Tuyển tập Điện xoay chiều Diễn đàn Vật lí phổ thông.pdf

TNG HI TUN (CH BIN)

TRN VN QUN - NGUYN MINH HIP

Tng hp

IN XOAY CHIU

Din n Vt l ph thng

2015

Din n Vt l ph thng http://vatliphothong.vn

Li ni u

Trong chng trnh Vt l 12 ca B gio dc, mng in xoay chiu c th ni l mt mng rtkh. Mi bi ton in xoay chiu u c nhng cch gii khc nhau, trong hi t nhng v pTon hc v Vt l m chc hn trong mi chng ta ai cng mun tm ra v thng thc n.

Tuyn tp in xoay chiu Din n Vt l ph thng ra i vi mong mun cung cp mt tiliu hu ch cho cc bn hc sinh ang n thi i hc v ng thi cng mong mun y l mt tiliu tham kho tt cho cc thy c gio ang dy ph thng.

Vi tinh thn lm vic nghim tc, ham hc hi, ban bin tp xin gi li chn thnh cm n titt c cc thnh vin tham gia ng bi, gii bi trn din n. Nu khng c s nhit tnh thamgia hc hi, trao i, bn lun ca cc bn th tuyn tp ny s khng c ngy xut bn nh hm nay.

Trong qu trnh bin tp, mc d rt c gng song khng trnh khi sai st, nn rt mong nhnc s thng cm, chia s, gp ca cc bn ti liu c hon thin hn.

Ti liu s c cp nht thng xuyn ti a ch http://vatliphothong.vn/download/59/

Mi kin ng gp xin gi v hm th [email protected]. Xin chn thnh cm n.

Thi Bnh, ngy 18 thng 2 nm 2015i din nhm bin son

Ch binTng Hi Tun

1


Cc thnh vin tham gia bin son

Ni dung

1. Tng Hi Tun - CLC K61 - Khoa Vt l - Trng i hc S phm H Ni

2. Trn Vn Qun - 12A1 - THPT L Li - Tn K - Ngh An

3. Nguyn Minh Hip - Thnh vin iHc2015 trn Din n Vt l ph thng

LATEX

H tr k thut Latex

1. Tng Hi Tun

2. Trn Vn Qun

3. Nguyn Minh Hip

Trnh by ba

Tng Hi Tun

2


Bi ton 1: Cho mch RLC mc ni tip theo th t LRC. t vo hai u 1 in p xoaychiu c hiu in th hiu dng U khng i, ! c th thay i c. Khi ! = !1 th on mchAB tiu th cng sut P1 = 100 W vi h s cng sut l 1. Khi ! = !2 th in p hiu dng URCt cc i v on mch AB tiu th cng sut l P2. Chng minh rng

P2 89P1:

Li gii

Khi ! = !1 th h s cng sut l 1, mch ang xy ra hin tng cng hng, suy ra

P1 =U2

R:

Khi ! = !2, l bi ton quen thuc hin nay, khng my kh khn c th thy c URC t gi tr

ln nht khi

!22 =2L

2

C2

L2

LC+q

L2

C2+ 2L

CR2 < 2

LC:

T ta suy ra ZC >1

2ZL v

Z2C =1

2ZLZC +

s1

2ZLZC

2+

1

2ZLZCR2;

Z2C 1

2ZLZC

2=

1

2ZLZC

2+

1

2ZLZCR

2;

R2 =2Z2C (ZC ZL)

ZL(1):

Cng sut lc ny lP2 = UI cos'

=U2

R RZ cos'

= P 21 cos2':

Ta c

cos2 ' =R2

R2 + (ZL ZC)2=

1

1 + (x y)2 :

Trong x =ZLR; y =

ZCR; x < 2y (vic t ny chnh l bn cht ca vic chun ha cho R = 1).

Vi php t nh th th ta c (1) tng ng vi

1 =2y2(y x)

x;

hay

(x y)2 = x(y x)2y2

=1

8 (2x y)

2

8y2 1

8:

(Ch rng, s dng iu kin c nghim ca tam thc bc hai, ta hon ton c th suy ra c

(x y)2 18, mnh vit vy cho gn).

3


T suy ra cos2 ' 89; hay

P2 89P1:

Bi ton c chng minh xong.

ng thc xy ra khi v ch khi 2x = y (tha mn x < 2y) tc l khi 2ZL = ZC hay tng ng

!0 =1p2LC

. Tuy nhin, n khc vi gi tr !2 nn ng thc trn khng th xy ra. V sao? Bi v

phng trnh (1) ch c c khi ! = !2.

Do ta khng th kt lun gi tr nh nht ca P2 l8

9P1:

Vy khi no th gi tr nh nht ca P2 l8

9P1? Khi bi ca chng ta sa li 1 cht nh sau:

Cho mch RLC mc ni tip theo th t L R C. t vo hai u 1 in p xoay chiu c hiuin th hiu dng U khng i, ! c th thay i c. Khi ! = !1 th on mch AB tiu th cng

sut P1 = 100 W vi h s cng sut l 1. Khi ! = !2 th in p hiu dng URC t cc i v

on mch AB tiu th cng sut l P2. Sau gi nguyn gi tr !2 v thay i L (hoc

C). Chng minh rng gi tr nh nht ca P2 l8

9P1. Tm L (hoc C) khi ng thc xy ra.

Bi ton 2: t in p u = 120p2 cos 100t (V ) vo hai u on mch ni tip gm bin tr

R, t in C =1

(4)mF v cun cm thun L =

1

H. Khi thay i gi tr ca bin tr th ng vi

hai gi tr ca bin tr l R1 v R2 th mch tiu th cng mt cng sut P v lch pha cain p hai u on mch so vi dng in trong mch tng ng l '1; '2 vi '1 = 2'2. Gi trcng sut P bng

A. 120 W. B. 240 W. C. 60p3 W. D. 120

p3 W.

Li gii

Ta tnh c ZC = 40;ZL = 100

Khi thay i gi tr ca bin tr th ng vi hai gi tr ca bin tr l R1 v R2 th mch tiu th

cng mt cng sut P v lch pha ca in p hai u on mch so vi dng in trong mch

tng ng l '1; '2 )'1+ '2 =

2(1)

M '1 = 2'2 (2). T (1) v (2) ) '1 = 3;'2 =

6

cos'1 = cos

3=

1

2) R1

Z1=

1

2. Thay s vo ta tnh c R1 = 20

p3) Z1 = 40

p3

) P = UI cos'1 = U2

Z1cos'1 =

1202

40p3:1

2= 60

p3W

Chn C.

Bi ton 3: Cho mch in ni tip gm in tr R = 50, cun cm thun L =1

H v t in

C =50

(F ). t vo hai u on mch in p u = 50+100

p2 cos (100t)+50

p2 cos (200t) (V ).

Cng sut tiu th ca mch in lA. 40W. B. 50W. C. 100W. D. 200W.

Li gii

Mch gm 3 mch nh

4


Mch 1 gm ngun 1 chiu U = 50V khng i qua C

Mch 2 gm ngun u1 = 100p2 cos (100t)

Mch 3 gm ngun u2 = 50p2 cos (200t)

Suy ra cng sut tiu th trn mch 2 + 3 l :

P =R (U21 + U

22 )

R2 + (ZL ZC)2= 50W

Chn B.

Bi ton 4: t in p xoay chiu c ga tr hiu dng khng i, f = 50Hz vo 2 u onmch mc ni tip gm in tr thun R, cun cm thun c t cm L thay i c v t c

in dung. iu chnh t cm L n gi tr1

5H hoc

4

5H th cng hiu dng nh nhau,

chnh L n3

5H hoc

6

5H th hiu in th 2 u cun cm bng nhau. Gi tr R gn

A. 53. B. 52. C. 54. D. 37.

Li gii

Khi I1 = I2 ! ZL1 + ZL22

= ZC , ZC = 50

Khi UL1 = UL2 ,1

ZL=

1

2

1

ZL1+

1

ZL2

, ZL = 80

Th tt c vo UL1 = UL2 , R 37Chn D.

Bi ton 5: on mch in gm RLC mc ni tip, cun dy thun cm vi CR2 < 2L; in p2 u on mch l u = U

p2cos(!t), U n nh v ! thay i. Khi ! = !L th in p 2 cun cm

L cc i v ULmax =41U

40. H s cng sut tiu th l?

Li gii

Gi '; l lch pha gia u vi uR v uL.

Ta c: ULmax nn + ' =

2) cos' = sin = p1 cos2 =

s1

U

ULmax

2=

9

41:

Bi ton 6: Mch RLC c in p xoay chiu u = 240p2 cos (314; 15t). Bit R = 80; I =

1; 732A;UCL = 138; 56 v in p uRC vung pha vi uCL. Tnh L?A. 0,47. B. 0,37. C. 0,68. D. 0,58.

Li gii

ZLC = 80; Z = 80p3. Gi ; l lch pha gia uLC , uRC v i.

Ta c:Z2 = Z2LC +R2 + 2RZLC cos) cos = 1

2= sin ) =

6) ZC = R tan

6=

80p3

cos =1

2) sin =

p3

2) ZL ZC = ZLC sin = 40

p3

) ZL = 115:47) L = 0:37Chn D.

5


Bi ton 7: Mch in xoay chiu gm AM v MB mc ni tip. on AM gm t in ni tipin tr R ,MB gm cuc dy c in tr r = R. t vo 2 u on mch AB mt in p xoaychiu c hiu in th hiu dng U v tn s thay i c, th in p tc thi trn AM v MB

lun lch gc .

2. Khi ! = !1 th in p hiu dng trn AM l U1 v tr pha hn AB gc 1. Khi

! = !2 th in p hiu dng trn AM l U2 v tr pha hn AB gc 2. Khi ! = !3 th in phiu dng trn AM l U3 v tr pha hn AB gc 3. Bit rng U1 + U2 + U3 > U v s chnh lch

gia 2 trong 3 i lng 1;2;3 khng vt qu

27. Hi h s cng sut ca mch AB ln nht

c th trong 3 ln thay i tn s t gi tr no?

A. 1. B.5

12. C. 0,73. D. 0,87.

Li gii

Ta xt vi mt bt ng thc: Cho 1 + 2 + 3 = th:

cos1 + cos2 + cos3 > 1

Do vy, ta gn 1 + 2 + 3 = .

Gi s : 1 < 2 < 3

Khi : 31 +2

27> ! 1 > 25

81. Kt hp 1 ZC1 . Khi C = C2 th UC = UAB +U

4; U2 = UR+r

gc lch in p hai u mch v dng in l '2; ZL < ZC2 . Tng (U1 + U2) gn gi tr no nht

trong cc gi tr sau? Bit '1 + '2 = 200;'1; '2 > 0;R =

7

9r v U 6= 0

A. 80p5V . B. 100

p5V . C. 120

p5V . D. 140

p5V .

Li gii

Lc u gc lch gia hiu in th AN v MN cc i nn Z2L = (R + r) r. Chn r = 9, R = 7, ZL

6


= 12. Gc hp bi uAN v i l arc tan3

4.

T gi thit ta suy ra c 4:UC2 UC1 = 3:UAB.V gin vector ra s thy

UAB0; 8

=UC1

sin

arc tan

3

4 '1

= UC2sin

arc tan

3

4 '1 + 20

.T suy ra 4: sin

arc tan

3

4 '1 + 20

sin

arc tan

3

4 '1

= 3:0; 8.

Bm my tnh tm c '1.

Vy U1 + U2 = UAB: cos'1 + UAB: cos (20 '1) 236; 2 (V ).Chn B.

Bi ton 10: Ngi dn Vit Nam ch yu s dng in xoay chiu mt pha c thng s 220 V- 50 Hz. Dy ngui c ni t c in th bng 0. Hi in th ca dy nng bin thin trongkhong no ?

A. -311 V 311 V. B. -311 V 11 V . C. -311 V 211 V . D. -311 V 111 V .

Li gii

220 V l in p hiu dng ca dng in xoay chiu

) u = p2U cos (!t+ ') ) 220p2 u 220p2Chn A.

Bi ton 11: Cho mch in xay chiu RLC mc ni tip. Bit L = 318(mH), C = 17(F ). in

p 2 u mch l u = 120p2 cos

100t

4

V , cng dng in trong mch c biu thc

i = 1; 2p2 cos

100t+

12

A. h s cng sut ca mch l 0,6 th phi ghp thm 1 in tr

Ro vi RA. ni tip, Ro =15; 25. B. ni tip, Ro = 20. C. ni tip, Ro = 30. D. ni tip, Ro = 45.

Li gii

ZL = 100;ZC = 187

Ban u: Z = 100 ) R1 = 50

Gi x l in tr tng ng sau khi ghp thm 1 in tr Ro

Theo bi ra: cos'0 =x

Z 0= 0; 6

) x = 0; 6Z 0 , x2 = 0:62: x2 + (ZL ZC)2) x = 65; 25

) Ro c mc ni tip vi Ro = 15; 25

Chn A.

Bi ton 12: Mch in gm c R = 100 , cun cm thun L =2

v t C bin i mc ni tip

vo 2 u A, B c in p u = 120p2 cos 100t (V ) . Khi C thay i t 0 n rt ln th in p

hiu dng gia 2 bn tA. tng t 0 n 120

p5V sau gim t 120

p5V n 0.

B. tng t 0 n 220p5V sau gim t 120

p5V n 0.

C. tng t 0 n 100p5V sau gim t 120

p5V n 0.

D. tng t 0 n 120p5V sau gim t 20

p5V n 0.

Li gii

7


ZL = 200

in p ln nht hai u t:

UCmax = U

pR2 + Z2LR

= 120p5V

Khi ta iu chnh C ! C0 !/ th in p hai u t s t 0 V n gi tr cc i sau gim dnv 0V

Do khi C = 0 th ZC rt ln nn dng in khng i qua t nn UC = 0V

Khi C =/ th ZC = 0 nn UC = 0VChn A.

Bi ton 13: Cho cun dy c r = 50; ZL = 50p3 mc ni tip vi hp en gm 2 trong 3

phn t R, L , C mc ni tip . t vo 2 u in p xoay chiu thy sau khi in p trn cundy t cc i 1/4 chu k th in p gia 2 u hp en t cc i. Trong hp en cha ccphn t l

A. L v R vi R = ZLp3. B. C v R vi R = ZC

p3.

C. C v R vi R = ZCp2. D. C v R vi R = ZC

p5.

