

2 + 5.10

3

310.125,0 .C -

2280

1

= 0

----> 5.103C

2 +

.625,0C -

6400

1 = 0

-----> C =

.25,1.10

-4 (F);

------>. = CZC

1 =

410.25,1

..100

1

= 80 rad/s. Chn p n A

Cch 2 ; ZLr = Z ------> r2 + ZL

2 = r

2 + (ZL ZC)

2 -------> ZC = 2ZL -----> ZL =

2

CZ = 50

Khi tng thm C = C + C th ZC = ZL = 2

CZ -------> C = 2C ----> C = C =

.25,110

-4F

------>. = CZC

1 =

410.25,1

..100

1

= 80 rad/s. Chn p n A

Cu 51 : t in p xoay chiu u U 2 cos(100 t)V vo on mch RLC. Bit R 100 2 , t in c

in dung thay i c. Khi in dung t in ln lt l

251 C (F) v

3

1252 C (F) th in p hiu

dng trn t c cng gi tr. in p hiu dng trn in tr R t cc i th gi tr ca C c th l:

A.

50C (F). B.

3

200C (F)., C.

20C (F). D.

3

100C (F)

Gii

Ta c 112 2

1( )

CC

L C

UZU

R Z Z

22

2 2

2( )

CC

L C

UZU

R Z Z

UC1 = UC2 --------->> 2 2

1 2

2 2 2 2

1 2( ) ( )

C C

L C L C

Z Z

R Z Z R Z Z

UL

Ud

UC

U

C





ZC1 = 400; ZC2 = 240

-----> R2 + ZL

2 =

21

212

CC

CCL

ZZ

ZZZ

=

240400

240.400.2

LZ = 300ZL

in p hiu dng trn in tr R t cc i th trong mch c cng hng ZL = ZC

Thay R =100 2 ; :

- ZC2

- 300ZC +20000 = 0

Phng trnh c hai nghim : ZC = 200 v ZC = 100

Khi ZC = 200 th C = 410 50

2F F

Khi ZC = 100 th C = 410 100

F F

Chn p n A

Cu 52: t vo hai u mch in RLC ni tip mt hiu in th xoay chiu c gi tr hiu dng khng i th hiu in th hiu dng trn cc phn t R, L v C u bng nhau v bng 20V. Khi t b ni tt th

in p dng hai u in tr R bng:

A. 10V. B. 10 2 V. C. 20V. D. 20 2 V.

Gii: Do UR = UL = UC trong mch c cng hng , nn U = UR = 20V

Khi t b ni tt UL = UR = 2

U = 10 2 (V). Chn p n B

Cu 53: Cho mch in xoay chiu khng phn nhnh RLC c tn s thay i c.Gi f0 ;f1 ;f2 ln lt

cc gi tr tn s lm cho hiu in th hiu dung hai u in tr cc i,hiu in th hiu dung hai u cun cm cc i,hiu in th hiu dung hai u t in cc i.Ta c :

A.f0 = 2

1

f

f B. f0 =

1

2

f

f C.f1.f2 = f0

2 D. f0 = f1 + f2

Gii: UR = Urmax khi trong mch c cng hng in ZL = ZC -----> f02 =

LC24

1

(1)

UC = UCmax khi ZC2 = 2

2

2

2

L

L

Z

ZR -----> R

2 = ZL2ZC2 ZL2

2 (*)

UL = ULmax khi ZL1 = 1

2

1

2

C

C

Z

ZR -----> R

2 = ZL1ZC1 ZC1

2 (**)

T (*) v (**) suy ra ZL1ZC1 ZC12 = ZL2ZC2 ZL2

2

ZL.ZC = C

L suy ra ZC1 = ZL2 ----->

Cf12

1

= 2f2L -----> f1f2 =

LC24

1

(2)

T (1) v (2) ta c f1f2 = f02 Chn p n C

Cu 54 : Mt mch in xoay chiu gm AM ni tip MB. Bit AM gm in tr thun R1, t in C1,

cun dy thun cm L1 mc ni tip. on MB c hp X, bit trong hp X cng c cc phn t l in tr thun, cun cm, t in mc ni tip nhau. t in p xoay chiu vo hai u mch AB c tn s 50Hz

v gi tr hiu dng l 200V th thy dng in trong mch c gi tr hiu dng 2A. Bit R1 = 20 v nu

thi im t (s), uAB = 200 2 V th thi im ( t+1/600)s dng in iAB = 0(A ) v ang gim. Cng sut ca on mch MB l:

A. 266,4W B. 120W C. 320W D. 400W





Gii:

Gi s in p t vo hai u mch c biu thc u = U 2 cost = 200 2 cos100t (V). Khi cng

dng in qua mch c biu thc i = 2 2 cos(100t -) vi gc lch pha gia u v i

Ti thi im t (s) u = 200 2 (V) -----> cost = 1. Do cng dng in ti thi im ( t+1/600)s

i = 0 ------> i = 2 2 cos[100(t + 600

1) -] = 0------> cos(100t +

6

-) = 0

----> cos100t.cos(6

-) - sin100t.sin(

6

-) = 0 -----> cos(

6

-) = 0 (v sin100t = 0 )--->

= 6

-

2

= -

3

----->

Cng sut ca on mch MB l: PMB = UIcos - I2R1 = 200.2.0,5 4. 20 = 120W. Chn p n B

Cu 55: Trong li in dn dng ba pha mc hnh sao, in p mi pha l u1 = 220 2 cos(100t) (V) ,

u2 = 220 2 cos(100t + 3

2) (V), u3 = 220 2 cos(100t -

3

2) (V), . Bnh thng vic s dng in

ca cc pha l i xng v in tr mi pha c gi tr R1=R2=R3 = 4,4. Biu thc cng dng in trong dy trung ho tnh trng s dng in mt cn i lm cho in tr pha th 1 v pha th 3 gim i mt na l:

A. i = 50 2 cos(100t +3

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

C. i = 50 2 cos(100t +3

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

3

) (A)

Gii: Do cc ti tiu th l cc in tr thun nn u v i lun cng pha

Khi mt cn i cc pha

I1 = I3 = 2,2

220 = 100 (A)

I2 = 4,4

220 = 50 (A). V gin vc t :

I0 = I1 + I2 + I3 = I13 + I2

I13 = I1 = I3 = 100A

I0 = I13 I2 = 50 (A)

0 = - 3

Do biu thc cng dng in trong dy trung ho

i = 50 2 cos(100t -3

) (A) Chn p n D

Cu 56: on mch AB gm cun dy thun cm c t cm L c th thay i mc gia A v M, in

tr thun mc gia M v N, t in mc gia N v B mc ni tip. t vo hai u A , B ca mch in mt in p xoay chiu c tn s f, in p hiu dng U n nh. iu chnh L c uMB vung pha vi

uAB, sau tng gi tr ca L th trong mch s c A. UAM tng, I gim. B. UAM gim, I gim. C. UAM gim, I tng. D. UAM tng, I tng.

Gii: V gin vect nh hnh v. Theo L hm sin

I1

-2/3

2/3

- /3

I0

I3

I2

I1

3





sinAMU =

sinABU ------> UAM =

sin

sinABU

Do gc , UAB xc nh nn UAM c gi tr ln nht khi = 900

Tc l khi uMB vung pha vi uAB th UAM c gi tr ln nht.

Do vy khi tng L th UAM gim Cng dng in qua mch

I = 22 )

1(

CLR

U AB

ta thy khi L tng th mu s tng do I gim

Chn p n B: UAM gim, I gim

Cu 57.t mt in p xoay chiu u = U0cos100t (V) vo hai u ca mt in tr thun R th trong

mch c dng in vi cng hiu dng I. Nu t in p vo hai u on mch gm in tr

thun R mc ni tip vi mt it bn dn c in tr thun bng khng v in trngc rtln th cng hiu dng ca dng in trong mch bng

A. 2I B.I 2 C.I D. I/ 2

Gii: Xt thi gian mt chu k

Lc ch c in tr thun R : P = I2R = 2

2

0 RI

Lc mc thm it, dng in qua Rch trong mt na chu ki P = I2R = 2

P=

4

2

0 RI

------> I

I ' =

2

1 -----> I =

2

I. Chn p n D

Cu 58: t mt in p xoay chiu c gi tr hiu dng U v tn s f khng i vo hai u on mch gm bin tr R mc ni tip vi t in c in dung C. Gi in p hiu dng gia hai u bin tr, gia

hai u t in v h s cng sut ca on mch khi bin tr c gi tr 1R ln lt l 1 1 1, , osR CU U c . Khi

bin tr c gi tr 2R th cc gi tr tng ng ni trn ln lt l 2 2 2, , osR CU U c bit rng s lin h:

1

2

0,75R

R

U

U v 2

1

0,75C

C

U

U . Gi tr ca 1osc l: A. 1 B.

