Các dạng bài tập Dòng điện xoay chiều VL 12

8/2/2019 Cc dng bi tp Dng in xoay chiu VL 12

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Phng php gii cc dng bi tp Vt l 12 Dao ng c - Sng c, sng m Trang 1

A - PHN M U

I. L DO CHN TI.Hin nay, khi m hnh thc thi trc nghim khch quan c p dng tron

k thi tt nghip v tuyn sinh i hc, cao ng th yu cu v vic nhn dgii nhanh v ti u cc cu trc nghim, c bit l cc cu trc nghim nhl rt cn thit c th t c kt qu cao trong k thi. Trong thi tuyH v C nm 2010, mn Vt L c nhng cu trc nghim nh lng khm cc thi trc cha c, nu cha gp v cha gii qua ln no th thkh m gii nhanh v chnh xc cc cu ny.

gip cc em hc sinh nhn dng c cc cu trc nghim nh lngc th gii nhanh v chnh xc tng cu, ti xin tp hp ra y cc bi tp itrong sch gio khoa, trong sch bi tp, trong cc thi tt nghip THPT, thi tsinh H C trong nhng nm qua v phn chng thnh nhng dng c bn a ra phng php gii cho tng dng. Hy vng rng tp ti liu ny gip cmt cht g cho cc qu ng nghip trong qu trnh ging dy v cc emsinh trong qu trnh kim tra, thi c.

II. I TNG V PHM VI P DNG 1) i tng s dng ti:Gio vin dy mn Vt l lp 12 tham kho hng dn hc sinh gii bHc sinh hc lp 12 luyn tp kim tra, thi mn Vt L.2) Phm vi p dng:

Phn dng in xoay chiu ca chng trnh Vt L 12 Ban C bn. III. PHNG PHP NGHIN CU Xc nh i tng p dng ti.Tp hp cc bi tp in hnh trong sch gio khoa, trong sch bi tp, tron

thi tt nghip THPT, thi tuyn sinh H C trong ba nm qua (t khi thay v phn chng thnh cc bi tp minh ha ca nhng dng bi tp c bn.

H thng cc cng thc, kin thc lin quan v phng php gii cho dng.

C li gii cc bi tp minh ha cc em hc sinh c th kim tra so sn bi gii ca mnh.

Cc cu trc nghim luyn tp l thi Tt nghip i hc Cao n ba nm qua.

Ngi vi t: Dng Vn ng Trng THPT Nguyn Vn Linh, Bnh Thun


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B - NI DUNGCC DNG BI TP V DNG IN XOAY CHIU

1. i cng v dng in xoay chiu* Cc cng thc:Biu thc ca i v u: I0cos(t +i); u = U0cos(t +u). lch pha gia u v i:=u -i.

Cc gi tr hiu dng: I =02 I ; U = 02

U ; E = 02 E . Chu k; tn s: T =2 ; f = 2

.

Trong 1 giy dng in xoay chiu c tn s f (tnh ra Hz) i chiu 2f ln.T thng qua khung dy ca my pht in:

= NBScos( ,n B

) = NBScos(t +) =0cos(t +); vi0 = NBS.Sut ng trong khung dy ca my pht in:e = - d

dt

= - = NBSsin(t +) = E0cos(t + - 2 ); vi E0 = 0 = NBS.

* Bi tp minh ha:1. Dng in xoay chiu c cng i = 4cos120t (A). Xc nh cng hiudng ca dng in v cho bit trong thi gian 2 s dng in i chiu bao ln?2. Mt n ng lm vic vi in p xoay chiu u = 2202cos100t (V). Tuy nhinn ch sng khi iu p t vo n c |u| = 155 V. Hi trung bnh trong 1 s cnhiu ln n sng?3. Dng in chy qua mt on mch c biu thc i = I0cos100t. Trong khongthi gian t 0 n 0,02 s, xc nh cc thi im cng dng in c gi thi c gi tr bng: a) 0,5 I0; b) 22 I0.

4. Ti thi im t, in p u = 2002cos(100t -2 ) ( u tnh bng V, t tnh bng s) c

gi tr l 100 2V v ang gim. Xc nh in p ny sau thi im 1300s.5. in p xoay chiu gia hai im A v B bin thin iu ha vi biu

u = 220 2cos(100t +6

) (trong u tnh bng V, t tnh bng s). Ti thi im t1 n cgi tr tc thi u1 = 220 V v ang c xu hng tng. Hi ti thi im t2 ngay sau t1 5 msth n c gi tr tc thi u2 bng bao nhiu?6. Mt khung dy dn phng dt hnh ch nht c 500 vng dy, din tch m54 cm2. Khung dy quay u quanh mt trc i xng (thuc mt phng ca ktrong t trng u c vect cm ng t vung gc vi trc quay v c B = 0,2 T. Tnh t thng cc i qua khung dy. sut in ng cm nghin trong khung dy c tn s 50 Hz th khung dy phi quay vi tc bao

vng/pht?7. Mt khung dy dn phng dt hnh ch nht c 500 vng dy, din tch ml 220 cm2. Khung dy quay u vi tc 50 vng/s quanh trc i xng trong mt phng khung dy, trong mt t trng u c vc t cm ng t B

vung Ngi vi t: Dng Vn ng Trng THPT Nguyn Vn Linh, Bnh Thun


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gc vi trc quay v c ln25 T. Tnh sut in ng cc i xut hin trokhung dy.8. Mt khung dy dn hnh ch nht c 1500 vng, din tch mi vng 100 2,quay u quanh trc i xng ca khung vi tc gc 120 vng/pht trong mtrng u c cm ng t bng 0,4 T. Trc quay vung gc vi cc ng sChn gc thi gian l lc vc t php tuyn ca mt phng khung dy cng vi vc t cm ng t. Vit biu thc sut in ng cm ng tc thi trong 9. T thng qua 1 vng dy dn l =

22.10

cos(100t - 4

) (Wb). Tm biu thcca sut in ng cm ng gia hai u cun dy gm 150 vng dy ny.* Hng dn gii

1. Ta c: I = 02 I

= 2 2 A; f = 2 = 60 Hz.

Trong 2 giy dng in i chiu 4f = 240 ln.2. n ch sng khi in p t vo n c |u| 155 V, do trong mt chu k s c

2 ln n sng. Trong 1 giy c1

2

= 50 chu k nn s c 100 ln n sng.

3. a) Ta c: 0,5I0 = I0cos100t cos100t = cos(3 ) 100t = 3

+ 2k

t = 1300 + 0,02k; vi k Z. Cc nghim dng nh hn hoc bng 0,02 s tro

2 h nghim ny l t =1

300s v t =160 s.

b) Ta c: 22 I0 = I0cos100t cos100t = cos(4 ) 100t = 4

+ 2k

t = 1400+ 0,02k; vi k Z. Cc nghim dng nh hn hoc bng 0,02 s tro

2 h nghim ny l t =1400s v t =7

400 s.

4. Ti thi im t: u = 1002= 200 2cos(100t -2 )

cos(100t -2 ) = 12= cos(3

). V u ang gim nn ta nhn nghim (+)

100t - 2 = 3

t = 1120 (s).

Sau thi im 1300s, ta c:

u = 200 2cos(100(1120+1

300) - 2 ) = 200 2cos23

= - 100 2 (V).

5. Ta c: u1 = 220 = 220 2cos(100t1 + 6

) cos(100t1 + 6

) =2

2 = cos( 4

) .V u ang tng nn ta nhn nghim (-)100t1 + 6

= - 4 t1 = -

1240s



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t2 = t1 + 0,005 =0,2240s u2 = 220 2cos(100t2 + 6

) = 220 V.

6. Ta c:0 = NBS = 0,54 Wb; n =60 f p = 3000 vng/pht.

7. Ta c: f = n = 50 Hz; = 2f = 100 rad/s; E0 = NBS = 220 2 V.

8. Ta c:0 = NBS = 6 Wb; = 60n

2 = 4 rad/s; = 0cos(

n B, ) =0cos(t +); khi t = 0 th (

n B, ) = 0 = 0.

Vy = 6cos4t (Wb); e = - = 24sin4t = 24cos(4t - 2 ) (V).

9. Ta c: e = - N = 150.10022.10

sin(100t - 4

) = 300cos(100t - 34 ) (V).

2. Tm cc i lng trn on mch xoay chiu c R, L, C * Cc cng thc:Cm khng, dung khng, tng tr: ZL = L; ZC = 1C ; Z =

2CL

2 )Z-(ZR + .

nh lut m: I =U Z = RU

R = L

L

U Z =

C

C

U Z .

Gc lch pha gia u v i: tan= L C Z Z

R .

Cng sut: P = UIcos= I2R =2

2U R Z . H s cng sut: cos=

R Z .

in nng tiu th mch in: W = A = Pt.* Phng php gii:

tm cc i lng trn on mch xoay chiu ta vit biu thc lin qucc i lng bit v i lng cn tm t suy ra v tnh i lng cn

Trong mt s trng hp ta c th dng gin vc t gii bi ton.

Trn on mch khuyt thnh phn no th ta cho thnh phn bng 0mch va c in tr thun R v va c cun dy c in tr thun r th ithun ca mch l (R + r).* Bi tp minh ha:1. Nu t vo hai u cun dy in p 1 chiu 9 V th cng dng incun dy l 0,5 A. Nu t vo hai u cun dy in p xoay chiu c gi trdng l 9 V th cng hiu dng ca dng in qua cun dy l 0,3 A. Xin tr thun v cm khng ca cun dy.



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2. Mt in tr thun R = 30v mt cun dy c mc ni tip vi nhau thmt on mch. Khi t in p khng i 24 V vo hai u on mch ndng in i qua n c cng 0,6 A; khi t mt in p xoay chiu tn svo hai u on mch, th dng in qua n lch pha 450 so vi in p ny. Tnh t cm ca cun dy, tng tr ca cun dy v tng tr ca c on mch3. Mt m in hot ng bnh thng khi ni vi mng in xoay chiu c hiu dng l 220 V, in tr ca m khi l 48,4. Tnh nhit lng do m ta ratrong thi gian mt pht.4. Mt on mch gm in tr thun R, cun cm thun L v t in C mtip. Cng dng in tc thi i qua mch c biu thc i = 0,284cos120t (A).Khi in p hiu dng gia hai u in tr, cun dy v t in c gi trng l UR = 20 V; UL = 40 V; UC = 25 V. Tnh R, L, C, tng tr Z ca on mch vin p hiu dng gia hai u on mch.5. t in p u = 100cos(t + 6

) (V) vo hai u on mch RLC th dng i

qua mch l i = 2 cos(t + 3 ) (A). Tnh cng sut tiu th v in tr thun c

on mch.6. t in p u = 2002cos(100t) (V) vo hai u on mch AB gm hai omch AM v MB mc ni tip. on AM gm in tr thun R mc ni tcun cm thun L, on MB ch c t in C. Bit in p gia hai u oAM v in p gia hai u on mch MB c gi tr hiu dng bng nhau lch pha nhau2

3

. Tnh in p hiu dng gia hai u on mch AM.7. Mt on mch AB gm hai on mch AM v MB mc ni tip. onAM c in tr thun R = 50 ni tip vi cun cm thun c L =1 H, on mchMB ch c t in vi in dung thay i c. t in p u = U0cos100t (V) vohai u on mch AB. iu chnh in dung ca t n gi tr C1 sao cho in phai u on mch AB lch pha2

so vi in p hai u on mch AM. Tnh C1.8. t in p xoay chiu c gi tr hiu dng khng i, tn s 50 Hz vo h

on mch mc ni tip gm in tr thun R, cun cm thun L v t indung C thay i c. iu chnh in dung C n gi tr

4104

F hoc

4102

F th

cng sut tiu th trn on mch u c gi tr bng nhau. Tnh t cm L.9. t in p xoay chiu c gi tr hiu dng 200 V vtn s khng i vo hai u A v B nh hnh v. Trong R l bin tr, L l cun cm thun v C l t in cin dung thay i. Cc gi tr R, L, C hu hn v khc khng. Vi C = C1 th in phiu dng gia hai u bin tr R c gi tr khng i v khc khng khi thay tr R ca bin tr. Tnh in p hiu dng gia A v N khi C =12C .



