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  • L L & KHO KH Phin bn 1.0

    Tuyn tp cc cu hi vt l kh nhn t cc thi th i hc trn ton quc km li gii chi tit v bnh lun.

    GSTT GROUP 11/12/2013

  • GSTT GROUP | 1

    ng bao gi b cuc cc em nh

    Anh ch yu cc em nhiu lm!

  • L l & kho kh 1.0 | 2

    Phn 1: bi Cu 1: Hai dao ng iu ha cng phng, cng tn s, bit phng trnh x1 = A1cos(t /6) cm v x2 = A2cos(t ) cm c phng trnh dao ng tng hp l x = 9cos(t + ). bin A2 c gi tr cc i th A1 c gi tr:

    A. 18 3 cm B. 7cm C. 15 3 cm D. 9 3 Christmas

    Cu 2: Mt vt dao ng iu ha vi phng trnh x = A.cos(t). T s gia tc trung bnh v vn tc trung bnh khi vt i c sau thi gian 3T/4 u tin k t lc bt u dao ng l A. 1/3 B. 3 C. 2 D. 1/2

    Cu 3: Mt con lc l xo nm ngang gm vt nng tch in q = 20C v l xo c cng k = 10N/m. Khi vt ang nm cn bng, cch in, trn mt bn nhn th xut hin tc thi mt in trng u trong khng gian bao quanh c hng dc theo trc l xo. Sau con lc dao ng trn mt on thng di 4cm. ln cng in trng E l: A. 2.10

    4 V/m. B. 2,5.10

    4 V/m. C. 1,5.10

    4 V/m. D.10

    4 V/m.

    Cu 4: Mt con lc n c chiu di l = 64cm v khi lng m = 100g. Ko con lc lch khi v tr cn bng mt gc 60 ri th nh cho dao ng. Sau 20 chu k th bin gc ch cn l 30. Ly g = 2 = 10m/s2. con lc dao ng duy tr vi bin gc 60 th phi dng b my ng h b sung nng lng c cng sut trung bnh l: A. 0,77mW. B. 0,082mW. C. 17mW. D. 0,077mW.

    Cu 5: Mt con lc l xo dao ng iu ha theo phng ngang vi nng lng dao ng 1J v lc n hi cc i l 10 N. Gi Q l u c nh ca l xo, khong thi gian ngn nht gia 2 ln lin

    tip Q chu tc dng ca lc ko 5 3 N l 0,1s. Qung ng ln nht m vt i c trong 0,4s l

    A. 60cm. B. 50cm. C. 55cm. D. 50 3 cm.

    Cu 6: Mt con lc l xo gm l xo c cng k = 2N/m, vt nh khi lng m = 80g, dao ng trn mt phng nm ngang, h s ma st trt gia vt v mt phng ngang l = 0,1. Ban u ko vt ra khi v tr cn bng mt on 10cm ri th nh. Cho gia tc trng trng g = 10m/s2. Tc ln nht m vt t c bng A. 0,36m/s B. 0,25m/s C. 0,50m/s D. 0,30m/s

    Cu 7: Mt si dy n hi c treo thng ng vo mt im c nh, u di ca dy t do.

    Ngi ta to sng dng trn dy vi tn s b nht l f1. c sng dng trn dy phi tng tn s

    ti thiu n gi tr f2. T s f2/f1 l:

    A. 1,5. B. 2. C. 2,5. D. 3.

    Cu 8: Cho mch in xoay chiu nh hnh v. in dung C c gi tr thay i c v cun dy thun cm. iu chnh gi tr ca C v ghi li s ch ln nht trn tng vn k th thy UCmax = 3ULmax. Khi UCmax gp bao nhiu ln URmax?

    A. 3

    8 B.

    8

    3 C.

    4 2

    3 D.

    3

    4 2

    Cu 9: Cho mch in xoay chiu nh hnh v. in dung C c gi tr thay i c v cun dy thun cm. iu chnh gi tr ca C th thy: cng thi im s, ch ca V1 cc i th s ch ca V1 gp i s ch ca V2. Hi khi s ch ca V2 cc i th s ch ca V2

    gp bao nhiu ln s ch V1?

    A. 2 ln. B. 1,5 ln. C. 2,5 ln. D. 2 2 ln Cu 10: Gi s ban u c mt mu phng x X nguyn cht, c chu k bn r T v bin thnh ht nhn bn Y. Ti thi im t1 t l gia ht nhn Y v ht nhn X l k. Ti thi im t2 = (t1 + 2T) th t l l

    Cu 11: Mt con lc l xo c cng k = 40N/m u trn c gi c nh cn pha di gn vt m. Nng m ln n v tr l xo khng bin dng ri th nh vt dao ng iu ha theo phng thng ng vi bin 2,5cm. Ly g = 10m/s2. Trong qu trnh dao ng, trng lc ca m c cng sut tc thi cc i bng

    A. 0,41W B. 0,64W C. 0,5W D. 0,32W

    Cu 12: Mt con lc l xo t trn mt phng nm ngang gm l xo nh c mt u c nh, u kia gn vi vt nh c khi lng m. Ban u vt m c gi v tr l xo b nn 9cm. Vt M c khi

  • GSTT GROUP | 3

    lng bng mt na khi lng vt m nm st m. Th nh m hai vt chuyn ng theo phng ca trc l xo. B qua mi ma st. thi im l xo c chiu di cc i ln u tin, khong cch gia hai vt m v M l: A. 9 cm. B. 4,5 cm. C. 4,19 cm. D. 18 cm.

    Cu 13: Mt CLLX nm ngang gm l xo c cng k = 20N/m va vt nng m = 100g .T VTCB

    ko vt ra 1 on 6cm ri truyn cho vt vn tc 20 14 cm/s hng v VTCB .Bit rng h s ma st gia vt v mt phng ngang l 0.4 ,ly g = 10m/s2. Tc cc i ca vt sau khi truyn vn tc bng :

    A. 20 22 cm/s B. 80 2 cm/s C. 20 10 cm/s D. 40 6 cm/s

    Cu 14: Mt vt dao ng iu ha vi phng trnh li : x = 4cos(8t 2/3) cm. Thi gian vt

    i c qung ng S = (2 + 2 2 ) cm k t lc bt u dao ng l: A. 1/12 B. 5/66 C. 1/45 D. 5/96

    Cu 15: Mt con lc l xo gm vt m1 (mng, phng) c khi lng 2kg v l xo c cng k = 100N/m ang dao ng iu ha trn mt phng nm ngang khng ma st vi bin A = 5cm. Khi vt m1 n v tr bin th ngi ta t nh ln n mt vt c khi lng m2. Cho h s ma st gia m2 v m1 l = 0,2 v g = 10m/s

    2. Gi tr ca m2 n khng b trt trn m1 l A. m2 0,5kg B. m2 0,4kg C. m2 0,5kg D. m2 0,4kg Cu 16: Mt con lc l xo treo thng ng, khi vt v tr cn bng l xo gin 4cm. Kch thch cho vt dao ng iu ha th thy thi gian l xo b nn trong mt chu k l T/3 (T l chu k dao ng ca vt). gin v nn ln nht ca l xo trong qu trnh vt dao ng l: A. 12 cm v 4 cm. B. 15 cm v 5 cm. C. 18 cm v 6 cm. D. 8 cm v 4 cm.