1

2 C. 0,49 D.

3

2

Gii:

2

1

R

R

U

U =

4

3------> UR2 =

9

16UR1 (*)

1

2

C

C

U

U =

4

3------> UC2 =

16

9UC1 (**)

U2 =

2

1RU + 2

1CU = 2

2RU + 2

2CU = (9

16)2 2

1RU + (16

9)

2 21CU -------->

(9

16)2 2

1RU - 2

1RU = 2

1CU - (16

9)2 2

1CU -------->2

1CU = (9

16)

2 21RU ------>

UAM

UMB

UA

B





U2 = 2

1RU + 2

1CU = [(1 + (9

16)2] 2

1RU --------> U = 9

169 22 UR1

cos1 = U

U R1 = 22 169

9

= 0,49026 = 0,49. Chn p n C

Cu 59: t mt in p u = U 2 cos(110t /3) (V) vo hai u on mch mc ni tip gm in tr R (khng i), cun dy thun cm c L = 0,3 H v mt t in c in dung C thay i c. Cn phi iu chnh in dung ca t n gi tr no in tch trn bn t in dao ng vi bin ln nht?

A. 26,9 F. B. 27,9 F. C. 33,77 F. D. 23,5 F

Gii: Gi s in tch gia hai bn cc t in bin thin theo phg trnh q = Q0 cos(t + )

Khi dng in qua mch c biu thc: i = q = -Q0sin(t + ) = I0cos((t + + 2

)

Vi I0 = Q0-----> Q0 c gi tr ln nht khi I0 c gi tr ln nht

---> I = Ic tc l khi trong mch c s cng hng ---- > ZC = ZL

Do C = L2

1

=

3,0.)110(

12

= 27,9 F. Chn p n B

Cu 60: Cho on mch xoay chiu RLC mc ni tip. Cho cc gi tr R = 60 m; ZC =600 m; ZL=140

m.t vo hai u on mch mt in p xoay chiu c tn s f = 50Hz. Bit in p gii hn (in p nh thng) ca t in l 400V. in p hiu dng ti a c th t vo hai u on mch t in khng b nh thng l

A. 400 2 V. B. 471,4 V. C. 666,67 V. D. 942,8 V.

Gii: Tng tr Z = 22 )( CL ZZR = 215200 = 464 ()

UC = Z

UZC =

464

600U UCmax = 400 2 (V) ------> U

600

464400 2 = 437,5 (V).

Chn p n A

Cu 61. Ni hai cc ca mt my pht in xoay chiu mt pha c 5 cp cc t vo hai u on mch

.AB gm in tr thun R=100, cun cm thun c t cm L=6

41H v t in c in dung C =

3

10 4F. Tc rto ca my c th thay i c. Khi tc rto ca my l n hoc 3n th cng

dng in hiu dng trong mch c cng gi tr I. Gi tr ca n bng

A. 10vng/s B. 15 vng/s C. 20 vng/s D. 5vng/s

Gii: Sut in ng cc i ca ngun in: E0 = N0 = 2fN0 => U = E = 2

0E (coi in tr trong

ca my pht khng ng k). Cng dng in qua mch I = Z

U

Vi f = np n tc quay ca roto, p s cp cc t Do I1 = I2 ta c:

2

1

1

2

2

1

)1

(C

LR

=2

2

2

2

2

2

)1

(C

LR

-------> ])1

([ 2

2

2

22

1C

LR

= ])1

([ 2

1

1

22

2C

LR





---> C

L

CLR 2122

2

2

122

2

2

1

22

1 2

=

C

L

CLR 2222

1

2

222

2

2

1

22

2 2

---> )2)(( 2222

1C

LR = )(

12

2

2

1

2

1

2

2

2

C =

2

2

2

1

2

1

2

2

2

1

2

2

2

))((1

C

-----> 2

2

2

1

11

= (2

C

L- R

2 )C

2 =

2

3

9

10.4

(*)