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10. t in p u = U2cost (V) vo hai u on mch gm cun cm thumc ni tip vi mt bin tr R. ng vi hai gi tr R 1 = 20v R 2 = 80ca bintr th cng sut tiu th trong on mch u bng 400 W. Tnh gi tr ca U11. t in p xoay chiu c gi tr hiu dng v tn s khng i vo hai mch gm bin tr R mc ni tip vi t in c in dung C. Gi in p higia hai u t in, gia hai u bin tr v h s cng sut ca on mchtr c gi tr R 1 ln lt l UC1, UR1 v cos1; khi bin tr c gi tr R 2 th cc gi trtng ng ni trn l UC2, UR2 v cos2. Bit UC1 = 2UC2, UR2 = 2UR1. Xc nh cos1v cos2.12. t in p u = U2cost vo hai u on mch AB gm hai on mch v NB mc ni tip. on AN gm bin tr R mc ni tip vi cun cm t

t cm L, on NB ch c t in vi in dung C. t1 =1

2 LC . Xc nhtn s gc in p hiu dng gia hai u on mch AN khng ph th

R.13. t in p u = 2cos2U ft (U khng i, tn s f thay i c) vo hai on mch mc ni tip gm in tr thun R, cun cm thun c t cmin c in dung C. Khi tn s l f 1 th cm khng v dung khng ca on mcc gi tr ln lt l 6v 8 . Khi tn s l f 2 th h s cng sut ca on mc bng 1. Tm h thc lin h gia f 1 v f 2.14. Mt on mch AB gm hai on mch AM v MB mc ni tip. onAM gm in tr thun R 1 mc ni tip vi t in c in dung C, on mch

gm in tr thun R 2 mc ni tip vi cun cm thun c t cm L. t ixoay chiu c tn s v gi tr hiu dng khng i vo hai u on mch A on mch AB tiu th cng sut bng 120 W v c h s cng sut bngni tt hai u t in th in p hai u on mch AM v MB c cng gi dng nhng lch pha nhau3

. Tnh cng sut tiu th trn on mch AB trontrng hp ny.15. on mch AB gm hai on mch AM v MB mc ni tip. on mgm in tr thun R 1 = 40mc ni tip vi t in c

310C F4

=

, on mch MB

gm in tr thun R 2 mc ni tip vi cun cm thun L. t vo A, B ixoay chiu c gi tr hiu dng v tn s khng i th in p tc thi hon mch AM v MB ln lt l:AM

7u 50 2cos(100 t )(V)12= v

MBu 150cos100 t (V)= . Tnh h s cng sut ca on mch AB.16. t mt in p xoay chiu c gi tr hiu dng v tn s khng i ln lhai u in tr thun R, cun cm thun c t cm L, t in c in ducng dng in hiu dng qua mch tng ng l 0,25 A; 0,5 A; 0,2 Acng dng in hiu dng qua mch nu t in p xoay chiu ny voon mch gm ba phn t trn mc ni tip.



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* Hng dn gii 1. Ta c: R = 1C U I = 18; Zd =

XC U I = 30; ZL =

22 R Z d = 24.2. Ta c: R + r =U I = 40 r = 10;

L Z R r + = tan= 1 ZL = R + r = 40

L = 2 L Z f = 0,127 H; Zd =

22 L Z r + = 41,2; Z = 22)( L Z r R ++ = 40 2.

3. Ta c: I =U R= 4,55 A; P = I2R =

2U R = 1000 W; Q = Pt = 60000 J = 60 kJ.

4. Ta c: I = 02 I

= 0,2 A; R = RU I = 100; ZL = LU

I = 200; L = L Z

= 0,53 H;

ZC = C U I = 125; C =

1C Z = 21,2.10

-6 F; Z = 2CL2 )Z-(ZR + = 125;

U = IZ = 25 V.5. Ta c:=u -i = - 6

; P = UIcos= 50 3W; R = 2 P I = 25 3.

6. Ta c: ABU = AM U

+ MBU U2 AB = U2 AM + U2 MB+ 2UAMUMBcos(U

AM, U

MB).

V UAM = UMBv ( AM U , MBU

) = 23 U2 AB = U2 AM UAM = UAB = 220 V.

7. Ta c: ZL = L = 100. V on mch AB c t in nn in p uAB tr pha hnin p uAN AB -AN = - 2

AN =AB + 2

tanAN = tan(AB + 2 ) = - cotanAB

tanAB.tanAN = R Z

R Z Z LC L .1

= tanAB.(- cotanAB) = - 1

ZC1 =1

L

R Z + ZL = 125 C1 = 1

1C Z =

58.10

F.

8. Ta c: ZC1 =1

12 fC = 400; ZC2 =

2

12 fC = 200. P1 = P2 hay 2

2

2

2

1

2

Z

RU

Z

RU =

Z21 = Z22 hay R 2 + (ZL ZC1)2 = R 2 + (ZL ZC2)2 ZL = 221 C C Z Z + = 300;

L = 2 L Z f =

3 H.

9. Khi C = C1 th UR = IR = 22 )(.

1C LZ Z R

RU

+ . UR khng ph thuc R th ZL = ZC1.

Khi C = C2 = 12

C th ZC2 = 2ZC1; ZAN = 22 L

Z R+

= 21

2

C Z R + ;

ZAB = 222 )( C L Z Z R + = 212

C Z R + = ZAN UAN = IZAN = UZAB = UAB = 200 V.



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10. Ta c: P = 221

12

L Z R RU

+= 22

2

22

L Z R RU + ZL = 21 R R = 40. U = 1

221 )( R

Z R P L+ = 200 V.

11. Ta c: UC1 = I1ZC = 2UC2 = 2I2ZC I1 = 2I2; UR2 = I2R 2 = 2UR1 = 2I1R 1 = 2.2I2R 1

R 2 = 4R 1; I1 = 221 C Z R

U + = 2I2 = 2 222 C Z R

U + R

22+ Z2C = 4R 21 + 4Z2C

16 R 21 + Z2C = 4R 21 + 4Z2C ZC = 2R 1 Z1 = 221 C Z R + = 5 R 1

cos1 = 11

R Z =

15 ; cos2 = 2

2

Z R

=1

1

24 Z R

=25 .

12. UAN = IZAN = 2222

)(.

C L

L

Z Z R

Z RU

+

+

khng ph thuc vo R th:

R 2 + Z2 L = R 2 + (ZL ZC)2 ZC = 2ZL hay1C = 2L

= LC 21 =

LC 22 = 1 2.

13. Ta c:21 1

11

1

2 6(2 )1 82

L

C

Z f L f LC Z f C

= = = = 34 v22 2

12

2

2 (2 )12

L

C

Z f L f LC Z f C

= = = 1

222

1

f f =

43 f 2 =

23 f 1.

14. Khi cha ni tt hai bn t, cos

= 1, on mch c cng hng in, do

PAB =2

1 2

U R R+ = 120 W. Khi ni tt hai bn t: tanMB = 2

L Z R = 3 ZL = 3 R 2;

UAM = UMB R 1 = 2 2 2 22 2 2( 3 ) L R Z R R+ = + = 2R 2

tan = 21 2 2

3 33 3

L R Z R R R= =+ = 6

; PAB =2

1 2

U R R+ =

2

23U R = 120

U2 = 360R 2;

Z =2 2 2 2

1 2 2 2( ) (3 ) ( 3 ) L R R Z R R+ + = + = 2 3 R 2. Vy: PAB =2

os ''U

c Z = 90 W.15. Ta c: ZC = 1C = 40; ZAM =

2 21 C R Z + = 40 2; I0 = AM

AM

U Z = 1,25;

tanAM =1

C Z R

= - 1 AM = - 4 ;i +AM = -

712

i = -712 -AM = -

712 + 4

= - 3 ;i +MB = 0 MB =i = 3

;

tanMB =2

L Z

R= 3 ZL = 3 R 2;

ZMB = 00

MBU I = 120=

2 2 2 22 2 2( 3 ) L R Z R R+ = + = 2R 2



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R 2 = 60; ZL = 60 3 . Vy: cos=1 22 2

1 2( ) ( ) L C

R R R R Z Z

++ + = 0,843.

16. Ta c: R = R

U I = 4U; ZL = L

U I = 2U; ZC = C

U I = 5U;

I = U

Z =

2 24 (2 5)

U

U + = 0,2 A.

3. Vit biu thc ca u v i trn on mch xoay chiu* Cc cng thc:Biu thc ca u v i: Nu i = I0cos(t +i) th u = (t +i +).

Nu u = U0cos(t +u) th i = I0cos(t +u -).Vi: I =U Z ; I0 =

0U Z ; I0 = I 2; U0 = U 2; tan=

L C Z Z R ; ZL > ZC th u nhanh

pha hn i; ZL < ZC th u chm pha hn i.

on mch ch c in tr thun R: u cng pha vi i; on mch ch c cuncm L: u sm pha hn i gc2 ; on mch ch c t in u tr pha hn i gc2

.Trng hp in p gia hai u on mch l u = U0cos(t +). Nu on mchch c t in th: i = I0cos(t ++ 2

) = - I0sin(t +) hay mch ch c cun cm

th: i = I0cos(t + - 2 ) = I0sin(t +) hoc mch c c cun cm thun v

in m khng c in tr thun R th: i = I0sin(t +). Khi ta c:2

20

i

I +

2

20

u

U

= 1.* Phng php gii: vit biu thc cng dng in chy qua on mhoc vit biu thc in p gia hai u mt on mch ta tnh gi tr cc cng dng in hoc in p cc i tng ng v gc lch pha gia icng dng in ri thay vo biu thc tng ng.Ch : Nu trong on mch c nhiu phn t R, L, C mc ni tip th trontnh tng tr hoc lch pha gia u v i ta t R = R 1 + R 2 + ...; ZL = ZL1 + ZL2+ ...; ZC = ZC1 + ZC2 + ... . Nu mch khng c in tr thun th ta cho R = 0; khc cun cm th ta cho ZL = 0; khng c t in th ta cho ZC = 0.* Bi tp minh ha:1. Mt t in c in dung C = 31,8F, khi mc vo mch in th dng in chqua t in c cng i = 0,5cos100t (A). Vit biu thc in p gia hai bn t2. Cho on mch RLC gm R = 80, L = 318 mH, C = 79,5F. in p gia haiu on mch l: u = 120 2cos100t (V). Vit biu thc cng dng in chtrong mch v tnh in p hiu dng gia hai u mi dng c.

