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TÀI LIỆU ÔN THI HỌC SINH GIỎI PHẦN I: NHIỆT HỌC I - CƠ … lieu on thi HSG lop 9... · TÀI LIỆU ÔN THI HỌC SINH GIỎI..... Sưu tầm bởi

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  • TI LIU N THI HC SINH GII

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    Su tm bi https://blogvatly.com

    PHN I: NHIT HC

    I - C S L THUYT:

    1/ Nguyn l truyn nhit:

    Nu ch c hai vt trao i nhit th:

    - Nhit t truyn t vt c nhit cao hn sang vt c nhit thp hn.

    - S truyn nhit xy ra cho n khi nhit ca hai vt bng nhau th dng li.

    -Nhit lng ca vt ny ta ra bng nhit lng ca vt khi thu vo.

    2/ Cng thc nhit lng:

    - Nhit lng ca mt vt thu vo nng ln: Q = mct (vi t = t2 - t1. Nhit cui tr

    nhit u)

    - Nhit lng ca mt vt ta ra lnh i: Q = mct (vi t = t1 - t2. Nhit u tr nhit

    cui)

    - Nhit lng ta ra v thu ca cc cht khi chuyn th:

    + S nng chy - ng c: Q = m ( l nhit nng chy)

    + S ha hi - Ngng t: Q = mL (L l nhit ha hi)

    - Nhit lng ta ra khi nhin liu b t chy:

    Q = mq (q nng sut ta nhit ca nhin liu)

    - Nhit lng ta ra trn dy dn khi c dng in chy qua: Q = I2Rt

    3/ Phng trnh cn bng nhit: Qta ra = Qthu vo

    4/ Hiu sut ca ng c nhit: H = %100tp

    ch

    Q

    Q

    5/ Mt s biu thc lin quan:

    - Khi lng ring: D = V

    m

    - Trng lng ring: d = V

    P

    - Biu thc lin h gia khi lng v trng lng: P = 10m

    - Biu thc lin h gia khi lng ring v trng lng ring: d = 10D

  • TI LIU N THI HC SINH GII

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    II - PHN BI TP.

    Bi 1: Ngi ta th mt thi ng 0,4kg nhit 800C vo 0,25kg nc nhit 18

    0C.

    Hy xc nh nhit khi cn bng nhit. Cho bit nhit dung ring ca ng l 380J/kg.k

    ca nc l 4200J/Kg.K.

    Hng dn gii:

    - Nhit lng do ming ng ta ra ngui i t 800C xung t

    0C:

    Q1 = m1.C1.(t1 - t) = 0,4. 380. (80 - t) (J)

    - Nhit lng nc thu vo nng ln t 180C n t

    0C:

    Q2 = m2.C2.(t - t2) = 0,25. 4200. (t - 18) (J)

    Theo phng trnh cn bng nhit:

    Q1 = Q2

    0,4. 380. (80 - t) = 0,25. 4200. (t - 18)

    t 260C

    Vy nhit xy ra cn bng l 260C.

    Bi 2: Trn ln ru v nc ngi ta thu c hn hp nng 140g nhit 360C. Tnh

    khi lng ca nc v khi lng ca ru trn. Bit rng ban u ru c nhit

    190C v nc c nhit 100

    0C, cho bit nhit dung ring ca nc l 4200J/Kg.K, ca

    ru l 2500J/Kg.k.

    Hng dn gii:

    - Theo bi ra ta bit tng khi lng ca nc v ru l 140

    m1 + m2 = m m1 = m - m2 (1)

    - Nhit lng do nc ta ra: Q1 = m1. C1 (t1 - t)

    - Nhit lng ru thu vo: Q2 = m2. C2 (t - t2)

    - Theo PTCB nhit: Q1 = Q2

    m1. C1 (t1 - t) = m2. C2 (t - t2)

    m14200(100 - 36) = m22500 (36 - 19)

    268800 m1 = 42500 m2

    42500

    268800 12

    mm (2)

    - Thay (1) vo (2) ta c: 268800 (m - m2) = 42500 m2

    37632 - 268800 m2 = 42500 m2

  • TI LIU N THI HC SINH GII

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    311300 m2 = 37632

    m2 = 0,12 (Kg)

    - Thay m2 vo pt (1) ta c:

    (1) m1 = 0,14 - 0,12 = 0,02 (Kg)

    Vy ta phi pha trn l 0,02Kg nc vo 0,12Kg. ru thu c hn hp nng

    0,14Kg 360C.

    Bi 3: Ngi ta m1(Kg) nc nhit 600C vo m2(Kg) nc nhit -5

    0C. Khi

    c cn bng nhit lng nc thu c l 50Kg v c nhit l 250C . Tnh khi lng

    ca nc v nc ban u. Cho nhit dung ring ca nc l 2100J/Kg.k. (Gii

    tng t bi s 2)

    Bi 4: Ngi ta dn 0,2 Kg hi nc nhit 1000C vo mt bnh cha 1,5 Kg nc ang

    nhit 150C. Tnh nhit cui cng ca hn hp v tng khi lng khi xy ra cn

    bng nhit. Bit nhit ha hi ca nc L =2,3.106J/kg, cn = 4200 J/kg.K.

    Hng dn gii:

    Nhit lng ta ra khi 0,2 Kg hi nc 1000C ngng t thnh nc 100

    0C

    Q1 = m1. L = 0,2 . 2,3.106 = 460000 (J)

    Nhit lng ta ra khi 0,2Kg nc 1000C thnh nc t

    0C

    Q2 = m1.C. (t1 - t) = 0,2. 4200 (100 - t)

    Nhit lng thu vo khi 1,5Kg nc 150C thnh nc t

    0C

    Q3 = m2.C. (t - t2) = 1,5. 4200 (t - 15)

    p dng phng trnh cn bng nhit: Q1 + Q2 = Q3

    460000 + 0,2. 4200 (100 - t) = 1,5. 4200 (t - 15)

    6780t = 638500

    t 940C

    Tng khi lng khi xy ra cn bng nhit.

    m = m1 + m2 = 0,2 + 1,5 = 1,7(Kg)

    Bi 5: C ba cht lng khng tc dng ha hc vi nhau v c trn ln vo nhau trong

    mt nhit lng k. chng c khi lng ln lt l m1=1kg, m2= 10kg, m3=5kg, c nhit

    dung ring ln lt l C1 = 2000J/Kg.K, C2 = 4000J/Kg.K, C3 = 2000J/Kg.K v c nhit

    l t1 = 60C, t2 = -40

    0C, t3 = 60

    0C.

    a/ Hy xc nh nhit ca hn hp khi xy ra cn bng.

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    b/ Tnh nhit lng cn thit hn hp c nng ln thm 60C. Bit rng khi trao i

    nhit khng c cht no b ha hi hay ng c.

    Hng dn gii:

    a/ Gi s rng, thot u ta trn hai cht c nhit thp hn vi nhau ta thu c mt hn

    hp nhit t < t3 ta c pt cn bng nhit:

    m1C1(t1 - t) = m2C2(t - t2)

    2211

    222111

    CmCm

    tCmtCmt

    (1)

    Sau ta em hn hp trn trn vi cht th 3 ta thu c hn hp 3 cht nhit t' (t < t'

    < t3) ta c phng trnh cn bng nhit:

    (m1C1 + m2C2)(t' - t) = m3C3(t3 - t') (2)

    T (1) v (2) ta c:

    332211

    333222111'CmCmCm

    tCmtCmtCmt

    Thay s vo ta tnh c t' -190C

    b/ Nhit lng cn thit nng nhit ca hn hp ln 60C:

    Q = (m1C1 + m2C2 + m3C3) (t4 - t') = 1300000(J)

    Bi 6: Mt thi nc c khi lng 200g -100C.

    a/ Tnh nhit lng cn cung cp nc bin thnh hi hon ton 1000C.

    b/ Nu b thi nc trn vo mt x nc bng nhm 200C. Sau khi cn bng nhit ta

    thy trong x cn li mt cc nc c khi lng 50g. tnh lng nc c trong x

    lc u. Bit x c khi lng 100g, c = 1800J/kg.k, = 3,4.105J/kg, cn = 4200 J/kg.K,

    cnh= 880J/kg.k, L =2,3.106J/kg .

    Hng dn gii:

    a/ Nhit lng nc thu vo tng nhit t -100C n 0

    0C

    Q1 = m1C1(t2 - t1) = 3600(J)

    Nhit lng nc thu vo nng chy hon ton 00C

    Q2 = m1. = 68000 (J)

    Nhit lng nc thu vo tng nhit t 00C n 100

    0C

    Q3 = m1C2(t3 - t2) = 84000(J)

    Nhit lng nc thu vo ha hi hon ton 1000C

  • TI LIU N THI HC SINH GII

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    Q4 = m1.L = 460000(J)

    Nhit lng cn cung cp trong sut qu trnh:

    Q = Q1 + Q2 + Q3 + Q4 = 615600(J)

    b/ Gi m' l lng nc tan: m' = 200 - 50 = 150g = 0,15Kg

    Do nc tan khng ht nn nhit cui cng ca hn hp l 00C.