Li gii

lch pha gia 2 u cun dy vi cng dng in: tan'1 =ZLr

=p3) '1 =

3Sau khi in p trn cun dy t cc i 1/4 chu k th in p gia 2 u hp en t cc i ) lch pha gia in p 2 u cun dy v in p hai u hp en:

' =T

4:2

T=

2) lch pha gia in p hai u hp en vi cng dng in: '2 =

2

3=

6Da vo gin vecto ta thy in p 2 u X chm pha so vi cng dng in :

6) Hp X cha C v R

C: tan'2 = tan

6=

1p3, ZC

R=

1p3) R = p3ZC

Chn B.

Bi ton 14: on mch ni tip gm cun dy thun cm L , in tr thun R, t in C c tdi in p xoay chiu n nh: u = Uo cos (!t+ '). Ban u CD qua mch li1 = 3 cos (!t).

Khi t C b ni tt th CD qua mch l i2 = 3 cos!t

3

. Tm '

A.

3. B.

3. C.

6. D.

6.

Li gii

Khi thay ni tt t C, t phng trnh dng in ta thy cng dng in hiu dng trong mch

khng thay i nn:

Z1 = Z2

,qR2 + (ZL ZC)2 =

qR2 + Z2L

) ZC = 2ZL

8


Li c:

tan'1 =ZL ZC

R

tan'2 =ZLR

Suy ra:

tan'1 + tan'2 =2ZL ZC

R= 0

Suy ra '1 = '2 ) 'u = 'i1 + 'i22

= 6

Chn D.

Bi ton 15: on mch ni tip theo th t gm R1; R2 v cun cm thun c t cm L thayi c. Nu t vo hai u on mch hiu in th khng i l 18 V th cng dng inqua cun cm l 20

p3 (mA) v hiu in th trn in tr l R1 l 40V . Nu t in p xoay

chiu c tn s 50Hz v iu chnh L n khi hiu in th tc thi gia hai u on mch chaR2 v L lch pha cc i so vi hiu in th hai u on mch. Gi tr t cm lc l:

A. L =2

H. B. L =

3

H. C. L =

4

H. D. L =

1

H.

Li gii

Nu t in p xoay chiu c tn s 50Hz v iu chnh L n khi hiu in th tc thi gia hai

u on mch cha R2 v L lch pha cc i so vi hiu in th hai u on mch.

, ZL =pR2 (R1 +R2)

I =Uq

(R1 +R2)2 + Z2L

, I = 18q(40 +R2)

2 +R2 (40 +R2)R2 = 337; 55

, ZL = 357 , L 1; 13Chn C.

Bi ton 16: t in p xoay chiu uAB = U0 cos (!t) vo 2 u on mch RLC ni tip viCbin thin. iu chnh in dung C ca t sao cho in p hiu dng ca t t gi tr cc i,khi in p tc thi cc i trn R l 12a. Bit khi in p tc thi hai u mch l 16a thin p tc thi hai u t l 7a. Chn h thc ng

A. 3R = 4ZL. B. R = 4ZL. C. 5R = 4ZL. D. R = ZL.

Li gii

Ta c um = uRL + uc ) uRL = 9aum = uRL + uc ) uRL = 9aVi C thay i in p 2 u t cc i nn:1

U2R=

1

U2RL+

1

U2m;

uRLURL

2+

umUm

2= 1 (cc gi tr u l tc thi; U l cc gi tr tc thi cc i)

Thay cc gi tr vo h trn ta c:

URL = 15a;Um = 20a) UL = 9a) 3R = 4ZLChn A.

9


Bi ton 17: Chn cu ng. Cho mch in AM cha in tr R, MN cha cun cm L, NB chat in C. Ngi ta o c cc in p uAN = uAB = 20V ; uMB = 12V . in p uAM ; uMN ; uNBln lt l

A. uAM = 26V; uMN = 12V; uNB = 24V .B. uAM = 16V; uMN = 12V; uNB = 34V .C. uAM = 16V; uMN = 12V; uNB = 24V .D. uAM = 20V; uMN = 12V; uNB = 24V .

Li gii

Do UAN = UAB ) ZAN = ZAB , R2 + Z2L = R2 + (ZL ZC)2 , 2ZL = ZCTa c: +

UMBUAN

=12

20=ZMBZAN

=ZLp

R2 + ZL=

12

20, R = 4

3ZL

+ I =UANZAN

=205

3ZL

) UMN = I:ZL = 12V

UAM = I:R =12

ZL:4

3ZL = 16V

UNB = 2UL = 2:UMN = 2:12 = 24V

Chn C.

Bi ton 18: Cho mch in RLC mc ni tip, trong RC2 < 2L. t vo hai u on mchin p u = U

p2 cos 2ft, trong U c gi tr khng i, f c th thay i c. Khi f = f1 th

in p hiu dng trn t c gi tr bng U , mch tiu th cng sut bng 75% cng sut cc i.Khi tn s ca dng in l f2 = f1 + 100Hz th in p hiu dng trn cun cm cng c gi trbng U . in p hiu dng trn t in cc i th tn s phi bng:

A. 100p2 Hz. B. 10

p2 Hz. C. 50

p2 Hz. D. 20

p2 Hz.

Li gii

Gi gi tr tn s UCmax v ULmax ln lt l fC v fL. Ta c hai b quan trng l: fC =f1p2

v fL = (f1 + 100)p2

Khi f = f1 th in p hiu dng trn t c gi tr bng U , mch tiu th cng sut bng 75% cng

sut cc i, do U2

R: cos2 ' =

U2

R) cos' =

p3

2Kt hp vi UC = U ) ZC = Z ta c ZL = 1; 5ZC hoc ZC = 2ZL+ ZL = 1; 5ZC ) ZC = 2Rp

3;ZL = R

p3

Ta c ZL:ZC =L

Ckhng ph thuc vo f. Trong trng hp ny ZL:ZC = 2R

2

Khi tn s ca dng in l f2 = f1 + 100Hz th in p hiu dng trn cun cm cng c gi tr

bng U nn ta c R2 + Z 02C 2Z 0L:Z 0C = 0) Z 0C = Rp3

Ta thy f1 < (f1 + 100) nn ZC > Z0C m ZC =

2Rp3< R

p3 nn trng hp ny loi.

+ ZC = 2ZL ) ZL = Rp3;ZC =

2Rp3, trng hp ny ta c ZL:ZC =

2R2

3

Tng t trng hp trn ta gii ra Z 0C =Rp3

Ta cf1 + 100

f1=ZCZ 0C

= 2) f1 = 100 (Hz)

in p hiu dng trn t in cc i th tn s phi bng fC =f1p2= 50

p2 (Hz)

10


Chn C.

Bi ton 19: t in p xoay chiu c gi tr hiu dng U = 120 V , tn s f thay i c vohai u on mch gm in tr thun, cun cm thun v t in mc ni tip. Khi tn s l f1th hai u on mch cha RC v in p gia hai u L lch pha nhau mt gc 135o. Khi tns l f2 th hai u on mch cha RL v in p gia hai u C lch pha nhau mt gc 135

o.

Khi tn s l f3 th mch xy ra cng hng. Bit rng

2f2f3

2f2f1

2=

96

25. iu chnh tn s

n khi in p hiu dng hai u t t cc i l Uo. Gi tr ca Uo gn gi tr no nht sauy?

A. 123 V . B. 130 V . C. 1803 V . D. 223 V .

Li gii

Li gii 1: f = f1 , R = ZC1 ; f = f2 , R = ZL2 ) ZC1 =f2f1ZL1 ) f 23 = f1f2

Thay vo cng thc

2f2f3

2f2f1

2=

96

25suy ra

f2f1

=12

5hoc

f2f1

=8

5

UCmax =2ULp

4L C2R4 tnh ra xp x 122(V)Chn A.

Li gii 2:

Theo bi suy ra f 23 = f1f2

UCmax =Us

1fCfL

2 = Us1

1 f2

2f1

2 = Usf2f1 1

4:

f2f1

2Thay s ta c UCmax =

120r24

25

122 (V )) A

Chn A.

Bi ton 20: Mt ngi nh cun mt bin th t hiu in th U1 = 110V ln 220V vi likhng phn nhnh, khng mt mt nng lng v cc cun dy c in tr rt nh , vi s vngcc cun ng vi 1,2 vng/Vn. Ngi cun ng hon ton cun th cp nhng li cun ngcchiu nhng vng cui ca cun s cp. Khi th my vi ngun th cp o c U2 = 264V so vicun s cp ng yu cu thit k, in p ngun l U1 = 110V . S vng cun sai l:

A. 6. B. 30. C. 11. D. 22.

Li gii

Gi s vng cc cun dy MBA theo ng yu cu l N1 v N2

Ta c:N1N2

=1

2) N2 = 2N1 (1) vi N1 = 110:1; 2 = 132 vng

Gi n l s vng dy b cun ngc. Ta c:

N1 2nN2

=e1e2

=E1E2

=U1U2

=110

264) N1 2n

2N1=

110

264(2)

T (1) ; (2)) n = 11. T ta chn C.

11


Bi ton 21: Mt mch in xoay chiu AB gm 2 hp kn X v Y ghp ni tip (mi hp chcha mt trong ba phn t: in tr thun R, cun thun cm c t cm L, t in c indung C). t vo hai u on mch AB mt ngun in mt chiu c hiu in th khng i 12V th hiu in th 2 u hp Y l 12 V. Khi t vo hai u on mch AB mt in p xoay

chiu uAB = 100p2 cos

100t

3

V th in p hai u hp X l uX = 50

p6 cos

100t

6

V

v cng dng in i = 2p2 cos

100t

6

A. Phn t trong hp X v Y ln lt l?

A. RX = 10p3; L =

0; 25

H.

B. RX = 20p3; L =

0; 25

H.

C. RX = 25p3; L =

0; 25

H.

D. RX = 25p3; L =

0; 24

H.

Li gii

+ Khi mc ngun mt chiu vo on mch ta o c hiu in th ca UY = UAB = 12V nn hp

Y cha cun cm thun L

+ Khi mc ngun xoay chiu vo on mch ta thy 'x = 'i do X phi cha in tr R.

Chn C.

Bi ton 22: t in p xoay chiu c gi tr hiu dng U khng i, tn s f thay i c vohai u on mch gm in tr thun, cun cm thun v t in mc ni tip. Khi f = f0 thin p hiu dng hai u t in UC = U . Khi f = f0 + 75 th in p hiu dng hai u cun

cm UL = U v h s cng sut ca ton mch lc ny l1p3. Hi f0 gn vi gi tr no nht

sau y?A. 10Hz. B. 16 Hz. C. 25 Hz. D. 30Hz.

Li gii

Li gii 1: Khi f = f0 ) ZC0 =qR2 + (ZL0 Zc0)2

) Z2L0 = 2ZL0ZC0 R2 =2L

CR2 (1)

Khi f = f0 + 75;UL = U ) ZL =qR2 + (ZL ZC)2

) Z2C = 2ZLZC R2 =2L

CR2 (2)

T (1) ; (2)) ZL0 = ZC ) !0L =1

C!) !!0 = 1

LC(3)

Li c:

cos' =Rq

R2 + (ZL ZC)2=R

Z=

1p3) R

Z=

!p3(4)

T (1)) Z2L0 =2L

CR2 ) !20L2 =

2L

CR2 ) !20 = 2LC

R2

L2(5)

Th (3) v (4) vo (5) ) !20 = 2!!0 !2

3) 3!20 6!!0 + !2 = 0

Hay 3f 20 6ff0+f 2 = 0) 3f 20 6 (f0 + f1) f0+(f0 + f1)2 ) 2f 20 +4f1f0f 21 = 0 (6) vi f1 = 75HzPhng trnh (6) c nghim f0 =

2f1 f1p6

2. Loi nghim m ta c f0 = 16; 86Hz.

T ta chn B.

12


Li gii 2:

Gi s f = f2 = f0+7 = kf0 (k > 1). Khi ta c tng tr cm khng v dung khng ca mch ln

lt l Z;ZL; ZC

T gi thit cos' =R

Z=

1p3ta chn R = 1; Z =

p3.

Mt khc UL = U ) ZL = Z =p3

Li c: Z =qR2 + (ZL ZC)2 )

ZC =

p3p2

ZC =p3 +

p2

Khi f = f0 th tng tr, cm khng, dung khng ca mch ln lt l: Z0; Z 0L =

ZLk; Z 0C = kZC

T gi thit Z 0C = Z0

,ZLk

2+ 1 = 2:

ZLk:kZC

, 3k2

+ 1 = 2ZLZC

Vi 2 gi tr ca ZC kt hp vi k > 1 ta suy ra k =

r3

5 2p6 Nn ta suy ra:

f0 + 75 =

s3

5 2p6f0 ) f0 16; 86

Chn B.

Bi ton 23: My bin p c s vng dy cun s cp l N1 = 400vng, th cp l N2 = 100vng.in tr cun s cp r1 = 4, in tr cun th cp r2 = 1. in tr mc vo cun th cpR = 10. Xem mch t l khp kn v b qua hao ph. t vo hai u cun s cp hiu in thxoay chiu c gi tr hiu dng U1 = 360V . Xc nh hiu in th hiu dng U2 ti hai u cunth cp v hiu sut my bin th

A. 80V, 88,8%. B. 80V, 99,8%. C. 80V, 78,8%. D. 50V, 88,8%.

Li gii

Xt vai tr cun s cp c vai tr l my pht, cun th cp c vai tr l my thu:

Sut in ng cun s cp:

E1 = U1 I1r1Sut in ng cun th cp:

E2 = U2 + I2r2

Mt khc:E1E2

=N1N2

=I2I1

= 4

Kt hp vi I2 =U2R

ta tnh c U2 = 80V

Khi h sut my bin th:

H =I22 :R

U1I1= 88; 8%

Chn C.

13


Bi ton 24: My pht in xoay chiu mt pha gm 1 cp cc; c 500 vng dy, din tch mivng l S = 60 (cm2) ; phn cm c B = 0; 05 (T ) . Tin hnh ni my ny vi on mch ABgm in tr R = 50 () ni tip vi t in c C = 104 (F ) ni tip vi mt cun thun cm c

L = 0; 4 (H). iu chnh s vng n t 3000

vng

pht

xung ti 1800

vng

pht

. Hi gi tr hiu dng

ln nht 2 u cun cm trong c qu trnh l bao nhiu?

Li gii

Ta c:

UL =E

Z:ZL =

!NBSsR2 +

!L 1

!C

2 :!L

UL =!2N:B:S:Ls

R2 +

!L 1

!C

2

UL =N:B:S:Lvuut 1

!6C2+R2 2L

C!4

+L2

!2

Xt mu s :

f

1

!2

=

1

!6C2+R2 2L

C!4

+L2

!2

f 0

1

!2

> 08!