    Cu 17. Mt si dy n hi cng ngang, ang c dng dng n nh. Trn dy A l mt nt, B l im bng gn A nht, AB = 14cm. C l mt im trn dy trong khong AB c bin bng mt na bin ca B. Khong cch AC l A. 14/3 cm B. 7 cm C. 3,5 cm D. 1,75 cm

    Cu 18. Hai im A, B nm trn cng mt ng thng i qua mt ngun m v hai pha so vi ngun m. Bit mc cng m ti A v ti trung im ca AB ln lt l 50 dB v 44 dB. Mc cng m ti B l A. 28 dB B. 36 dB C. 38 dB D. 47 dB

    Cu 19: Ti O c 1 ngun pht m thanh ng hng vi cng sut khng i. Mt ngi i b t A n C theo mt ng thng v lng nghe m thanh t ngun O th nghe thy cng m tng t I n 4I ri li gim xung I. Khong cch AO bng:

    A. AC 2

    2 B.

    AC 3

    3 C.

    AC

    3 D.

    AC

    2

    Cu 20: Cho hai mch dao ng l tng L1C1 v L2C2 vi C1 = C2 = 0,1F, L1 = L2 = 1 H. Ban du tch in cho t C1 n hiu in th 6V v t C2 n hiu in th 12V ri cho mch dao ng. Thi gian ngn nht k t khi mch dao ng bt u dao ng th hiu in th trn 2 t C1 v C2 chnh lch nhau 3V?

    A. 610

    6

    s B. 610

    3

    s C. 610

    2

    s D. 610

    12

    s

    Cu 21: Mc vo hai u on mch RLC ni tip gm mt ngun in xoay chiu c tn s f thay i C. Khi tn s f1 = 60Hz, h s cng sut t cc i cos1 = 1. Khi tn s f1 = 120Hz, h s

    cng sut nhn gi tr cos2 = 2

    2. Khi tn s f3 = 90Hz th h s cng sut ca mch bng

    A. 0,874 B. 0,486 C. 0,625 D. 0,781

    Cu 22: t in p u = U 2 cos(t + ) (V) vo hai u mch RLC ni tip, cun dy thun cm, in dung C thay i C. Khi in dung c C = C1, o in p hai u cun dy, t in v in

    tr ln lt UL = 310V v UC = UR = 155V. Khi thay i C = C2 UC2 = 155 2 V th in p hai u cun dy khi bng A. 175,3V. B. 350,6V. C. 120,5V. D. 354,6V

    Cu 23: Cho on mch RLC ni tip, cun dy thun cm v in tr R thay i C. t vo hai u on mch in p xoay chiu c gi tr hiu dng U = 200V. Khi R = R1 v R = R2 th

    mch c cng cng sut. Bit R1 + R2 = 100. Cng sut ca on mch khi R = R1 bng A. 400W. B. 220W. C. 440W D. 880W

  • L l & kho kh 1.0 | 4

    Cu 24: Mt on mch xoay chiu gm 3 phn t mc ni tip: in tr thun R, cun dy c (L; r) v t in c in dung C. t vo hai u on mch mt in p xoay chiu, khi in p tc

    thi hai u cun dy v hai u t in ln lt l: ud = 80 6 cos(t +

    6) V, uC = 40 2 cos(t

    2

    3)V, in p hiu dng hai u in tr l UR = 60 3 V. H s cng sut ca on mch trn

    l

    A. 0,862. B. 0,908. C. 0,753. D. 0,664.

    Cu 25: Cho mt mch in xoay chiu AB gm in tr thun R = 100, cun dy thun cm L, t in c in dung C. t vo hai u on mch mt hiu in th xoay chiu u =

    220 2 cos100t (V), bit ZL = 2ZC. thi im t hiu in th hai u in tr R l 60(V), hai u t in l 40(V). Hi hiu in th hai u on mch AB khi l:

    A. 220 2 (V) B. 20 (V) C. 72,11 (V) D. 100 (V)

    Cu 26: t in p u = U 2 cos(2ft) vo hai u on mch gm in tr thun R, cun cm thun L v t in C mc ni tip. Bit U, R, L, C khng i, f thay i C. Khi tn s l 50Hz th dung khng gp 1,44 ln cm khng. cng sut tiu th trn mch cc i th phi iu chnh tn s n gi tr bao nhiu?

    A. 72Hz B. 34,72Hz C. 60Hz D. 50 2 Hz

    Cu 27: t in p xoay chiu c u = 100 2 cos(t) V vo hai u mch gm in tr R ni tip vi t C c ZC = R. Ti thi im in p tc thi trn in tr l 50V v ang tng th in p tc thi trn t l

    A. 50V. B. 50 3 V. C. 50V. D. 50 3 V.

    Cu 28. Cho mch in xoay chiu RLC c CR2 < 2L. t vo hai u on mch mt in p

    xoay chiu c biu thc u = U. 2 cos(t) , trong U khng i, bin thin. iu chnh gi tr

    ca in p hiu dng gia hai bn t t cc i. Khi UL = 0,1UR. Tnh h s cng sut ca mch khi .

    A. 1

    17 B.

    1

    26 C.

    2

    13 D.

    3

    7

    Cu 29. Cho mch in AB gm in tr thun R, cun thun cm L v t C ni tip vi nhau theo th t trn., v c CR2 < 2L. t vo hai u on mch mt in p xoay chiu c biu thc u =

    U. 2 cos(t) , trong U khng i, bin thin. iu chnh gi tr ca in p hiu dng

    gia hai bn t t cc i. Khi Cmax

    5UU

    4 . Gi M l im ni gia L v C. H s cng sut

    ca on mch AM l:

    A. 2

    7 B.

    1

    3 C.

    5

    6 D.

    1

    3

    Cu 30. Cho mch in xoay chiu RLC c CR2 < 2L. t vo hai u on mch mt in p

    xoay chiu c biu thc u = U. 2 cos(t) , trong U khng i, bin thin. iu chnh gi tr

    ca in p hiu dng gia hai u ca cun cm t cc i. Khi Lmax

    41UU

    40 . Tnh h s

    cng sut ca mch khi .

    A. 0,6 B. 0,8 C. 0,49 D. 3

    11

    Cu 31. Cho mch in AB gm in tr thun R, cun thun cm L v t C ni tip vi nhau theo th t trn., v c CR2 < 2L. t vo hai u on mch mt in p xoay chiu c biu thc u =

    U. 2 cos(t) , trong U khng i, bin thin. iu chnh gi tr ca in p hiu dng gia hai bn t t cc i. Gi M l im ni gia cun cm v t. Ngi ta dng vn k V1 theo di gi tr ca UAM, vn k V2 theo di gi tr ca UMN gi tr ln nht m V2 ch l 90V.

    Khi V2 ch gi tr ln nht th V1 ch gi tr 30 5 V. Tnh U.

  • GSTT GROUP | 5

    A. 70,1V. B. 60 3 V C. 60 5 D. 60 2 V

    Cu 32. Cho mch in RLC mc ni tip, trong RC2 < 2L. t vo hai u on mch in p

    xoay chiu u = U 2 cos 2ft, trong ng 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 cc i, mch tiu th cng sut bng 3

    4 cng sut cc i.

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

    a. Tnh tn s ca dng in khi in p hiu dng ca t cc i.

    A. 125Hz B. 75 5 Hz C. 50 15 Hz D. 75 2 Hz.

    b. Tnh h s cng sut ca mch khi in p hiu dng gia hai u cun cm cc i.