    Nhit lng m m' (Kg) nc thu vo nng chy:

    Q' = m' = 51000 (J)

    Nhit lng do m'' Kg nc v x nhm ta ra gim xung t 200C n 0

    0C

    Q" = (m"C2 + mnhCnh)(20 - 0)

    p dng phng trnh cn bng nhit:

    Q" = Q' + Q1 hay:

    (m"C2 + mnhCnh)(20 - 0) = 51000 + 3600

    m" = 0,629 (Kg)

    Bi 7: Khi thc hnh trong phng th nghim, mt hc sinh cho mt lung hi nc

    1000C ngng t trong mt nhit lng k cha 0,35kg nc 10

    0C. Kt qu l nhit ca

    nc tng ln 420C v khi lng nc trong nhhit k tng thm 0,020kg. Hy tnh nhit

    ha hi ca nc trong th nghim ny? Bit nhit dung ring v nhit hoa hi ca nc l

    cn = 4200 J/kg.K, L =2,3.106J/kg

    Hng dn gii:

    Nhit lng m 0,35kg nc thu vo:

    Q Thu vo = m.C.(t2 - t1) 46900(J)

    Nhit lng m 0,020Kg hi nc 1000C ngng t thnh nc

    Q1 = m.L = 0,020L

    Nhit lng m 0,020Kg nc 1000C ta ra khi h xung cn 42

    0C

    Q 2 = m'.C.(t3 - t2) 4860(J)

    Theo phng trnh cn bng nhit:

    Q Thu vo = Q1 + Q 2 hay:

    46900 = 0,020L + 4860

    L = 21.105 (J/Kg)

  • TI LIU N THI HC SINH GII

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    Bi 8: C hai bnh cch nhit, bnh th nht cha 2Kg nc 200C, bnh th hai cha 4Kg

    nc 600C. Ngi ta rt mt ca nc t bnh 1 vo bnh 2. Khi bnh 2 cn bng nhit

    th ngi ta li rt mt ca nc t bnh 2 sang bnh 1 lng nc trong hai bnh nh lc

    u. Nhit bnh 1 sau khi cn bng l 21,950C.

    a/ Xc nh lng nc rt mi ln v nhit cn bng bnh 2.

    b/ Nu tip tc thc hin ln th hai, tm nhit cn bng mi bnh.

    Hng dn gii:

    a/ Gi s khi rt lng nc m t bnh 1 sang bnh 2, nhit cn bng ca bnh 2 l t nn

    ta c phng trnh cn bng:

    m.(t - t1) = m2.(t2 - t) (1)

    Tng t ln rt tip theo nhit cn bng bnh 1 l t' = 21,950C v lng nc trong

    bnh 1 lc ny ch cn (m1 - m) nn ta c phng trnh cn bng:

    m.(t - t') = (m1 - m).(t' - t1) (2)

    T (1) v (2) ta c pt sau:

    m2.(t2 - t) = m1.(t' - t1)

    2

    122 '

    m

    tttmt

    (3)

    Thay (3) vo (2) tnh ton ta rt phng trnh sau:

    11122

    121

    '

    '.

    ttmttm

    ttmmm

    (4)

    Thay s vo (3) v (4) ta tm c: t = 590C v m = 0,1 Kg.

    b/ Lc ny nhit ca bnh 1 v bnh 2 ln lt l 21,950C v 59

    0C by gi ta thc hin rt

    0,1Kg nc t bnh 1 sang bnh 2 th ta c th vit c phng trnh sau:

    m.(T2 - t') = m2.(t - T2)

    Cmm

    tmtmT 0

    2

    212 12,58

    '

    By gi ta tip tc rt t bnh 2 sang bnh 1 ta cng d dng vit c phng trnh sau:

    m.(T1 - T2) = (m1 - m).(t - T1)

    Cm

    tmmmTT 0

    1

    121 76,23

    ')(

  • TI LIU N THI HC SINH GII

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    Bi 9: Bp in c ghi 220V-800W c ni vi hiu in th 220V c dng un si

    2lt nc 200C. Bit hiu sut ca bp H = 80% v nhit dung ring ca nc l

    4200J/kg.K.

    a/ Tnh thi gian un si nc v in nng tiu th ca bp ra Kwh.

    b/ Bit cun dy c ng knh d = 0,2mm, in tr sut m 710.5 c qun trn mt

    li bng s cch in hnh tr trn c ng knh D = 2cm. Tnh s vng dy ca bp in

    trn.

    Hng dn gii:

    a/ Gi Q l nhit lng m nc thu vo nng ln t 200C n 100

    0: Q = m.C.t

    Gi Q' l nhit lng do dng in ta ra trn dy t nng Q' = R.I2.t = P. t

    Theo bi ra ta c: sHP

    tCmt

    tP

    tCm

    Q

    QH 1050

    .

    ..

    .

    ..

    '

    in nng tiu th ca bp: A = P. t = 233,33 (Wh) = 0,233 (Kwh)

    b/ in tr ca dy: 22

    4

    4

    d

    Dn

    d

    Dn

    S

    lR

    (1)

    Mt khc: P

    UR

    2

    (2)

    T (1) v (2) ta c: P

    U

    d

    Dn 2

    2

    4

    Vng

    DP

    dUn 5,60

    4

    22

    Bi 10: Cu ch trong mch in c tit din S = 0,1mm2, nhit 27

    0C. Bit rng khi

    on mch th cng dng in qua dy ch l I = 10A. Hi sau bao lu th dy ch t?

    B qua s ta nhit ra mi trng xung quanh v s thay i in tr, kch thc dy ch

    theo nhit . cho bit nhit dung ring, in tre sut, khi lng ring, nhit nng chy v

    nhit nng chy ca ch ln lt l: C = 120J/kg.K; m 610.22,0 ; D = 11300kg/m3;

    kgJ /25000 ; tc=3270C.

    Hng dn gii:

    Gi Q l nhit lng do dng in I ta ra trong thi gian t, ta c:

    Q = R.I2.t = tI

    S

    l 2 ( Vi l l chiu di dy ch)

    Gi Q' l nhit lng do dy ch thu vo tng nhit t 270C n nhit nng chy tc

    = 3270C v nng chy hon ton nhit nng chy, ta c

  • TI LIU N THI HC SINH GII

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    Q' = m.C.t + m = m(C.t + ) = DlS(C.t + ) vi (m = D.V = DlS)

    Do khng c s mt mt nhit nn:

    Q = Q' hay: tIS

    l 2 = DlS(C.t + )

    stCI

    DSt 31,0.

    2

    2

    PHN II: CHUYN NG C HC - VN TC

    I - C S L THUYT

    1. VN TC L MT I LNG VC - T:

    a. Th no l mt i lng vc t:

    - Mt i lng va c ln, va c phng v chiu l mt i lng vec t.

    b. Vn tc c phi l mt i lng vc t khng:

    - Vn tc l mt i lng vc t, v:

    + Vn tc c phng, chiu l phng v chiu chuyn ng ca vt.

    + Vn tc c ln, xc nh bng cng thc: v = t

    s.

    c. K hiu ca vc t vn tc: v (c l vc t v hoc vc t vn tc )

    2. MT S IU CN NH TRONG CHUYN NG TNG I:

    a. Cng thc tng qut tnh vn tc trong chuyn ng tng i :

    v13 = v12 + v23

    v = v1 + v2

    Trong : + v13 (hoc v ) l vc t vn tc ca vt th 1 so vi vt th 3

    + v13 (hoc v) l vn tc ca vt th 1 so vi vt th 3

    + v12 (hoc v1 ) l vc t vn tc ca vt th 1 so vi vt th 2

    + v12 (hoc v1) l vn tc ca vt th 1 so vi vt th 2

    + v23 (hoc v2 ) l vc t vn tc ca vt th 2 so vi vt th 3

    + v23 (hoc v2) l vn tc ca vt th 2 so vi vt th 3

  • TI LIU N THI HC SINH GII

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    b. Mt s cng thc tnh vn tc tng i c th:

    b.1. Chuyn ng ca thuyn, can, xung trn sng, h, bin:

    B sng ( vt th 3)

    Nc (vt th 2)

    Thuyn, can (vt th 1)

    * KHI THUYN, CA N XUNG CHUYN NG XUI DNG:

    Vn tc ca thuyn, can so vi b c tnh bng 1 trong 2 cp cng thc sau:

    vcb = vc + vn

    t

    ABS )( = vc + vn ( Vi t l thi gian khi can i xui dng )

    Trong :

    + vcb l vn tc ca can so vi b

    + vcn (hoc vc) l vn tc ca can so vi nc

    + vnb (hoc vn) l vn tc ca nc so vi b

    * Lu : - Khi can tt my, tri theo sng th vc = 0

    vtb = vt + vn

    t

    ABS )( = vc + vn ( Vi t l thi gian khi

    thuyn i xui dng )

    Trong :

    + vtb l vn tc ca thuyn so vi b

    + vtn (hoc vt) l vn tc ca thuyn so vi nc

    + vnb (hoc vn) l vn tc ca nc so vi b

    * KHI THUYN, CA N, XUNG CHUYN NG NGC DNG:

  • TI LIU N THI HC SINH GII

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    Tng qut: v = vln - vnh

    Vn tc ca thuyn, can so vi b c tnh bng 1 trong 2 cp cng thc sau:

    vcb = vc - vn (nu vc > vn)

    '

    )(

    t

    ABS = vc - vn ( Vi t l thi gian khi can i ngc

    dng )

    vtb = vt - vn (nu vt > vn)

    '

    )(

    t

    ABS = vc - vn ( Vi t l thi gian khi can i ngc

    dng )

    b.2. Chuyn ng ca b khi xui dng:

    vBb = vB + vn

    t

    ABS )( = vB + vn ( Vi t l thi gian khi

    can i xui dng )

    Trong :

    + vBb l vn tc ca b so vi b; (Lu : vBb =

    0)

    + vBn (hoc vB) l vn tc ca b so vi nc

    + vnb (hoc vn) l vn tc ca nc so vi b

    b.3. Chuyn ng xe (tu ) so vi tu:

    Tu (vt th 3) Tu th 2 (vt th 3)

    ng ray ( vt th 2) ng ray ( vt th 2)

    Xe ( vt th 1) tu th 1 ( vt th 1)

  • TI LIU N THI HC SINH GII

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    * KHI HAI VT CHUYN NG NGC CHIU:

    vxt = vx + vt

    Trong :

    + vxt l vn tc ca xe so vi tu

    + vx (hoc vx) l vn tc ca xe so vi ng ray

    + vt (hoc vt) l vn tc ca tu so vi ng

    * KHI HAI VT CHUYN NG CNG CHIU:

    vxt = vx - vt hoc vxt = vx - vt ( nu vx > vt ; vx > vt)

    vxt = vt - vx hoc vxt = vt - vx ( nu vx < vt ; vx < vt)

    b.4. Chuyn ng ca mt ngi so vi tu th 2:

    * Khi ngi i cng chiu chuyn ng vi tu th 2: vtn = vt + vn

    * Khi ngi i ngc chiu chuyn ng vi tu th 2: vtn = vt - vn ( nu vt > vn)

    Lu : Bi ton hai vt gp nhau:

    - Nu hai vt cng xut pht ti mt thi im m gp nhau th thi gian chuyn ng bng

    nhau: t1= t2=t

    - Nu hai vt chuyn ng ngc chiu th tng qung ng m mi vt i c bng

    khong cch gia hai vt lc ban u: S = S1 + S2

    - Nu hai vt chuyn ng cng chiu th qung ng m vt th nht (c vn tc ln hn)

    i tr i qung ng m vt th hai i bng khong cch ca hai vt lc ban u: S =

    S1 - S2

    II - BI TP VN DNG.

    Bi 1: Lc 7h mt ngi i b khi hnh t A n B vi vn tc 4km/h. Lc 9h mt ngi

    i xe p cng khi hnh t A v B vi vn tc 12km/h.

    a. Hai ngi gp nhau lc my gi? Lc gp cch A bao nhiu?

    b. Lc my gi hai ngi cch nhau 2km?

  • TI LIU N THI HC SINH GII

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    Hng dn gii:

    a/ Thi im v v tr lc hai ngi gp nhau:

    - Gi t l khong thi gian t khi ngi i b n khi hnh khi n lc hai ngi gp nhau

    ti C.

    - Qung ng ngi i b i c: S1 = v1t = 4t (1)

    - Qung ng ngi i xe p i c: S2 = v2(t-2) = 12(t - 2) (2)

    - V cng xut pht ti A n lc gp nhau ti C nn: S1 = S2

    - T (1) v (2) ta c: 4t = 12(t - 2) 4t = 12t - 24 t = 3(h)

    - Thay t vo (1) hoc (2) ta c: (1) S1 = 4.3 =12 (Km)

    (2) S2 = 12 (3 - 2) = 12 (Km)

    Vy: Sau khi ngi i b i c 3h th hai ngi gp nhau v cch A mt khong 12Km

    v cch B 12Km.

    b/ Thi im hai ngi cch nhau 2Km.

    - Nu S1 > S2 th:

    S1 - S2 = 2 4t - 12(t - 2) = 2 4t - 12t +24 =2 t = 2,75 h = 2h45ph.

    - Nu S1 < S2 th:

    S2 - S1 = 2 12(t - 2) - 4t = 2 12t +24 - 4t =2 t = 3,35h = 3h15ph.

    Vy: Lc 7h + 2h45ph = 9h45ph hoc 7h + 3h15ph = 10h15ph th hai ngi cch nhau

    2Km.

    Bi 2: Lc 9h hai t cng khi hnh t hai im A v B cch nhau 96km i ngc chiu

    nhau. Vn tc xe i t A l 36km/h, vn tc xe i t A l 28km/h.

    a. Tnh khong cch ca hai xe lc 10h.

    b. Xc nh thi im v v tr hai xe gp nhau.

    Hng dn gii:

    a/ Khong cch ca hai xe lc 10h.

    - Hai xe khi hnh lc 9h v n lc 10h th hai xe i c trong khong thi gian t = 1h

    - Qung ng xe i t A: S1 = v1t = 36. 1 = 36 (Km)

    - Qung ng xe i t B: S2 = v2t = 28. 1 = 28 (Km)

    - Mt khc: S = SAB - (S1 + S2) = 96 - (36 + 28) = 32(Km)

    Vy: Lc 10h hai xe cch nhau 32Km.

  • TI LIU N THI HC SINH GII

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    Su tm bi https://blogvatly.com

    b/ Thi im v v tr lc hai xe gp nhau:

    - Gi t l khong thi gian t khi ngi i b n khi hnh khi n lc hai ngi gp nhau

    ti C.

    - Qung ng xe i t A i c: S1 = v1t = 36t (1)

    - Qung ng xe i t B i c: S2 = v2t = 28t (2)

    - V cng xut pht mt lc v i ngc chiu nhau nn: SAB = S1 + S2

    - T (1) v (2) ta c: 36t + 28t = 96 t = 1,5 (h)

    - Thay t vo (1) hoc (2) ta c: (1) S1 = 1,5.36 = 54 (Km)

    (2) S2 = 1,5. 28 = 42 (Km)

    Vy: Sau khi i c 1,5h tc l lc 10h30ph th hai xe gp nhau v cch A mt khong

    54Km v cch B 42Km.

    Bi 3: Cng mt lc hai xe gn my cng xut pht t hai im A v B cch nhau 60km,

    chng chuyn ng thng u v i cng chiu nhau t A n B. Xe th nht xut pht t A

    vi vn tc 30km/h, xe th hai khi hnh t B vi vn tc 40km/h.

    a. Tnh khong cch ca hai xe sau khi chng i c 1h.

    b. Sau khi xut pht c 1h, xe th nht bt u tng tc v t vn tc 60km/h. Hy Xc

    nh thi im v v tr hai ngi gp nhau.

    Hng dn gii:

    a/ Khong cch ca hai xe sau 1h.

    - Qung ng xe i t A: S1 = v1t = 30. 1 = 30 (Km)

    - Qung ng xe i t B: S2 = v2t = 40. 1 = 40 (Km)

    - Mt khc: S = S1 + S2 = 30 + 40 = 70 (Km)

    Vy: Sau 1h hai xe cch nhau 70Km.

    b/ Thi im v v tr lc hai ngi gp nhau:

    - Gi t l khong thi gian t khi ngi i b n khi hnh khi n lc hai ngi gp nhau

    ti C.

    - Qung ng xe i t A i c: S1 = v1t = 60t (1)

    - Qung ng xe i t B i c: S2 = v2t = 40t (2)

    - V sau khi i c 1h xe th nht tng tc nn c th xem nh cng xut mt lc v n

    lc gp nhau ti C nn: S1 = 30 + 40 + S2

    - T (1) v (2) ta c: 60t = 30 +40 +40t t = 3,5 (h)

  • TI LIU N THI HC SINH GII

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    - Thay t vo (1) hoc (2) ta c: (1) S1 = 3,5. 60 = 210 (Km)

    (2) S2 = 3,5. 40 = 140 (Km)

    Vy: Sau khi i c 3,5 h th hai ngi gp nhau v cch A mt khong 210 + 30 =

    240Km v cch B 140 + 40 = 180Km.