= 2f = 2np

2

2

2

1

11

=

224

1

p(

2

2

2

1

11

nn ) =

224

1

p(

2

1

n +

29

1

n) =

22236

10

np =

222 536

10

n (**)

-------> 222 536

10

n =

2

3

9

10.4

------> n2 =

22 536

10

3

2

10.4

9

= 25----->

n = 5 vng /s. Chn p n D

Cu 62: Cho on mch R,L,C ni tip, in p gia hai u on mch

u = 220 2 cos2ft (V); R =100; L l cun cm thun, L = 1/(H); T in c in dung C v tn s f thay i c. iu chnh C= CX, sau iu chnh tn s, khi f = fX th in p hiu dng gia hai bn t

C t cc i; gi tr ln nht ny gp 5/3 ln in p hiu dng gia hai u on mch. Gi tr CX, v tn s fX bng

Gii:

UC = 22 )_( CL

C

ZZR

UZ

=

C

1

22 )1

(C

LR

U

= C

1

2

2242 1)2(CC

LRL

U

UC = UCmax khi 2 =

2

2

2

2

L

RC

L

v UCmax = C

1

2

42

4

4

L

RC

LR

U

= 224

2

CRLCR

LU

=

3

5U

----> 6L = 5R 224 CRLC ------> R2C2 4LC +

2

2

25

36

R

L

-----> C = 2

6,12

R

LL = (21,6).

410F ------> c 2 gi tr ca C: C1 =

410..6,3 F v C2 =

510.4 F

2 = 2

2

2

2

L

RC

L

= LC

1 -

2

2

2L

R > 0 -----> C <

2

2

R

L =

410.2 F.-----> loi nghim C1

CX = C2 =

510.4 F ----> 2 =

2

1

LC -

2

2

2L

R =

4

10 25 -

2

100 22= 2.10

42 -----> = 100 2 rad/s

Do fX = 50 2 Hz p s CX =

510.4 F v fX = 50 2 Hz

Cu 63: Cho mch in gm cun dy c in tr hot ng R ni tip t C. t vo hai u mch in

mt in p xoay chiu n nh u = U 2 cost. Khi C = C0 th in p hiu dng gia hai u cun dy ln nht bng 2U. Vi gi tr no ca C th UC t cc i?

A. C = 03C

4. B. C = 0

C

2. C. C = 0

C

4. D. C = 0

C

3.





Gii:

Ta c Ud = I22

LZR ; Ud = Udmax khi I = Imax mch c cng hng ZL = ZC0 (*)

Udmax = 2U----> Zd = 2Z = 2R ( v ZL = ZC0)-----> R2 + ZL

2 = 4R

2 ----> R =

3

LZ = 3

0CZ (**)

UC = UCmax khi ZC = L

L

Z

ZR 22 =

0

2

0

2

0

3

C

CC

Z

ZZ

= 3

4 0CZ -

---> ZC = 3

4 0CZ -----> C = 4

3 0C Chn p n A

Cu 64: t mt in p xoay chiu )(cos0 VtUu vo hai u mch in AB mc ni tip theo th t

gm in tr R, cun dy khng thun cm (L, r) v t in C vi rR . Gi N l im nm gia in tr

R v cun dy, M l im nm gia cun dy v t in. in p tc thi uAM v uNB vung pha vi nhau

v c cng mt gi tr hiu dng l V530 . Gi tr ca U0 bng:

A. 2120 V. B. 120V. C. 260 V. D. 60 V. Gii: Do R = r ---> UR = Ur

Ta c :(UR + Ur)2 + 2LU =

2

AMU

----> 4 2RU + 2

LU = 2

AMU (1)

2RU + (UL UC)2 = 2NBU (2)

UAM = UNB -----> ZAM = ZNB ------>

4R2 + ZL

2 = R

2 + (ZL ZC)

2

3R2 + ZL

2 = (ZL ZC)