T BBT ta thy hm f ng bin nn UL max khi ! = !2 = 60

T ta tnh cc gi tr E = 9; ZL = 24; ZC =500

3T ta tnh c: Z 57; 66; I 0; 52A

UL = I:ZL = 38; 9V

Bi ton 25: t hiu in th xoay chiu u = U0 cos (100t+ ') vo hai u on mch ni tiptheo th t gm R, C v cun cm thun c t cm L. T in c in dung C thay i c.Ban u iu chnh C hiu in th hiu dng gia hai u on mch cha R v C t cci. Sau , phi gim gi tr in dung i ba ln th hiu in th hai u t mi t cc i. T

sR

ZLca on mch xp x bng?

A. 1. B. 2. C. 3,2. D. 4.

14


Li gii

Li gii 1: Ta c: URC =Us

R2 + (ZL ZC)2R2 + Z2C

=Us

1 +Z2L 2ZCZLR2 + Z2C

t y =Z2L 2ZCZLR2 + Z2C

; y0 = 0 th URCmax

ZC1 =ZL +

p4R2 + Z2L2

, t x =R

ZL) ZC1

R=

0; 5

x+ 0; 5

r4 +

1

x2

ZC2 =R2 + ZL

2

ZL) ZC2

R=

1

x+ x = 3

0; 5

x+ 0; 5

r4 +

1

x2

!) x = 3; 2. Chn C.Li gii 2:

Thay i C URCmax : ZC1 =ZL +

p4R2 + Z2LZL

Gim C i 3 ln UCmax ! ZC2 = 3ZC1 ,R2 + Z2L

ZL= 3

ZL +p4R2 + Z2L2

, 4R4 28R2Z2L 5Z4L = 0Gii PT ta c R =

r7 +

3p6

2' 3; 2. Chn C.

Bi ton 26: t in p u = U0 cos (!t) (V ) vo 2 u cun dy ni tip vi 1 t in C c indung thay i c. Ban u t in c dung khng 100, cun dy c dung khng 50. Gim

in dung mt lng C =103

8(F ) th tn s gc dao ng ring ca mch l 80 (rad=s). Tn

s gc ! ca dng in trong mch l:A. 40 (rad=s). B. 20 (rad=s). C. 10 (rad=s). D. 50 (rad=s).

Li gii

ZL = !L = 50! L = 50!

ZC =1

!C= 100! LC = 1

2!2

!0 = 80 =1s

L

C 10

3

8

Th LC, L vo phng trnh trn tm c ! = 40

Bi ton 27: t in p u = U0 cos!t vo hai u on mch mc ni tip gm in tr R, tin c in dung C, cun cm thun c t cm L thay i c. Khi L = L1 in p hai ucun cm c gi tr cc i ULmax v in p hai u on mch sm pha hn dng in trong

mch l 0; 24 (0 < :tg1 =

4

3

tg2 =3

4

C tg1 =4

3) UMB

UAM=

4

3, r

2 + Z2LR2 + Z2C

=16

9

R2 = ZL:ZC )

8>:ZL =

4R

3

ZC =3R

4

, 16Z2C + 7ZLZC 9Z2L = 0 ) ZC =9

16ZL

cos' =R + rq

(R + r)2 + (ZL ZC)2= 0; 96

T ta chn D.

16


Bi ton 29: t in p xoay chiu c gi tr hiu dng khng i, tn s thay i c vo hai

u on mch gm in tr thun R, cun cm thun L v t in C

R =

rL

C

!. Thay i tn

s n cc gi tr f1 v f2 th cng dng in trong mch l nh nhau v cng sut ca mchlc ny l P0. Thay i tn s n gi tr f3 th in p hai u t in cc i v cng sut mch

lc ny l P . Bit rng

f1f3

+f2f3

2=

25

2. Gi =

P0P. Gi tr gn gi tr no nht sau y?

A. 0,45. B. 0,57. C. 0.66. D. 2,2.

Li gii

D c !1!2 =1

LC, !23 =

1

LC R

2

2L2

M theo R2 =L

C) !23 =

1

2LC) f 23 =

f1f22

Mt khc

f1f3

+f2f3

2=

25

2suy ra f1 = 2f3

p2 = 2f0

Nn =Z23Z21

=6

13:

Chn A.

Bi ton 30: Mt on mch gm in tr thun R = 32 mc ni tip vi mt cun cm thun.t vo hai u on mch mt in p xoay chiu n nh c tn s f = 50Hz. Gi uR v uL lin p tc thi hai u in tr thun v hai u cun dy. Bit 65u2R + 256u

2L = 1600, t

cm ca cun dy l?

A.0; 1

H. B.

0; 36

H. C.

0; 26

H. D.

0; 16

H.

Li gii

T gi thit ta cu2R

8p65

13

!2 + u2L(2; 5)2 = 1. Vy U0R = 8p65

13V; U0L = 2; 5V

Lp t sU0LU0R

=ZLR) ZL = 2

p65 ) L =

p65

50 0; 16

.

Chn D.

Bi ton 31: Mt mch in AB gm on AM ni tip vi on MB. on AM gm in tr Rmc ni tip vi cun cm thun c t cm L thay i, on MB ch c t in C. in p tcthi uAB = 100

p2 cos1 00t V. iu chnh L = L1 th cng dng in hiu dng I = 0,5 A,

UMB = 100V , dng in i tr pha so vi uAB mt gc

3. iu chnh L = L2 in p hiu dng

UAM t cc i. Tnh t cm L2?

A. L2 =1 +

p2

H. B. L2 =

1 +p3

H. C. L2 =

1 +p6

H. D. L2 =

2 +p2

H.

Li gii

Ta c: ZC =100

0; 5= 200; tg' =

ZL ZCR

= tan 600 =p3 ) ZL ZC = R

p3

Z =U

I=

100

0; 5= 200

Z =qR2 + (ZL ZC)2 = 2R

9>=>;) R = 100

17


UAM = I:ZAM =UpR2 + Z2Lq

R2 + (ZL ZC)2=

UsR2 + Z2L + Z

2C 2ZLZC

R2 + Z1L

=Us

1 +400 (100 ZL)

1002 + Z2L

UAMmin ,100 ZL1002 + Z2L

= y = ymax

y = ymax khi o hm y0= 0) Z2L 200ZL 100 = 0

) ZL = 1001 +

p2

) L = 1 +

p2

(H)

Chn A.

Bi ton 32: Cho on mch xoay chiu gm in tr thun R, cun dy thun cm L, t in Cmc ni tip. t vo hai u on mch in p xoay chiu n nh th cng hiu dng cadng in l I1. Nu ni tt t in th cng dng in hiu dng l I2 = 2 I1, ng thi haidng in i1; i2 vung pha vi nhau. H s cng cut ca on mch khi khng ni tt t in l:

A. 0,1p2. B. 0,2

p3. C. 0,1

p5. D. 0,2

p5.

Li gii

Gi s h s cng sut ca mch lc trc v sau ln lt l cos'1; cos'2

Khi ta c

cos'2 = sin'12 cos'1 = cos'2

) cos'21 + 4 cos'21 = 1) cos'1 =1p5

Chn D.

Bi ton 33: t 1 in p xoay chiu uAB = U0: cos (100t)V vo hai u on mch ni tipth t gm AM cha cun dy thun cm c ZL = 200

p3, MB cha in tr R = 100 v t

in c ZC = 100p3. Ti thi im t, uAB =

U02

v ang gim th cng dng in c gi tr

i =

p3

2A. thi im t+

1

200s in p tc thi gia 2 u on mch MB c gi tr:

A. uMB = 150 V. B. uMB = 250 V. C. uMB = 50 V. D. uMB = 350 V.

Li gii

Ta c: tan' =p3) ' =

3vy i chm pha hn u

3Ti thi im t:

!U 0 hp vi trc honh mt gc

3v i > 0 nn

!I 0 nm trn trc honh. Ta c:

I0 = i1 =

p3

2A

thi im t+1

200= t+

T

4ta c

!I 0 quay thm c mt gc

2nn n hp vi trc honh mt

gc

2. uMB chm pha hn i mt gc

3nn

!U MB hp vi trc honh mt gc

6. Vy in p tc

thi gia 2 u on mch MB c gi tr bng: uMB = U0MB : cos

6= I0

pR2 + Z2C : cos

6= 150V .

Chn A.

Bi ton 34: Mt my pht in xoay chiu mt pha c roto l mt nam chm in c mt cpcc quay u vi tc n (b qua in tr thun cc cun dy phn ng). Mt on mch RLCc mc vo hai cc ca my. Khi roto quay vi tc n1 = 30 (vng/s) th dung khng t inbng R; cn khi roto quay vi tc n2 = 40 (vng/s) th in p hiu dng trn t in t gitr cc i. cng hiu dng qua mch t gi tr cc i th roto phi quay vi tc :

A. 120 vng/s. B. 220 vng/s. C. 10 vng/s. D. 50 vng/s.

Li gii

18


n = n1 = 30 vng/dy ! R = ZC1 ! !1:R:C = 1

n = n2 : UC =

p2:!:

1

!:Cq(ZL ZC)2 +R2

=

p2:1

Cq(ZL ZC)2 +R2

UCmax , ZL = ZC ! LC =1

!22

I =

p2:!q

(ZL ZC)2 +R2

n = n3: Thay i ! Imax trong trng hp ny tng t thay i ! ULmax trong mch

RLC mc ri tip c U khng i ! 1!3:C

=

rL

C R

2

2! !3 = 1

C:

rL

C R

2

2

=1r

LC R2C2

2

=1vuuut 1

!22

1

!21

2

Vy: n3 =1r

1

n22 1

2:n21

= 120 (vng/dy)

Chn A.

Bi ton 35: Mt my tng p c t l s vng dy 2 cun l 0,5. Nu ta t vo 2 u cun scp mt in p xoay chiu c gi tr hiu dng l 130V th in p o c 2 u cun th cp h l 240V . Hy lp t l gia in tr thun r ca cun dy s cp v cm khng ZL ca cundy ?

A. 5/12. B. 1/12. C. 1/3. D. 5/12.

Li gii

Ta c:N1N2

=ULU2

= 0; 5) UL = 0; 5U2 = 120VLi c: U2L + U

2r = U

21 = 130

2 ) Ur = 50VUrr

=ULZL

) rZL

=UrUL

=5

12Chn D.

Bi ton 36: Mt my bin p c li st gm n nhnh i xng nhng ch c hai nhnh l ccun dy.Coi hao ph ca my l rt nh. Khi in p xoay chiu mc vo cun 1( c s vng N1)th in p o cun 2 (C s vng N2) h l U2. Tnh U2 theo U, N1,N2 v n

A. U2 = U1N2

(n 2) :N1 . B. U2 = U1N3

(n 1) :N1 .

C. U2 = U2N1

(n 1) :N1 . D. U2 = U1N2

(n 1) :N1 .

Li gii

Sut in ng cun s cp: E1 = N101Th cp nhn c: E2 = N2

02

n 1Li c:

E1E2

=U1U2, U1

U2= (n 1) N1

N2. Chn D.

19


Bi ton 37: Cho on mch xoay chiu RLC1 mc ni tip (cun dy thun cm). Bit tn s

dng in l 50Hz, L =1

5(H) C1 =

103

5(F ). Mun dng din cc i th phi ghp thm vi t

in C1 mt t in dung C2 bng bao nhiu v ghp th no?

A. Ghp ni tip v C2 =3:104

. B. Ghp ni tip v C2 =

5:104

.

C. Ghp song song v C2 =3:104

. D. Ghp song song v C2 =

5:104

.

Li gii

Mun dng in cc i th C =1

L!2=

5:104

(F )

Thy C > C1 nn phi mc thm mt t in dung C2 =3:104

(F ) v mc song song. Chn C.

Bi ton 38: t mt in p u = U0: cos!:t (vi ! thay i) vo on mch R;L;C ni tiptha mn CR2 2L. Gi V1; V2; V3 ln lt l cc Vn k t vo ln lt hai u R;L;C. Khitng dn tn s th thy trn mi Vn k u c mt gi tr cc i, tm th t ln lt cc Vnk ch gi tr cc i khi tng dn tn s.(Gi s cc Vn k chy bnh thng)

Li gii

URmax , !21 =1

LC, ULmax , !22 =

1

C2

0B@ 1L

C R

2

2

1CA = 1LC C

2R2

2

> !21

UCmax , !23 =1

L2

L

C R

2

2

=

1

LC R

2

2L2< !21

Vy th t l V3; V1; V2

Bi ton 39: Mt m-t in s dng in p xoay chiu 220V 50Hz, h s cng sut ca m-tbng 0; 9 v coi tn hao m-t ch yu do s ta nhit. Cho in tr dy cun ca m-t l 10; 5.Ngi cng nhn dng m-t in nng mt kin hng c khi lng 100kg t mt t ln cao 36m trong thi gian 1 pht. Coi kin hng chuyn ng u. Ly g = 10 (m=s2). Cng ng in qua m-t gn gi tr no nht sau y?

A. 15 A. B. 8 A. C. 3 A. D. 4 A.

Li gii

Ta c:

Ptoan:phan = Pco:ich + Ptoa:nhiet , UI cos' = mght

+ I2R ) I ' 3; 8A: Chn D.Bi ton 40: t in p xoay chiu n nh vo 2 u on mch gm bin tr, cun cm thunv t in. Thay i bin tr n khi cng sut trn t cc i th dng in trong mch l

i = 2p2 cos

!t+

3

(A). Thay i bin tr n gi tr RX th cng sut trn mch lc ny l P

v dng in trong mch l i =p2 cos

!t+

2

(A). Thay i bin tr n gi tr RY th lc ny

l cng sut trn mch lc ny l P , dng in trong mch lc ny l?

A. i = 2 cos!t+

6

(A). B. i = 2 cos

!t+

3

(A).

C. i =p14 cos

!t+

4

(A). D. i =

p14 cos

!t+

6

(A).

Li gii

20


R thay i P t cc i khi v ch khi: R2 = (ZL ZC)2Khi lch pha ca u so vi i l ' =

4) 'u =

12; hoc 'u =

7

12;

Khi R = RX hoc R = RY th cng sut bng nhau nn ta c: RXRY = R2 = (ZL ZC)2

, ~iX ? ~iY, 'iY =

6Mt khc ta c P = UI cos' , IY =

p14 cos

!t+

6

(A)

Chn D.

Bi ton 41: t in p xoay chiu u = Up2 cos2 ft vo 2 u on mch gm R v C

mc ni tip. Khi tn s l f1 hoc f2 = 3f1 th h s cng sut tng ng ca on mch l

cos'2 =p2 cos'1. Khi tn s l f3 =

f1p2h s cng sut ca on mch cos'3 bng.