    Bi 4: Mt ngi d nh i b mt qung ng vi vn tc khng i l 5km/h, nhng

    khi i c 1/3 qung ng th c bn o bng xe p i tip vi vn tc 12km/h do

    n xm hn d nh l 28 pht. Hi nu ngi i b ht qung ng th mt bao lu?

    Hng dn gii:

    Gi S1, S2 l qung ng u v qung ng cui.

    v1, v2 l vn tc qung ng u v vn tc trn qung ng cui

    t1, t2 l thi gian i ht qung ng u v thi gian i ht qung ng cui

    v3, t3 l vn tc v thi gian d nh.

    Theo bi ra ta c: v3 = v1 = 5 Km/h; S1 = 3

    S; S2 = S

    3

    2; v2 = 12 Km

    Do i xe nn ngi n xm hn d nh 28ph nn: 21360

    28ttt (1)

    Mt khc: 33

    3 55

    tSS

    v

    St (2)

    v: 155

    3

    1

    11

    S

    S

    v

    St

    1836

    2

    12

    3

    2

    2

    22

    SS

    S

    v

    St

    Thay (2) vo (3) ta c: 18

    5

    3

    33

    21

    tttt

    So snh (1) v (4) ta c: httt

    t 2,118

    5

    360

    283

    33

    3

    Vy: nu ngi i b th phi mt 1h12ph.

    Bi 5: Mt can chy trn hai bn sng cch nhau 90km. Vn tc ca can i vi nc l

    25km/h v vn tc ca dng nc l 2km/h.

    a. Tnh thi gian can ngc dng t bn n n bn kia.

    181521

    SStt (3)

  • TI LIU N THI HC SINH GII

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    Su tm bi https://blogvatly.com

    b.Gi s khng ngh bn ti. Tnh thi gian i v v?

    Hng dn gii:

    a/ Thi gian can i ngc dng:

    Vn tc ca can khi i ngc dng: vng = vcn - vn = 25 - 2 = 23 (Km)

    Thi gian can i: 3,91( ) 3 54 36ng ngng ng

    S Sv t h h ph giy

    t v

    b/ Thi gian can xui dng:

    Vn tc ca can khi i ngc dng: vx = vcn + vn = 25 + 2 = 27 (Km)

    3,33( ) 3 19 48x xx x

    S Sv t h h ph giy

    t v

    Thi gian c i ln v: t = tng + tx = 7h14ph24giy

    Bi 6: Hai bn l ng c hai hng dc cc vn ng vin chuyn ng theo cng mt

    hng: Hng cc vn ng vin chy v hng cc vn ng vin ua xe p. Cc vn ng

    vin chy vi vn tc 6 m/s v khong cch gia hai ngi lin tip trong hng l 10 m; cn

    nhng con s tng ng vi cc vn ng vin ua xe p l 10 m/s v 20m. Hi trong

    khong thi gian bao lu c hai vn ng vin ua xe p vt qua mt vn ng vin

    chy? Hi sau mt thi gian bao lu, mt vn ng vin ua xe ang ngang hng mt vn

    ng vin chy ui kp mt vn ng vin chy tip theo?.

    Hng dn gii:

    - Gi vn tc ca vn ng vin chy v vn ng vin ua xe p l: v1, v2 (v1> v2> 0).

    Khong cch gia hai vn ng vin chy v hai vn ng vin ua xe p l l1, l2 (l2>l1>0).

    V vn ng vin chy v vn ng vin ua xe p chuyn ng cng chiu nn vn tc

    ca vn ng vi ua xe khi chn vn ng vin chy lm mc l:

    v21= v2 - v1 = 10 - 6 = 4 (m/s).

    - Thi gian hai vn ng vin ua xe vt qua mt vn ng vin chy l:

    21

    21

    205

    4

    lt

    v (s)

    - Thi gian mt vn ng vin ua xe p ang ngang hng mt vn ng vin chy ui

    kp mt vn ng vin chy tip theo l:

    12

    21

    102,5

    4

    lt

    v (s)

  • TI LIU N THI HC SINH GII

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    Su tm bi https://blogvatly.com

    Bi 7: Xe 1 v 2 cng chuyn ng trn mt ng trn vi vn tc khng i. Xe 1 i ht

    1 vng ht 10 pht, xe 2 i mt vng ht 50 pht. Hi khi xe 2 i mt vng th gp xe 1 my

    ln. Hy tnh trong tng trng hp.

    a. Hai xe khi hnh trn cng mt im trn ng trn v i cng chiu.

    b. Hai xe khi hnh trn cng mt im trn ng trn v i ngc chiu nhau.

    Hng dn gii:

    - Gi vn tc ca xe 2 l v vn tc ca xe 1 l 5v

    - Gi t l thi gian tnh t lc khi hnh n lc 2 xe gp nhau.

    (C < t 50) C l chu vi ca ng trn

    a/ Khi 2 xe i cng chiu.

    - Qung ng xe 1 i c: S1 = 5v.t; Qung ng xe 2 i c: S2 = v.t

    - Ta c: S1 = S2 + n.C

    Vi C = 50v; n l ln gp nhau th n

    5v.t = v.t + 50v.n 5t = t + 50n 4t = 50n t = 4

    50n

    V C < t 50 0 < 4

    50n 50 0 <

    4

    n 1 n = 1, 2, 3, 4.

    - Vy 2 xe s gp nhau 4 ln

    b/ Khi 2 xe i ngc chiu.

    - Ta c: S1 + S2 = m.C (m l ln gp nhau th m, m N*)

    5v.t + v.t = m.50v 5t + t = 50m 6t = 50m t = 6

    50m

    V 0 < t 50 0 V1 nn ngi i xe p phi hng v pha A.

    V ngi i xe p lun cch u hai ngi u nn h phi gp nhau ti im G cch B

    150km lc 9 gi. Ngha l thi gian ngi i xe p i l: t = 9 - 7 = 2gi

    Qung ng i c l: DG = GB - DB = 150 - 125 = 25 km

    Vn tc ca ngi i xe p l. V3 = ./5,122

    25hkm

    t

    DG

    PHN III: CNG, CNG SUT - NH LUT V CNG

    I - C S L THUYT:

    1/ Cng c hc:

    - Mt lc tc dng ln vt chuyn di theo phng ca lc th lc thc hin mt cng

    c hc ( gi tt l cng).

    - Cng thc tnh cng c hc:

    A = F.S

    2/ Cng sut:

    - Cng sut c xc nh bng cng thc hin c trong mt n v thi gian.

    - Tng thc tnh cng sut:

    t

    AP

    3/ My c n gin:

    RNG RC

    C NH

    RNG RC

    NG N BY

    MT PHNG

    NGHING

    Trong :

    A: Cng c hc (J)

    F: Lc tc dng (N)

    S: Qung ng vt dich chuyn (m)

    Trong :

    A: Cng c hc (J)

    P: Cng sut (W)

    t: Thi gian thc hin cng (s)

  • TI LIU N THI HC SINH GII

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

    Su tm bi https://blogvatly.com

    C

    U T

    O

    T

    C D

    N

    G

    BI

    N

    I

    L

    C

    Ch c tc dng

    bin i phng

    chiu ca lc:

    F = P

    Bin i v ln

    ca lc:

    2

    PF

    Bin i v phng, chiu v ln

    ca lc.

    1

    2

    l

    l

    F

    P

    l

    h

    P

    F

    C

    NG

    C

    C

    H

    Aich = P.S1 Aich = P.S1 Aich = P.h1 Aich = P.h

    C

    NG

    T.P

    H

    N

    Atp = F.S2 Atp = F.S2 Atp = F.h2 Atp = Fl

    TN

    H

    CH

    T

    CH

    UN

    G

    Asinh ra = Anhn c

    ( Khi cng hao ph khng ng k)

    HI

    U

    SU

    T

    %100tp

    ch

    A

    AH

    4/ nh lut v cng:Khng mt my c n gin no cho ta li v cng. c li bao

    nhiu ln v lc th thit by nhiu ln v ng i v ngc li.