3. Cho on mch xoay chiu RLC c R = 503; L =1 H; C =

3105

F . in pgia hai u on mch c biu thc uAB = 120cos100t (V). Vit biu thc cng dng in trong mch v tnh cng sut tiu th ca mch.



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4. Mt mch in AB gm in tr thun R = 50, mc ni tip vi cun dy c t cm L =1 H, in tr R 0 = 50. t vo hai u on mch in p xoay chiuAB = 100 2cos100t (V). Vit biu thc in p tc thi hai u cun dy.

5. t mt in p xoay chiu u = U0cos(100t - 3

) (V) vo hai u mt t in cin dung

42.10

(F). thi im in p gia hai u t in l 150 V th c

dng in trong mch l 4 A. Vit biu thc cng dng in chymch.6. t in p xoay chiu u = U0cos(100t + 3

) (V) vo hai u mt cun cm

thun c t cm L =12 H. thi im in p gia hai u cun cm l

2 V th cng dng in qua cun cm l 2 A. Vit biu thc cng in chy qua cun cm.7. Mch RLC gm cun thun cm c L =2 H, in tr thun R = 100v t in

c C =410

F. Khi trong mch c dng in i =2cost (A) chy qua th h s

cng sut ca mch l22 . Xc nh tn s ca dng in v vit biu thc ingia hai u on mch.8. Cho mch in xoay chiu gm in tr thun R = 10, cun dy thun cm Lv t in C =

3102

F mc ni tip. Biu thc ca in p gia hai bn t

uC = 50 2cos(100t 0,75) (V). Xc nh t cm cun dy, vit biu thcng dng in chy trong mch.* Hng dn gii 1. Ta c: ZC = 1C = 100; U0C = I0ZC = 50 V; uC = 50cos(100t - 2

) (V).

2. Ta c: ZL = L = 100; ZC = 1C = 40;Z = 2CL

2 )Z-(ZR + = 100; I =U Z = 1,2 A; tan= L C Z Z R = tan370

= 37180 rad; i = 1,2 2cos(100t - 37180

) (A); UR = IR = 96 V;UL = IZL = 120 V; UC = IZC = 48 V.3. Ta c: ZL = L = 100; ZC = 1C = 50; Z =

2CL

2 )Z-(ZR + = 100;

tan= L C Z Z R = tan300 = 6 rad; I0 = 0U Z = 1,2 A; i = 1,2cos(100t - 6 ) (A);P = I2R = 62,4 W.



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4. Ta c: ZL = L = 100; Z = 220)( L Z R R ++ = 100 2;

I = U Z =12 A; tan= 0

L Z R R+ = tan4

= 4 ; Zd = 220 L Z R + = 112; Ud = IZd = 56 2V;

tand =0 R

Z L = tan630 d = 63180 .

Vy: ud = 112cos(100t - 4 + 63180

) = 112cos(100t + 10 ) (V).

5. Ta c: ZC = 1C = 50; i = I0cos(100t - 3 + 2

) = - I0sin(100t - 3 ).

Khi :2

20

i I +

2

20

uU = 1 hay 22

0

2

20

2

C Z I u

I i + = 1 I0 = 22 )(

C Z u

i + = 5 A.

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

6. Ta c: ZL = L = 50; i = I0cos(100t + 3 - 2

) = I0sin(100t + 3 ).

Khi :2

20

i I +

2

20

uU = 1 hay 22

0

2

20

2

L Z I u

I i + = 1 I0 = 22 )(

L Z u

i + = 2 3 A.

Vy: i = 23cos(100t - 6 ) (A).

7. Ta c: cos= R Z Z = os Rc = 100 2; ZL ZC = 22 R Z = 100

2fL -1

2 fC = 4f -410

2 f = 102 8f 2 2.102f - 104 = 0

f = 50 Hz hoc f = 25 Hz; U = IZ = 1002V.Vy: u = 200cos(100t + 4

) (A) hoc u = 200cos(25t - 4 ) (A).

8. Ta c: ZC = 1C = 20; - - 2 = - 34

= 4 ; tan= L C Z Z R

ZL = ZC + R.tan= 30 L = L Z =3

10 H; I =C

C U Z = 2,5 A.

Vy: i = 2,52 cos(100t - 4 ) (A).

4. Bi ton cc tr trn on mch xoay chiu* Cc cng thc:

Khi ZL = ZC hay =1

LC th Z = Zmin = R; Imax =U R ; Pmax =

2U R ;= 0 (u cng pha

vi i). l cc i do cng hng in.Cng sut: P = I2R =

2

2U R Z .



12/44


in p hiu dng gia hai u cun thun cm: UL = IZL = LUZ Z .

in p hiu dng gia hai bn t: UC = IZC = C UZ Z .

* Phng php gii:+ Vit biu thc i lng cn xt cc tr (I, P, UL, UC) theo i lng cn tm (R, L,C,).+ Xt iu kin cng hng: nu trong mch xy ra hin tng cng hnlun suy ra i lng cn tm.+ Nu khng c cng hng th bin i biu thc a v dng ca bt Csi hoc dng ca tam thc bc hai c cha bin s tm cc tr.

Sau khi gii cc bi tp loi ny ta c th rt ra mt s cng thc sau skhi cn gii nhanh cc cu trc nghim dng ny:

Cc i P theo R: R = |ZL ZC|. Khi Pmax =2

2| | L C

U Z Z =

2

2

U

R.

Cc i UL theo ZL: ZL =C

C

Z Z R 22 + .

Khi ULmax= R

Z RU C 22 + ; U2max L = U2 + U2 R + U2C

Cc i ca UC theo ZC: ZC = L

L

Z Z R 22 + .

Khi UCmax= R

Z RU L22 + ; U2 maxC = U2+ U2 R + U2 L

Cc i ca UL theo: UL = ULmaxkhi = 2222

C R LC .

Cc i ca UC theo: UC = UCmax khi = 22

21

L R

LC .

* Bi tp minh ha:1. Cho mch in nh hnh v. Trong R = 60, cundy thun cm c t cm L =12 H, t in c indung C thay i c. t vo gia hai u on mch mt in p xoay chnh: uAB = 120 2cos100t (V). Xc nh in dung ca t in cho cng sutiu th trn on mch t gi tr cc i. Tnh gi tr cc i .2. Mt on mch gm R = 50, cun thun cm c t cm L v t in c dung C =

42.10

F mc ni tip. t vo hai u on mch mt in p xoay

c in p hiu dng 110 V, tn s 50 Hz. Th thy u v i cng pha vi nhau. t cm ca cun cm v cng sut tiu th ca on mch.



13/44


3. Cho mch in nh hnh v. Trong in tr thunR = 50, cun dy thun cm c t cm L = 159 mH,t in c in dung C = 31,8F, in tr ca ampe k v dy ni khng ng kt vo gia hai u on mch mt in p xoay chiu uAB = 200cost (V). Xcnh tn s ca in p ampe k ch gi tr cc i v s ch ca ampe k l4. t in p u = 1002cost (V), c thay i c vo hai u on mcgm in tr thun R = 200, cun cm thun c t cm L =2536 H v t in

c in dung C =410

F mc ni tip. Cng sut tiu th ca on mch l 50

Xc nh tn s ca dng in.5. Cho mch in xoay chiu gm bin tr R, cun thun cm L =12 H, t in

C =410

F mc ni tip vi nhau. t vo hai u on mch in p xoay

u = 220 2cos100t (V). Xc nh in tr ca bin tr cng sut tiu th on mch t gi tr cc i. Tnh gi tr cc i .6. Cho mch in nh hnh v. Trong cun dy c in tr thun r = 90, c t cm L =1,2 H, R l mt bin tr. t vo hai u onmch mt in p xoay chiu n nh uAB = 200 2cos100t (V). nh gi tr ca bin tr R cng sut to nhit trn bin tr t gi tr cc i. Tnh cng si .7. Cho mch in nh hnh v. Trong R = 1003; C = 4102

F; cun dy thuncm c t cm L thay i c. in p gia hai uon mch l u = 200cos100t (V). Xc nh t cm cacun dy in p hiu dng trn cun cm L l cc i.Tnh gi tr cc i .8. Cho mch in nh hnh v. Trong R = 60, cun dy thun cm c tcm L = 1

2 H, t in c in dung C thay i c. t

vo gia hai u on mch mt in p xoay chiu nnh: uAB = 120 2cos100t (V). Xc nh in dung ca t in in p gia bn t t gi tr cc i. Tnh gi tr cc i .9. Cho mt mch ni tip gm cun thun cm L =2 H, in tr R = 100, t

in C =410

F. t vo mch in p xoay chiu u = 2002cost (V). Tm gi

tr ca :a) in p hiu dng trn R t cc i. b) in p hiu dng trn L t cc i.c) in p hiu dng trn C t cc i.



14/44


10. t in p u = U2cost vi U khng i vo hai u on mch AB gm on mch AN v NB mc ni tip. on AN gm bin tr R mc ni tip v

cm thun L, on NB ch c t in, in dung C. Vi = 0 =1

LC th cng dng in qua on mch t gi tr cc i. Tnh tn s gc theo 0 inp hiu dng gia hai u on mch AN khng ph thuc vo R.11. Ln lt t cc in p xoay chiu u1 = 12 cos(100 )U t + ;u2 = 22 cos(120 )U t + v u3 = 32 cos(110 )U t + vo hai u on mch gmin tr thun R, cun cm thun L v t in C mc ni tip th cng in trong on mch c biu thc tng ng l: i1 = 2cos100 I t ;i2 = 22 cos(120 )3 I t

+ v i3 = 2' 2 cos(110 )3 I t . So snh I v I.

12. t in p xoay chiu 2 cos100u U t = vo hai u on mch mc n

tip gm in tr thun R, t in c in dung C v cun cm thun c L thay i c. iu chnh L in p hiu dng hai u cun cm cc i th thy gi tr cc i bng 100 V v in p hiu dng hai u bng 36 V. Tnh U.13. t in p xoay chiu u = U0cost (U0 khng i v thay i c) vo haiu on mch gm in tr thun R, cun cm thun c t cm L v tin dung C mc ni tip, vi CR 2 < 2L. Khi = 1 hoc = 2 th in p hiudng gia hai bn t in c cng mt gi tr. Khi = 0 th in p hiu dng giahai bn t in t cc i. Tm h thc lin h gia1, 2 v0.14. t in p xoay chiuu U 2 cos100 t= (U khng i, t tnh bng s) vo haiu on mch mc ni tip gm in tr thun R, cun cm thun c t15H v t in c in dung C thay i c. iu chnh in dung ca t in p hiu dng gia hai bn t in t gi tr cc i v bngU 3. Tnh R.* Hng dn gii

1. Ta c: ZL = L = 50. P = Pmax th ZC = ZL = 50 C =1

C Z =42.10

F.