2 (*)

uAM v uBN vung pha ----> tanAM.tanNB = -1

R

Z L

2 R

ZZ CL = -1---->(ZL ZC)2 = 2

24

LZ

R (**)

T (*) v (**) 3R2 + ZL2 =

2

24

LZ

R

------> ZL4 + 3R

2ZL

2 4R2 = 0 -----> ZL

2 = R

2

Do UL2 = UR

2 (3). T (1) v (3)----> 5UR

2 = 2AMU = (30 5 )

2 -----> UR = 30 (V)

UR = UL =30 (V) (4)

2

RU + (UL UC)2 = 2NBU ------>(UL UC)

2 = (30 5 )2 302 = 4.302

UAB2 = :(UR + Ur)

2 + (UL UC)

2 = 4UR

2 + (UL UC)

2 = 2.4.30

2

---> UAB = 60 2 (V)-------> U0 = UAB 2 = 120 (V). Chn p n B

Cu 65: Cho mch in RL ni tip, cun dy thun cm, L bin thin t 0 in p hiu dng t vo hai u on mch l U. Hi trn gin vc t qu tch ca u mt vc t I l ng g?

A. Na ng trn ng knh R

U B. on thng I = kU, k l h s t l.

C. Mt na hiperbol 22

LZR

U

D. Na elip

2

0

2

U

u +

2

0

2

I

i=1

UC

UA

M

UL

UAB

Ur

UR 2U

R

UN

B





Gii Ta c I = 22

LZR

U

------> Trn gin vc t qu tch ca u mt vc t I l mt na hiperbol 22

LZR

U

Chn p n C

Cu 66. Stato ca mt ng c khng ng b ba pha gm 9 cun dy, cho dng in xoay chiu ba pha tn s 50Hz vo ng c. Rto lng sc ca ng c c th quay vi tc no sau y? A. 1000vng/min. B. 900vng/min. C. 3000vng/min. D. 1500vng/min.

Gii: p dng cng thc f = np. vi p l s cp cc t. ng c khng ng b 3 pha mi cp cc t ng vi 3 cun dy stato. Do p = 3. n l tc quay ca t trng.

--------> n = p

f=

3

50 vng/s =

3

50.60 vng/min = 1000 vng/min.

Tc quay ca roto ng c n < n nn c th l n = 900 vng /min. Chn p n B

Cu 67: t in p xoay chiu u=U0cost (U0 khng i v thay i c) vo hai u on mch

gm in tr thun R,cun cm thun c t cm L v t in c in dung C mc ni tip,vi CR2<

2L.Khi = 1 hoc = 2 th in p hiu dng gia hai u cun cm c cng mt gi tr.Khi = 0

th in p hiu dng gia hai u cun cm c gi tr cc i.H thc lin h gia 1,2 v 0 l :

A. )(2

1 22

2

1

2

0 B. )(2

1210 B. 2

0

1

=

2

1 (

2

1

1

+

2

2

1

) C. 0 = 21

Gii: UL = 22 )( CL

L

ZZR

UZ

. Do UL1 = UL2 ----->

2

1

1

2

2

1

)1

(C

LR

= 2

2

2

2

2

2

)1

(C

LR

-----> 2

1

2 2

C

LR

+ 24

1

1

C=

2

2

2 2

C

LR

+ 24

2

1

C

------> (2C

L- R

2)(

2

2

1

-

2

1

1

) =

24

2

1

C-

24

1

1

C -----> (2

C

L- R

2) =

2

1

C 22

2

1

2

2

2

1

-----> 2

1

1

+

2

2

1

= C

2 (2

C

L- R

2) (*)

UL = ULmax khi 2

2 2

C

LR

+ 24

1

C + L

2 c gi tr cc tiu.----->

2

0

1

=

2

2C(2

C

L- R

2) (**)

T(*) v (**) suy ra:2

0

1

=

2

1 (

2

1

1

+

2

2

1

) . Chn p n C. Vi iu kin CR2< 2L

Cu 68: Cho mch in AB c hiu in th khng i gm c bin tr R, cun dy thun cm L v t in C mc ni tip. Gi U1, U2 , U3 ln lt l hiu in th hiu dng trn R, L v C. Bit khi U1 =