A.

p7

4. B.

p7

5. C.

p5

4. D.

p7

5.

Li gii

Theo ra:f2 = 3f1 ) !2 = 3!1 ) ZC1 = 3ZC2cos'2 =

p2 cos'1 , R2 = Z2C1 2Z2C2 =

7

9ZC1

f3 =f1p2) ZC3 =

p2ZC1

Nn: cos'3 =Rq

R2 + Z2C3

=Rr

R + 2:9

7R

=

p7

5

Chn D.

Bi ton 42: Mch in xoay chiu gm bin tr, cun dy khng thun cm v t in ghp nitip. t vo 2 u on mch in p c biu thc thc u = U

p2 sin!t (V ). Trong U v !

khng i. Khi bin tr R = 75 th cng sut tiu th trn bin tr t gi tr ln nht. Xc nhin tr thun ca cun dy v tng tr ca mch AB, bit chng c gi tr nguyn.

A. r = 21 v Z = 120. B. r = 21 v Z = 100.C. r = 21 v Z = 105. D. r = 15 v Z = 120.

Li gii

Cng sut tiu th trn c xc nh bi:

P = I2:R =U2

Z2:R =

U2

(R + r)2 + (ZL ZC)2:R =

U2

2r +R +r2 + (ZL ZC)2

R

cng sut trn R ln nht th 2r +R +r2 + (ZL ZC)2

Rmin

p dng BT Cauchy ta c Pmax , R2 = r2 + (ZL ZC)2Mt khc Z2 = R2 + 2Rr + r2 + (ZL Zc)2 nn:

Z2 = 2R2 + 2Rr = 2R (R + r) = 150 (R + r)

Theo ra Z; r 2 Z nn (R + r) 2 Z v Z2 chia ht cho 150So snh vi p n ta suy ra Z = 120, t ta c r = 21.

21


Chn A.

Bi ton 43: t in xoay chiu n nh c gi tr hiu dng 100p3 vo 2 u on AB gm

2 on mch AM v MB mc ni tip. on AM l cun dy c in tr thun, on MB gmin tr thun R mc ni tip vi t in. in p hiu dng trn MB gp i in p trn R vcng dng in hiu dng trn R v cng dng in hiu dng trn mch 0; 5A. in pMB vung pha vi in p AB. Cng sut tiu th ca ton on mch l?

A. 150W . B. 90W . C. 50W . D. 100W .

Li gii

Ta c: Z =U

I= 200

p3 ()

+UMB = 2UR ) UMB chm pha hn i mt gc 600+u nhanh pha hn UMB mt gc 90

0 ) u nhanh pha hn i mt gc 300) cos (300) = R

Z) R = 300 ()

Vy: P = 2RI2 = 2:300:0; 52 = 150 (W ) Chn A.

Bi ton 44: Mt on mch AB gm AM v MB mc ni tip. on mch AM gm R v C nitip vi on mch MB (gm L v r = R), t vo AB mt in p xoay chiu ch c tn s gcthay i th AM lun vung pha vi MB.Khi ! = !1 th in p trn AM c gi tr hiu dng lU1 v tr pha vi in p trn AB mt gc 1 Khi ! = !2 th in p trn AM c gi tr hiu dng

l U2 v tr pha vi in p trn AB mt gc 2. Bit 1+2 =

2v U1 = U2

p3. H s cng sut

ca AM trong 2 trng hp ! = !1 V ! = !2 Ln lt l:

A.

p3

2v

p3

2. B. 0,45 v 0,75. C.

p3

2v 0,5. D. 0,5 v

p3

2.

Li gii

Do uAM ? uMB ) tan'AM : tan'MB = 1, ZLZC = R2Ta c: U1 = U cos1; U2 = U cos2 = U sin1 (do 1 + 2 =

2)

) tan1 = U2U1

=p3 =

UMBUAM

AMC MBD ) ZLr

=R

ZC=BM

AM=p3

Vy tan' =ZL ZCR + r

=1p3) cos' =

p3

2Chn A.

Dng bi ny b c nh bi cc gi thit v ch thay i t l U1; U2 nn c th a ra cng thc

tng qut l viU1U2

= k th tan' =k 1

k2

, n SHIFT + tan+Ans) ') cos' =

22


Bi ton 45: Cho mch RLC ni tip. in tr thun R, L thay i c, t in c in dung C.

in p xoay chiu t vo 2 u mch u = Uo cos (!t). Khi thay i t cm n L1 =1

(H)

th cng dng in hiu dng qua mch cc i, lc cng sut ca mch bng 200W. Khi

thay i L n L2 =2

(H) th in p hiu dng gia 2 cun cm cc i bng 200V. in dung

C c gi tr:

A. C =200

(F ). B. C =

50

(F ). C. C =

150

(F ). D. C =

100

(F ).

Li gii

Ta c: khi L = L1 th Imax tc l cng hng nn: Pmax =U2

R= 200 v ZC = ZL1

Khi L = L2 = 2L1 ) ZL2 = 2ZL1 = 2ZC khi :ULmax ) ZL2 =

Z2C +R2

ZC, 2ZC = Z

2C +R

2

ZC) R = ZC

Khi : ULmax =UpR2 + Z2CR

, thay R = ZC vo ta c: Up2 = 200) U = 100p2

thay vo Pmax ta c R = 200 v t R = ZC = ZL1 , t ta s tnh c C

Chn C.

Bi ton 46: Mt nh my thy in cung cp in cho 1 thnh ph cch n 80 km bng ngdy ti in 1 pha, h s cng sut ca ng dy bng 1.ng dy ti lm tiu hao 5% cng sutcn ti v thnh ph cn nhn c cng sut 47500 kW vi in p hiu dng 190 kV. ng

dy lm bng ng c in tr sut 1; 6:108:m v khi lng ring l8800kg

m3. Khi lng ng

dng lm ng dy ny bng.A. 190 tn. B. 90 tn. C. 180 tn. D. 100 tn.

Li gii

Ta c: Cng sut hao ph trn ng dy:

P =P 2R

U2= 0; 05P ) PR = 0; 05U2 ) R = 0; 05U

2

PTrong : 0; 95P = 47500kW ) P = 50000kWU = U0 +U = U0 + IR = U0 +

PR

U) U = U0 + 0; 05U

) U = U00; 95

= 200kV

) R = 0; 05U2

P=

0; 05:4:1010

5:107= 40

Li c: R =l

S) S = l

R

m = V D = SlD =lDl

R=:l2D

R= 1; 6:108:

162:108:8; 8:103

40 90 tn.

Chn B.

Bi ton 47: t in p u = Up2 cos!t vo 2 u on mch RLC v R l bin tr. Khi R c

gi tr R1 v R2 th cng sut tiu th ca mch l P1 = P2 = P v ng vi R1 in p 2 u onmch lch pha so vi dng in 1 gc . Cng sut ca mch ln nht khi R = Ro l.

A. Ro =U2

2Psin 2. B. Ro =

U2

2Pcos2 . C. Ro =

U2

2Psin. D. Ro =

U2

Psin.

Li gii

23


Ta c: thay i R P cng gi tr ta s c nh sau:

P =U2R

R2 + (ZL ZC)2=

U2

R +(ZL ZC)2

R

U2

2jZL ZC j

T ta s thu c nh sau: khi R = R0 th Pmax =U2

2Rv khi c hai gi tr R1; R2 P bng

nhau th tg'1:tg'2 = 1) '1 + '2 = 2

Bi ton 48: t in p xoay chiu vo 2 u R, L, C (C thay i) L =3

(H). Khi C = C1 =

0; 1

2(mF) th dng in tr pha

4so vi in p 2 u on mch. Khi C =

3C15

th in p gia

2 bn t cc i. Tn s trong mch l?A. 100. B. 200. C. 300. D. 150.

Li gii

Ta c: C = C1 ) tg'1 = ZL ZC1R

= tg

4(1)

Li c: C = C2 ) ZC2 =5

3ZC1 =

R2 + Z2LZL

=(ZL ZC1)2 + Z2L

ZL

) 5ZC13

=(ZL ZC1)2 + Z2L

ZL=Z2C1 2ZLZC1 + 2Z2L

ZL(2)

Chia hai v ca (2) cho ZL ta c:

5ZC13ZL

=

ZC1ZL

2 2ZC1

ZL+ 2) X2 11X

3+ 2 = 0 vi X l

ZC1ZL

Ta s c 2 nghim:

Nghim th nht:

X = 3) ZC1 = 3ZL khng tho (1)Nghim th hai:

X =2

3) ZC1 =

2

3ZL ) 1

!C1=

2

3!L) ! =

r3

2LC1= 100 (rad=s)

Chn A.

Bi ton 49: t hiu in th xoay chiu u = Uo cos (100t+ ') (V ) hai u on mch ni tiptheo th t R1, R2 v cun thun cm c t cm L thay i c. Bit R1=2R2=200

p3. iu

chnh L cho n khi hiu in thc thi gia 2 u mch cha R2 v L lch pha cc i so vi hiuin th gia 2 u on mch, gi tr ca t cm lc l.

A. L =2

. B. L =

3

. C. L =

4

. D. L =

1

.

Li gii

tg'1 =ZLR1

tg' =ZL

R1 +R2

) tg'1 = 32tg'

lch pha uLR2 vi U mch l ln:

tg ('1 ') = tg'1 tg'1 + tg'1:tg'

=0; 5:tg'

1 + 1; 5:tg2'S dng BT Cauchy ta s c kt qu.

24


Bi ton 50: X l hp en cha trong 3 phn t L1; R1; C1 mc ni tip UAN =

100 cos (100t) ; UMB = 200 cos100t

3

v ! = 100 =

1pLC

. Vit biu thc UX?

A. UX = 25p28 cos

100t 41

180

. B. UX = 16

p28 cos

100t+

36

180

.

C. UX = 25p28 cos

100t+

36

180

. D. UX = 16

p28 cos

100t 41

180

.

Li gii

ZL = L!;ZC =1

C!;ZL = ZC

) !UAL = !UL +!UX!U MB =

!U0 +

!UX

Vi UMB = 2YAN = 100p2

Ly trc (), biu din!UAL;

!UMB

Xt tam gic OHK: HK = 2U2 = 2UC

) HK =r

50p22

+100p22 2:50:100: cos

3= 50

p6

) UL = UC = 25p6 (V )

Dng nh lut hm s sin:HK

sin

3

=CK

sin=

50p6p3

2

=100p2

sin) = 900 ) !UL ? ()

Li c: tg'x =HE

OH=

25p6

50p2

=

p3

2) 'X 410 UX =

pOH2 +HE2 =

p252:6 + 502:2 =

25p14 (V )

uX = UXp2 cos (100t 'x) = 25

p28 cos

100t 41

180

25


Chn A.

Bi ton 51: Trn on mch xoay chiu khng phn nhnh c bn im theo ng th t A, M,N v B. Gia hai im A v M ch c in tr thun, gia hai im M v N ch c cun dy, gia2 im N v B ch c t in. t vo hai u on mch AB mt in p xoay chiu 175 V 50Hz th in p hiu dng trn on AM l 25V, trn on MN l 25V v trn on NB l 175V.H s cng sut ca on mch AB l

A.3

5. B.

1

7. C.

17

25. D.

7

25.

Li gii

Li gii 1: Theo bi, ta c: UAB = UC = 175 (V ) ; UR = Ud = 25 (V )

Cun dy c in tr r.

Ta c: U2d = U2r + U

2L = 25

2 (1)

U2AB = (UR + Ur)2 + (UL UC)2

= U2R + (U2L + U

2r ) + U

2C + 2URUr 2ULUC

) U2AB = U2R + U2d + U2C + 2URUr 2ULUC = 1752 (2)T (1) v (2) ) UL = 7 (V ) ; Ur = 7UL 25 = 24 (V )Li c: cos' =

UR + UrUAB

=25 + 24

175=

7

25Chn D.

Li gii 2:

Tam gic ABN v AMN cn ti B v M. Ta c: NB=HB+NH

) 175: sin'+ 25: cos' = 175) 1752 (1 sin')2 = 2) sin' = 2425

Vy: cos' =7

25Chn D.

26


Bi ton 52: Khi t vo hai u cun dy mt in p xoay chiu 120V 50Hz th thy dngin chy qua cun dy c gi tr hiu dng l 2A v tr pha

3so vi in p hai u mch. Khi

ni tip cun dy trn vi mt mch in X ri t vo 2 u on m ch in p xoay chiu nh

trn th thy dng in chy qua mch c gi tr hiu dng 1A v sm pha

6so vi in p hai

u on mch X. Cng sut tiu th trn ton mch khi ghp thm X l:A. 120W. B. 300W. C. 200

p2 W. D. 300

p3W.

Li gii

Ta c: Zd =U

I=

120

2= 60

Khi mc ni tip vi X th: Ud = I:Zd = 1:60 = 60 (V )

Da vo gin vect ta thy: Tam gic AMB vung ti M nn: cos =AM

AB=

1

2) = 600 )

!AB !I ) P = UI cos = UI = 120 (W )Chn A.

Bi ton 53: Mt on mch gm mt cun cm c t cm L v in tr thun r mc ni tipvi mt t in c in dung C thay i c. t vo 2 u mch mt in p xoay chiu c gitr hiu dng U v tn s f khng i. Khi iu chnh in dung t in c gi tr C = C1 thin p gia hai u t v cun cm c cng gi tr bng U, cng dng in trong mch khi

c biu thc i1 = 2p6 cos

100t+

4

(A). Khi iu chnh C = C2 th in p gia 2 bn t in

c gi tr cc i. Cng dng in tc thi trong mch c biu thc l:

A. i2 = 2p2 cos

100t+

5

12

(A). B. i2 = 2

p3 cos

100t+

5

12

(A).

C. i2 = 2p2 cos

100t+

3

(A). D. i2 = 2

p3 cos

100t+

3

(A).

Li gii

Khi C = C1: r2 + Z2L = Z

2C = r

2 + (ZL ZC)2) ZC = 2ZL; r = ZL

p3) tan' = 1p

3Khi C = C2 ta c: ZC2 = 4ZL ) Z2 = Z1

p3; tan' = p3

Vy i = 2p2 cos

100t+

5

12

Chn A.

27


Bi ton 54: t in p u = 200 cos(100t)V vo hai u on mch gm AM cha in trthun R mc ni tip vi mt t in, MB cha cun dy. in p trn AM v BM lch pha so vi

dng in ln lt l '1 v '2 sao cho '2 '1= 2. Biu thc in p trn AM c th l

A. u = 50p2 cos

100t

3

. B. u = 50

p2 cos

100t

6

.

C. u = 100 cos100t

6

. D. u = 100 cos

100t

3

.