    II - BI TP P DNG:

    F

    P

    S1 S2

    F

    P

    S1

    S2

    P

    F

    h2

    h1

    l1 l2

    P

    F

    l h

  • TI LIU N THI HC SINH GII

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    Su tm bi https://blogvatly.com

    Bi 1: Mt ngi ko mt gu nc t ging su 10m. Cng ti thiu ca ngi phi

    thc hin l bao nhiu? Bit gu nc c khi lnh l 1Kg v ng thm 5lt nc, khi

    lng ring ca nc l 1000kg/m3.Hng dn gii:

    Th tch ca nc: V = 5l = 0,005 m3

    Khi lng ca nc: mn = V.D = 0,005 . 1000 = 5 (Kg)

    Lc ti thiu ko gu nc ln l: F = P

    Hay: F = 10(mn + mg) = 10(5 + 1) = 60(N)

    Cng ti thiu ca ngi phi thc hin: A = F.S = 60. 10 = 600(J)

    Bi 2: Ngi ta dng mt rng rc c nh ko mt vt c khi lng 10Kg ln cao 15m

    vi lc ko 120N.

    a/ Tnh cng ca lc ko.

    b/ Tnh cng hao ph thng lc cn.

    Hng dn gii:

    a/ Cng ca lc ko: A = F.S = 120.15 = 1800(J)

    b/ Cng c ch ko vt: Ai = P.S = 100.15 =1500(J)

    Cng hgao ph: Ahp = A - Ai = 1800- 1500 = 300 (J)

    Bi 3: a mt vt coa khi lng 200Kg ln cao 10m ngi ta dng mt trong hai

    cch sau:

    a/ Dng h thng mt rng rc c nh, mt rng rc ng. Lc ny lc ko dy nng vt

    ln l F1 = 1200N.

    Hy tnh:

    - Hiu sut ca h thng.

    - Khi lng ca rng rc ng, Bit hao ph nng rng rc bng 4

    1hao ph tng cng do

    ma st.

    b/ Dng mt phng nghing di l = 12m. Lc ko lc ny l F2 = 1900N. Tnh lc ma st

    gia vt v mt phng nghing, hiu sut ca c h.

    Hng dn gii:

    a/ Cng dungd nng vt ln 10m: A1 = 10.m.h = 20 000 (J)

    - Khi dng h thng rng rc trn th khi vt ln cao mt on h th phi ko dy mt on

    S = 2h. Do cng dng ko vt: A = F1 . S = F1 . 2h = 24000(J)

  • TI LIU N THI HC SINH GII

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    Su tm bi https://blogvatly.com

    - Hiu sut ca h thng: %33,83%10024000

    20000%1001

    A

    AH

    - Cng hao ph: Ahp = A - A1 = 4000(J)

    - Cng hao ph nng rng rc ng: )(10004

    .' J

    hAA

    hp

    hp

    - Khi lng ca rng rc ng: )(1010

    '''..10' Kg

    h

    AmhmA

    hp

    hp

    b/ Cng c ch dng ko vt l A1 = 20000(J)

    - Cng ton phn ko vt lc nay: A = F2. l = 22800(J)

    - Cng hao ph do ma st: Ahp = A - A1 = 2800(J)

    - Lc ma st gia vt v mt phng nghing: )(33,233. Nl

    AFlFA

    hp

    msmshp

    - Hiu sut ca mt phng nghing: %72,87%1001 A

    AH

    Bi 4: Mt u tu ko mt toa tu chuyn ng t ga A ti ga B trong 15pht vi vn tc

    30Km/h. Ti ga B on tu c mc thm toa v do on tu i t ga B n ga C vi

    vn tc nh hn 10Km/h. Thi gian i t ga B n ga C l 30pht. Tnh cng ca u tu

    sinh ra bit rng lc ko ca u tu khng i l 40000N.

    Hng dn gii:

    - Qung ng i t ga A n ga B: S1 = v1.t1 = 7,5 (Km) = 7500m

    - Qung ng i t ga B n ga C: S2 = v2.t2 = 10 (Km) = 10000m

    - Cng sinh ra: A = F (S1 + S2) = 700000000 (J) = 700000(KJ)

    Bi 5: Ngi ta dng mt mt phng nging c chiu di 3m ko mt vt c khi lng

    300Kg vi lc ko 1200N . Hi vt c th ln cao bao nhiu? Bit hiu sut ca mt phng

    nghing l 80%.

    Hng dn gii:

    - Cng ca l ko vt: A = F.l = 3600(J)

    - Cng c ch: A1 = P.h = 10.m.h = 3000h (J)

    - cao vt c th ln c:

    )(96,03000.100

    3600.80

    %1003600

    3000%80%1001

    mh

    h

    A

    AH

  • TI LIU N THI HC SINH GII

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    Su tm bi https://blogvatly.com

    Bi 6: Ngi ta dng h thng rng rc trc mt vt c bng ng c

    trng lng P = 5340N t y h su H = 10m ln (hnh v). Hy tnh:

    1) Lc ko khi:

    a. Tng pha trn mt nc.

    b. Tng chm hon ton di nc.

    2) Tnh cng tng cng ca lc ko tng t y h ln

    pha trn mt nc h = 4m. Bit trng lng ring ca ng v

    ca nc ln lt l 89000N/m3, 10000N/m

    3. B qua trng lng ca cc rng rc.

    Hng dn gii:

    1a/ Dng rng rc ng c li hai ln v lc, nn lc ko khi vt ln khi mt nc:

    )(26702

    NP

    F

    1b/ Khi vt cn di nc th th tch chim ch: 306,089000

    5340m

    d

    PV

    - Lc y Acsimet tc dng ln vt: FA= V.d0 = 0,06.10000 = 600(N)

    - Lc do dy treo tc dng ln vt: P1 = P - FA = 5340 - 600 = 4740 (N)

    - Lc ko vt khi cn trong nc: )(23702

    1 NP

    F

    2/ Do dng rng rc ng nn b thit hai ln v ng i nn cng tng cng ca lc ko:

    A =F1.2H + F. 2h = 68760 (J)

    Bi 7: Ngi ta ln 1 ci thng theo mt tm vn nghing ln t. Sn xe t cao 1,2m, vn

    di 3m. Thng c khi lng 100Kg v lc y thng l 420N.

    a/ Tnh lc ma st gia tm vn v thng.

    b/ Tnh hiu sut ca mt phng nghing.

    Hng dn gii:

    - Nu khng c ma st th lc y thng l: )(400.

    ' Nl

    hPF

    - Thc t phi y thng vi 1 lc 420N vy lc ma st gia vn v thng: Fms = F - F' =

    20(N)

    - Cng c ch a vt ln: Ai = P . h = 1200(J)

    - Cng ton phn a vt ln: A = F. S = 1260 (J)

    - Hiu sut mt phng nghing: %95%1001 A

    AH

  • TI LIU N THI HC SINH GII

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    Su tm bi https://blogvatly.com

    Bi 8: Ngi ta dng mt palng a mt kin hng ln cao 3m. Bit qung ng dch

    chuyn ca lc ko l 12m.

    a/ Cho bit cu to ca palng ni trn.

    b/ Bit lc ko c gi tr F = 156,25N. Tnh khi lng ca kin hng ni trn.

    c/ Tnh cng ca lc ko v cng nng vt khng qua palng. T rt ra kt lun g?

    a/ S cp rng rc 26

    12

    2

    '

    S

    Sn (Cp)Vy palng c cu to bi 2 r

    2 c nh v 2 r

    2

    ng.

    b/ Ta c: 26

    12

    2

    '

    2

    S

    S

    F

    Pn

    - Trng lng ca kin hng: P = 4F = 4. 156,25 = 625(N)

    - Khi lng ca kin hng: )(5.6210

    10 KgP

    mmP

    c/ cng ca lc ko: Ak = FK.S' = 156,25.12 = 1875 (J)

    - Cng ca lc nng vt: An = P.S = 625.3 = 1875(J)

    - H thng palng khng cho li v cng.

    Bi 9: Cho h ging nh hnh v. vt m1 c khi lng 10Kg, vt m2 c khi lng 6Kg.

    Cho khong cch AB = 20cm. Tnh chiu di ca thanh OB h cn bng.

    Hng dn gii:

    - Trng lng ca vt m1:

    P1 = F1 = 10.m1 = 100N

    P2 = F2 P1 = F1

    1 2

    B A O

    F'

  • TI LIU N THI HC SINH GII

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    Su tm bi https://blogvatly.com

    - Trng lng ca vt m2: P2 = F2 = 10.m2 = 60N

    - Do vt m1 nng hn m2 nn m1 i xung vy u B c xu th i ln:

    - ln lc tc dng ln u B: NF

    F 502

    100

    2'

    - p dng h thc cn bng ca n by ta c:

    CMOA

    OAOA

    OA

    OA

    ABOA

    OA

    OB

    OA

    F

    F

    100

    .6205

    2060

    50

    '

    2

    - Chiu di thanh OB: OB = OA + AB = 100 + 20 = 120 (cm)

    Bi 10: Thanh AB di 160cm, u A ngi ta treo mt vt c khi lng m1 = 9Kg, im

    ta O nm cch A mt on 40cm.

    a/ Hi phi treo vo u b mt vt m2 c khi lng bao nhiu thanh cn bng?

    b/ Vt m2 gi nguyn khng i, bay gi ngi ta dch chuyn im O v pha u B v

    cch B mt on 60cm. Hi vt m1 phi thay i nh th no thanh vn ccn bng?