Khi : Pmax = 2U R = 240 W.

2. Ta c: ZC =1

2 fC = 50. u v i cng pha th ZL = ZC = 50

L = 2 L Z f =

12 H. Khi : P = Pmax =

2U R = 242 W.

3. Ta c: I = Imax khi ZL = ZC hay 2fL =1

2 fC f =1

2 LC = 70,7 Hz.

Khi I = Imax = U R= 2 2 A.

4. Ta c: P = I2R I = P R = 0,5 A =U R = Imax do c cng hng in.



15/44


Khi c cng hng in th = 2f =1

LC f =1

2 LC = 60 Hz.

5. Ta c: ZL = L = 50; ZC = 1C = 100;

P = I2R =2 2 2

22 2 2 ( )( ) L C L C

U R U R U

Z Z Z R Z Z R

R

= =+

+. V U, ZL v ZC khng i

nn P = Pmax th R =2( ) L C Z Z

R

(bt ng thc Csi)

R = |ZL ZC| = 50. Khi : Pmax =2

2U R = 484 W.

6. Ta c: ZL = L = 120; PR = I2R = 222

)( L Z r R RU

++ = R

Z r r R

U

L22

2

2 +++ .

V U, r v ZL khng i nn PR = PRmax khi: R = R

Z r L22 + (bt ng thc Csi)

R = 22 L Z r + = 150. Khi : PRmax=2

2( )

U

R r += 83,3 W.

7. Ta c: ZC =1

C Z = 200;

UL = IZL = 22 )( C L L

Z Z R

UZ + = 1121)( 222 ++

LC

LC Z

Z Z

Z R

U

.

V U, R v ZC khng i nn UL = ULmax khi1 L Z

= -)(2

222

C

C

Z R

Z

+

(khi x = -2ab )

ZL =C

C

Z

Z R 22 += 350 L = 3,5 H. Khi ULmax= R

Z RU C 22 + = 216 V.

8. ZL = L = 50; UC = IZC = 22 )( C LC

Z Z R

UZ + = 1121)( 222 ++

C L

C L Z

Z Z

Z R

U

;

UC = UCmaxkhi1C Z = - )(2

222

L

L

Z R

Z

+

ZC = L

L

Z

Z R 22 += 122

C =1

C Z =

22,110 4 F. Khi : UCmax =

R Z RU L

22 += 156 V.

9. a) Ta c: UR = IR = URmaxkhi I = Imax; m I = Imax khi =1

LC = 70,7 rad/s.

b) UL = IZL = 22 )1(C

L R

LU Z

UZ L

+

== 2

22

421).2(1.1

.

L RC L

C

LU

+

.



16/44


UL = ULmaxkhi 21

= -

2

2

12

)2(

C

RC L

= 2222

C R LC = 81,6 rad/s.

c) UC = IZC =22 )1(

1

C L R

C U

Z

UZ C

+

= =2

2242 1

)2(

.

C RC

L

L

LU

+ .

UC = UCmax khi2 = -2

2

2

)2(

L

RC L

= 22

21

L R

LC = 61,2 rad/s.

10. Ta c: UAN = I.ZAN = AN UZ

Z =2 2 2

2 21( )

U R L

R L C

++ =

2 2 2

2 2 22 212

U R L L R L C C

++ +

= 2 22 2 2

1 21

U LC C

R L

+ +

.

V U khng i nn UAN khng ph thuc vo R th2 21C - 2

LC = 0 hay

=1

2 LC =02

.

11. V I1 = I2 = I Z1 = Z2 hay R 2 + (100L - 1100 C

)2 = R 2 + (120L - 1120 C

)2

100L - 1100 C = - (120L -1

120 C ) 220L =22

1200 C 120002 = 1 LC

ch = 212000 110 = 3 I3 = Imax = I > I.Qua bi ny c th rt ra kt lun: Vi1 2 (1 < 2) m I1 = I2 = I, th khi

1 < 3 < 2 ta s c I3 = I > I.12. Vi UL = ULmax theo L ta c: U2 L = U2 + U2 R + U2C (1).Mt khc U2 = U2 R + (UL UC)2 U2 R = U2 - (UL UC)2 (2).

Thay (2) vo (1) ta c: U2 L = U2 + U2 - (UL UC)2 + U2C 2U2 = U2 L - U2C + (UL UC)2 = 128000 U = 80 (V).

13. Khi 1= hoc 2= th UC1 = UC2

hay 2 21

1

1( )

U

R L C + .

1

1C = 2 22

2

1( )U

R L C + .

2

1C

21 (R 2 + 21 L2 - 2 LC + 2 21

1C ) =

22(R 2 + 22L2 - 2

LC + 2 22

1C )

21 R 2 + 41 L2 - 21 2 LC + 21

C = 22R 2 + 42L2 - 222 LC + 2

1C



17/44


(21 - 22 )(R 2 - 2 LC ) = - (

41 - 42)L2 21 + 22 = 2

1 LC -

2

2 R L (1) (vi CR

2 ZL.

Ta c tan= L C Z Z

R = tan(-6

) = -13 R = 3 (ZC ZL).

3. Khi = 1 = 100 hay = 2 = 50 th u v i u lch pha nhau gc2 . Vy

on mch ch c L v C m khng c R.

4. V uMB tr pha hn uR tc l tr pha hn i nn uMB c tnh dung khng tc l hpen cha t in. Ta c: UAB = IZ = I 2 2C R Z + U2 AB = U2 R + U2C

UC = 2 2 R ABU U = 160 V ZC = C C R

U RU I U = =

2003 .

5. lch pha gia u v i l:= 4 3 12 = , do hp en cha R v C.

6. Ta c: tanAN = C Z R

= - 1 = tan(-4 ) AN = -4

;MA -AN= - 12

MA =AN - 12 = - 3 . Vy, hp en cha in tr thun R x v t in Cx.Ta li c: ZAN = 2 2C R Z + = 100 2v UMA = I.ZMA = 3UAN = 3.I.ZAM

ZMA = 3ZAN = 300 2. V tanMA = Cx x

Z R

= tan(-3 ) = - 3 ZCx = 3R x

R x = 2 MA Z = 150 2v ZCx = 150 6.

7. V uAB cng pha vi i nn hp en Y cha in tr thun R v R = ABU

I = 100.

V uAN tr pha4 so vi i nn on mch AN cha R v C tc l hp en Z ch

in v ZAN = AN U I = 100 2 ZC = 100. V u v i cng pha nn on mch c

cng hng in, do X l cun cm thun v ZL = ZC = 100.6. Dng gin vc t gii mt s bi ton v on mch xoay chiu* Kin thc lin quan:

Trn on mch RLC ni tip th uR cng pha vi i, uL sm pha hn i gc2

, uCtr pha hn i gc2

. on mch gm cun thun cm v intr thun hoc cun dy c in tr thun th u sm pha hn i.on mch gm t in v in tr thun th u tr pha hn i.

on mch RLC ni tip c: u = uR + uL + uC.Biu din bng gin vc t:U

= RU + LU

+ C U .

Khi v gin vc t cho on mch in gm cc phn t

mc ni tip th chn trc gc trng hng vi vc t biu din cng n din I (v I ging nhau vi mi phn t mc ni tip).



20/44


* Phng php gii:Cn c vo iu kin bi ton cho v gin vc t cho on mch. C th

t tngU bng cch p dng lin tip qui tc hnh bnh hnh. Nhng nn s dcch v thnh hnh a gic th thun li hn.

Nu gin c dng hnh hc c bit, ta c th da vo nhng cng thhc gii bi tp mt cch ngn gn.* Bi tp minh ha:1. Cho on mch xoay chiu nh hnh v.

Trong uAB = 50 2cost (V) ;UAN = 50 V ; UC = 60 V.Cun dy L thun cm. Xc nh UL v UR .2. Cho on mch in xoay chiu nh hnh v.

Trong UAB = 40 V; UAN = 30 V; U NB = 50 V. Cun dy Lthun cm. Xc nh UR v UC.3. Cho on mch xoay chiu nh hnh v.

Cun dy L thun cm. Cc in p hiu dng o c lUAB = 180 V; UAN = 180 V; U NB = 180 V. Xc nh h s cngsut ca on mch.4. Mt on mch in xoay chiu gm cun dy mc ni tip vi in tr t biu thc ca in p hai u mch c dng u = 300cos100t (V). o in p hiudng gia hai u cun dy v hai u in tr c cc gi tr ln lt l 5010 Vv 100 V, cng sut tiu th trn cun dy l 100 W. Tnh in tr thun v cm ca cun dy.5. t mt in p xoay chiu c gi tr hiu dng U vo hai u on mgm cun cm thun c t cm L, in tr thun R v t in c in mc ni tip theo th t trn. Gi UL, UR v UC ln lt l cc in p hiu dnggia hai u mi phn t. Bit in p gia 2 u on mch AB lch pha2 so viin p gia hai u on mch NB (on mch NB gm R v C ). H thdi y ng?

A. 2 2 2 2R C LU U U U= + + . B. 2 2 2 2C R LU U U U= + + . C. 2 2 2 2L R CU U U U= + + . D. 2 2 2 2R C LU U U U= + + .6. Cho on mch xoay chiu nh hnh v. Trong cun dyl thun cm. t vo hai u on mch AB in p xoaychiu uAB = U0cos(100t +) th ta c in p trn cc onmch AN v MB l uAN = 100 2cos100t (V) v uMB = 100 6cos(100t - 2

) (V).Tnh U0.



21/44


7. Cho on mch xoay chiu nh hnh v. Trong cun dyL l thun cm. t vo hai u on mch AB in p xoaychiu uAB = 50 2cos(100t - 3

) (V) th in p gia hai uon mch AM c biu thc l uL = 100 2cos100t (V). Tm biu thc in pgia hai u on mch MB.* Hng dn gii:1. Ta c: UAB = 50 V = UAN.

Gin Fre-nen c dng l mt tam gic cn m y l UC.Do ta c: UL = 12UC = 30 V; UR =

2 2 AN LU U = 40 V.

2. V U2 NB = U2 AB + U2 AN nn trn gin Fre-nen tam gic ABN ltam gic vung ti A; do ta c:12UAB.UAN =

12UL.UR

UR =. AB AN L

U U U = 24 V; UC =

2 2 AN RU U = 18 V.

3. Gin Fre-nen c dng l mt tam gic u vi UR l ng cao

trn cnh y UC nn: cos= cos( ABU ; RU

) = cos(-6 ) = 32 .

4. Ta c: U = 1502 V. Da vo gin vc t ta thy:

U2 = U2d + U2 R + 2UdUR cosd cosd =2 2 2

2d R

d R

U U U U U

=

110 .

Pd = UdIcosd I osd

d d

P U c = 2 A; R d = 2

d P

I = 25;Zd = d U

I = 25 10; ZL = 2 2d Z R = 75 L = L

Z

= 34

H.