    A. 3

    2 B.

    1

    3 C.

    5

    7 D.

    2

    5

    Cu 33. Dng d kin sau tr li cc cu hi:

    Cho mch in nh hnh v. C ba linh kin : in tr, t, cun thun cm c ng trong ba hp kn, mi hp cha mt linh kin, v mc ni tip vi nhau. Trong : RC2 < 2L.

    t vo hai u on mch mt in p xoay chiu c biu thc u = U. 2 .cos t, trong U

    khng i, c th thay i c. Tng dn gi tr ca t 0 n v theo di s ch ca cc vn k v am pe k, ri ghi li gi tr cc i ca cc dng c o th thy gi tr cc i ca V1 l 170V, ca V2 l 150V, ca V3 l 170V, ca A l 1A. Theo trnh t thi gian th thy V3 c s ch cc i u tin.

    a Theo th t t tri sang phi l cc linh kin: A. R, L, C B. L, R, C C. R, C, L D. C, R, L

    b. Theo trnh t thi gian, cc dng c o c s ch cc i ln lt l: A. V3, V2, A, V1

    B. V3, sau V2 v A ng thi, cui cng l V1 C. V3 sau l V1, cui cng l V2 v A ng thi. D. V3 v V1 ng thi, sau l V2 v A ng thi.

    c. Tnh cng sut tiu th trong mch khi V1 c s ch ln nht. A. 150W B. 170W C. 126W D. 96W

    Cu 34. Cho mch in RLC mc ni tip, trong RC2 < 2L. t vo hai u on mch in p

    xoay chiu u = U 2 cos 2ft, trong ng 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 3

    4 cng sut cc i.

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

    a. Tnh tn s ca dng in khi in p hiu dng ca t cc i.

    A. 50Hz B. 75Hz C. 50 2 Hz D. 75 2 Hz.

    b. Tnh h s cng sut ca mch khi in p hiu dng gia hai u cun cm cc i.

    A. 6

    7 B.

    1

    3 C.

    5

    7 D.

    2

    5

    Cu 35. Cho mch in nh hnh v:

    t vo hai u on mch in p xoay chiu c biu thc u = U0cos t (V) trong , U0 c gi tr khng i, c th thay i c. iu chnh in p hiu dng trn t c gi tr cc i, khi uAN lch pha gc 71,57

    0 (tan 71,57

    0 =3) so vi uAB, cng sut tiu th ca mch khi

    l 200W. Hi khi iu chnh cng sut tiu th ca mch t cc i th gi tr cc i bng bao nhiu? Bit rng h s cng sut ca on mch AN ln hn h s cng sut ca on mch AB.

  • L l & kho kh 1.0 | 6

    Cu 36. Cho mch in nh hnh v:

    t vo hai u on mch in p xoay chiu c biu thc u = U0cos t (V) trong , U0 c gi tr khng i, c th thay i c. iu chnh in p hiu dng trn t c gi tr cc i, khi uAN lch pha gc so vi uAB. Tm gi tr nh nht ca .

    Cu 37. Cho mch in xoay chiu nh hnh v, trong cun dy c in tr thun r. t vo hai u on mch in p xoay chiu c biu thc u = U0cos t (V), trong U0 khng thay i, c th thay i c. iu chnh gi tr ca in p hiu dng ca on MB t cc i th gi tr cc i ng bng U0, cng sut tiu th ca on mch khi l 182W, in p hiu dng ca on AM khi l 135,2V.

    a. Tnh r.

    b. Tnh U0.

    Cu 38. Cho mch in xoay chiu RLC ni tip, trong L l cun thun cm, RC2 > 2L. t vo hai u on mch in p xoay chiu

    c biu thc 0

    u U .cos t V trong

    U0 khng i, cn c th thay i c. Ban u tn s gc ca dng in l , h s cng sut ca on mch MB bng 0,6. Khi tng tn s ca dng in ln gp i th in p gia hai u cun cm t cc i. Hi t gi tr , phi thay i tn s ca dng in th no :

    a. Cng sut tiu th trn on mch t cc i.

    b. in p hiu dng trn t t cc i.

    Cu 39: Trong th nghim Y-ng v giao thoa nh sng, Ngun pht ng thi 2 bc x n sc

    1 =0,64m () v 2 =0,48m (lam). Trn mn hng vn giao thoa, trong on gia 3 vn

    sng lin tip cng mu vi vn trung tm c s vn sng v vn lam l:

    A. 4 vn , 6 vn lam. B. 6 vn , 4 vn lam. C. 7 vn , 9 vn lam.

    D. 9 vn , 7 vn lam

    Cu 40: Mt cht im dao ng iu ha trn trc Oy. chnh gia khong thi gian ngn

    nht khi vt i t v tr bin n v tr cn bng th tc l 40m/s. Khi vt c li 10cm th

    tc ca vt l 30m/s. Chu k dao ng l:

    A. B. C. D.

    Cu 41: Ni hai cc ca mt my pht in xoay chiu mt pha vo hai u on mch ngoi

    RLC ni tip. B qua in tr dy ni, coi t thng cc i gi qua cc cun dy ca my pht

    khng i. Khi Rto ca my pht quay vi tc n0 (vng/pht) th cng sut tiu th

    mch ngoi t cc i. Khi Rto ca my pht quay vi tc n1 (vng/pht) v n2

    (vng/pht) th cng sut tiu th mch ngoi c cng mt gi tr. H thc quan h gia n0,

    n1, n2 l:

    A. B. C. 2 2

    2 1 20 2 2

    1 2

    n .nn =

    n +n D.

    2 22 1 20 2 2

    1 2

    n .nn = 2

    n +n

    Cu 42: Ni hai cc ca mt my pht in xoay chiu mt pha vo hai u on mch AB gm

    in tr thun mc ni tip vi mt cun dy thun cm. B qua in tr ca my pht. Khi

    roto quay u vi tc n vng/pht th cng dng in hiu dng trong on mch l

    1(A). Khi roto quay vi tc 3n vng/pht th cng dng in hiu dng trong on

    mch l 3(A) . Nu roto quay u vi tc 2n vng/pht th cm khng ca on mch l:

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

    Cu 43: Mt con lc n gm mt qu cu khi lng m = 250g mang in tch q = 10-7C c treo bng mt si dy khng dn, cch in, khi lng khng ng k chiu di 90cm trong

    L R C

    M N A B

    C L,r

    A B M

  • GSTT GROUP | 7

    in trng u c E = 2.106 V/m ( c phng nm ngang). Ban u qu ng yn v tr cn bng. Ngi ta t ngt i chiu ng sc in trng nhng vn gi nguyn ln ca E, ly g = 10m/s2. Chu k v bin dao ng ca qu cu l:

    A. 1,878s;14,4cm B. 1,887s; 7,2cm C. 1,883s; 7,2cm D. 1,881s;14,4cm Cu 44: Trong giao thoa Y-ng c a = 0,8mm, D = 1,2m. Chiu ng thi hai bc x n sc

    = 0,75m v = 0,45m vo hai khe. V tr trng nhau ca cc vn ti ca hai bc x trn

    mn l:

    A. 0,225(k+1/2)mm (k = 0; 1; 2; 3....) B. 0,375(k+1/2)mm (k = 0; 1; 2;

    3....)

    C. 2(2k+1)mm (k = 0; 1; 2; 3....) D. 1,6875(2k+1)mm (k = 0; 1; 2;

    3....)