    Hng dn gii:

    a/ Ta c: OA = 40cm cmOAABOB 12040160

    Trng lng ca vt m1: P1 = F1 = 10.m1 = 90N

    p dng h thc cn bng ca n by: OA

    OB

    l

    l

    F

    F

    1

    2

    2

    1

    Lc tc dng vo u B: NOB

    OAFF 30

    .12

    Vy thanh AB cn bng th phi treo vo u B vt m2 = 3Kg.

    b/ Ta c: OB = 60cm

    cmOBABOA 10060160

    p dng h thc cn bng ca n by, thanh AB cn bng th lc tc dng vo u A:

    NOA

    OBF

    l

    lFF 18

    100

    60.30..' 2

    1

    22

    Vy vt m1 = 1,8Kg tc l vt m1 phi bt i 7,2Kg.

    PHN IV: P SUT - P SUT CHT LNG

    P SUT KH QUYN - LC Y AC-SI-MET

    I - C S L THUYT:

  • TI LIU N THI HC SINH GII

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

    Su tm bi https://blogvatly.com

    1/ p sut:

    - Cng thc tnh p sut: S

    FP

    P

    FS

    SPF .

    - n v p sut l paxcan(Pa): 21

    11

    m

    NPa

    2/ p sut cht lng:

    - Cht lng ng trong bnh s gy p sut theo mi phng ln y bnh, thnh bnh v

    mi vt t trong n.

    - Cng thc tnh p sut cht lng: P = d.h

    d

    Ph

    h

    Pd

    ( Vi d l trng lng ring ca cht lng; h l chiu cao ( su) ca ct cht lng tnh

    t mt thong cht lng)

    Ch :

    Trong ct cht lng ng yn, p sut ca mi im trn cng mt phng nm ngang c

    ln nh nhau (cng su)

    Mt vt nm trong lng cht lng, th ngoi p sut cht lng, vt cn chu thm p sut

    kh quyn do cht lng truyn ti.

    3/ Bnh thng nhau:

    - Trong bnh thng nhau cha cng cht lng ng yn, cc mt thong ca cht lng

    cc nhnh khc nhau u mt cao.

    - Trong bnh thng nhau cha hai hay nhiu cht lng khng ha tan, th mc mt thong

    khng bng nhau, trong trng hp ny p sut ti mi im trn cng mt phng nm

    ngang c gi tr bng nhau.

    - Bi ton my dng cht lng: p sut tc dng ln cht lng c cht lng truyn i

    nguyn vn theo mi hng.

    + Xc nh ln ca lc: Xc nh din tch ca pittng ln, pittng nh.

    + i n v thch hp.

    f

    FsS

    F

    fSs

    S

    Fsf

    s

    SfF

    s

    S

    f

    F

    .

  • TI LIU N THI HC SINH GII

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    4/ p sut kh quyn:

    - Do khng kh c trng lng nn Tri t v mi vt trn Tri t chu tc dng ca

    p sut kh quyn. Ging nh p sut cht lng p sut ny tc dng theo mi phng.

    - p sut kh quyn c xc nh bng p sut ct thy ngn trong ng T-ri-xe-li.

    - n v ca p sut kh quyn l mmHg (760mmHg = 1,03.105Pa)

    - Cng ln cao p sut kh quyn cng gim ( c ln cao 12m th gim 1mmHg).

    5/ Lc y Acsimet:

    - Mi vt nhng trong cht lng u b cht lng y thng ng t di ln vi mt lc

    c ln bng trng lng ca phn cht lng m vt chim ch. Lc ny c gi l

    lc y Acsimet.

    - Cng thc tnh: FA = d.V

    - iu kin vt ni, chm, l lng:

    + FA > P Vt ni

    + FA = P Vt l lng

    + FA < P Vt chm

  • TI LIU N THI HC SINH GII

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    II - BI TP P DNG:

    Bi 1: Bnh thng nhau gm hai nhnh hnh tr tit

    din ln lt l S1, S2 c cha nc nh hnh v. Trn mt nc

    c t cc pittng mng, khi lng m1, m2 . Mc nc hai nhnh

    chnh nhau mt on h = 10cm.

    a. Tnh khi lng m ca qu cn t ln pittng ln

    mc nc hai nhnh ngang nhau.

    b. Nu t qu cn sang pittng nh th mc nc hai nhnhlc by gi s chnh nhau

    mt on H bng bao nhiu?

    Cho khi lng ring ca nc D = 1000kg/m3, S1 = 200cm

    2, S2 = 100cm

    2 v b qua p

    sut kh quyn. Hng dn gii:

    a -p sut mt di pittng nh l : 2 1

    2 1

    10 1010

    m mDh

    S S

    2 1

    2 1

    m mDh

    S S (1)

    - Khi t qu cn m ln pittng ln mc nc hai bn ngang nhau nn:

    2 1 2 1

    2 1 2 1

    10 10( )m m m m m m

    S S S S

    (2)

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

    1 1

    10m m m

    DhS S

    1

    .m

    D hS

    => m = DS1h = 2kg

    b. Khi chuyn qu cn sang pittng nh th ta c : 2 1

    2 1

    10( ) 1010

    m m mDH

    S S

    2 1

    2 1

    m m mDh

    S S

    2 1

    2 1

    m m mDh

    S S

    (3)

    Kt hp (1), (3) v m = DhS1 ta c : H = h( 1 +1

    2

    S

    S)

    hS2

    S1

  • TI LIU N THI HC SINH GII

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    H = 0,3mBi 2:Trong mt bnh nc hnh tr c mt khi nc ni c gi bng

    mt si dy nh, khng gin (xem hnh v bn). Bit lc u sc cng ca si dy l

    10N. Hi mc nc trong bnh s thay i nh th no, nu khi nc

    tan ht? Cho din tch mt thong ca nc trong bnh l 100cm2 v

    khi lng ring ca nc l 1000kg/m3.

    Hng dn gii:

    Nu th khi nc ni (khng buc dy) th khi nc tan ht, mc nc trong

    bnh s thay i khng ng k.

    Khi buc bng dy v dy b cng chng t khi nc chm su hn so vi khi

    th ni mt th tch V, khi lc y Ac-si-met ln phn nc ngp thm ny to

    nn sc cng ca si dy.

    Ta c: FA = 10.V.D = F

    10.S.h.D = F (vi h l mc nc dng cao hn so vi khi khi nc th ni)

    => h = F/10.S.D = 0,1(m)

    Vy khi khi nc tan ht th mc nc trong bnh s h xung 0,1m

    Bi 3: Mt khi g hnh hp y vung ,chiu cao h=19cm, nh hn cnh y, c khi

    lng ring Dg=880kg/m3c th trong mt bnh nc. thm vo bnh mt cht

    du (khi lng ring Dd=700kg/m3), khng trn ln c vi nc.

    a/ Tnh chiu cao ca phn chm trong nc.Bit trng lng ring ca nc

    dn=10000N/m3

    b/ xc nh nhit dung ring ca du Cx ngi ta thc hin th nghim nh sau:

    khi lng nc mn vo mt nhit lng k khi lng mk.Cho dng in chy qua

    nhit lng k nung nng nc.Sau thi gian T1 nhit ca nhit lng k v nc

    tng ln 1t (0C).Thay nc bng du vi khi lng md v lp li cc bc th nghim

    nh trn. Sau thi gian nung T2 nhit ca nhit lng k v du tng ln 2t (0C).

  • TI LIU N THI HC SINH GII

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    tin tnh ton c th chn mn=md=mk=m.B qua s mt mt nhit lng trong qu trnh

    nung nng.Hy tnh cx.