Li gii

Do '2 '1= 2nn uAM v uMB vung pha

Mt khc: cos'AM =R

ZAM=

p3

2) ZAM = 2Rp

3

Tng t: cos'MB =R

ZMB=

1

2) ZMB = 2R

UAM = I:ZAM =2IRp

3=UMBp

3) UAB =

pU2AM + U

2MB = 2UAM ) UAM =

UAB2

= 50V

T ta chn p n B v uAM chm pha hn uAB

Chn B.

Bi ton 55: Cun s cp ca my bin p mc qua ampe k vo in p xoay chiu c gi trhiu dng 100V th ampe k ch 0; 0125A. Bit cun th cp mc vo mch gm 1 cun dy khngthun cm c r = 1, t cm L v mt in tr R = 9 mc ni tip. T s gia vng dy cuns cp v cun th cp bng 20. B qua hao ph. lch pha gia cng dng in v in p cun th cp l.

A.

3. B.

2. C.

4. D.

3 .

Li gii

Ta c:U1U2

=I2I1

=N1N2

= 20

) U2 = 5V ; I2 = 0; 25A

) Z = U2I2

= 20 Khi :

cos' =R + r

Z=

1

2) ' =

3

Chn A.

Bi ton 56: Mt phn xng c kh s dng mt ng c in xoay chiu c hiu sut 80% . Khing c hot ng n sinh ra mt cng sut c l 7; 5kW . Bit mi ngy ng c hot ng 8 giv gi tin ca mt s in cng nghip l 2000 ng. Trong mt thng (30 ngy) , s tin mphn xng phi tr cho ngnh in l

A. 1.350.000. B. 5.400.000.C. 2.700.000. D. 4.500.000.

Li gii

Ta c hiu sut: H =PdcPcc

Vi Pdc l cng sut ng c v Pcc l cng sut in cn cung cp

) Pcc = PdcH

= 9; 375kW

28


in nng tiu th in nng cung cp trong thng:

A = P:T = 9; 375:30:8 = 2250kWh

Vy s tin tr: 4.500.000.

Chn D.

Bi ton 57: t in p u = U0 cos!t vo hai u on mch AB gm hai on mch AM vMB mc ni tip. on AM gm cun cm thun L mc ni tip vi in tr R, on MB ch ct in C. Bit in p gia hai u on mch AM v in p gia hai u MB c gi tr hiu

dng bng nhau v cng dng in chy trong mch lch pha

12so vi in p hai u on

mch. H s cng sut hai u on mch AM bng

A.

p3

2. B.

p2

2. C.

1

2. D. 0,26.

Li gii

T gin vec-t ta thy ngay tam gic AMB cn ti M do UAM = UMB Khi :

\MAB =\MBA =

2

12=

5

12

Suy ra: =\MAB 12

=4

12=

3

Nn: cos'AM = cos =1

2Chn C.

Bi ton 58: t in p U = Uo cos (2ft)V vo hai u on mch RLC ni tip. Ln ltthay i tn s f = f1Hz; f = f1 + 150Hz; f = f1 + 50Hz th h s cng sut ca on mch ln

lt l 1;3

5;15

17. Tn s mch xy ra cng hng gn gi tr no nht sau y ?

A. 60 Hz. B. 150 Hz. C. 120 Hz. D. 100 Hz.

Li gii

ZL1 = ZC1 ;ZL2 ZC2 =4R

3;ZL3 ZC3 =

8R

15

29


Do c biu thc:(!22 !21)!3!2 (!23 !21)

=5

2! f1 60 (Hz).

Chn A.

Bi ton 59: t in p u = Up2 cos!t vo 2 u on mch mc ni tip gm 1 cun dy v

t in C thay i c. Ban u C = C1 in p hiu dng 2 u t in t cc i, dng insm pha 1 gc '. Khi C = C2 in p 2 u t in l 473; 2V dng in tr pha hn in p 2u 1 on mch l '. C = C3 in p hiu dng 2 u t in p l 473; 2V , in p hiu dng 2 u cun dy gim 100

p2 so vi khi C = C2, gi tr U l.

Li gii

Chn O lm gc, Oi l trc th hin dng in.

Ban u C = C1 in p hiu dng 2 u t in t cc i, dng in sm pha 1 gc '. Th

hin iu ny vi OB(th hin in p gia hai u mch) hp vi Oi mt gc '; OD th hin in

p hai u cun dy, OD vung gc vi OB, lc ny DB th hin in p hiu dng gia hai u t

in.

Khi C = C2 in p 2 u t in l 473,2V dng in tr pha hn in p 2 u 1 on mch l '.

Th hin iu ny OC l in p hiu dng hai u on mch, OC hp vi Oi gc '

C = C3 in p hiu dng 2 u t in p l 473,2V, in p hiu dng 2 u cun dy gim

100p2 so vi khi C = C2

OA th hin in p gia hai u on mch. T A k ng vung gc vi Oi ct OD ti E, lc

30


ny AE th hin in p hiu dng gia hai u t in, OE th hin in p hiu dng gia hai

u cun dy.

Nh vy OA = OB = OC = U ;AE = CD = 473; 2 (V ) ;ED = AC = 100p2 Tam gic AOC cn ti

O nn[AOC = 2\BOC = 4' Trong tam gic y ta c AC = 2OC: sin 2') 50p2 = U: sin'T ta tnh c U 100p2Bi ton 60: Cho mch in R, L, ,! khng i mc ni tip vi in dung C thay i c. iuchnh C (URL + UC)max. Tm ZC theo R v ZL.

Li gii

Ta c: U 0 =Uq

R2 + (ZL ZC)2p

R2 + Z2L + ZC

Xt: A =

rZ2LR2

+ 1 +ZCRs

1 +

ZLR ZC

R

2Ta chn: ZL = 2

p2R;

ZCR

= x. Khi : A2 =9 + x2 + 6x

x2 + 9 + 4p2x

= 1 +

6 4p2

x+9

x+ 4p2

Thy: x+9

x 6! Amax ! U 0max khi : ZC =

pR2 + Z2L

Bi ton 61: Cho on mch xoay chiu R, L, C mc ni tip. t vo 2 u mch 1 in pxoay chiu c tn s thay i c. Khi tn s ca in p 2 u mch l fo = 60Hz th inp hiu dng 2 u cun cm thun t cc i. Khi tn s ca in p 2 u l f = 50Hz thin p 2 u cun cm l uL = UL

p2 cos (100t+ '1). Khi f = f

0 th in p 2 u cun cm l

uL = ULp2 cos (100t+ '2). Bit UL =

UOLp2, gi tr !0 bng.

A. 160. B. 100. C. 200. D. 150.

Li gii

Theo ra khi ! = ! hoc ! = !0 th in p hai u cun cm bng nhau v bng: UL nn ta c:

1

!2+

1

!02= 2LC R2C2 (1)

Khi ! = !0:

!0 =

r2

2LC R2C2 ) 2LC R2C2 =

2

!20(2)

T (1) v (2) ta suy ra:1

!2+

1

!02=

2

!20

) !0 = 160

Chn A.

31


Bi ton 62: on mch mc ni tip AB gm t in c in dung C =1

6mF , cun cm c

t cm L =0; 3

(H) v in tr trong r = 10, v mt bin tr R. t vo hai u on mch

mt in p xoay chiu c in p hiu dng khng i cn tn s thay i c. Khi f = 50Hz,thay i R th in p hiu dng trn t t gi tr cc i l U1. Khi R = 30, thay i f th by

gi in p hiu dng trn t t gi tr cc i l U2. T sU1U2

gn gi tr no nht sau y:

A. 0,79. B. 6,29. C. 1,58. D. 3,15.

Li gii

Khi f = 50Hz, thay i R th in p hiu dng trn t t gi tr cc i l U1 lc ny l cng

hng.

U1 =U

ZZC =

3p10

5U

Khi R = 30, thay i f th by gi in p hiu dng trn t t gi tr cc i l U2

U2 = UC max =2UL

(r +R)q4LC C2: (R + r)2

, U2 = 1; 2U ) U1U2

= 1; 58

Chn C.

Bi ton 63: t in p u = Uo cos (!t+ ') v 2 u on mch gm cun dy khng thuncm mc ni tip vi t in C thay i c. Khi C = Co th cng dng in sm pha hnin p '1(0 < '1 < 90

o), in p hiu dng 2 u cun dy l 45V . Khi C = 2Co th cng dng in tr pha hn in p '2 = 120

o '1 in p hiu dng 2 u cun dy l 135V , gi trUo gn gi tr no nht.

A. 40 V. B. 80 V. C. 100 V. D. 200 V.

Li gii

Khi C = C0 t hnh v ta c:U

sin=

Ud1sin (900 '1)

32


Khi C = 2C0 t hnh v ta c:U

sin=

Ud2sin (900 + '2)

NnUd1

sin (900 '1) =Ud2

sin (900 + '2)

Suy ra:sin (900 '1)sin (900 + '2)

=1

3

cos'1sin (900 + '2)

=1

3

M '2 = 1200 '1 ta suy ra c '1 760; '2 440 Li c:8>>>:

tan (760) = ZL ZC1r

tan (440) =ZL ZC1

2r

Suy ra: 8>>>:ZC1 ZL

r 4

ZL ZC12

r 1

Suy ra:

8>:ZC1 =

5

3ZL

r =1

6ZL

M Ud1 =Uq

r2 + (ZL ZC1)2:pr2 + Z2L = 45V C s ng bc, n gin ta c th chn

ZL = 1; ZC1 =5

3; r =

1

6ta suy ra c U0 = U

p2 43V

Chn A.

Tnh chnh xc bng my tnh th ta tnh c: Ud1 =

p19

3U

Bi ton 64: ho on mch xoay chiu AB ni tip gm: AM cha bin tr R, on MN cha r,on NP cha cun cm thun, on PB cha t in c in dung bin thin. Ban u thay it in sao cho UAP khng thuc vo bin tr R. Gi nguyn gi tr in dung khi v thay ibin tr. Khi uAP lch pha cc i so vi uAB th UPB = U1. Khi (UAN :UNP ) cc i th UAM = U2.Bit rng U1 = 2

p6 +

p3U2, lch pha cc i gia uAP v uAB gn gi tr no nht sau

y:

A.3

7. B.

4

7. C.

5

7. D.

6

7.

Li gii

+ Khi thay i C UAP khng thuc vo bin tr R. D c:

ZC = 2ZL

+ Khi R thay i ta lun c APB lun l tam gic cn ti A (Hnh v)

33


Ta thy khi R thay i, nu ta di chuyn im A!M th gc 2' chnh l lch pha ca UAP vUAB cng ln. Vy lch pha cc i ca UAP v UAB khi im A trng vi im M hay lc

R = 0.

Khi , U1 = UPB =U

Z1:ZC =

Upr2 + Z2L

:2ZL

+ Khi R = R0 :

UAN :UNP U2AN + U

2NP

2=U2

2

Vy UAN :UNP ln nht khi UAN = UNB hay khi tam gic APB l tam gic vung cn

Lc ny: U2 = UAM = U: cos

4 Ur

Suy ra: U2 =Up2 Ur

T hnh v ta cng suy ra :ZL = R + r; Z2 =p2 (R + r) Nn:

U2 =Up2 I:r = Up

2 UZ2:r =

Up2 Up

2 (R + r):r

Hay: U2 =U: (ZL r)p

2ZLLi c. T bi: U1 = 2

p6 +

p3U2 nn ta c:

Upr2 + Z2L

:2ZL = 2p

6 +p3 U: (ZL r)p

2ZL

, ZLpZ2L + r

2=

p6 +

p3p

2:ZL rZL

) Z2L =p6 +

p3p

2: (ZL r)

qr2 + Z2L

,ZLr

2=

p6 +

p3p

2:

ZLr 1s

1 +

ZLr

2

34


t x = tan' =ZLr

ta c PT:

x2 =

p6 +

p3p

2: (x 1)

px2 + 1

Bng my tnh ta rt ra c x 1; 37672 suy ra ' = arc tanx 540Suy ra lch pha cc i bng 2' = 1080

Chn B.

Bi ton 65: t in p xoay chiu u = Up2 cos!t (V ) vo u on mch R, L, C (2L > R2C).

Khi ! = !1 th Uc = U h s cng sut lc ny l x. Khi ! = !1 + y th UL = U . Chn h thcng.

A. x2 =2!1

!1 + y

!1!1 + y

2. B. x2 =

!1!1 + y

!1!1 + y

2.

C. x2 =3!1

!1 + y

!1!1 + y

2. D. x2 =

4!1!1 + y

!1!1 + y

2.

Li gii

Ta c: UC1 = U ! Z2C1 = R2 + (ZL1 ZC1)2 ! R2 = 2ZL1 :ZC1 Z2L1Tng t, c: R2 = 2ZL2 :ZC2 Z2C2Nn: ZL1 = ZC2

C: x2 =R2

R2 + (ZL1 ZC1)2, x2 = R

2

Z2C1, x2 = 2ZL1 :ZC1

Z2C1 Z

2L1

Z2C1, x = 2!1

!2 !

21

!22Chn A.

Bi ton 66: t in p xoay chiu c biu thc u = Uo cos 100t (V )(trong t tnh bng s)vo hai u on mch cha R;L;C mc ni tip. Trong mt chu k, khong thi gian in p haiu on mch sinh cng m bng 5; 9ms. H s cng sut ca mch bng bao nhiu?

A. 0; 2. B. 0; 1. C. 0; 3. D. 0; 6.

Li gii

Gi s i = I0 cos (100t+ ')

Khi cng sut tc thi l: p = U0 cos (100t) I0 cos (100t+ ')

in p sinh cng m p < 0 khi hai ci cos biu thc trn tri du. lch pha ca hai ci l '

Dng ng trn lng gic, ta s thy p < 0 khi vt qut c mt gc bng 2'

Suy ra t =2'T

2= 5; 9ms) ' = 0; 295 ) cos' = 0; 6

Chn D.

Bi ton 67: Cho dng in xoay chiu c cng hiu dng I chy qua mt in tr R = 500.in tr c nhng trong mt bnh nhit lng k cha 2,5 lt nc 27 ( C), sau 10 pht nc

si. Cho khi lng ring, nhit dung ring ca nc l D = 1000kg

m3v C = 4; 2

kJ

kg:. B qua s

mt mt nng lng. Cng hiu dng I c gi tr lA. 1,6. B. 2. C. 3. D. 1.

Li gii

Q = mCT = Pt = I2R) I = 1:6Chn A.