5. Theo gin Fre-nen ta c:U2 L = U2 + U2 NB= U2 + U2 R + U2C .

6. Theo gin Fre-nen ta c:UL + UC = 2 2 AN MBU U + = 200 V; UR =

. AN MB L C

U U U U + = 50 3 V ;

U2 AN = U2 R + U2 L v U2 MB= U2 R + U2C U2 MB - U2 AN = U2C - U2 L = (UC + UL)(UC - UL)

UC UL =2 2

MB AN

C L

U U U U

+ = 100 V UL UC = - 100 V

U = 2 2( ) R L C U U U + = 50 7 V U0 = U 2 = 50 14 V.7. Trn gin Fre-nen ta thy: AB =1

2AM v =

3

= 6

ABM l tam gic vung ti B

Ngi vit: Dng Vn ng Trng THPT Nguyn Vn Linh, Bnh Thun


22/44


UMB = 2 2 AM ABU U = 50 3 V; v uMB tr pha hn uAB gc 2 nn:

uMB = UMB 2cos(100t - 3 - 2

) = 50 6cos(100t - 56 (V).

7. My bin p Truyn ti in nng * Cc cng thc:

My bin p: 21

U U = 12 I I =21

N N .

Cng sut hao ph trn ng dy ti: Php = rI2 = r 2 P

U

= P2 2r

U .

gim in p trn ng dy ti in:U = Ir.Hiu sut ti in: H = hp

P P P .

* Phng phi gii: tm cc i lng trn my bin p hoc trn ng d

in ta vit biu thc lin quan n cc i lng bit v i lng cn tsuy ra v tnh i lng cn tm.* Bi tp minh ha:1. Mt my bin p c s vng dy trn cun s cp v s vng dy ca cucp l 2000 vng v 500 vng. in p hiu dng v cng hin dng th cp l 50 V v 6 A. Xc nh in p hiu dng v cng hiu dng s cp.2. Cun s cp v th cp ca mt my bin p c s vng ln lt l N1 = 600vng, N2 = 120 vng. in tr thun ca cc cun dy khng ng k. Ni hacun s cp vi in p xoay chiu c gi tr hiu dng 380 V.

a) Tnh in p hai u cun th cp. b) Ni 2 u cun th cp vi bng n c in tr 100. Tnh cng dng

in hiu dng chy trong cun s cp. B qua hao ph my bin p.3. Mt my pht in c cng sut 120 kW, in p hiu dng gia hai cc c pht l 1200 V. truyn n ni tiu th, ngi ta dng mt dy ti in ctr tng cng 6.

a) Tnh hiu sut ti in v in p hai u dy ni tiu th.

b) tng hiu sut ti in, ngi ta dng mt my bin p t ni my t s vng dy cun th cp v s cp l 10. B qua mi hao ph trong my bitnh cng sut hao ph trn dy v hiu sut ti in lc ny.4. in nng c ti t trm tng p ti trm h p bng ng dy ti i pha c in tr R = 30. Bit in p hai u cun s cp v th cp ca mp ln lt l 2200 V v 220 V, cng dng in chy trong cun th cmy h p l 100 A. B qua tn hao nng lng cc my bin p. Tnh inhai cc trm tng p v hiu sut truyn ti in. Coi h s cng sut bng 1.5 . t vo 2 u cun s cp ca mt my bin p l tng (b qua hao ph) in pxoay chiu c gi tr hiu dng khng i th in p hiu dng gia hcun th cp h l 100 V. cun th cp, nu gim bt n vng dy th hiu dng gia hai u h ca n l U, nu tng thm n vng dy th in



23/44


2U. Tnh in p hiu dng gia hai u cun th cp h khi tng thm 3n v cun th cp.6. T ni sn xut n ni tiu th l hai my bin p. My tng p A c h s bK A = 120, my h p B c h s bin i K B = 15. Dy ti in gia hai bin p c itr tng cng R = 10. B qua hao ph trong hai bin p v gi s ng dy c hcng sut l cos= 1. m bo ni tiu th, mng in 120 V 36 kW hot bnh thng th ni sn xut in nng phi c I1A v U1A bng bao nhiu? Tnh hiusut ca s ti in.7. Mt hc sinh qun mt my bin p vi d nh s vng dy ca cun s chai ln s vng dy ca cun th cp. Do s sut nn cun th cp b thiuvng dy. Mun xc nh s vng dy thiu qun tip thm vo cun th c, hc sinh ny t vo hai u cun s cp mt in p xoay chiu c gi dng khng i, ri dng vn k xc nh t s in p cun th cp h s cp. Lc u t s in p bng 0,43. Sau khi qun thm vo cun th cvng dy th t s in p bng 0,45. B qua mi hao ph trong my bin p. Tvng dy m hc sinh ny phi tip tc qun thm vo cun th cp bin p ng nh d nh.* Hng dn gii

1. Ta c: 21

U U =

1

2

I I =

2

1

N N . U1 =

1

2

N N U2 = 200 V; I1 =

2

1

N N I2 = 1,5 A.

2. a) Ta c: U2 = 21

N N U1 = 76 V.

b) Ta c: I2 = 2U R = 0,76 A v I1 =2

1 N N I2 = 0,152 A.

3. a) Ta c:P = RI2 = R 2

2 P U = 60000 W = 60 kW; H =

P P P = 0,5 = 50%;

U = IR = P U R = 600 V U1 = U U = 600 V.

b) U = 10U = 12000V;P = RI2 = R 2

'2 P U

= 600 W; H =' P P

P

= 0,995 = 99,5%.

4. Ta c: I1 =1

22

U I U = 10 A;U = I1R = 300 V; U = U1 + U = 2500 V.

5. Ta c: U U

N N 2

1

2 = ; vi U2 = 100 V. V:12

2

1

2

N n

N N

N n N = = 2

1

U U - 1

n N = 1U

U (1)

1

n N = 1

2

U U U

(1). Tng t:12

2

1

2

N n

N N

N n N +=+ = 2

1

U U + 1

n N = 1

2U

U

(2).

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

22U U

= 13U U

U = 32 2U = 3200 V.



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Mt khc:12

2

1

2 33 N

n

N

N

N

n N +=

+

= 21

U U + 1

3n N =

3

1

U U (3).

T (1) v (3) ta c:1

2 34U

U U = 3

1

U U U3 = 4U2 3U = 200 V.

6. Ti B: U2B = 120 V; I2B =2

B

B

P

U = 300 A; U1B = K B.U2B = 1800 V; I1B = 2 B

B

I

K = 20 A.

Ti A: I2A = I1B = 20 A; I1A = 2 A A

I K = 400 A; U2A = U1B + I1BR = 2000 V;

U1A = K AU2A = 100 V.Cng sut truyn ti: PA = I1AU1A = 40000 W = 40 kW.

Hiu sut ti in: H = B A

P P = 90%.

7. Ta c: 21

N

N = 0,43 v 2

1

24 N

N

+ = 0,45 N2 = 516; N1 = 1200.

Ta li c: 21

24 N N N

+ + = 0,5 N = 60 (vng).8. My pht in ng c in* Cc cng thc:Tn s dng in do my pht in xoay chiu mt pha pht ra (tnh ra Hz):

My c 1 cp cc, rto quay vi tc n vng/giy: f = n.My c p cp cc, rto quay vi tc n vng/giy: f = pn.

My c p cp cc, rto quay vi tc n vng/pht: f =60 pn .Cng sut tiu th trn ng c in: I2r + P = UIcos.* Bi tp minh ha:1. Mt my pht in xoay chiu mt pha c phn cm l rto gm 8 cp ccnam v 8 cc bc). Rto quay vi tc 300 vng/pht.

a) Tnh tn s ca sut in ng cm ng do my pht ra. b) tn s ca sut in ng cm ng do my pht ra bng 50 Hz th r

quay vi tc bng bao nhiu?2. Mt my pht in xoay chiu mt pha c 4 cp cc. Biu thc ca sung do my pht ra l: e = 2202cos(100t 0,5) (V). Tnh tc quay ca rtotheo n v vng/pht.3. Mt my pht in xoay chiu mt pha c phn ng gm bn cun dy nhau mc ni tip. Sut in ng xoay chiu do my pht sinh ra c tn sv gi tr hiu dng100 2 V. T thng cc i qua mi vng dy ca phn ng5 mWb. Tnh s vng dy trong mi cun dy ca phn ng.

4. Ni hai cc ca mt my pht in xoay chiu mt pha vo hai u on mgm in tr thun R mc ni tip vi cun cm thun. B qua in tr cc cca my pht. Khi rto ca my quay u vi tc n vng/pht th cng in hiu dng trong on mch l 1 A. Khi rto ca my quay u vi tc



25/44


vng/pht th cng dng in hiu dng trong on mch l3 A. Tnh cmkhng ca on mch AB theo R nu rto ca my quay u vi tc 2n vng5. Trong gi hc thc hnh, hc sinh mc ni tip mt qut in xoay chiu vtr R ri mc hai u on mch ny vo in p xoay chiu c gi tr hiu dV. Bit qut in ny c cc gi tr nh mc: 220 V - 88 W v khi hot ng cng sut nh mc th lch pha giain p hai u qut v cng dngin qua n l , vi cos = 0,8. Tnh R qut chyng cng sut nh mc.6. Mt ng c in xoay chiu c in tr dy cun l 32, khi mc vo mch cin p hiu dng 200 V th sn ra cng sut 43 W. Bit h s cng sut l 0,cng dng in chy qua ng c.7. Mt ng c in xoay chiu khi hot ng bnh thng vi in p hi220 V th sinh ra cng sut c hc l 170 W. Bit ng c c h s cng suv cng sut ta nhit trn dy qun ng c l 17 W. B qua cc hao ph kTnh cng dng in cc i qua ng c.

* Hng dn gii 1. a) f =

60 pn = 40 Hz. b) n =

p f 60

= 375 vng/pht.

2. Ta c: f =2

= 60 pn n =

602 p

= 750 vng/pht.

3. E0 = E 2 = 2fN0 N =0

22 E

f = 400 vng. Mi cun: N1c = 4

N = 100 vng.

4. Tn s ca dng in xoay chiu do my pht ra: f =60 pn

.Sut in ng cc i do my pht ra: E0 = NBS = 2fNBS.in p hiu dng t vo 2 u on mch: U = E =

2

0 E

= 2fNBS.

Cm khng ca on mch: ZL = L = 2fL.+ Khi rto ca my quay u vi tc n1 = n th: f 1 = 60

pn ;

U1 = 2f 1 NBS; ZL1 = 2f 1L; I1 = 2121

L Z R

U

+ = 1 (1).+ Khi rto ca my quay u vi tc n3 = 3n th: f 3 = 360

pn= 3f 1;

U3 = 2f 3 NBS = 3U1; ZL3 = 2f 3L = 3ZL1; I3 = 23

23

L Z R

U

+ = 2121

9

3

L Z R

U

+ = 3 (2).

T (1) v (2) suy ra: 3 21

2

21

2

9 L L

Z R Z R

+

+= 3 ZL1 = 3

R.