    (Bit 1t =9,2

    0C

    2t =16,20C. cn=4200J/KgK; ck=380J/KgK. Cho

    rng T1 = T2)

    Hng dn gii:

    a/ Gi h1 v h2ln lct l phn g chm trong nc v trong du:

    h=h1+h2=19(cm) (1)

    Khi g chu tc dng ca ba lc cn bng nhau:

    -Trng lc: P=dg.V=dg.S.h

    -Lc y Ac-si-met ca nc: Fn=dnS.h1

    -Lc y Ac-si-met ca du : Fd=ddS.h2

    Ta c: Fn+Fd=P

    ddS.h2+dnS.h1=dg.S.h

    dd.h2+dn.h1=dg.h

    7000h2+10000h1=8000.19

    7h2+10h1=167,2 (2)

    Thay (1) vo (2),suy ra: 3h1=34,2 =>h1=11,4(cm) :

    h2=19-11,2=7,6 (cm)

    Vy :-phn chm trong nc l 11,4(cm) v phn chm trong du l 7,6(cm)

    b/ Nhit lng m nc v nhit lng k hp thu: Q1=(mn.cn+mk.ck) 1t

    =m(4200+380)9,2=42136m

    Nhit lng m du v nhit lng k hp thu: Q2=(md.cd+mk.ck) 2t

    =m(cd+380)16,2

    Dng mt loi dy nung do cng sut nh nhau v thi gian T1=T2 nn

    Q1=Q2

    42136m=m(cd+380)16,2

    => cd=2221J/Kg.K

  • TI LIU N THI HC SINH GII

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    Bi 4: Mt qu cu c trng lng ring d1=8200N/m3, th tch V1=100cm

    3, ni trn

    mt mt bnh nc. Ngi ta rt du vo ph kn hon ton qu cu. Trng lng ring

    ca du l d2=7000N/m3 v ca nc l d3=10000N/m

    3.

    a/ Tnh th tch phn qu cu ngp trong nc khi du.

    b/ Nu tip tc rt thm du vo th th tch phn ngp trong nc ca qu cu

    thay i nh th no?

    Hng dn gii:

    a/ Gi V1, V2, V3ln lt l th tch ca qu cu, th tch ca qu cu ngp trong du v

    th tch phn qu cungp trong nc. Ta c V1=V2+V3 (1)

    Qu cu cn bng trong nc v trong du nn ta c: V1.d1=V2.d2+V3.d3 . (2)

    T (1) suy ra V2=V1-V3, thay vo (2) ta c:

    V1d1=(V1-V3)d2+V3d3=V1d2+V3(d3-d2)

    V3(d3-d2)=V1.d1-V1.d2 23

    2113

    )(

    dd

    ddVV

    Tay s: vi V1=100cm3, d1=8200N/m

    3, d2=7000N/m

    3, d3=10000N/m

    3

    b/T biu thc: 23

    2113

    )(

    dd

    ddVV

    . Ta thy th tch phn qu cu ngp trong nc (V3) ch

    ph thuc vo V1, d1, d2, d3 khng ph thuc vo su ca qu cu trong du, cng

    nh lng du thm vo. Do nu tip tc thm du vo th phn qu cu ngp

    trong nc khng thay i

    Bi 5: Mt khi nc hnh lp phng cnh 3cm, khi lng ring 0.9 g /cm 3 . Vin

    ni trn mt nc. Tnh t s gia th tch phn ni v phn chm ca vin , t

    suy ra chiu cao ca phn ni. Bit khi lng ring ca nc l 1g /cm 3 .

    Hng dn gii:

    D 1 = 1g/cm3 => d 1 = 10N/ g/cm

    3 ;

    D 2 = 0.9g/cm3 => d 2 = 9N/ g/cm

    3 ;

    Gi d 1 v d 2 l trng lng ring cu nc v

    V 1 v V 2 l th tch phn nc b chm v ni

    Khi vin ni th lc y c simet bng trng lng ca vt ta c

  • TI LIU N THI HC SINH GII

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    d 1 . V 1 = d 2 ( V 1 + V 2 )

    1(

    2

    1

    1

    )21

    2

    1

    V

    V

    V

    VV

    d

    d

    Hay 11,09

    1

    9

    9101

    9

    101

    2

    1

    1

    2

    d

    d

    V

    V

    Vy 11,01

    2 V

    V

    cao ca phn ni l:

    h 2 = 0,11.3 = 0,33 cm

    Bi 6: Mt thanh ng cht, tit din u c chiu di

    AB = l = 40cm c ng trong mt chu (hnh v )

    sao cho OB3

    1OA . Ngi ta nc vo chu cho

    n khi thanh bt u ni (u B khng cn ta trn

    y chu). Bit thanh c gi cht ti O v ch c th quay quanh O.

    a. Tm mc nc cn vo chu. Cho khi lng ring ca thanh v nc ln lt l :

    D1 = 1120kg/m3 ; D2 =1000kg/m

    3.

    Hng dn gii:

    a. Gi x = BI l mc nc vo chu thanh bt u ni, S l tit din ca thanh.

    Thanh chu tc dng ca trng lc P t ti im M ca AB v lc y Archimede t

    ti trung im N ca BI. Theo iu kin cn bng ta c :

    P.MH = F.NK

    Trong P = 10D1Sl

    F = 10D2Sx

    Suy ra : D1l.MH = D2x.NK

    1

    2

    .D l MH

    xD NK

    (1)

    Xt hai tam gic ng dng : OMH ONK ta c

    B

    A

    O

    A O

    B

    I

    N

    P

    K

    F

    H

    M

  • TI LIU N THI HC SINH GII

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    MH OM

    NK ON

    Vi OM = MA OA = 20 10 = 10cm

    ON = OB NB = 60

    302 2

    x x

    T : 1

    2

    20(2)

    60

    Dx l

    D x

    1120

    60 .40.20 8961000

    x x

    2 60 896 0x x

    2

    32

    28

    x cm

    x cm

    Loi nghim x1 = 32cm v ln hn OB. Phi ngp nc mt on 28cm.

    b. T phng trnh (2) ta suy ra ;

    1

    2

    20

    60

    DD l

    x x

    Mc nc ti a vo chu l x = OB = 30cm, ng vi trng hp ny, cht lng

    phi c khi lng ring l

    31

    2

    20 20.1120.40995,5 /

    60 30 60 30

    D lD kg m

    x x

    Vy, thc hin c th nghim, cht lng vo chu phi c khi lng ring

    3

    2995,5 /D kg m

    Bi 7: Mt cc nc ang tan trong n c cha mt mu ch c th vo trong

    nc. Sau khi c 100g tan chy th th tch phn ngp trong nc ca cc gim i

    mt na. Khi c thm 50g na tan chy th cc nc bt u chm. Tnh khi

    lng ca mu ch. Cho bit khi lng ring ca nc , nc v ch ln lt l

    0,9g/cm3 , 1g/cm

    3 v 11,3g/cm

    3

    Hng dn gii:

    Trng lng ca nc v ch l P = (mc + md).10

    Trc khi tan 100g nc tan P = (mc + md).10 = Vc. Dn.10

  • TI LIU N THI HC SINH GII

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    Sau khi 100g nc tan chy: P, = (mc + md -0,1 ).10 =

    1

    2. Vc. Dn.10

    Bin i v => mc + md = 0,2

    Th tch ca khi nc sau khi tan chy 150 g l:

    c

    d

    c

    c

    D

    m

    D

    mV

    15,0 khi cc bt u chm (mc + md - 0,15 ).10 = V. Dn.10

    => mc + md - 0,15 =

    d

    d

    c

    c

    D

    m

    D

    m 15,0Dn bin i v thay s vo ta c h pt

    mc + md = 0,2

    103 1 0,05

    113 9 3c dm m

    gii h phng trnh ta c

    mc 5,5 g ; md 194,5g

    Bi 8: Trong bnh hnh tr, tit din S cha nc c chiu cao H = 15cm. Ngi ta th

    vo bnh mt thanh ng cht, tit din u sao cho n ni trong nc th mc nc

    dng ln mt on h = 8cm.

    a) Nu nhn chm thanh hon ton th mc nc s cao bao nhiu? (Bit khi

    lng ring ca nc v thanh ln lt l D1 = 1g/cm3; D2 = 0,8g/cm

    3

    b) Tnh cng thc hin khi nhn chm hon ton thanh, bit thanh c chiu di l =

    20cm; tit din S = 10cm2. Hng dn gii:

    a) Gi tit din v chiu di thanh l S v l. Ta c trng lng ca thanh:

    b) P = 10.D2.S.l

    Th tch nc dng ln bng th tch phn chm trong nc : V = ( S S).h

    Lc y Acsimet tc dng vo thanh : F1 = 10.D1(S S).h

    H

    h l

    P

    F1

    S

    h

    S

    F

    l

    Do thanh cn bng nn: P = F1

    10.D2.S.l = 10.D1.(S S).h

    hS

    SS

    D

    Dl .

    '

    '.

    2

    1 (*)

    Khi thanh chm hon ton trong nc, nc dng ln mt lng

    bng th tch thanh.

    Gi Vo l th tch thanh. Ta c : Vo = S.l

    Thay (*) vo ta c: hSSD

    DV ).'.(

    2

    10

    Lc mc nc dng ln 1 on h ( so vi khi cha th

    thanh vo) hD

    D

    SS

    Vh .