35


Bi ton 68: Cho mch R, L, C mc ni tip. Tn s ca hiu in th thay i c. Khi tn sl f1 v 4f1 th cng sut trong mach nh nhau v u bng 80% cng sut cc i m mch cth t c. Khi tn s l 3f1 th h s cng sut l?

A. 0,2. B. 0,5. C. 0,96. D. 0,6.

Li gii

Khi tn s l f1 v 4f1 th cng sut trong mach nh nhau) jZ1L Z1C j =4Z1L Z1C4

) Z1L =Z1C4

Cng sut =80% cng sut cc i ) U2R

R2 + (z1L 4z1L)2=

4

5

U2

R) R = 6z1L

cos'3 =Rs

R2 +

3Z1L Z1C

3

2 = 0:96Chn C.

Bi ton 69: on mch xoay chiu AB theo th t R, C, L khng thun cm. Bit R2 = r2 =L

C.

t in p xoay chiu vo AB, th h s cng sut ca mch AB l 0,96. TmURCUrL

A.3

8. B.

3

5. C.

1

4. D.

3

4.

Li gii

Dng ton ny ty chn lng cht l nhanh nht. Ta c cos' =R + r

Z= 0; 96 =

24

25Chn R + r = 24) R = r = 12 v Z = 25m Z2 = (R + r)2 + (ZL ZC)2 = 252 ) ZL ZC = 7Theo bi ta c h: 8>:

!L 1!C

= 7

L

C= 122 = 144

) 144!2C2 7!C 1 = 0) !C = 19

T suy ra ZL = 16;ZC = 9

Khi URCUrL

=

pR2 + Z2Cpr2 + Z2L

=3

4Chn D.

Bi ton 70: Mt cun dy c in tr R v t cm L ghp ni tip vi t in c in dungC ri mc vo mch in xoay chiu c tn s f. Dng vn k nhit o hiu in th ta thy giahai u mch in l U = 37; 5V ; gia hai u cun cm UL = 50V ; gia hai u bn t inUC = 17; 5V . Dng ampe k nhit o cng dng in ta thy I = 0; 1A. Khi tn s f thay in gi tr fm = 330Hz th cng dng in trong mch in t gi tr cc i. Tn s f lcban u l:

A. 70Hz. B. 50Hz. C. 100Hz. D. 500Hz.

Li gii

36


Ta c: Z =U

I= 375; Zd =

UdI

= 500; ZC =UCI

= 175

Z2 = r2 + (ZL ZC)2 v Z2d = r2 + Z2L ) ZL =Z2C + Z

2d Z2

2ZC= 400

ZL = 2fL;ZC =1

2fC) ZL

ZC= 42f 2LC ) 42LC = 400

175:1

f 2(1)

Khi I = Imax ) 2fmaxL = 12fmaxC

) 42LC = 1f 2max

(2)

T (1) ; (2)) f = 4fmaxp7

=4:330p

7 500Hz

Chn D.

Bi ton 71: Mch gm in tr R = 100 mc ni tip C=104

F , in p t vo hai u on

mch c dng u = 400 cos2 (50t). Cng hiu dng qua mch c gi tr bng?

Li gii

Ta c u = 200 + 200 cos (100t)t s cn dng 1 chiu. Nn ta c Uhd=100p2 mt khc tnh c

Zc=100 ) Z =pR2 + Z2c t tnh c I = 1

Nu bi ton ch c R v L mi c I =p5 nh v lc khng c s cn dng 1 chiu.

Bi ton 72: Mt bin tr c mc vo mt ngun pin 3V c in tr trong r = 10. Ngita iu chnh bin tr t gi tr R1 th cng sut tiu th trn bin tr l ln nht. Gi nguynbin tr, sau mc vo cnh in tr ca mt mch gm t in v cun cm mc ni tip c mt mch gm 3 phn t cun cm, t in, v in tr mc ni tip. Ni hai u mch vomt ngun dng xoay chiu c biu thc dng in tc thi i = 0; 2

p2 cos 100t (V ). Ngi ta iu

chnh cun cm v t in sao cho in p gia hai u cun cm vung pha vi in p gia haiu mch c t v tr. Bit khi iu chnh cun cm ch thay i c thnh phn cm khng can. Khi in p gia hai u mch t gi tr cc tiu th ngi ta thy cng sut tiu th trnmch l 1,2 W. in p s o c gia hai u cun dy l:

A. 1V. B. 2V. C. 2p3V . D. 2

p6V .

Li gii

Li gii 1:!U1 =

!UL +

!Ur!

U2 =!UC +

!UR

Mch c cp mt dng xoay chiu c nh; ng thi thnh phn R v r c c nh nn thnh

phn UR v Urcng c nh khng thay i. Khi thay i L v C sao cho U1 vung pha vi U2. Ta

d thy hiu in th hai u mch!U =

!U1 +

!U2 s nh nht khi UL = UC =

pUR:Ur

T thng tin th nht ta c R = 10 (= r).

Cng sut ca mch P = I2 (R + r)

) r = PI2R = 20:

Lc ny:

U1 = Iqr2 + Z2L = I

pR2 +Rr = 0; 2

p202 + 20:10 = 2

p6 (V ) :

37


Li gii 2:

Khi mc bin tr vo pin ta c cng sut ca bin tr:

P1 = RI21 =

E2R

(R + r)2 E

2R

4Rr

Cng sut trn cc i khi v ch khi:

R = r = 10

in cm ca cun cm v ca on RC vung pha khi:

ZLZC = RR0 (1)

in p hai u mch cc tiu khi tng tr mch cc tiu. Nn ta c:

ZL = ZC (2)

Li c: cng sut ca mch:

P = I2 (R +R0)) R0 = 20

T (1) ; (2) ta c:

ZL =pRR0 = 10

p2) Zd =

qR20 + Z

2L = 10

p6

Vy: Ud = I:Zd = 2p6 (V )

Chn D.

Bi ton 73: Mch in xoay chiu AB gm 2 on AN v NB mc ni tip. on AN cha intr R = 50 v Cun dy thun cm c t cm L. on NB cha t in vi in dung C c

th thay i c. Bit rng khi UAN + 2UNB t cc tiu th h s cng sut l1p2. Tm cos'd ?

Li gii

UAN + 2UMB =

pR2 + Z2L + 2ZCqR2 + (ZL ZC)2

U:

) (UAN + 2UMB)min, f (ZC) =pR2 + Z2L + 2ZCqR2 + (ZL ZC)2

min

, f 0 (ZC) = 0, ZC = 2R2

2ZL +pR2 + Z2L

+ ZL:

Kt hp vi ZC ZL = R ta c 2R 2ZL =pR2 + Z2L

) R = 4 +p7

3ZL ) ZC = 7 +

p7

3ZL:

) cos'd =4 +

p7

3vuut 4 +p73

!2+ 1

0; 9114:

38


Comment: Gi lch pha ca uAB v i l . Ta c:UAN

sin2

= UNBsin ( + 'd)

=UAB

sin2 'd

) UAN + 2UNB = UAB

cos'd[cos + 2 sin (+ 'd)]

=UABcos'd

[(1 + 2 sin'd) cos + 2 cos'd sin]

Vi 2h0;

2

ith hm f () = (1 + 2 sin'd) cos+2 cos'd sin s t cc tiu ti bin nn khng

th no khi =

4c. Em xem li xem bi l cc i hay cc tiu?

Nu bi l cc i gi tr cc i t c ti f 0 () = 0

, tan = 2 cos'd1 + 2 sin'd

, 2 cos'd = 1 + 2 sin'd , sin'd

4

= 1

2p2, 'd = 24; 3o ) cos'd = 0; 91

Hoc c th dng bt ng thc BCS on nh gi:

(1 + 2 sin'd) cos + 2 cos'd sin q[(1 + 2 sin'd)

2 + (2 cos'd)2]sin2 + cos2

=q(1 + 2 sin'd)

2 + (2 cos'd)2

Du " = " xy ra khi tan =2 cos'd

1 + 2 sin'd

Bi ton 74: Cho mch in xoay chiu AB, AN cha cun dy, NB cha t in. t 1 hiu inth khng i vo 2 u on mch AB. Bit gi tr ca t in c th thay i c v in p

hai u mch AN lun sm pha hn cng dng in mt gc

7. iu chnh gi tr ca t

gi tr UAN + UNB t gi tr cc i. H s cng sut t khi l.

Li giiUAB

sin5

14

=UAN

sin2 '

= UNBsin7+ '

= UAN + UNBsin2 '

+ sin

7+ '

) UAN + UNB = UAB5

14

[sin2 '

+ sin

7+ '

= 2 sin

9

28cos

'+

5

28

(UAN + UNB)max , ' = 5

28) cos' 0; 85

Bi ton 75: Cho mch RLC mc ni tip, tn s gc c th thay i. Khi iu chnh cho !nhn cc gi tr ln lt l 60,66,70,72,80,90 th in p gia hai u t c gi tr tng ng lUc : 1; 2; 3; 4; 5; 6. Bit UC2 = UC6 . Gi tr ln nht trong cc gi tri Uc l

A. 1. B. 2. C. 5. D. 4.

Li gii

Bi ny ch cn p CT l xong

! thay i c hai gi tr cho UC bng nhau th

!2 =1

2(!22 + !

26) ) ! 78; 9

Vy UC5 c gi tr ln nht.

39


Bi ton 76: Khi t dng in p xoay chiu vo hai u mch gm in tr thun R mc ni

tip t in C th biu thc dng in c dng i1 = I0 cos!t+

6

(A). Mc ni tip thm vo

mch in cun dy thun cm L ri mc vo in p ni trn th biu thc dng in c dng

i2 = I0 cos!t

3

(A). Biu thc hai u on mch c dng?

Li gii

Gi biu thc ca mch l. u = Uo cos (!t+ 'u) do c cng Io nn cng gi tr hiu dng.

Ta c I1 = I2 , (ZL ZC)2 = Z2C , ZL = 2ZCM tan'1 =

ZCR

, tan'2 =ZL ZC

R, '1 + '2 = 0

'1 = 'u 6

'2 = 'u +

3

, 'u = 12

Bi ton 77: Trong 1 hp en c hai trong ba linh kin: cun cm, in tr thun, t in. Khit vo mch u = 100

p2 cos (!t) (V ) th i =

p2 cos (!t). Khi gi nguyn U, tng ! ln

p2 ln th

mch c h s cng sut l1p2. Hi nu t gi tr ba u ca ! gim i hai ln th h s cng sut

mi l bao nhiu?A. 0; 26. B. 0; 326. C. 0; 426. D. 0; 126.

Li gii

D thy lc u mch xy ra cng hng

Do Z0L = Z0C

Khi !p2 th ZL = 2ZC

M h s cng sut l1p2nn R = ZC =

Z0Cp2

Khi ! gim i hai ln th Z1L = 0; 5Z0C ;Z1C = 2Z0C

T tnh c h s cng sut lc l 0,426

p n C.

Bi ton 78: in nng c truyn t mt nh my in n khu cng nghip (KCN) bng

ng dy mt pha. Nu in p truyn i l U th KCN phi lp mt my h p vi t s54

1

p ng12

13nhu cu in nng ca KCN. Nu mun cung cp in nng cho KCN th in p

truyn i l 2U , khi phi dng my h p nh th no? Coi h s cng sut lun bng 1

A.117

5. B.

117

3. C.

117

1. D.

17

1.

Li gii

Gi in p cun th cp ca my h p l Uo th in p cun s cp ca my h p trong 2

trng hp l 54U0;nUo.

Gi s khu cng nghip tiu th 1 cng sut 39W.

Ban u cng sut cn truyn i, hao ph trn ng dy ln lt l P ; P

Khi in p truyn i l 2U th hao ph trn ng dy l:P4

40


Ta c h phng trnh: 8>:P P = 12

1339

P P4

= 39)P = 40 (W )P = 4 (W )

Mt khc: 8>:H1 =

54UoU

=36

40

H2 =nUo2U

=39

40

) n = 117

Chn C.

Bi ton 79: t mt in p u = Up2 cos!t vo on mch AB ni tip. Gia hai im AM l

mt bin tr R, gia MN l mt cun dy c r v gia NB l t C. Khi R = 75 th ng thi cbin tr R tiu th cng sut cc i v thm bt k mt t C' no vo NB d l ni tip hay songsong vi t C vn thy U trn NB gim. Bit cc gi tr r; ZL; ZC ; Z nguyn. Gi tr ca r v ZCl?

A. 45;12. B. 17;32. C. 29;86. D. 21;200.

Li gii

Ta c: PR = I2R =

U2R

(R + r)2 + (ZL ZC)2=

U2

R +r2 + (ZL ZC)2

R+ 2r

PR = PRmax khi R2 = r2 + (ZLZC)2 (1)

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

Do ta c: ZC =(R + r)2 + Z2L

ZL=

(R + r)2

ZL+ 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 + n2 = R2 = 752 (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 = 72Thay R; r; n vo (3) ) ZL = 128. Thay vo (4) ) ZC = 200

T ta chn p n D.

Bi ton 80: t in p u = U0 cos (!t) vi U0 v ! khng i vo hai u on mch mc nitip in tr R, t in C v cun cm thun L thay i c. Khi L = L1 v L = L2: in phiu dng hai u cun cm c cng gi tr; lch pha ca in p hai u on mch so vicng dng in ln lt l 0,32 rad v 1,28 rad. Khi L = L0: in p hiu dng gia hai ucun cm t cc i; lch pha gia hai u dng in v in p hai u mch l '. Gi trca ' gn gi tr no nht sau y?

A. 0,75 rad. B. 1,7 rad. C. 0,5 rad. D. 1,2 rad.

Li gii

41


Ta c: UL =U:ZLq

R2 + (ZL ZC)2=

UR:ZL

R:qR2 + (ZL ZC)2

= U: cos':ZLR

) ULZL

=U cos'

R)

8>:ULZL1

=U: cos'1

RULZL2

=U: cos'2

R

) ULZL1

+ULZL2

=U (cos'1 + cos'2)

R= UL

1

ZL1+

1

ZL2

= ULmax :

2

ZLmax=

2U

Rcos'

) cos' = cos'1 + cos'22

) ' = 0; 933.Do gn gi tr no nht nn ta chn p n A.