100V, U2 = 200V, U3 = 100 V. iu chnh R U1 = 80V, lc y U2 c gi tr

A. 233,2V. B. 100 2 V. C. 50 2 V. D. 50V.

Gii:





U = 2322

1 )( UUU = 2

32

2

1 )''(' UUU = 100 2 (V)

Suy ra : (U2 U3)2 = U

2 U1

2 = 13600

U2 U3 = I(Z2 Z3) =100 (V) (*)

U2 U3 = I(Z2 Z3) = 13600 (V) (**)

T (*) v (**) ------> I

I '=

100

13600------>

2

2'

U

U =

2

2'

IZ

ZI=

I

I '=

100

13600---->

U2 = 100

13600U2 = = 233,2 V. Chn p n A

Cu 69 Mc vo on mch RLC khng phn nhnh gm mt ngun in xoay chiu c tn s thay i

c. tn s 1 60f Hz , h s cng sut t cc i cos 1 . tn s 2 120f Hz , h s cng sut

nhn gi tr cos 0,707 . tn s 3 90f Hz , h s cng sut ca mch bng

A. 0,872 B.0,486 C. 0,625 D. 0,781

Gii: Ta c ZL1 = ZC1 -----> 1L = C1

1

-----> LC =

2

1

1

(1)

cos2 = 0,707 -----> 2 = 450 ---->

tan2 = R

ZZ CL 22 =1 -----> R = ZL2 - ZC2

tan3 = R

ZZ CL 33 = 22

33

CL

CL

ZZ

ZZ

=

CL

CL

2

2

3

3

1

1

= 3

2

1

1

2

1

2

2

2

1

2

3

= 3

2

2

1

2

2

2

1

2

3

=

3

2

f

f2

1

2

2

2

1

2

3

ff

ff

tan3 = 3

2

f

f2

1

2

2

2

1

2

3

ff

ff

=

3

422

22

60120

6090

=

3

4

12

5 =

9

5 -----> (tan3)

2 = 25/91---->

81

106

81

251

cos

1

3

2

-------> cos

23 = 81/106 ------> cos3 = 0,874. p n A

Cu 70: Mt on mch AB gm hai on mch AM v MB mc ni tip. on mch AM gm in tr

thun R mc ni tip vi t in C c in dung thay i c, on mch MB l cun dy thun cm c t cm L. Thay i C in p hiu dng ca on mch AM t cc i th thy cc in p hiu

dng gia hai u in tr v cun dy ln lt l UR = 100 2 V, UL = 100V. Khi in p hiu dng

gia hai u t in l:

A. UC = 100 3 V B. UC = 100 2 V C. UC = 200 V D. UC = 100V

Gii:

Ta c UAM = 22

22

)( CL

C

ZZR

ZRU

=

22

22 )(

1

C

CL

ZR

ZZR

=

22

2 21

1

C

CLL

ZR

ZZZ

UAM = UAMmax th biu thc y = 22

2 2

C

CLL

ZR

ZZZ

= ymin ------> o hm y = 0

---> (22

CZR )(-2ZL) ( CLL ZZZ 22 )2ZC = 0 ZC

2 ZLZC R2 = 0

Hay UC2 ULUC UR

2 = 0 UC

2 100UC 20000 = 0

UC = 200(V) (loi nghim m). Chn p n C





Cu 71: Mch in R1L1C1 c tn s cng hng 1 v mch R2L2C2 c tn s cng hng 2 , bit

1=2. Mc ni tip hai mch vi nhau th tn s cng hng ca mch s l . lin h vi 1v

2theo cng thc no? Chn p n ng:

A. =21. B. = 31. C. = 0. D. = 1.

Gii:

2 = LC

1 =

21

2121 )(

1

CC

CCLL

2

1 = 11

1

CL ---> L1 =

1

2

1

1

C ; 22 =

22

1

CL----->L2 =

2

2

2

1

C

L1 + L2 = 1

2

1

1

C +

2

2

2

1

C =

2

1

1

(

1

1

C +

2

1

C) =

2

1

1

21

21

CC

CC

( v 1=2.)