+ Khi rto ca my quay u vi tc n2 = 2n th: f 2 = 260 pn = 2f 1;



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ZL2 = 2f 2L = 2ZL1=3

2 R.

5. Ta c: PQ = UQIcos I = cosQ

Q

U

P = 0,5 A; ZQ =

I

U Q = 440;

R Q = ZQcos= 352; Z = I U

= 760; Z2

- Z2Q

= 384000(R + R Q)2+(ZLQ - ZCQ)2- (R 2Q + (ZLQ- ZCQ)2) = (R + R Q)2 - R 2Q = 384000(R + R Q)2 = 384000 + R 2Q = 712,672 R = 712,67 R Q = 360,67 361 ().

6. Ta c: I2r + P = UIcos 32I2 - 180I + 43 = 0 I =438 A (loi v cng sut

hao ph qu ln, khng ph hp thc t) hoc I = 0,25 A (nhn).

7. Ta c: Ptp = Pci + Php = 187 W; Ptp = UIcos I = cosU

P tp = 1 A; I0 = I 2= 2 A

MT S CU TRC NGHIM LUYN TP thi TN H C nm 20091. Mt my pht in xoay chiu mt pha c phn cm l rto gm 4 cp ccnam v 4 cc bc). sut in ng do my ny sinh ra c tn s 50 Hz th phi quay vi tc

A. 750 vng/pht. B. 75 vng/pht. C. 25 vng/pht. D. 480 vng/pht.2. t mt in p xoay chiu c gi tr hiu dng 50 V vo hai u on min tr thun R mc ni tip vi cun cm thun L. in p hiu dng gi

R l 30 V. in p hiu dng gia hai u cun cm bngA. 10 V. B. 20 V. C. 30 V. D. 40 V.3. t mt in p xoay chiu tn s f = 50 Hz v gi tr hiu dng U = 80 V vu on mch gm R, L, C mc ni tip. Bit cun cm thun c L =0,6 H, t

in c in dung C =410

F v cng sut ta nhit trn in tr R l 80 W. Gi

ca in tr thun R lA. 80. B. 30. C. 20. D. 40.

4. t mt in p xoay chiu vo hai u on mch ch c t in thA. cng dng in trong on mch sm pha/2 so vi in p gia hai uon mch.

B. dng in xoay chiu khng th tn ti trong on mch.C. cng dng in trong on mch tr pha/2 so vi in p gia hai u

on mch.D. tn s ca dng in trong on mch khc tn s ca in p gia h

on mch.5. Khi ng c khng ng b ba pha hot ng n nh vi tc quay ctrng khng i th tc quay ca rto

A. lun bng tc quay ca t trng.B. ln hn tc quay ca t trng.



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C. nh hn tc quay ca t trng.D. c th ln hn hoc bng tc quay ca t trng, ty thuc ti.

6. in p gia 2 u on mch xoay chiu c biu thc u = 2202cos100t (V).Gi tr hiu dng ca in p ny l

A. 110 2 V. B. 220 2 V. C. 110 V. D. 220 V.7. Mt my bin p l tng c cun s cp gm 1000 vng, cun th c50 vng. in p hiu dng gia hai u cun s cp l 220 V. B qua hain p hiu dng gia hai u cun th cp h l

A. 440 V. B. 44 V. C. 110 V. D. 11 V.8. t mt in p xoay chiu u = 1002cos100t (V) vo hai u on mch cR, L, C mc ni tip. Bit R = 50, cun thun cm c t cm L =1 H v t

in c in dung C =42.10

F. Cng hiu dng ca dng in trong o

mch lA. 2 A. B. 2 A. C. 2 2 A. D. 1 A.9. Khi t hiu in th khng i 12 V vo hai u mt cun dy c in trR v t cm L th dng in qua cun dy c cng 0,15 A. Nu t vu cun dy ny mt in p xoay chiu c gi tr hiu dng 100 V th cdng in hiu dng qua n l 1 A. Cm khng ca cun dy l

A. 50. B. 30. C. 40. D. 60.10. t in p xoay chiu c gi tr hiu dng 120 V, tn s 50 Hz vo ha

on mch mc ni tip gm in tr thun R = 30, cun cm thun L =0,4 H

v t in c in dung thay i c. iu chnh in dung ca t in th hiu dng gia hai u cun cm t gi tr cc i bng

A. 150 V. B. 160 V. C. 100 V. D. 250 V.11. t in p u = U0cost vo hai u on mch mc ni tip gm in tr thR, t in C v cun cm thun c L thay i c. Bit dung khng ca t

3 . iu chnh L in p hiu dng gia hai u cun cm t cc i, kA. in p 2 u in tr R lch pha

6

so vi in p 2 u on mch.

B. in p 2 u t in C lch pha6 so vi in p 2 u on mch.

C. trong mch c cng hng in.D. in p 2 u cun cm L lch pha6

so vi in p 2 u on mch.12. t mt in p xoay chiu c gi tr hiu dng U vo hai u on mgm cun cm thun L, in tr thun R v t in C mc ni tip theo thGi UL, UR v UC ln lt l cc in p hiu dng gia hai u mi phn t.

in p gia 2 u on mch AB lch pha2 so vi in p gia hai u onmch NB (on mch NB gm R v C ). H thc no di y ng?



28/44


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

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

13. Mt on mch in xoay chiu gm in tr thun, cun cm thun vmc ni tip. Bit cm khng gp i dung khng. Dng vn k xoay chiu (rt ln) o in p gia hai u t in v in p gia hai u in tr thca vn k l nh nhau. lch pha ca in p gia hai u on mchcng dng in trong on mch l

A. 4 . B. 6

. C. 3 . D. - 3

.14. in p gia hai u mt on mch l u = 150cos100t (V). C mi giy c bao nhiu ln in p ny bng khng?

A. 100 ln. B. 50 ln. C. 200 ln. D. 2 ln.15. My bin p l thit b

A. bin i tn s ca dng in xoay chiu.B. c kh nng bin i in p ca dng in xoay chiu.C. lm tng cng sut ca dng in xoay chiu.D. bin i dng in xoay chiu thnh dng in mt chiu.

16. Khi t hiu in th khng i 30V vo hai u on mch gm in trR mc ni tip vi cun cm thun L =14 H th dng in trong on mch ldng in mt chiu c cng 1 A. Nu t vo hai u on mch ny u 150 2 cos120 t= (V) th biu thc ca cng dng in trong on mch

A. i 5 2cos(120 t )4= (A). B. i 5cos(120 t )4

= + (A).

C. i 5 2 cos(120 t )4

= + (A). D. i 5cos(120 t )4

= (A).17. t in p xoay chiu c gi tr hiu dng khng i vo hai u ongm bin tr R mc ni tip vi t in. Dung khng ca t in l 100. Khi iuchnh R th ti hai gi tr R 1 v R 2 cng sut tiu th ca on mch nh nhau. Bin p hiu dng gia hai u t in khi R = R 1 bng hai ln in p hiu dng

gia hai u t in khi R = R 2. Cc gi tr ca R 1 v R 2 lA. R 1 = 50, R 2 = 100. B. R 1 = 40, R 2 = 250.C. R 1 = 50, R 2 = 200. D. R 1 = 25, R 2 = 100.

18. t in pu 100 2 cos t= (V), c thay i c vo hai u on mcgm in tr thun R = 200, cun cm thun L =2536 H v t in c in dung

410

F mc ni tip. Cng sut tiu th ca on mch l 50 W. Gi tr ca l

A. 150 rad/s. B. 50 rad/s. C. 100 rad/s. D. 120 rad/s.19. t in p xoay chiu u = U0cost c U0 khng i v thay i c vo haiu on mch c R, L, C mc ni tip. Thay i th cng dng in hiu



29/44


dng trong mch khi = 1 bng cng dng in hiu dng khi = 2. Hthc ng l

A. 1+ 2 = 2 LC . B. 1.2 =1

LC . C. 1 + 2 =2

LC . D. 1.2 =2

LC .

20. t in p u = U0cos(100t - 3 ) (V) vo hai u mt t in c in dun

42.10

(F). thi im in p gia hai u t in l 150 V th cng in trong mch l 4 A. Biu thc ca cng dng in trong mch l

A. i = 4 2cos(100t + 6 ) (A). B. i = 5cos(100t + 6

) (A).

C. i = 5cos(100t - 6 ) (A). D. i = 4 2cos(100t - 6

) (A).21. Trong on mch in xoay chiu gm in tr thun mc ni tip vi cuthun, so vi in p hai u on mch th cng dng in trong mch

A. tr pha2 . B. tr pha4 . C. sm pha2 . D. sm pha 4 .22. Khi truyn i mt cng sut 20 MW trn ng dy ti in 500 kV m dy ti in c in tr 20th cng sut hao ph l

A. 320 W. B. 500 W. C. 50 kW. D. 32 kW.23. t in p u = U0cos(100t + 4

) vo hai u on mch ch c t in tcng dng in trong mch l i = I0cos(t +i);i bng

A. -2 . B. - 3

4 . C.

2 . D. 3

4 .

24. Khi ng c khng ng b ba pha hot ng n nh, t trng quay ng c c tn s

A. bng tn s ca dng in chy trong cc cun dy ca stato.B. ln hn tn s ca dng in chy trong cc cun dy ca stato.C. c th ln hn hay nh hn tn s ca dng in, ty vo ti.D. nh hn tn s ca dng in chy trong cc cun dy ca stato.

25. t in p u = U0cos(100t + 3 ) (V) vo hai u cun cm thun c t

cm L = 12 (H). thi im in p gia hai u cun cm l 1002 V thcng dng in qua cun cm l 2 A. Biu thc ca cng dng icun cm l

A. 2 3 cos 100 ( )6

i t A

= . B. 2 3 cos 100 ( )

6i t A

= +

.

C. 2 2 cos 100 ( )6

i t A

= + . D. 2 2 cos 100 ( )

6i t A

=

.

26. t in p u = 100cos(t + 6 ) (V) vo hai u on mch RLC th dng i

qua mch l i = 2cos(t + 3 ) (A). Cng sut tiu th ca on mch l



30/44


A. 100 3 W. B. 50 W. C. 50 3 W. D. 100 W.

27. T thng qua 1 vng dy dn l =

210.2 cos(100t - 4 ) (Wb). Biu thc

ca sut in ng cm ng xut hin trong vng dy ny l

A. e = 2cos(100t - 4

) (V) B. e = 2cos(100t + 4

) (V).C. e = 2cos(100t + 2

) (V). D. e = 2cos100t (V).28. Mt my bin p l tng c cun s cp 2400 vng dy, cun th cp 800dy. Ni hai u cun s cp vi in p xoay chiu c gi tr hiu dng in p hiu dng gia hai u cun th cp khi bin p hot ng khng t

A. 0. B. 105 V. C. 630 V. D. 70 V.29. t in p xoay chiu u = U0cos2ft, c U0 khng i v f thay i c vo

hai u on mch c R, L, C mc ni tip. Khi f = f 0 th trong on mch c cnghng in. Gi tr ca f 0 l

A.2LC

. B.2LC

. C.1LC

. D.1

2 LC .30. t in p xoay chiu c gi tr hiu dng 60 V vo hai u on mchni tip th cng dng in chy qua on mch l i1 = I0cos(100t + 4

) (A).