Bi ton 81: t mt in p xoay chiu u = U0 cos (!t+ ') vo hai u on mch ni tip gmin tr, t in v cun cm thun c t cm L thay i c. Thay i L th thy in phiu dng hai u cun dy t cc i bng U0

p2. ng thi trong qu trnh thay i c hai gi

tr ca L cho in p hiu dng hai u cun dy nh nhau m cng sut tiu th ca mch tronghai trng hp ny tng ng l P v 4P . Hi cng sut tiu th cc i ca mch l bao nhiu

A. P . B. 4P . C. 2P . D. P .

Li gii

Do UL = U0p2 = 2U nn tam gic OHK l na tam gic u v ZC = R

p3

Mt khc khi P tng ln 4 ln th I tng ln 2 ln, m UL khng i nn cm khng gim hai ln:

ZL1 = 2ZL2

Vy: P =U2R

R2 +2ZL2 R

p32 (1) v 4P = U2R

R2 +ZL2 R

p32 (2)

Ly (2) chia (1) ta c: ZL2 = Rp3 (3)

Li c: Pmax =U2

R(4)

Thay (3) vo (2) ri ly (4) chia (2) sau khi thay (3) vo

) PmaxP

= 4) Pmax = 4P

T ta chn c p n B.

Bi ton 82: Khu chung c Times City tiu th mt cng sut in 14289W , trong cc dngc in khu ny u hot ng bnh thng hiu in th hiu 220V . in tr dy ti in tni cp in n khu chung c l r. Khi khu chung c khng dng my bin p h th, cc dngc in ca khu nh ny hot ng bnh thng th hiu in th hiu dng ni cp in l 359 V,

khi hiu in th tc thi hai u dy ca khu nh nhanh pha

6so vi dng in tc thi

chy trong mch. Khi khu nh ny dng my bin p h th l tng c t sN1N2

= 15, cc dng

c in ca khu chung c vn hot ng bnh thng ging nh khi khng dng my bin p hth th hiu in th hiu dng ni cp in l (bit h s cng sut mch s cp my bin p l1):

A. 2236. B. 1234. C. 3309. D. 2014.

Li gii

42


Ta c: khi khng dng my bin p: cng dng in chy qua mch cung cp cho khu tp th

cng chnh l dng in chy qua ng dy ti I =P

U cos'=

14289

220: cos

6

= 75 (V )

st p trn ng dy: U1 = U1 U = Ir ) r = U1 UI

=359 220

75=

139

75

Khi dng my bin p: in p hiu dng ni cung cp l: U2 = U2 + U0

U0l in p hai u cun s cp: U

0= U

N1N2

= 15U = 3300 (V )

st p trn ng dy: U2 = I0r

Vi I0l cng dng in chy qua cun s cp:

I0= I

N2N1

=I

15= 5 (A)) U2 = I 0r = 5:139

75= 9; 27 9; 3 (V )

Do : U2 = U2 + U0= 3309; 3 (V ) 3309 (V )

T ta chn p n C.

Bi ton 83: t in p u = 120p2 cos 2ft (V ) (f thay i c) vo hai u on mch mc

ni tip gm cun cm thun c t cm L, in tr R v t in c in dung C, vi CR2 < 2L.Khi f = 25Hz th in p hiu dng hai u t in cc i. Khi f= 50Hz th in p hai u intr cc i. Khi f= 100Hz th in p hai u cun cm cc i. Gi tr cc i ca cun cm gngi tr no nht sau y?

A. 100V. B. 124 V. C. 200V. D. 180V.

Li gii

Cch 1:

Ta c:

U

ULmax

2+

!21!23

2= 1 ) U

2

U2Lmax=f 21f 23) U

2

U2Lmax+

252

1002= 1

) 1516U2Lmax = 120

2 ) ULmax = 32p15 (V ) 124 (V )

Cch 2:

Cng thc tnh ULmax l:

ULmax =Us

1fCfL

2 = 120s1

25

100

2 = 32p15 (V ) 124 (V )

Cch ny s d hn v y l cng thc quen thuc trong chng ny.

Chn B.

Bi ton 84: Cho mch in xoay chiu mc ni tip RLC c R = 100 (). t vo hai u mchin p xoay chiu u = 200

p2 cos (!t+ ') (V ) c tn s gc thay i c. Khi ! thay i ti !1

v !2 th1

f 21+

1

f 22=

1

122. Bit ZL1ZC1 = 31200 v ZL1 =

13

8ZC2 . Tm tng tr Z ca on mch?

A. 106 (). B. 100 (). C. 200 (). D. 150 ().

Li gii

Ta c: ZL1 :ZC1 = 31200)L

C= 31200) ZL2ZC2 = 31200) L2!1!2 =

31200

8

Li c:1

f 21+

1

f 22=

1

122) 1

!21+

1

!22=

1

(24)2

) !0 = 24 l tn s ULmax

43


) LC R

2

2=

1

C!0

2T hai trn ta lp c h:

8>:!1!1 =

12:31200

8L21

!21+

1

!22=

1

576

Vi: C =

r1

26200:242:2v L = 31200:

r1

26200:242:2

T ta tnh !1; !2 tm c Z

Ta c th (n may) d on c a l ng cao ca tam gic vung vi 2 cnh gc vung l a1 v

a2.

Nu chn Ni i cht v tam gic Ai-Cp -Tam gic vung vi s o 3 cnh t l: 3 : 4 : 5. Gi a1;a2

l 2 cnh gc vung;h l ng cao k t nh gc vung th ta c:

a1 : a2 : h = 3 : 4 : 2; 4 = 15 : 20 : 12

Ti y. Ta c th d on f1 = 15; f2 = 20 ho f1 = 20; f2 = 15.

Chn A.

Bi ton 85: Cho mch in xoay chiu AB gm in tr thun R = 100, cun dy thuncm L, t in c in dung C. t vo hai u on mch mt in p xoay chiu u =

220p2 cos (100t) (V ), bit ZL = 2ZC . thi im t in p hai u in tr R l 60 (V ), hai

u t in l 40 (V ). Hi in p hai u on mch AB khi l:A. 100 (V ). B. 20 (V ). C. 10 (V ). D. 30 (V ).

Li gii

Ta c: Hi in p thi im t chnh l gi tr tc th:

u = uR + uL + uC = uR + (2uC) + uC = 60 40 = 20 (V )Do: ZL = 2ZC , v uL ngc pha uC ) uL = 2uC .Chn B.

Bi ton 86: Cho mch in RLC, vi C thay i c v tng in tr thun ca ton mchl 100. in p gia hai u on mch l u = 100

p2 cos (100t) (V ). Khi C tng ln 2 ln th

cng dng in vn nh c nhng pha ban u ca dng in thay i mt gc

2. Biu thc

cng dng in trc khi tng in dung l

A. i = 1 cos100t

4

(A). B. i = 1 cos

100t

3

(A).

C. i = 2 cos100t

4

(A). D. i = 2 cos

100t

3

(A).

Li gii(C = C1

C = C2: I1 = I2 = I ! Z1 = Z2 ! ZL = 3

4ZC1

(C = C1 : '1

C = C2 : '2;

8>:tan'1 =

ZL ZC1R + r

=ZC1

4 (R + r)tan'2 =

ZL ZC2R + r

=ZC1

4 (R + r)

j'1 '2j = 2! tan'1 tan'2 = 1! ZC1 = 4 (R + r) = 400 v ZL = 300

44


Suy ra '1 = 4v I0 = 1A

Chn A.

Bi ton 87: Ni hai cc ca my pht in xoay chiu mt pha vo hai u on mch AB gmin tr R, cun cm thun L, t in C mc ni tip theo th t. iu chnh tc quay caroto mch AB tiu th cng sut cc i th thy in p hiu dng hai u in tr ln hnin p hiu dng hai u t in 35 V . Gi M l im ni gia in tr v cun cm, hi in phiu dng hai u on mch MB lc ny c th l:

A. 23 V . B. 50 V . C. 70 V . D. 73 V .

Li gii

iu chnh tc quay r to PAB max ng vi trng hp UL max khi UAB khng i

Khi : U2R = 2UC : (UL UC) = 2 (UR 35) : (UL UC), U2R 2UR:UMB + 70UMB = 0; 0 = U2MB 70UMB 0) UMB 70Chn D.

Bi ton 88: t in p xoay chiu c gi tr hiu dng U khng i, tn s thay i c vohai u on mch gm in tr thun R, t in v cun cm thun mc ni tip. Khi tn s lf1 Hz th dung khng bng in tr R. Khi tn s l f2 Hz in p hiu dng hai u cun cmt cc i. Khi tn s l fo th mch xy ra cng hng in. Biu thc lin h gia f1; f2 v fol

A.5

f 2o 1f 22

=1

2f 21. B.

2

f 2o 1f 22

=1

2f 21. C.

1

f 2o 1f 22

=1

2f 21. D.

1

f 2o 1f 22

=1

3f 21.

Li gii

Ta c:1

!21= (RC)2;

1

(!2C)2 =

L

C R

2

2! 1

!22= LC (RC)

2

21

!2

0

= LC

Do :1

2!21+

1

!22=

1

!20p n C.

Bi ton 89: t in p xoay chiu c gi tr hiu dng v tn s khng i vo hai u onmch ni tip gm in tr R, cun cm thun L v t in C c in dung thay i c. Lnlt thay i in dung C in p hiu dng hai u t in c gi tr cc i v na cc ith thy dng in trong trng hp sau nhanh pha hn dng in trong trng hp trc mt gc'. Hi ' gn gi tr no nht sau y:

A. 3 rad. B. 1 rad. C. 1; 1 rad. D. 1; 4 rad.

Li gii

Hng ch c l t cng thc kh c ny:

UC = UCmax cos (' 'o) :

p dng cng thc trn:

UC1 = UCmax cos ('1 'o) :

UC2 = UCmax cos ('2 'o) :

45


Kt hp vi bi ta c '1 = 'o

nn cos' = cos ('2 '1) = cos ('2 'o) = 12) ' =

3Chn C.

Bi ton 90: Mc mt cun cm thun v mt t in mc song song ri mc vo in xoaychiu th dung khng gp i cm khng. Nu cng dng in qua t in c biu thci = 2 cos (!t) (A) th cng dng in qua cun cm c biu thc:

A. i = 4 cos (!t ) (A). B. i = 2 cos (!t ) (A).C. i = 3 cos (!t ) (A). D. i = 4 cos (!t+ ) (A).

Li gii

Ta c: Do on mch mc song song nn ta c: uL = uC v ZL = 2ZC ) uCZC

= 2uLZL

nn ta thu

c:1

iC=

2

iL) iL = 4

v iC ; iL ngc pha nhau nn s lch pha nhau mt gc . Vy biu thc s l: iL = 4 cos (!t ) (A)Chn A.

Bi ton 91: Cho dng in xoay chiu i = sin (100t) (A) (t o bng giy)chy qua bnh inphn cha dung dch H2SO4 vi cc in cc tr. Tnh th tch kh H2 (ktc) thot ra trong 16pht 5 giy l:

A. 0.1 lt. B. 0.224 lt. C. 0.12 lt. D. 0.222 lt.

Li gii

Ta c: nH2 =It!

F=

1

10p10:! =

1

10:!.

Vy th tch kh H2 l VH2 = nH2 :22; 4 = 10:22; 4 = 224 (ml) = 0; 224 (l).

Chn B.

Bi ton 92: Cho on mch RLC ni tip tha mn iu kin 2L > CR2. t vo hai u onmch in p xoay chiu u = U

p2 cos 2ft (V ) trong U khng i cn f c th thay i c.

Khi f = f1 th in p hiu dung trn t bng

k2 +

1

2

U , dng in chy qua mch c cng I,

ng thi mch tiu th cng sut bng (1 k22) cng sut cc i. Khi f = f2 = f1+f (f > 0),cng dng in trong mch cng l I. Khi f = f3 =

f

2th in p hai u t in ln nht.k2

gn gi tr no nht sau y?A. 0,4. B. 0,5. C. 0,2. D. 0,1.

Li gii

Chn Z = 1, ta c ngay ZL1 =1

2hoc ZL1 = 2k +

1

2) f1

f2=

4k + 1

2k + 1(loi) hoc

f1f2

=1

2k + 1

Li c cos2 ' =1

f1f2

+f2f1 1

= 1 k2

46


Bi ton 93: Cho mch in xoay chiu AB gm hai on mch AM v MB mc ni tip nhau.on mch AM cha in tr R1, cun cm thun L1 v t in C1 mc ni tip. on mch MBgm in tr R2, cun cm thun L2 v t in C2 mc ni tip. Bit in dung ca t in C1v t cm ca cun cm L2 c th thay i c. t in p u = 100 cos (100:t+ ') (V) vohai u on mch AB. Khi C1 = C v L2 = L th biu thc in p hai u on mch AM lu1AM = U cos 100:t (V). Khi C1 = C

0 v L2 = L0 th biu thc in p hai u on mch AM l

u2AM = 2U cos

100:t+

2

3

(V). Bit trong hai trng hp th in p hai u on mch AM

v MB u vung pha vi nhau. Gi tr ca U gn vi gi tr no nht trong cc gi tr sau?A. 5V. B. 10V. C. 20V. D. 30 V.

Li gii

uAM + uMB = u

Do uAM vung pha vi uMB nn v hnh th ta s c gc lch gia uAM v u mch:

cos' =U

100; cos

' 2

3

=

2U

100Gii ra c U = 32:7V

Chn D.

Bi ton 94: t in p xoay chiu c gi tr hiu dng U khng i, tn s thay i c vohai u on mch gm in tr thun, t in v cun cm thun mc ni tip theo th t .Khi tn s l 50 Hz th in p hiu dng hai u t in l U . Khi tn s l 125 Hz th in phiu dng hai u cun cm l U . Thay i tn s n khi in p gia hai u on mch chaRC v in p gia hai u L lch pha nhau mt gc 135o th tn s lc ny l

A. 100 Hz. B. 50 Hz. C. 60 Hz. D. 62; 5 Hz.

Li gii

Ta c: f = 50Hz;UC = U ) Z2C = R2 + (ZL ZC)2 (1)f = 125Hz;UL = U ) (2; 5ZL)2 = R2 +

2; 5ZL ZC

2; 5

2(2)

T (1) v (2) suy ra: ZC = 2; 5ZL = 1; 25R

Khi in p gia hai u on mch cha RC v in p gia hai u L lch pha nhau mt gc 135o

th ZC = R

Suy ra f0= 1; 25f = 62; 5Hz

Chn D.

Bi ton 95: t in p xoay chiu c gi tr hiu dng U khng i, tn s thay i c vohai u don mch gm in tr thun R, t in v cun cm thun mc ni tip. Khi tn s l62; 5 Hz th dung khng ca t bng in tr R. Khi tn s l 100 Hz th cm khng ca cuncm bng in tr R. Thay i tn s n khi in p gia hai u on t in t cc i th hs cng sut trong mch lc ny gn gi tr no nht sau y?