    Cu 45: on mch AB theo th t gm cc on mch AM, MN v NB mc ni tip. on

    mch AM cha in tr thun R, on mch MN cha t in C, on mch NB cha cun dy

    khng thun cm r, L. t vo A, B in p xoay chiu Bit in p

    hiu dng , , in p gia 2 im M, B lch pha 90o so vi in p

    gia 2 im A, N. H s cng sut ca on mch AB l:

    A. 0,642 B. 0,5 C. 0,923 D. 1

    Cu 46: Cho on mch RLC ghp ni tip, cun cm thun c t cm L thay i: R = 120, 410

    C F0,9

    , in p hai u on mch u = Uocos100t(V). iu chnh L = L1 th ULmax = 250V.

    Tm gi tr ca L LU 175 2 (V)?

    A. 3,09

    L H

    B. 0,21

    L H

    C. 3,1

    L H

    D. 2,5

    L H

    Cu 47: Khi thc hin giao thoa vi hai ngun kt hp O1O2 cch nhau 12 cm v c phng trnh

    1

    u =3cos(40t + )cm

    6;

    2

    5u =3cos(40t - )cm

    6. Vn tc truyn sng 60cm/s. Tm s im dao

    ng vi bin 3 cm trn on O1O2?

    A. 16 B. 8 C. 9 D. 18

    Cu 48: on mch xoay chiu AB gm mt cun dy mc ni tip vi mt in tr R, UAB =

    150 2 V. in p hiu dng gia hai u in tr v hai u cun dy ln lt l 70V; 170V.

    Cng sut tiu th l 75W, gi tr ca R l:

    A. 65,3 B. 140 C. 160 D. 115,7

    Cu 49: Mt on mch ni tip gm cun dy c in tr thun r = 100 3 v t cm L =

    0,191 H, t in c in dung C = 1/4(mF), in tr R c gi tr thay i c. in p t vo

    hai u on mch u = 200 2 cos(100t) V. Thay i gi tr ca R cng sut tiu th trong

    mch t cc i. Xc nh gi tr cc i ca cng sut trong mch.

    A. 200 W B. 228W C. 100W D. 50W

    Cu 50: Trong mch dao ng c T=0,12s. Ti thi im gi tr in tch v cng dng

    in l 01Q 3

    q =2

    , . Ti thi im (trong gi tr mi ca

    chng l 02

    Qq =

    2, Gi tr ln nht ca l:

    A. 240,12s B. 240,24s C. 241,33s D. 241,45s

    Cu 51: Cho mch in xoay chiu AB nh hnh v, trong in tr R =

    20, cun dy c in tr thun r =10, t cm L = 1/ H, t in c

    in dung C thay i c. in p gia hai u on mch c biu thc

    uAB = 120 cos100t (V). Ngi ta thy rng khi C = Cm th in p

    hiu dng gia hai im M v B t cc tiu l U1min. Gi tr U1min khi l:

  • L l & kho kh 1.0 | 8

    A. 40 V B. 40 V C. 60 V D. 60 V

  • GSTT GROUP | 9

    p n Cu 1:

    V gin vect nh hnh v v theo nh l hm s sin:

    22

    A A Asin= A =

    sinsin sin

    6 6

    , A2 c gi tr cc i khi sin c

    gi tr cc i bng 1 = /2

    A2max = 2A = 18cm A1 = 2 2 2 2

    2A A = 18 9 = 9 3

    Cu 2:

    Vn tc trung bnh: 2 1tb2 1

    x xv =

    t t

    , 2 1x = x x l di. Vn tc trung bnh trong mt chu k lun

    bng khng

    Tc trung bnh lun khc 0: tb

    2 1

    Sv =

    t t trong S l qung ng vt i c t t1 n t2.

    Tc trung bnh: tocdoS 3A 4A

    v = = =3Tt T

    4

    (1); 3T

    4 chu k u vt i t x1 = + A (t1 = 0) n x2 = 0

    (t2 = 3T

    4 ) (VTCB theo chiu dng)

    Vn tc trung bnh: 2 1van toc tb2 1

    x x 0 A 4Av = = =

    3Tt t 3T0

    4

    (2). T (1) v (2) suy ra kt qu bng 3.

    Cu 3:

    V chiu di on thng dao ng l 4cm nn suy ra bin A = 2cm. Khi vt m dao ng hp ca lc in trng v lc n hi gy gia tc a cho vt. Ti v tr bin, vt c gia tc cc i. Khi ta c: F Fh = m.amax

    qE kA = m.2.A = m.k

    m.A qE = 2kA E = 2.104 V/m

    Cu 4:

    0 = 60 = 0,1047rad v T = 2

    g

    l = 2

    2

    0,64

    = 1,6 (s)

    C nng ban u W0 = mgl(1 cos0) = 2mglsin2

    2

    0 mgl

    2

    2

    0

    C nng sau t = 20T: W = mgl(1 cos) = 2mglsin2 2

    mgl2

    2=mgl

    8

    2

    0

    gim c nng sau 20 chu k: W = mgl(2

    2

    0 8

    2

    0 ) = mgl8

    3 20 = 2,63.103J

    Cng sut trung bnh cn cung cp con lc dao ng duy tr vi bin gc l 60

    Ptb = 3

    3W 2,63.10= = 0,082.1020T 32

    W = 0,082mW.

    Cu 5:

    21 k = 50 N / mkA = 12

    A = 20 cmkA = 10

    v kx = 5 3 x =10 3cm

    max

    Tt = 0,1= T = 0,6s S = 2A + A = 60cm

    6

    Cu 6:

    Vt c tc cc i khi gia tc bng 0; tc l

    lc hl dh msF = F + F = 0 ln u tin ti N

  • L l & kho kh 1.0 | 10

    ON = x kx = mg x = mg/k = 0,04m = 4cm Khi vt i c qung ng S = MN = 10 4 = 6cm = 0,06m

    Theo L bo ton nng lng ta c: 2 2 2

    maxmv kx kA+ = mgS2 2 2

    (Cng ca lc ma st Fms =

    mgS)

    2 2 2

    maxmv kA kx= mgS2 2 2

    06,0.10.08,0.1,02

    04,0.2

    2

    1,0.2

    2

    08,0 222max v

    = 0,0036 2maxv = 0,09 vmax = 0,3(m/s) =

    30cm/s.

    Cch 2:

    gim bin sau na chu k 1 22mg 2.0.1.0,08.10

    A A = = = 0,08m = 8cmk 2

    Sau na chu k u tin bin cn li A2 = 2cm Tc ln nht t c ti v tr cn bng mi

    1 2 1 2max

    A + A A + Ak 2 10 + 2v = = = = 30

    2 m 2 0,08 2cm/s

    Cu 7:

    Si dy 1 u c nh, 1 u t do nn v

    (2k 1) f (2k 1).4 4

    ll

    1

    vk 1 f

    4l v 22 1

    1

    fvk 2 f 3. 3f 3

    4 fl Ch :

    Tn s ti thiu bng k 1 kf f

    2

    Cu 8:

    V C bin thin nn: 2 2Cmax LU

    U R ZR

    (1)

    Lmax max L L L

    min

    U UU I .Z .Z .Z

    Z R (2) (cng hng in)

    v RmaxU U (3) (cng hng in)

    2 2

    LCmaxL

    Lmax L

    R + ZU(1)= 3 = R = Z 8

    (2) U Z (4)

    2 2

    LCmax

    Rmax

    R + ZU(1)=

    (3) U R (5)

    T (4) v (5) 8

    3

    U

    U

    maxR

    maxC

    Cu 9:

    Khi V1 cc i th mch cng hng: UR = U = 2UC = 2UL hay R = 2ZL (1)

    Khi V2 cc i ta c:R

    ZRUU

    2L

    2

    maxC

    theo (1)

    2 2

    L L

    Cmax

    L

    U 4Z + Z U 5U =

    2Z 2 (2)

    Khi li c: L

    2L

    2

    CZ

    ZRZ

    theo (1) ta c: ZC = 5ZL = 2,5R Z = R 5 (3)

    Ch s ca V1 lc ny l RUR U

    U = IR = =Z 5

    (4)

  • GSTT GROUP | 11

    T (3) v (4) ta c: Cmax

    R

    U 5= = 2,5

    U 2

    Cu 10:

    p dng cng thc nh lut phng x ta c: 1

    1 1

    1

    1

    tY t01

    t

    1X 1 0

    N N (1 e )N 1= = = k e =

    N N N e k +1

    (1)

    2 1

    2

    2 1 1

    2

    t (t +2T)Y 02

    2 t (t +2T) t 2T

    1X 2 0

    N N (1 e )N (1 e ) 1k = = = = = 1

    N N N e e e e

    (2)

    Ta c ln2

    2 T2T 2ln2T

    1e = e = e =

    4

    (3). Thay (1), (3) vo (2) ta c t l cn tm:

    2

    1k = 1 = 4k + 3

    1 1.

    1+ k 4

    .

    Cng sut tc thi ca trng lc Pcs = F.v = mg.v vi v l vn tc ca vt m

    Pmax = mg.vmax = mg.2kA

    m = gA mk = gA

    kAk

    g; (v A = l0)

    Pmax = kA Ag = 40.2,5.102

    10.10.5,2 2 = 0,5W.

    Cu 12:

    Khi qua v tr cn bng, vn tc 2 vt l v p dng nh lut bo ton c nng cho qu trnh hai vt chuyn ng t v tr l xo b nn

    l n khi hai vt qua v tr cn bng:

    2 21 1 kk( ) = (m + M)v v = 2 2 m + M

    l l (1)

    n v tr cn bng, vt m chuyn ng chm dn, M chuyn ng thng u, hai vt tch ra, h con lc l xo ch cn m gn vi l xo.

    Khi l xo c di cc i th m ang v tr bin, thi gian chuyn ng t v tr cn bng n v tr bin l T/4

    Khong cch ca hai vt lc ny: 2 1

    Tx = x x = v. A

    4 (2), vi

    mT = 2

    k;

    mA = v

    k,

    T (1) v (2) ta c: k 2 m m k 1 1

    x = . . . . = . = 4,19cm1,5m 4 k k 1,5m 2 1,5 1,5

    l l l l

    Cch 2

    Khi h vt chuyn ng t VT bin ban u n VTCB: CLLX (m + M = 1,5m): vmax =

    kA = A

    1,5m

    Khi n VTCB, hai vt tch khi nhau do m bt u chuyn ng chm dn, lc ny M chuyn ng thng u vi vn tc vmax trn. Xt CLLX c vt m (vn tc cc i khng thay i):

    vmax = k

    A'' = A'm

    = k A 9

    A A' = = cm1,5m 1,5 1,5

  • L l & kho kh 1.0 | 12

    T khi tch nhau (qua VTCB) n khi l xo c chiu di cc i th m n v tr bin A, thi gian

    dao ng l T' 2

    t = = =4 4' 2'

    ; vi k

    ' = = 1,5 t =m .2 1,5

    . Trong thi gian ny, M

    i c qung ng:

    s = vmax.t = 4,5

    A. = cm.2 1,5 1,5

    khong cch hai vt: d = s A 4,19 cm

    Cch 3

    Sau khi th h con lc l xo dao ng iu ha, sau khi hai vt t vn tc cc ai th M tch ra chuyn ng thng u, cn m dao ng iu ha vi bin A

    22

    max(m + M)vk( ) =2 2

    l maxv = l

    k

    m + M = l

    k

    1,5m

    22

    maxmvkA =2 2

    A = maxvm

    k = l

    k

    1,5m

    m

    k =

    1,5

    l = 7,348 cm

    Sau khi tch nhau vt m dng li v tr bin sau thi gian t = T

    4 =

    2

    4

    m

    k khi M i c

    qung ng S2 = maxv t = lk

    1,5m .

    2

    4

    m

    k =

    .

    2 1,5

    l = 11,537 cm

    Khong cch gia hai vt khi l S = S2 A = 11,537 7,348 = 4,189 = 4,19 cm

    Cu 13: Vt c tc cc i khi gia tc bng 0;

    tc l lc hl dh msF = F + F = 0 ln u tin ti N

    ON = x kx = mg x = mg/k = 0,02m = 2cm Khi vt i c qung ng S = MN = 6 2 = 4cm = 0,04m

    Ti t = 0 x0 = 6cm = 0,06m, v0= 20 14 cm/s = 0,2 14 m/s Theo L bo ton nng lng ta c:

    2 2 22

    max 0 0mv mv kxkx+ = + mgS2 2 2 2

    (Cng ca Fms = mgS)

    2 2 2 2

    max 0 0mv mv kx kx= + mgS2 2 2 2

    2 2 2 2

    max0,1v 0,1(0,2 14) 20.0,06 20.0,02= + 0,4.0,1.10.0,042 2 2 2

    = 0,044 2maxv = 0,88

    vmax = 2204,088,0 = 0,2. 22 (m/s) = 20 22 cm/s.

    Cu 14:

    Vt xut pht t M n N th i c qung ng S = 2 + 2 2 . Thi gian:

    T T 5t = + = (s)

    12 8 96

    Cu 15: vt m2 khng trt trn m1 th lc qun tnh cc i tc dng ln m2 c ln khng vt qu lc ma st ngh gia m1 v m2 tc l

    msn qtmaxF F 2 2 maxm g m a 2

    2

    1 2

    kg A g A m 0,5(kg)

    m + m

    Cch 2

    Sau khi t m2 ln m1 h dao ng vi tn s gc = 1 2

    k

    m + m 2 =

    1 2

    k

    m + m

  • GSTT GROUP | 13

    m2 khng trt trn m1 th gia tc chuyn ng ca m2 c ln ln hn hoc bng ln gia

    tc ca h (m1 + m2); vi a = 2x. Lc ma st gia m2 v m1 gy ra gia tc ca m2 c ln: a2 =

    g = 2m/s2 iu kin m2 khng b trt trong qu trnh dao ng l

    amax = 2A a2; suy ra

    1 2

    kAg

    m + m g(m1 + m2) kA 2(2 + m2) 5 m2 0,5kg.

    Cu 16: Thi gian l xo nn l T/3. Thi gian khi l xo bt u b nn n lc nn ti a l T/6. nn ca l xo l A/2, bng gin ca l xo khi vt v tr cn bng. Suy ra A = 8cm. Do gin ln nht ca l xo A/2 + A = 4cm + 8cm = 12cm, cn nn ln nht A/2 = 4cm

    Cu 17.