Ngt b t in C th cng dng in qua mch l i2 = I0cos(100t - 12 ) (A).

in p hai u on mch lA. u = 60 2cos(100t - 12

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

C. u = 60 2cos(100t + 12 ) (V). D. u = 60 2cos(100t + 6

) (V).31. Mt khung dy dn phng dt hnh ch nht c 500 vng dy, din tchvng 54 cm2. Khung dy quay u quanh mt trc i xng (thuc mt phngkhung), trong t trng u c vect cm ng t vung gc vi trc quay vln 0,2 T. T thng cc i qua khung dy l

A. 0,27 Wb. B. 1,08 Wb. C. 0,81 Wb. D. 0,54 Wb.32. Trong on mch in xoay chiu gm R, L (thun cm) v C mc ni tiA. in p gia hai u t in ngc pha vi in p gia hai u on mB. in p gia hai u cun cm cng pha vi in p gia hai u t inC. in p gia hai u t in tr pha so vi in p gia hai u on mD. in p gia hai u cun cm tr pha so vi in p gia hai u on m

33. Cho on mch xoay chiu RLC mc ni tip. Bit cc in p hiu dnUR = 10 3 V, UL = 50 V, UC = 60 V. in p hiu dng gia hai u on mch

lch pha gia in p hai u on mch v cng dng in chy tronc gi tr lA. U = 20 2V;= 6

. B. U = 20 2V;= 3 .



31/44


C. U = 20 V;= - 6 . D. U = 20 V;= - 3

. thi H C nm 201034. t in p u = U2cost vo hai u on mch AB gm hai on mch AN NB mc ni tip. on AN gm bin tr R mc ni tip vi cun cm thun

cm L, on NB ch c t in, in dung C. t

1 = LC 2

1. in p hiu dng

gia hai u on mch AN khng ph thuc vo R th tn s gc bng

A. 21 . B.

221 . C. 21. D. 1 2 .

35. t in p xoay chiu c gi tr hiu dng v tn s khng i vo hai mch gm bin tr R mc ni tip vi t in c in dung C. Gi in p higia hai u t in, gia hai u bin tr v h s cng sut ca on mchtr c gi tr R 1 ln lt l UC1, UR1 v cos1; khi bin tr c gi tr R 2 th cc gi tr

tng ng ni trn l UC2, UR2 v cos2. Bit UC1 = 2UC2, UR2 = 2UR1. Gi tr ca cos1v cos2 l:

A. cos1 = 51

, cos2 = 31

. B. cos1 = 31

, cos2 = 51

.

C. cos1 = 51

, cos2 = 52

. D. cos1 = 221

, cos2 = 21

.

36. t in p xoay chiu c gi tr hiu dng 200 V v tn s khng i vu A v B ca on mch mc ni tip theo th t gm bin tr R, cun cc t cm L v t in c in dung C thay i. Gi N l im ni gia cuthun v t in. Cc gi tr R, L, C hu hn v khc khng. Vi C = C1 th in phiu dng gia hai u bin tr R c gi tr khng i v khc khng khi thay tr R ca bin tr. Vi C =12

C th in p hiu dng gia A v N bngA. 200 2 V. B. 100 V. C. 200 V. D. 100 2 V.

37. Ti thi im t, in p u = 2002cos(100t -2 ) (u tnh bng V, t tnh bng s) c

gi tr 100 2V v ang gim. Sau thi im 1300s, in p ny c gi tr lA. - 100 2 V. B. 100 V. C. 100 3 V. D. 200 V.

38. Mt on mch AB gm hai on mch AM v MB mc ni tip. onAM c in tr thun 50 mc ni tip vi cun cm thun c t cm1 H,on mch MB ch c t in c in dung thay i c. t iu = U0cos100t (V) vo hai u on mch AB. iu chnh in dung ca tn gi tr C1 sao cho in p hai u on mch AB lch pha2

so vi in p haiu on mch AM. Gi tr ca C1 l

A.

510.8 F. B.

510 F. C.

510.4 F. D.

510.2 F.



32/44


39. t in p xoay chiu c gi tr hiu dng khng i, tn s 50 Hz vo hon mch mc ni tip gm in tr thun R, cun cm thun L v t indung C thay i c. iu chnh in dung C n gi tr

4104

F hoc

4102

F th

cng sut tiu th trn on mch u c gi tr bng nhau. Gi tr ca L bngA. 13 H. B. 12 H. C. 3 H. D. 2 H.

40. t in p u = U0cost vo hai u on mch gm in tr thun R, cun thun c t cm L v t in c in dung C mc ni tip. Gi i l cng in tc thi trong on mch; u1, u2, u3 lnlt l in p tc thi gia hai u itr, gia hai u cun cm v gia hai u t in. H thc ng l

A. i = 2u L . B. i =

1u R . C. i = u3C. D. i = 2 21( )

u R L C +

.

41. Trong gi hc thc hnh, hc sinh mc ni tip mt qut in xoay chiu vtr R ri mc hai u on mch ny vo in p xoay chiu c gi tr hiu dV. Bit qut in ny c cc gi tr nh mc: 220 V - 88 W v khi hot ng cng sut nh mc th lch pha giain p hai u qut v cng dngin qua n l , vi cos = 0,8. qut in ny chyng cng sut nh mc thR bng

A. 354 . B. 361 . C. 267 . D. 180 .42. t in pu = U0cost vo hai u cun cm thun c t cm L thcng dngin qua cun cm l

A. i = 0U L cos(t + 2 ). B. i = 0 2

U L cos(t + 2

).

C. i = 0U L cos(t - 2 ). D. i = 0 2

U L cos(t - 2

).43. t vo hai u cun s cp ca mt my bin p l tng (b qua hao mt in pxoay chiu c gi tr hiu dng khng i th in p hiu dng giu cun th cp h l 100 V. cun th cp, nu gim bt n vng dy p hiu dng gia hai u h ca n l U, nu tng thm n vng dy th inl 2U. Nu tng thm 3n vng dy cun th cp th in p hiu dng gia haih ca cun ny bng

A. 100 V. B. 200 V. C. 220 V. D. 110 V.44. Ni hai cc ca mt my pht in xoay chiu mt pha vo hai u on mgm in tr thun R mc ni tip vi cun cm thun. B qua in tr cc cca my pht. Khi rto ca my quay u vi tc n vng/pht th cng dng trong on mch l 1 A. Khi rto ca my quay u vi tc 3n vng/pcng hiu dng trong on mch l3 A. Nu rto ca my quay u vi tc

2n vng/pht th cm khng ca on mch AB lA. 3

R. B. R 3 . C.

23

R. D. R 3 .



33/44


45. Mt ng c in xoay chiu khi hot ng bnh thng vi in p hi220 V th sinh ra cng sut c hc l 170 W. Bit ng c c h s cng suv cng sut ta nhit trn dy qun ng c l 17 W. B qua cc hao ph kcng dng in cc i qua ng c l

A. 2 A. B. 3 A. C. 1 A. D. 2 A.46. t in p xoay chiu u = U0cost vo hai u on mch ch c in trthun. Gi U l in p hiu dng gia hai u on mch; i, I0 v I ln lt l gitr tc thi, gi tr cc i v gi tr hiu dng ca cng dng in tronmch. H thc no sau ysai ?

A.0 0

0U I U I

= . B.0 0

2U I U I

+ = . C. 0u iU I

= . D.2 2

2 20 0

1u iU I

+ = .

47. t in p u = U0cost c thay i c vo hai u on mch gm cucm thun c t cm L, in tr thun R v t in c in dung C mc

Khi I.67. Mt khung dy dn phng quay u vi tc gc quanh mt trc c nhnm trong mt phng khung dy, trong mt t trng u c vect cm vung gc vi trc quay ca khung. Sut in ng cm ng trong khung c

thc e = E0cos(t + 2

). Ti thi im t = 0, vect php tuyn ca mt phng khdy hp vi vect cm ng t mt gc bngA. 450. B. 1800. C. 900. D. 1500.

68. Mt on mch AB gm hai on mch AM v MB mc ni tip. onAM gm in tr thun R 1 mc ni tip vi t in c in dung C, on mchgm in tr thun R 2 mc ni tip vi cun cm thun c t cm L. t ixoay chiu c tn s v gi tr hiu dng khng i vo hai u on mch A on mch AB tiu th cng sut bng 120 W v c h s cng sut bngni tt hai u t in th in p hai u on mch AM v MB c cng gi dng nhng lch pha nhau3

, cng sut tiu th trn on mch AB trong trhp ny bng

A. 75 W. B. 160 W. C. 90 W. D. 180 W.69. Mt hc sinh qun mt my bin p vi d nh s vng dy ca cun sgp hai ln s vng dy ca cun th cp. Do s sut nn cun th cp b this vng dy. Mun xc nh s vng dy thiu qun tip thm vo cun thcho , hc sinh ny t vo hai u cun s cp mt in p xoay chiu chiu dng khng i, ri dng vn k xc nh t s in p cun th cp cun s cp. Lc u t s in p bng 0,43. Sau khi qun thm vo cun t24 vng dy th t s in p bng 0,45. B qua mi hao ph trong my bin



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c my bin p ng nh d nh, hc sinh ny phi tip tc qun thm vth cp

A. 40 vng dy. B. 84 vng dy. C. 100 vng dy. D. 60 vng dy.70. t in p xoay chiu u =U 2cos100 t vo hai u on mch mc ni tigm in tr thun R, t in c in dung C v cun cm thun c tthay i c. iu chnh L in p hiu dng hai u cun cm t gii th thy gi tr cc i bng 100 V v in p hiu dng hai u t bng 36 V. Gi tr ca U l

A. 80 V. B. 136 V. C. 64 V. D. 48 V.71. t in p 2cosu U t = vo hai u mt t in th cng dng in qn c gi tr hiu dng l I. Ti thi im t, in p hai u t in l u v dng in qua n l i. H thc lin h gia cc i lng l

A.2 2

2 2u i 1

4U I+ = . B.2 2

2 2u iU I+ = 1. C.

2 2

2 2u iU I+ = 2. D.

2 2

2 2u i 1

2U I+ = .

72. t in p xoay chiu u = U0cost (U0 khng i v thay i c) vo haiu on mch gm in tr thun R, cun cm thun c t cm L v tin dung C mc ni tip, vi CR 2 < 2L. Khi = 1 hoc = 2 th in p hiudng gia hai bn t in c cng mt gi tr. Khi = 0 th in p hiu dng giahai bn t in t cc i. H thc lin h gia1, 2 v0 l

A. 0 = 12(1+ 2). B. 20 =

12(

21 + 22). C. 0 1 2 = . D. 2 2 2

0 1 2

1 1 1 1( )2= + .73. t in p xoay chiuu U 2 cos100 t= (U khng i) vo hai u on mchmc ni tip gm in tr thun R, cun cm thun L =15 H v t in c indung C thay i c. iu chnh in dung ca t in in p hiu dhai bn t in t gi tr cc i. Gi tr cc i bngU 3. in tr R bng

A. 10. B. 10 2. C. 20 2. D. 20.74. on mch AB gm hai on mch AM v MB mc ni tip. on mgm in tr thun R 1 = 40mc ni tip vi t in c din dng

310C F4

= , onmch MB gm in tr thun R 2 mc ni tip vi cun cm thun. t vo Ain p xoay chiu c gi tr hiu dng v tn s khng i th in p tc thu on mch AM v MB ln lt l:AM

7u 50 2 cos(100 t )(V)12= v

MBu 150cos100 t (V)= . H s cng sut ca on mch AB lA. 0,86. B. 0,95. C. 0,84. D. 0,71.