----> 21 =

21

2121 )(

1

CC

CCLL

= 2 --------> = 1. p n D

Cu 72. Dng in xoay chiu c chu k T, nu tnh gi tr hiu dng ca dng in trong th gian T/3 l 3(A), trong T/4 tip theo gi tr hiu dng l 2(A) v trong 5T/12 tip theo na gi tr hiu dng l

2 3 (A). Tm gi tr hiu dng ca dng in:

A. 4 (A). B. 3 2 (A). C. 3 (A). D. 5(A).

Gii: Nhit lng ta ra trn in tr R ca mch trong thi gian: t1 = T/3: Q1 = I1

2Rt1 = 9RT/3 = 3RT

t2 = T/4: Q2 = I22Rt2 = 4RT/4 = RT

t3 = 5T/12: Q3 = I32Rt3 = 12R.5T/12 = 5RT

t = t1 + t2 + t3 = T l Q = I2Rt = I

2RT

M Q = Q1 + Q2 + Q3 = 9RT-------> I2 = 9 -----> I = 3 (A). Chn p n C

Cu 73 : t in p xoay chiu c gi tr hiu dng khng i 150 V vo on mch AMB gm on AM ch cha in tr R, on mch MB cha t in c in dung C mc ni tip vi mt cun cm thun c t cm L thay i c. Bit sau khi thay i t cm L th in p hiu dng hai u mch

MB tng 2 2 ln v dng in trong mch trc v sau khi thay i lch pha nhau mt gc 2

. Tm in

p hiu dng hai u mch AM khi cha thay i L?

A. 100 V. B. 100 2 V. C. 100 3 V. D. 120 V.

Gii:

tan1 = 1

11

R

CL

U

UU ; tan2 =

2

22

R

CL

U

UU

1 - 2 = /2 -------> tan1 tan2 = 1

11

R

CL

U

UU

2

22

R

CL

U

UU = -1

(UL1 UC1)2 .(UL2 UC2)

2 =

2

1RU2

2RU .-------> 2

1MBU2

2MBU = 2

1RU2

2RU .------>

84

1MBU =2

1RU2

2RU .(*) (v UMB2 = 2 2 UMB1)

Mt khc 21RU + 2

1MBU = 2

2RU + 2

2MBU (= U2) ----->

2

2RU = 2

1RU - 72

1MBU (**)

T (*) v (**): 8 4 1MBU =2

1RU2

2RU = 2

1RU (2

1RU - 72

1MBU )

-----> 4

1RU - 72

1MBU2

1RU - 84

1MBU = 0 ------> 2

1RU = 82

1MBU





21RU + 2

1MBU = U2 ------> 2

1RU + 8

2

1RU = U2

----> UR1 = 3

22U = 100 2 (V). Chn p n B

Cu 74: 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: Ta thy I1 = I2 ----> (ZL ZC)

2 = ZL

2 ----. ZC = 2ZL

tan1 = R

ZZ CL = -R

Z L (*) tan1 = R

Z L (**) ----> 1 + 2 = 0

1 = u - 4

; 2 = u +

12

-------> 2u -

4

+

12

= 0 ---> u =

12

Do u 60 2 cos(100 t )12

, Chn p n C

Cu 75: Ba in tr ging nhau u hnh sao v ni vo ngun n nh cng u hnh sao nh cc ng dy dn. Nu i cch u ba in tr thnh tam gic (ngun vn u hnh sao) th cng dng in hiu dng qua mi ng dy dn:

A. tng 3 ln. B. tng 3 ln. C. gim 3 ln. D. gim 3 ln.

Gii:

Khi cc in tr u sao: Id = Ip = R

U p

Khi cc in tr u tam gic: Id = 3 Ip = 3R

U p'= 3

R

U d = 3R

U

R

UPp 3

3 = 3I

Tng ln gp 3 ln. Chn p n A

Cu 76 : Cho on mch xoay chiu ni tip RLC, in dung C = 2F. t vo hai u on mch mt

in p xoay chiu th in p gia hai bn t in c biu thc 100cos(100 / 3)( )u t V . Trong

khong thi gian 5.10-3(s) k t thi im ban u, in lng chuyn qua in tr R c ln l

A. 4( 3 2).10 ( )C B. 4(1 3).10 ( )C

C. 4(

Download - Tuyenchondien Xoay Chieu Kho Diem 10hocmaivncom.thuvienvatly.com.4f260.39098

Top Related