    = 4.AB = 46 cm Dng lin h gia H v chuyn ng trn

    u: AC = 30

    .360

    = 14/3 cm

    Cu 18. T cng thc I = P/4d2

    Ta c: 2A M

    M A

    I d= ( )

    I d v LA LM = 10.lg(IA/IM) dM =

    0,6

    A10 .d

    Mt khc M l trung im cu AB, nn ta c: AM = (dA + dB)/2 = dA + dM; (dB > dA) Suy ra dB = dA + 2dM

    Tng t nh trn, ta c: 2 0,6 2A B

    B A

    I d= ( ) = (1+ 2 10 )

    I d v LA LB = 10.lg(IA/IB)

    Suy ra LB = LA 10.lg0,6 2(1 2 10 ) = 36dB

    Cch 2

    Cng m ti im cch ngun m khong R; I = 2

    P

    4R = 10

    L.I0; vi P l cng sut ca ngun;

    I0 cng m chun, L mc cng m R = 0

    P

    4.I L1

    10

    M l trung im ca AB, nm hai pha ca gc O nn: RM = OM = B AR R

    2

    (1)

    Ta c RA = OA v LA = 5 (B) RA = 0

    P

    4.I LA1

    10=

    0

    P

    4.I 51

    10 (2)

    Ta c RB = OB v LB = L RB = 0

    P

    4.I LB1

    10=

    0

    P

    4.I L1

    10 (3)

    Ta c RM = OM v LM = 4,4 (B) RM = 0

    P

    4.I LM1

    10=

    0

    P

    4.I 4,41

    10 (4)

    T ta suy ra 2RM = RB RA 2 4,410

    1=

    L10

    1

    510

    1

    L10

    1 =

    510

    1 + 2

    4,410

    1

    L10 = 9,4

    4,4 5

    10

    10 + 2 10

    L

    210 = 5,22,2

    7,4

    10.210

    10

    = 63,37 8018,1

    2

    L L = 3,6038 (B) = 36

    (dB)

  • L l & kho kh 1.0 | 14

    Cu 19: Do ngun pht m thanh ng hng. Cng m ti im cch

    ngun m R l 2

    PI =

    4R. Gi s ngi i b t A qua M ti C IA

    = IC = I OA = OC Gi thuyt: IM = 4I OA = 2.OM. Trn ng thng qua AC IM t gi tr ln nht, nn M gn O nht OM vung gc vi AC v l trung im ca AC

    AO2 = OM

    2 + AM

    2 =

    2 2AO AC+

    4 4 3AO2 = AC2 AO =

    AC 3

    3

    Cu 20:

    Hai mch dao ng c1 2 1 2

    C = C ; L = L nn 1 2

    1

    1 = = =

    L C1

    Khi cho hai mch bt u dao ng cng mt lc th hiu in th gia hai bn t ca mi mch dao ng bin thin cng tn s gC.

    Ta biu din bng hai ng trn nh hnh v Ti thi im t k t lc bt u dao ng, hiu in th trn mi t l u1, u2

    Theo bi ton: u2 u1 = 3V (1) T hnh v, ta c: 02 2

    01 1

    U u= = 2

    U u (2)

    T (1) v (2), ta c: 6

    01

    1

    U 10u = 3V = = t = = = (s)

    2 3 3 3

    .

    Cch 2: Phng trnh hiu in th: 1 2

    u = 6cos(t); u =12cos(t)

    V hiu in th bin thin cng tn s, c ngha l khi u1 gim v 0 th u2 cng gim v 0.

    Do , ta c: 2 1

    u u = 3 12cost 6cost = 3 1

    cost = t = + k22 3

    V hiu in th trn mi t ang gim nn ta chn h nghim

    t = + k23

    Thi gian ngn nht nn ta chn k = 0. Vy 6 10

    t = t = = (s)3 3 3

    Cu 21:

    Khi cos1 = 1 ZL1 = ZC1120L = 1

    120.C LC =

    2

    1

    (120) (1)

    Khi cos2 = 2

    2 2 = 45

    0 tan2 =

    L2 C2Z Z

    R

    = 1 R = ZL2 ZC2

    tan3 = 2

    L3 C3 L3 C3

    2

    L2 C2

    1180L

    Z Z Z Z 4 (180) LC 1180C= = = .1R Z Z 3 (240) LC 1

    240L240C

    tan3 =

    2

    2

    2

    2

    (180)1

    4 4 5 5(120)= =

    (240)3 3 4.3 91

    (120)

    (tan3)2

    = 25/91 2

    3

    1 25 1061

    cos 81 81

    cos3 =

    0,874.

    Cch 2

  • GSTT GROUP | 15

    T/h 1: ZL1 = ZC1

    T/h 2: f2 = 2f1 ZL2 = 4ZC2 v cos2 = 2

    2 2 = 45

    0 R = ZL2 ZC2 ZC2 = R/3

    2

    3C =

    2f R

    T/h 3: f3 = 1,5.f1 ZL3 = 2,25.ZC3

    32 2 2 2 2

    2C 2

    2

    3

    R Rcos 0,874

    R (1,25) Z (2 f ) RR 1,5625

    (2 f )

    Cu 22:

    2

    2 2L LL

    Z = 2R U155 2 = + U 155 2

    2U =155 2

    LU = 350,6V

    Cu 23:

    P1 = P2 1

    2 2

    1 L C

    R

    R + (Z Z )= 2

    2 2

    2 L C

    R

    R + (Z Z ) (ZL ZC)

    2 = R1.R2

    P1 = 2

    1

    2 2

    1 L C

    U R

    R + (Z Z )=

    2

    1

    2

    1 1 2

    U R

    R + R R=

    2

    1 2

    U

    R + R = 400W.

    Cu 24:

    d C

    2 5 = + =

    6 3 6 uC chm so vi i mt gc /2 vy ud nhanh pha so vi i mt gc /2

    tand = tan

    3= L

    r

    U

    U nn L rU = 3U m

    2 2 2 2

    d r L rU = U + U = 4U

    r LU = 40 3 (V) v U =120 (V)R rU + Ucos = = 0,908U

    Cu 25: Ta c hiu in th hai u on mch thi im t l: uAB = uR + uC + uL = 20(V); (v uCv uL ngc pha nhau) Cu 26:

    Khi f = f1 = 50 (Hz): ZC1 = 1,44.ZL1

    1

    1

    2f .C= 1,44.2f1L LC = 2 2

    1

    1

    1,44.4 f (1)

    Gi f2 l tn s cn iu chnh cng sut tiu th trn mch cc i. Khi f = f2 th trong mch xy

    ra cng hng: ZC2 = ZL2 2

    1

    2f .C = 2f2.L LC = 2 2

    2

    1

    4 f (2)

    So snh (1) v (2), ta c: 2 2

    2

    1

    4 f =

    2 2

    1

    1

    1,44.4 f f2 = 1,2.f1 = 1,2.50 = 60 (Hz)

    Cu 27:

    T ZC = R U0C = U0R = 100V m Ru 50i = =

    R R cn 0R0

    UI =

    R

    p dng h thc c lp trong on ch c t C:

    2R2 22

    C C

    2 2 220R0C 0

    u( )

    u ui R+ = 1 = 1UU I 100

    ( )R

    2

    C Cu = 7500 u = 50 3V ; v ang tng nn chn Cu = 50 3V

    Cch 2 R = ZCUR = UC.

    Ta c: U2 = UR

    2 + Uc

    2 = 2UR

    2 UR = 50 2 V = UC. Mt khc:

    CZtan =R

    = 1

    =

    4

    T ta suy ra pha ca i l (

    t +4

    ).