75. Mt my pht in xoay chiu mt pha c phn ng gm bn cun dy nhau mc ni tip. Sut in ng xoay chiu do my pht ra c tn s 50 Hztr hiu dng100 2V. T thng cc i qua mi vng ca phn ng l5

mWb. S

vng dy trong mi cun dy ca phn ng lA. 71 vng. B. 200 vng. C. 100 vng. D. 400 vng.



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76. t mt in p xoay chiu c gi tr hiu dng v tn s khng i ln lhai u in tr thun R, cun cm thun c t cm L, t in c in ducng dng in hiu dng qua mch tng ng l 0,25 A; 0,5 A; 0,2 A. Nin p xoay chiu ny vo hai u on mch gm ba phn t trn mc ni cng dng in hiu dng qua mch l

A. 0,2 A. B. 0,3 A. C. 0,15 A. D. 0,05 A.77. Mt khung dy dn phng, hnh ch nht, din tch 0,025 m2, gm 200 vng dyquay u vi tc 20 vng/s quanh mt trc c nh trong mt t trng trc quay l trc i xng nm trong mt phng khung v vung gc vi phca t trng. Sut in ng hiu dng xut hin trong khung c l222V. Cm ng t c ln bng:

A. 0,50 T. B. 0,60 T. C. 0,45 T. D. 0,40 T.78. Khi ni v h s cng sut cos ca on mch xoay chiu, pht biu no sauysai ?

A. Vi on mch ch c t in hoc ch c cun cm thun th cos= 0.B. Vi on mch c in tr thun th cos= 1.C. Vi on mch c R, L, C mc ni tip ang xy ra cng hng th cos= 0.D. Vi on mch gm t in v in tr thun mc ni tip th 0 < cos< 1.

79. t in p u = U0cost ( U0 v khng i) vo hai u on mch xoaychiu ni tip gm in tr thun, cun cm thun v t in c in duchnh c. Khi dung khng l 100th cng sut tiu th ca on mch t ci l 100 W. Khi dung khng l 200th in p hiu dng gia hai u t in

100 2 V. Gi tr ca in tr thun lA. 100. B. 150. C. 160. D. 120.80. t in p u = 220 2cos100t (V) vo hai u on mch gm mt bng dy tc loi 110V 50W mc ni tip vi mt t in c in dung C thc. iu chnh C n sng bnh thng. lch pha gia cng dnv in p hai u on mch lc ny l:

A. 2 . B. 3

. C. 6

. D. 4

.

81. Mt my tng p c cun th cp mc vi in tr thun, cun s cp mngun in xoay chiu. Tn s dng in trong cun th cpA. c th nh hn hoc ln hn tn s trong cun s cp.

B. bng tn s dng in trong cun s cp.C. lun nh hn tn s dng in trong cun s cp.D. lun ln hn tn s dng in trong cun s cp.

82. Mt on mch in xoay chiu gm mt t in v mt cun cm thunni tip. lch pha gia in p hai u t in v in p hai u o bng

A. 2

. B. 2

. C. 0 hoc . D. 6

hoc 6

.



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83. t in p 150 2 os100u c t = (V) vo hai u on mch gm in tr thucun cm thun v t in mc ni tip th in p hiu dng gia hai u thun l 150 V. H s cng sut ca mch l

A. 32 . B. 1. C.12 . D.

33 .

84 . Trong my pht in xoay chiu ba pha ang hot ng, sut in ngchiu xut hin trong mi cun dy ca stato c gi tr cc i l E0. Khi sut inng tc thi trong mt cun dy bng 0 th sut in ng tc thi trong mdy cn li c ln bng nhau v bng

A. 0 32

E . B. 023 E . C. 0

2 E . D. 0 2

2 E .

85 . Cho dng in xoay chiu c tn s 50 Hz chy qua mt on mch. Khothi gian gia hai ln lin tip cng dng in ny bng 0 l

A. 1100

s. B. 1200

s. C. 150

s. D. 125

s.86 . Khi truyn in nng c cng sut P t ni pht in xoay chiu n nth th cng sut hao ph trn ng dy l P. cho cng sut hao ph trn dy ch cn l P n

(vi n > 1), ni pht in ngi ta s dng mt my bin

tng) c t s gia s vng dy ca cun s cp v s vng dy ca cun thA. n . B.

1n . C. n. D.

1n .

p n cc cu trc nghim luyn tp1 A. 2 D. 3 D. 4 A. 5 C. 6 D. 7 D. 8 A. 9 D. 10 B. 11 A. 12 C. 13 A. 14 A. 15 B.D. 17 C. 18 D. 19 0 B. 20 B. 21 B. 22 D. 23 D. 24 A. 25 A. 26 C. 27 B. 28 D. 2930 C. 31 D. 32 C. 33 C. 34 D. 35 B. 36 C. 37 A. 38 A. 39 C. 40 B. 41 B. 42 C. 4344 C. 45 D. 46 D. 47 B. 48 D. 49 D. 450C. 51 A. 52 B. 53 A. 54 A. 55 B. 56 AB. 58 C. 59 D. 60 B. 61 A. 62 A. 63 B. 64 A. 65 A. 66 C. 67 B. 68 C. 69 D. 7071 C. 72 B. 73 B. 74 C. 75 C. 76 A. 77 A. 78 C. 79 A. 80 B. 81 B. 82 C. 83 B. 8485 A. 86 B.



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C - KT LUN

Thc t ging dy v kt qu cc bi kim tra, bi thi trong cc nm hni cc trng ti ging dy (THPT Bi Th Xun, THPT Nguyn Vn LBnh Thun) cho thy nu cc em hc sinh nhn c dng cc cu hi trc nnh lng trong cc thi th vic gii cc cu ny s cho kt qu kh tt.

Trong thi tuyn sinh H v C cc nm 2010, 2011 c mt s cu nghim nh lng kh di v kh nn nhiu th sinh khng lm kp. gip cnhn dng gii nhanh mt s cu trc nghim nh lng, ti a vo tliu ny mt s dng bi tp c xem l mi vi cch gii c coi l ngnht (theo suy ngh ch quan ca bn thn ti) cc ng nghip v cc emsinh tham kho. t c kt qu cao trong cc k thi th cc em hc singii nhiu luyn tp rn luyn k nng nhn dng t a ra phnu gii nhanh v chnh xc tng cu. Nu c nhng cu kh v di qu tdnh li gii sau cng. Nu sp ht gi m cha gii ra mt s cu nocng ng b trng, hy la chn mt phng n m mnh cho l kh thi nhvo la chn (d sao vn cn xc sut 25%).

Ti liu ch trnh by c mt phn ca chng trnh Vt L 12. Cccc bi tp theo suy ngh ch quan ca ti cho l ngn gn nhng cha chc lgn lm v chc chn s khng trnh khi nhng thiu st trong cch phncng nh cch gii cc bi tp minh ha. Rt mong nhn c nhng nhn x ca cc qu ng nghip xy dng c mt tp ti liu hon ho hn.

Xin chn thnh cm n.

Hm Thun Bc, thng 01 nm 2012 Ngi vit

Dng Vn ng



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MC LC

STT NI DUNG TRANG1 A PHN M U 12 B NI DUNG 23 1. i cng v dng in xoay chiu 24 2. Tm cc i lng trn on mch xoay chiu c R, L, C 45 3. Vit biu thc ca u v i trn on mch xoay chiu 96 4. Bi ton cc tr trn on mch xoay chiu 117 5. Bi ton nhn bit cc thnh phn trn on mch xoay chiu 18 6. Dng gin vc t gii mt s bi ton v on mch

xoay chiu19

9 7. My bin p Truyn ti in nng 2210 8. My pht in ng c in 24

11 Mt s cu trc nghim luyn tp 2612 thi TN H C nm 2009 2613 thi H C nm 2010 3114 thi TN H C nm 2011 3515 p n cc cu trc nghim luyn tp 3916 C . KT LUN 40

TI LIU THAM KHO

1. Vt l 12 - V Quang (ch bin) - NXB GD - Nm 2011.2. Bi tp vt l 12 - V Quang (ch bin) - NXB GD - Nm 2011.3. Vt l 12 - Nng cao - V Thanh Khit (ch bin) - NXB GD - Nm 2011.4. Bi tp vt l 12 - Nng cao - V Thanh Khit (ch bin) - NXB GD - Nm 25. Ni dung n tp mn Vt l 12 - Nguyn Trng Su - NXB GD - Nm 20106. Hng dn n tp thi tt nghip THPT nm hc 2010 - 2011 - Nguyn Trn- NXB GD - Nm 2011.

7. Vt l 12 - Nhng bi tp hay v in hnh - Nguyn Cnh He - NXB HQ Ni 2008.8. Bi ging trng tm chng trnh chun Vt l 12 - V Thanh Khit - HQG H Ni - 2010.9. Cc thi tt nghip THPT v tuyn sinh H - C cc nm 2009, 2010 v 210. Cc ti liu truy cp trn cc trang web thuvienvatly.com v violet.vn.



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CNG HA X HI CH NGHA VIT NAMc lp T do Hnh phc

----- -----

PHIU NH GI, XP LOI SNG KIN KINH NGHIMNm hc 2011 2012

I. nh gi, xp loi ca HKH trng THPT Nguyn Vn Linh1. Tn ti:

PHNG PHP GII CC DNG BI TP VT L 12PHN DNG IN XOAY CHIU.

2. H v tn ngi vit: Dng Vn ng.3. Chc v: Ph Hiu trng. T: Vn phng.4. Nhn xt ca Ch tch HKH v ti:a) u im:................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................b) Hn ch:...........................................................................................................................................................................................................................................................................................................................................................................................5. nh gi, xp loi:

Sau khi thm nh, nh gi ti trn, HKH trng THPT Nguyn Vn

thng nht xp loi: ...............................................................................................Nhng ngi thm nh: Ch tch HKH C S

(K, ghi r h tn) (K, ng du, ghi r h tn).....................................................................................................................................................................................................................................................................................................................................

II. nh gi, xp loi ca HKH S GD&T Tnh Bnh ThunSau khi thm nh, nh gi ti trn, HKH S GD&T Bnh Thun thngxp loi: ...................................................................................................................

Nhng ngi thm nh: Ch tch HKH NGNH GD(K, ghi r h tn) (K, ng du, ghi r h tn)

..................................................................................................................................

.................................................................



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S GIAO DUC VA AO TAO BNHTHUAN

TRNG THPT NGUYEN VAN LINH---------- ----------

Sang kien kinhnghiem :

PHNG PHAP GIAICAC DANG BAI TAP

VAT LY 12PHAN DONG IENXOAY CHIEU



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Ngi viet : Dng Van ong----- -----

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