A. 0; 1:. B. 0; 8:. C. 0; 5:. D. 0; 2:.

Li gii

Ta c w1 = 125; ZC1 = R = 1; 6ZL1 ) w0 = 0; 4sqrt10w1 = 50sqrt10Khi UC t max ti w3 th R =

p2:ZL:: (ZC ZL)! R

ZL

=

s2:

ZCZL

1

47


, 1; 6!1!3

=

vuut2:"!0!3

2 1#) !3 = 100

) ZC3 = 1; 25R;ZL3 = 0; 5R) cos' =Rq

R2 + (0; 5R 1; 25R)2= 0; 8

Chn B.

Bi ton 96: t vo 2 u on mch AB in p xoay chiu c gi tr hiu dng v tn skhng i gm RLC trong L thay i c. Thay i L sao cho in p 2 u cun cm c gitr hiu dng 120V thy c 2 gi tha mn l L1 v L2. Tip tc thay i L sao cho ZL = 100 ththy in p gia 2 u in tr t gi tr cc i U. Tnh U bit ZL1 + ZL2 = 400

Li gii

Ta c: ZC = 100

2

ZLo=ZL1 + ZL2ZL1ZL2

=400

ZL1ZL2) ZL0 =

ZL1ZL2200

Li c:

ZL0 =R2 + Z2C

ZC

! ZL1ZL2200

=R2 + Z2C

100$ R

2 + Z2CZL1ZL2

=1

2

M:

1U

UL

2=R2 + Z2CZL1ZL2

=1

2

! U = ULp2= 60

p2 (V )

Ta c UL = UZLq

R2 + (ZL ZC)2! 1

U

UL

2!Z2L 2ZCZL +R2 = 0

Xem y l phng trnh bc hai n ZL, p dng nh l Viet ta c : ZL1 + ZL2 =2ZCU

UL

2 1

Khi cng hng th ZC = ZL = 100

Li c ZL1 + ZL2 = 400! 1U

UL

2=

1

2! U = ULp

2= 60

p2 (V )

Bi ton 97: Cho on mch AB gm hai on mch AM v BM mc ni tip. on mch AMgm cun thun cm ni tip vi t in theo th t , on mch MB ch c in tr thun R.

in p t vo AB c biu thc u = 80p2 cos 100t(V), h s cng sut ca mch AB l

p2

2. Khi

in p tc thi gia hai im A v M l 48V th in p tc thi gia hai im M v B c lnl:

A. 64 V. B. 60 V. C. 50 V. D. 100 V.

Li gii

Ch mu cht ca bi l uR(hay uMB)v uLC (hay uAM) vung pha nh

Do ta c h thc :

48


uRU0R

2+

uLCU0LC

2= 1:

M theo bi ta c

U0LC = jU0L U0C j = U0R:

V

U20LC + U20R = 2:80

2:

Thay s vo ra p n A.

Bi ton 98: t in p xoay chiu c tn s ! thay i c vo hai u on mch RLC nitip. Khi thay i ! th cng hiu dng trong mch t cc i l Imax v khi t hai gi tr

!1 v !2 th cng hiu dng trong mch t gi tr u bngImaxp

5. Cho

!1 !2C:!1!2

=60. Tnh

RA. 30. B. 60. C. 20. D. 50.

Li gii

Ta c:1p5=

I

Imax=

RqR2 + (ZL ZC)2

V c hai gi tr !1; !2 cng I nn !1!2 =1

LC

C!1 !2C:!1!2

=

1

LC!2 !2

C

LC

=1 !22LC

!2C=

1 ZLZC1

ZC

= ZC ZL

Suy raRp

R2 + 602=

1p5. Suy ra R = 30

Chn A.

Bi ton 99: My pht in xoay chiu c 2 cp cc ni vi on mch RLC. R = 1. Khi cho

Roto quay vi tc 750

vng/pht hoc

1200

th cng sut ca mch u bng 50W . Bit t

thng cc i qua tt c cc vng dy l o = 0; 8Wb. Gi tr !o mch cng hng gn vi gitr no nht sau y? Bit !o 30 (rad=s)

A. 60 rad/s. B. 50 rad/s. C. 100 rad/s. D. 70 rad/s.

Li gii

Ta c: !1 = 50 ( (rad=s)) ;!2 = 80 ( (rad=s))

P =(o!)

2R

R2 +

L! 1

!C

2) 1

!4C2 2

L

C R

2

2

1

!2+ L2 (o!)

2R

P= 0

49


S dng nh l Viete ta c:8>>>>>: L2 (o!)

2R

P

!C2 =

1

!21!22

2

L

C R

2

2

C2 =

1

!21+

1

!22T bng cch th LC t phng trnh (2) vo phng trnh (1) ta tm c 2 cp nghim nhng

loi 1 nghim:L ' 0; 2559C ' 1; 089:103 ) !o =

1pLC

' 60 (rad=s)Chn A.

Bi ton 100: Cho on mch in AB khng phn nhnh theo th t t C, cun cm L cin tr trong r, in tr thun R. Gi M l im gia t v cun dy, N l im gia R vcun dy. in p hai u on mch khn thay i nhng tn s thay i. Khi tn s dng inf = f1 = 50Hz th cng sut trn ton mch cc i. h s cng sut trn mch MB v trnton mch tng ng l k1 = 0; 8 v k2 = 0; 6 th tn s ca dng in phi bng bao nhiu?

Li gii

Ta c tan'MB =ZL

R + r=

3

4, tan'AB =

ZC ZLR + r

=4

3

) ZCR + r

=25

12

Thit lp c t l:ZLZC

=9

25. M vi Pmax , LC = 1

(!1)2

) LC (!1)2 = 925! f = 30Hz

Bi ton 101: Mt vng dy kn, phng c din tch S t trong mt t trng u vi cm ngt B vung gc vi mt phng vng dy. Nu cho vng dy quay mt gc 180 xung quanh mttrc nm trong mt phng ca n th trong vng dy c mt in lng Q di chuyn. B qua t cm ca vng dy. Nu cho vng dy quay u xung quanh trc ny vi tc gc khng i!, th cng dng in cc i trong vng dy l:

A.!Q

5. B.

!Q

3. C.

!Q

4. D.

!Q

2.

Li gii

Ta c

Q = 2Qo ! Qo = Q2! Io = !Q

2:

Bi ton 102: in nng c truyn t mt trm pht in n mt h tiu th in bng ngdy mt pha vi hiu sut 82 %. Coi hao ph in nng ch do to nhit trn ng dy v khngvt qu 30%. Nu cng sut s dng in ca h tiu th tng 25% v gi nguyn in p nipht th hiu sut truyn ti in nng trn chnh ng dy l:

A. 77,6%. B. 75,6%. C. 70,6%. D. 65,6%.

Li gii

Cng sut cn truyn i ban u l P1, th cng sut nhn c ni tiu th l 0; 82P1.

Cng sut cn truyn i lc sau l P2, th cng sut nhn c ni tiu th l 1; 25:0; 82P1 = 1; 025P1.

Ta c:

50


8>:0; 82 = 1 P1R

U2

H2 = 1 P2RU2

$ 0; 181H2 =

P1P2

(1)

Li c:H2P2 = 1; 025P1 $ P1P2

=H2

1; 025(2)

Th (2) vo (1):$ 0; 181H2 =

H21; 025

! H2 = 0; 7559 do H2 > 0; 7.Chn B.

Bi ton 103: Cho mch in xoay chiu RLC trong L l cun thun cm v c t cm thayi c. in p t vo hai u on mch c biu thc u = U0 cos (wt+ ') trong U0;w; lcc hng s. iu chnh gi tr ca L th thy, khi L = L1 th in p hiu dng gia hai u t tcc i v khi mch tiu thu cng sut 120W. Khi L = L2 th in p gia hai u cun cm

c gi tr hiu dng ln nht l ULmax khi mch tiu th cng sut 43.2W. T s giaULmaxUCmax

l

A.5

3. B.

5

2. C.

4

3. D.

1

3.

Li gii

Khi L = L2 th UL max ) ZL = Z2C +R

2

ZC) ZL ZC = R

2

ZC

P1 =U2

R

P2 =U2

R: cos'2

) cos2 ' = R2

R2 + (ZL ZC)2) ZC = 4

3R

UCmax =U

R:ZC

ULmax =U

R

pR2 + Z2L

Thay s ) ULmaxUCmax

=5

3Chn A.

Bi ton 104: Mt on mch AB gm hai on mch AM v MB mc ni tip. on mch AMgm in tr thun R1 mc ni tip vi t in c in dung C, on mch MB gm in tr R2mc ni tip vi cun cm thun c t cm L. t in p xoay chiu c tn s v gi tr hiudng khng i vo hai u on mch AB. Khi on mch AB tiu th cng sut bng 120Wv c h s cng sut bng 1 . Nu ni tt hai u t in th in p hai u on mch AM

v MB c cng gi tr hiu dng nhng lch pha nhau

3, cng sut tiu th trn on mch AB

trong trng hp ny bng :A. 120W. B. 30W. C. 90W. D. 50W.

Li gii

Ban u mch xy ra cng hng nn cng sut tiu th: Pmax = 120 (W )

Sau khi ni tt hai u t in, th h s cng sut ca mch l:cos' =

p3

2

51


) P = Pmax cos2 ' = 120:34= 90 (W )

Chn C.

Bi ton 105: Cho on mch AB gm RLC mc ni tip theo th t. Cun cm thun, intr R = 50. t vo hai u on mch mt in p xoay chiu u = U

p2: cos (2ft) (V ), U

khng i, tn s f thay i c. iu chnh f in p hai u cun cm t gi tr cc iULmax = U

p3. Khi in p hiu dng ca on mch RC c gi tr 150V . Cng sut ca mch

khi c gi tr

Li gii

UpR2 + Z2CR

= Up3 ) ZC =

p2R

M ta li c R2 = 2ZC (ZL ZC)T ta s tm c ZL

URC = I:ZRC

URC =UpR2 + Z2Cq

R2 + (ZL ZC)2

S tm c U. P = RI2 = RU2q

R2 + (ZL ZC)2

Bi ton 106: t hiu in th xoay chiu u = U0 cos (100t+ ') hai u on mch ni tiptheo thc t gm R1; R2 v cun thun cm c t cm L c th thay i c/ Bit R1 = 2R2 =200p3 (). iu chnh L cho n khi hiu in th tc thi gia hai u on mch cha R2 v L

lch pha cc i so vi hiu in th hai u on mch . Gi tr ca t cm lc l

A. L =1

(H). B. L =

3

(H). C. L =

2

(H). D. L =

5

(H).

Li gii

Bi ny bn c th thay cc p n vo th trc tip lun t cc pha trong nh sau R1L v m

Ta c nhn xt cc pha trn u nm trn gc phn t th nht v R1L lun ln hn m

T ta c tha mn ycbt th R1L m t max hay tan (R1L m) t maxTa c.

tan (R1L m) =tanR1L tanm1 + tanR1L: tanm

Bn gii phng trnh trn vi n l ZL sau bn dng o hm theo ZL bng 0

Tm c ZL = 300

Bi ton 107: t in p xoay chiu c gi tr hiu dng khng i 150 V vo on mch AMBgm on AM ch cha in tr R, on mch MB cha t in c in dung C mc ni tip vimt cun cm thu cn c t cm L thay i c. Bit sau khi thay i t cm L th inp hiu dng hai u mch MB tng 2

p2 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. 100p5. B. 100

p2. C. 100. D. 100

p3.

Li gii

52


UR l in p hiu dng hai u in tr ban u

! ULC = UR2p2

! U2R + U2LC = U2 ! U2R +UR

2p2

2= 1502

! UR = 100p2V

Chn B.

Bi ton 108: Mch in RLC gm 3 phn t mc ni tip vi R l bin tr thay i c. KhiR = R1 th cng sut mch l P1, dng in chm pha so vi ip p '1, khi R = R2 th cng sutmch l P2, dng in chm pha so vi in p '2. Khi R = Ro th cng sut mch cc i. Bit

R1 + R2 = Ro, P1 + P2 = Po,'2 = '1 12

. lch pha nh nht gia dng in v in p gia

hai u mch khi R = R1 gn gi tr no nht sau y?

A.5

12. B.

3. C.

4. D.

2

5.

Li gii

Ta c: P =U2

Rcos2 '

P1 + P2 = Po

) U2 cos2 '1R1

+U2 cos2 '2

R2=

U2

2Ro

) 12Ro

=cos2 '1R1

+cos2 '2R2

(cos'1 + cos'2)2

R1 +R2=

(cos'1 + cos'2)2

Ro

) cos'1 + cos'2 p2

2

, 2 cos'1

6

cos

6p2

2) ' 96oBi ton 109: Cho mch in RLC mc ni tip trong cun dy thun cm, hai u mch tmt in p xoay chiu c in p hiu dng U=150V tn s f. Bit in p hiu dng on RL l200V, hai u t in l 100V v cng sut tiu th l 72,6V. Nu mc thm cun dy thun cmc t cm l 0,73H th cng sut tiu th ca on mch cng khng i. Tn s f ca dngin l

A. 20Hz. B. 50Hz. C. 100Hz. D. 70Hz.

Li gii

T cc d kin u, chng ta s x l c

ZL = 275;ZC = 200

Do cng sut tiu th ca mch khng i:

! Z1 = Z2 $ ZL ZC = ZC Z 0L ! Z 0L = 125

Do cm khng gim, nn cun cm s c mc song song:

125 =275:ZL1275 + ZL1

! ZL1 =1375

6

) f = ZL12L

=1375

6:2:0; 73= 50Hz

53


Chn B.

Bi ton 110: bm nc ngoi mt cnh ng xa, ngi ta dng mt my bm nc m ngc ca n l mt ng c in loi 220V 704W . Cch xa ng c, c mt ngun in xoay chiuc in p hiu dng U. ng c hot ng, ngi ta dng dy truyn ti c in tr tng cng2; 5 ni t ngun in ti ng c. Bit h s cng sut ca ng c khi chy ng cng sutnh mc l 0.8. ng c chy ng cng sut nh mc th in p hiu dng U c gi tr xpx bng:

A. 100V. B. 120V. C. 228V. D. 220V.

Li gii

Cng dng in hiu dng: I =P

U cos'=

704

220:0; 8= 4 (A)

! UR = 4:2; 5 = 10 ()

Ta c:

U =qU2d + U

2R + 2UdUR cos'

=p2202 + 102 + 2:220:10:0; 8 = 228 (V )

Chn C.

Bi ton 111: Ni hai cc ca 1 my pht in xoay chiu mt pha vo 2 u on AB gm 1

in tr thun R = 100 , cun cm c t cm L =41

6(H) v t in c in dung C =

104

3F .

Tc roto ca my c th thay i c. Khi tc roto ca my l n hoc 3n th cng sut tiuth in ca mch l nh n

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