    R L, r = 0 C A B

    M N

  • L l & kho kh 1.0 | 16

    Xt on cha R: uR = U0Rcos(

    t +4

    ) = 50cos(

    t +4

    ) = 2

    1

    V uR ang tng nn u'R > 0 suy ra sin(

    t +4

    ) < 0 vy ta ly sin(

    t +4

    ) = 2

    3(1)

    v uC = U0C.cos(

    t +4

    2) = U0C.sin(

    t +

    4) (2) Th U0C = 100V v th (1) vo (2) ta c uC =

    50 3 V

    Cu 28:

    Ta c: L1 2

    R 1

    U 0,5tan 0,1 tan 5

    U tan

    H s cng sut ca mch l : 2 2

    2

    1 1cos

    1 tan 26

    Cu 29:

    Ta c: Cmax C

    5U 5ZU Z

    4 4 .

    Khng lm nh hng n kt qu bi ton, c th gi s ZC = 5, Z =

    4. Khi : 2 2LZ 5 4 3

    L C LR 2.Z . Z Z 2.3. 5 3 2 3 . Suy ra: ZAM =

    2 2

    LR Z 12 9 21

    H s cng sut ca on mch AM

    1AM

    R 2 3 2cos

    Z 21 7

    Cu 30:

    Tng t trn, c th gi s: Z = 40, ZL = 41.

    Khi : 2 2CZ 41 40 9

    C L CR 2.Z . Z Z 2.9. 41 9 24

    H s cng sut ca mch khi : R 24

    cos 0,6Z 40

    Cu 31:

  • GSTT GROUP | 17

    Bn gin vc t, ta c:

    22y 90 30 5 60V

    x = 90 y = 30V

    2 2 2 2U 90 x 90 30 60 2V

    Lu : Nu cn tnh UR khi th ta c:

    RU v 2.x.y 2.60.30 60V

    H s cng sut ca mch khi l: RU 1

    U 2

    Cu 32:

    a. Hai tn s f1 v f2 tho mn cng thc: 2 2 2

    1 2 Rf .f f . Vy tn s ca dng in in p hiu

    dng trn in tr t cc i l: R 1 2f f .f (*)

    Khi iu chnh f cng sut tiu th trn mch cc i th trong mch xy ra cng hng. H s

    cng sut khi bng 1. V cng sut tiu th ca mch c tnh bng biu thc: 2

    max

    UP

    R

    Trong cc trng hp khc th cng sut ca mch c tnh bng biu thc: 2 2 2 2

    2 2 2

    max2 2

    U U R UP I .R .R . .cos P .cos

    R RZ Z

    ng vi tn s f1, cng sut tiu th trn mch bng 3

    4 Pmax. Vy ta

    suy ra h s cng sut khi Ucmax l 3 3

    4 2 ( trn hnh v, h s

    cng sut ca mch khi ny c gi tr bng 1

    cos .

    Khng lm nh hng n kt qu, c th gi s v = 3 , z = 2. Khi ta suy ra y = 1.

    Theo cng thc ca phn l thuyt trn th ta c: 2v 3

    x 1,52.y 2

    Theo t l trn hnh v th khi tn s dng in l f1 th t s gia dung khng v cm

    khng ca mch l : C1

    L1

    Z x y 2,5 5

    Z x 1,5 3

    V khi tn s ca dng in tng t f1 n f2 th in p ca t v ca cun cm i gi tr cho nhau, nn cm khng v dung khng trong mch cng i gi tr cho nhau. Nn tn s

    f2 th ta c: L2

    C2

    Z 5

    Z 3 . Hay L2 2

    L1 1

    Z f 5

    Z f 3

  • L l & kho kh 1.0 | 18

    Mt khc: f2 = f1 + 100 (Hz)

    Gii h phng trnh ta suy ra: f1 = 150Hz, f2 = 250Hz

    Thay hai gi tr f1 v f2 trn vo(*) ta c: Rf 150.250 50. 15 Hz

    b. H s cng sut ca mch khi in p gia hai u cun cm t cc i cng bng h s

    cng sut ca mch khi in p gia hai u t in t cc i v bng 3

    2

    Cu 33:

    a. Khi tng dn t 0 n th UC t cc i u tin. Theo , V3 c s ch cc i u tin. Vy Z l hp cha t.

    Do Lmax CmaxU U . M s ch cc i ca V1 v V3 bng nhau. Nn ta suy ra X l hp cha

    cun cm.

    Cui cng, Y l hp cha in tr thun.

    Vy theo th t t tri sang phi l cc linh kin: L, R, C. Chn p n B.

    b. Khi I t cc i th UR cng t cc i nn A v V2 ng thi c s ch cc i. Theo trnh t thi gian, cc dng c o c s ch cc i ln lt l: V3 , sau V2 v A

    ng thi, cui cng l V1. Chn B.

    c. V2 c s ch cc i Rmax ABU U . Vy ta c UAB = 150V.

    Khi V2 (v ng thi A) c s ch cc i th cng sut tiu th

    trn mch ln nht v bng: max maxP U.I 150.1 150W

    Khi V1 c s ch cc i th ta c gin vc t nh hnh bn:

    Ta c: 2 2CU 170 150 80V

    RU 2.80. 170 80 120V

    H s cng sut ca mch l 2

    120cos cos 0,8

    150

    Cng sut tiu th ca mch khi l: 2

    2 2 2

    max

    UP .cos P .cos 150.0,8 96W

    R

    Cu 34:

    a. Cng sut tiu th ca on mch c tnh bng cng

    thc: 2max

    P P .cos

    Theo , khi f = f1 th UC = U v c

    2 3 3cos cos4 2

    . Gin vc t ca mch khi

    c dng nh hnh v:

    trn hnh v: ta c = 300, = 600, OB = MB. Suy ra tam gic OMB l tam gic u. Vy UC = 2UL.

    Suy ra: 1

    1

    12 f L

    2 f C

    ng vi hai tn s f1 v f2 th UL v UC i gi tr cho nhau nn ZL v ZC cng i gi tr cho nhau, ta c:

    UC

    UAB

    UL

    UR O

    B

    M

  • GSTT GROUP | 19

    ZL2 = ZC1 = 2ZL1. Suy ra f2 = 2f1.

    Mt khc, f2 = f1 + 100 Hz

    Suy ra: f1 = 100Hz, f2 = 200Hz.

    Tn s ca dng in khi UC = U gp 2 ln tn s ca dng in khi Ucmax. Vy khi Ucmax th tn s ca dng in l:

    1C

    f 100f 50 2 Hz

    2 2

    b. ng vi tn s f2, UL = U, gin vc t ca mch nh hnh v:

    Khng lm nh hng n kt qu, c th gi s: ZL = ZAB

    = 2 . Khi , ZC = 1 , R = 3 .

    ng vi tn s fL = f2. 2 th in p trn t t gi tr cc

    i. Lc , cm khng ca mch tng ln 2 ln, dung khng

    ca mch gim i 2 ln. Gin vc t nh hnh v c.

    Trn gin ny, ta c: OH = 3 , HM = 1 3

    2 22 2

    Suy ra: MO = 9 15

    32 2

    H s cng sut ca mch khi l:

    OH 3 6 2cos

    MO 15 515

    2

    Cu 35:

    Khi UC t cc i th gin vc t ca mch nh hnh v.

    Ta c:

    01 21 2

    1 2

    tan tantan tan71,57 3

    1 tan . tan

    (1)

    Mt khc, ta c: 1 2

    tan .tan 0,5 (2)

    V v h s cng sut ca on mch AN ln hn h s

    cng sut ca on mch AB nn ta c: 1 2

    (3)

    T (1),(2),(3) ta suy ra: 1 2

    1tan , tan 1

    2

    H s cng sut ca on mch AB l

    2

    2cos cos cos

    4 2

    Cng sut tiu th ca on mch c tnh bi cng thc:

    2max max

    1P P .cos P .

    2

    Theo th P = 200W. Suy ra Pmax = 400W.

    ZC

  • L l & kho kh 1.0 | 20

    Cu 36:

    Gii:

    Khi UC t cc i th gin vc t ca mch nh hnh v.

    Ta c: 1 2

    tan .tan 0,5

    1 2 1 21 2 1 2

    1 2

    tan tan tan tantan tan 2. tan tan

    1 tan .tan 1 0,5

    V 1, 2 l nhng gc nhn, nn tan ca chng l nhng s dng.

    Theo bt ng thc Cosi ta c:

    1 2 1 2

    1tan tan 2. tan . tan 2. 2

    2

    Vy thay vo biu thc trn ta c:

    0tan 2 2 70,53

    Vy khi UC t gi tr cc i th uRL sm pha hn uAB mt gc ti thiu bng 70,530.

    Cu 37:

    a. Khi tn s gc l , h s cng sut ca on MB l 0,6. Khng lm nh hng n kt qu c th gi s khi : R = 6, ZMB = 10. Suy ra ZC = 8.

    Khi tng tn s ca dng in ln gp i (n = 2) th dung khng ca mch l 'CZ 4 ,

    in p hiu dng trn cun cm t cc i. Lc gin vc t ca mch nh hnh v.

    Ta c: 2 2

    C

    R 6x 4,52.Z 2.4

    Cm khng ca mch khi ny l : 'LZ 4 4,5 8,5

    T l gia cm khng v dung khng ca mch l:

    '2L

    'C

    Z 8,5 172 L.2 C 4 .LC

    4 8Z (1)

    Khi iu chnh cng sut tiu th ca mch t gi tr cc i th trong mch xy ra cng hng. Lc t s gia cm khng v dung khng ca mch l:

    ''" " "2L

    "C

    ZL. C .LC 1

    Z (2)

    Chia hai v ca (1) cho (2) ta c: ""

    2 17 32.

    8 17 .

    Vy t tn s gc , mun cho cng sut ca mch t cc i th phi tng tn s gc ln 32

    17

    ln.

    b. Gi l tn s gc khi in p trn t t cc i. Ta c:

    x

    y

    v

    ZRL

    2

    1

    O

    Z

    x

    6 O H

    Q

    4

    Z

    ZRC

  • GSTT GROUP | 21

    2"2

    "'

    '

    32. 1617 .2. 17

    Vy t gi tr tn s gc , mun cho in p hiu dng trn t t cc i th phi gim tn s

    gc xung n gi tr 16.17

    ( tc l gim bt i mt lng 17

    )

    Cu 38:

    a. iu chnh Ucmax th gin vc t ca mch nh hnh v:

    Ta c: 2 2 2 20

    x U U 2U U U

    0y U x U 2 1

    v 2xy 2U.U 2 1 U. 2 2 2 (*)

    in p hiu dng ca on AM l:

    2 2 2 2rLU x v U U 2 2 2 U 2 2 1 =135,2 (V)

    Suy ra: U = 100(V). Thay vo (*) suy ra v = 91(V)

    Ta c: 2 2v 91

    P 182 r 45,5r r

    b. Gi tr ca U0

    0U U. 2 100 2 V

    Cu 51:

    Cu 50:

    Khong cc ca 2 vt:

    Khi 2 vt gp nhau:

    Ban u vt (v tr gp nhau). Gc quay:

    Vy vt qua 6 ln (k c ban u)

    Cu 49:

    Ta c

    x

    y

    v

    UrL

    2

    1

    O

    U

    U0

  • L l & kho kh 1.0 | 22

    Cu 48:

    Cu 47:

    Gi s bin dao ng ca phn t M l 3 cm, ta s c phng trnh sng ti M l:

    Cu 46:

    Ta c R=120 v ZC=90

  • GSTT GROUP | 23

    Cu 45:

    Ta c gin vecto:

    C MB AB NBU 50 2(V);U U 130(V) OE 50 2;OP OQ 130

    t R C ANU x U x U x 2 EP EQ x;PQ x 2

    Gi F l trung im ca PQ ta c x 2

    OF PQ;EF PF QF2

    C

    2 2

    2 2 2 2x 2 x 2PF OF OP 50 2 130 x 70(V)2 2

    Vy 2 2 2OP PE OE

    cos 0,9232OP.PE

    Cu 44:

    Khi vn ti ca 2 bc x trng nhau th

    Ta c th vit:

    Ch : Vi cc bi ton c cc i lng thay i v mi lin h gia chng, ta c th th loi tr cc p n sai, nh vy c th rt ngn thi gian lm bi.

    Cu 43:

    uR

    uC

    ur

    uL

    uAB

    uAN

    uMB

    O

    P

    Q

    E

    F

  • L l & kho kh 1.0 | 24

    Khi i chiu th v tr cn bng mi i sang pha bn kia (hnh v).

    Bin chnh bng:

    Cu 42:

    Khi roto quay vi tc n (vng/pht) th 2 2

    L

    UI 1(A)

    R Z

    Khi roto quay vi tc 3n (vng/pht) th 2 2

    L

    3UI 3(A)

    R 9Z

    2 2 2 2L L

    L2 22 2LL

    R 9Z R 9Z13 R 3Z

    R Z33 R Z

    Khi roto quay vi tc 2n (vng/pht) th L1 L

    2RZ 2Z

    3

    Ch : Khi thay i tc quay ca roto th tn s ca dng in thay i hiu in th gia 2 u on mch cng thay i.

    Cu 41:

    Gii theo phong cch t lun nh:

    Cc em cn nm c cc tr tam thc bc 2 mi hiu r li gii bi ton ny

    Cu 40:

    Ti A: A AA 2

    x = ;v = 40 (m / s)2

    Ti B: B Bx =0,1 (m);v =30 (m/s)

  • GSTT GROUP | 25

    2 2

    2

    A 40 40 2(1) = A =

    2 thay vo (2) ta c:

    22

    2300 2300 2 0,2= 0,1 = = T = (s)

    0,1 T 2300

    Cu 39:

    Cc im c mu ging vn trung tm (hay c cc vn sng trng nhau) th tha mn

    1 21 1 2 2

    2 1

    k 3x k k

    k 4

    V 1 2 1 2k ,k Z k 3;k 4

    Trong khong gia 3 vn sng lin tip cng mu vi vn trung tm, ly k1 l 3, 6, 9 th

    k2 l 4, 8, 12.

    Cc vn sng n sc nm trong khong gia 3 vn sng lin tip ng vi k1 l : 4,

    5, 7, 8

    Cc vn sng n sc lam nm trong khong gia 3 vn sng lin tip ng vi k2 l :

    5, 6, 7, 9, 10, 11

    Vy c 4 vn v 6 vn lam

    Phin bn 1.0 tm thi 51 cu , cc em ch i phin bn 1.1 sau y mt tun na nh (d kin l 20/11 ra mt)

    Mt ln na, anh ch mong cc em lun lun n lc phn u, ng bao gi nn ch!

    Ngoi ra, xem thm cc hot ng, bi ging ca anh ch GSTT GROUP, cc em vo nhng knh sau: Website : gstt.vn

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