    ' 2

    10

  • TI LIU N THI HC SINH GII

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    T chiu cao ct nc trong bnh l: H = H +h =H + hD

    D.

    2

    1

    H = 25 cm

    Lc tc dng vo thanh lc ny gm : Trng lng P, lc y Acsimet F2 v lc tc

    dng F. Do thanh cn bng nn F = F2 - P = 10.D1.Vo 10.D2.S.l

    F = 10( D1 D2).S.l = 2.S.l = 0,4 N

    T pt(*) suy ra : 2

    1

    2 30'.3'.1. cmSSh

    l

    D

    DS

    Do khi thanh i vo nc thm 1 on x c th tch V = x.S th nc dng thm

    mt on:

    2'2'

    x

    S

    V

    SS

    Vy

    Mt khc nc dng thm so vi lc u:

    cmhD

    Dhh 2.1

    2

    1

    ngha l : 42

    2 x

    x

    Vy thanh c di chuyn thm mt on: x + cmxxx

    3

    84

    2

    3

    2 .

    V lc tc dng tng u t 0 n F = 0,4 N nn cng thc hin c:

    JxFA 32 10.33,510.3

    8.4,0.

    2

    1.

    2

    1

    Bi 9: Ti y ca mt ci ni hnh tr tit din S1 = 10dm2,

    ngi ta khot mt l trn v cm vo mt ng kim loi

    tit din S2 = 1 dm2. Ni c t trn mt tm cao su nhn,

    y ln ngc ln trn, rt nc t t vo ng pha trn.

    Hi c th rt nc ti cao H l bao nhiu nc khng

    thot ra t pha di.

    (Bit khi lng ca ni v ng kim loi l m = 3,6 kg.

    Chiu cao ca ni l h = 20cm. Trng lng ring ca nc dn = 10.000N/m3).

    h

    S1

    S2

    H

  • TI LIU N THI HC SINH GII

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    Hng dn gii: Nc bt u chy ra khi p lc ca n ln y ni cn bng vi

    trng lc:

    P = 10m ; F = p ( S1 - S2 ) (1)

    *Hn na: p = d ( H h ) (2)

    T (1) v (2) ta c:10m = d ( H h ) (S1 S2 )

    H h = 1 2 1 2

    10m 10mH h

    d(S S ) d(S S )

    *Thay s ta c: H = 0,2 + 10.3,6

    0,2 0,04 0,24(m) 24cm

    10000(0,1 0,01)

    PHN V: IN HC

    I - C S L THUYT:

  • TI LIU N THI HC SINH GII

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    1/ nh lut m:

    Cng dng in chy qua dy dn t

    l thun vi hiu in th t vo hai u

    dy v t l nghch vi in tr ca dy .

    I =R

    U I : Cng dng in ( A )

    .

    U : Hiu in th ( V ) ; R : in tr (

    ) .

    2/ on mch ni tip :

    Cng dng in : I = I1 = I2 .

    Hiu in th : U = U1 + U2 .

    in tr tng ng : Rtd = R1 + R2 .

    Hiu in th t l thun vi in tr :

    2

    1

    2

    1

    R

    R

    U

    U

    3/ on mch song song :

    I = I1 + I2 U = U1 = U2 .

    21

    111

    RRRtd =>

    21

    21.

    RR

    RRRtd

    Cng dng in t l nghch vi in

    tr

    1

    2

    2

    1

    R

    R

    I

    I

    4/ on mch hn hp :

    R1 nt ( R2 // R3 ) .

    I = I1 = I 23 = I3 + I2 .

    U = U1 + U23 (m U23 = U2 = U3 ) .

    Rtd = R1 + R23 ( m 32

    32

    23

    .

    RR

    RRR

    )

    ( R1 nt R2 ) // R3 .

    IAB = I12 + I3 ( m I12 = I1 = I2 ) .

    UAB = U12 = U3 (m U12 = U1 + U2 ) .

    312

    312.

    RR

    RRRtd

    ( m R12 = R1 + R2 ) .

    1K = 1000

    1M = 1000 000

    in tr dy dn t l thun vi chiu

    di dy dn :

    2

    1

    2

    1

    R

    R

    l

    l .

    in tr ca dy dn t l nghch vi tit

    din ca dy :

    2

    1

    1

    2

    R

    R

    S

    S

    Cng thc tnh in tr :S

    lR

    : in tr sut ( m) .

    l : chiu di ca dy ( m )

    S : tit din ca dy dn ( m2 ) .

    1mm= 1 .10-6

    m2 ; d = 2r =>

    2

    dr

    S = 3,14 .r2 ;

    d : ng knh

    r :bn knh ca dy .

    V

    mD

    D : khi lng ring ( kg / m3 )

    m: khi lng ca dy ( kg ) .

    V : th tch ca dy ( m3 )

    S

    Vl

  • TI LIU N THI HC SINH GII

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    l: chiu di ca dy ( m ) .

    V : th tch ca dy ( m3 ) .

    S : tit din ca dy (m2 ) .

    Chu vi ng trn :2 r (vi =3,14)

    Cng sut in :P = U .I = I2 . R =

    R

    U 2

    P : cng sut ( W ) .

    Hiu sut :H = Atp

    Qi ; H : hiu sut ( %

    )

    Ai = Qi : in nng c ch ( J )

    (Qi =m.C. t)

    Atp : in nng ton phn ( J )

    5/Cng ca dng in :

    A = P . t = U.I.t = I2.R.t =

    R

    U 2.t

    A : cng ca dng in ( J )

    P : cng sut in ( W )

    t: thi gian ( s )

    1kW = 1000 W .

    1 h = 3600 s .

    1kWh = 3,6 .10-6

    J

    nh lut Jun Len-X : Nhit lng

    ta ra dy dn khi c dng in chy

    qua t l thun vi bnh phng cng

    dng in, vi in tr ca dy v thi

    gian dng in chy qua .

    Q = I2 . R . t .

    Nu o nhit lng Q bng n v calo

    th h thc ca nh lut Jun Len-X

    l

    Q = 0,24 . I2 .R. t

    S vng dy r

    ln

    .2

  • TI LIU N THI HC SINH GII

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  • TI LIU N THI HC SINH GII

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    II - MT S VN CN LU KHI GII BI TP IN:

    1/. Mun duy tr mt dng in lu di trong mt vt dn cn duy tr mt in

    trng trong vt dn . Mun vy ch cn ni 2 u vt dn vi 2 cc ca ngun

    in thnh mch kn.

    Cng gn cc dng ca ngun in th cng cao. Quy c in th ti cc

    dng ca ngun in, in th l ln nht , in th ti cc m ca ngun in

    bng 0.

    Quy c chiu dng in l chiu chuyn di c hng ca cc ht mang in tch

    dng, Theo quy c bn ngoi ngun in dng in c chiu i t cc

    dng, qua vt dn n cc m ca ngun in (chiu i t ni c in th cao

    n ni c din th thp).

    chnh lch v in th gia 2 im gi l hiu in th gia 2 im : VA -

    VB = UAB. Mun duy tr mt dng in lu di trong mt vt dn cn duy tr mt

    HT gia 2 u vt dn ( U = 0 I = 0)

    2/. Mch in:

    a. on mch in mc song song:

    *c im: mch in b phn nhnh, cc nhnh c chung im u v im cui.

    Cc nhnh hot ng c lp.

    *Th cht: 1. Uchung

    2. Cng dng in trong mch chnh bng trng cng dng

    in trong cc mch r: I=I1+I2+...+In

    3. Nghch o ca in tr tng ng bng tng cc nghch o ca

    cc in tr thnh phn: nRRRR

    1...

    111

    21

    -T t/c 1 v cng thc ca nh lut m .I1R1 = I2R2 =....= InRn = IR

    - T t/c 3 on mch gm n in tr c gi tr bng nhau v bng r th in tr

    ca on mch mc song song l R = n

    r

    - T t/c 3 in tr tng ng ca on mch mc song song lun nh hn

    mi in tr thnh phn.

    b. on mch in mc ni tip:

  • TI LIU N THI HC SINH GII

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

    Su tm bi https://blogvatly.com

    *c im:cc b phn (cc in tr) mc thnh dy lin tc gia 2 cc ca ngun

    in ( cc b phn hot ng ph thuc nhau).

    *tnh cht: 1.I chung

    2. U = U1 + U2 +....+ Un.

    3. R = R1 + R2 +,...+ Rn.

    *T t/c 1 v cng thc ca nh lut m I=U/R U1/R1=U2/R2=...Un/Rn. (trong

    on mch ni tip, hiu in th gia 2 u cc vt dn t l thun vi in tr

    ca chng) Ui=U Ri/R...

    T t/s 3 nu c n in tr ging nhau mc ni tip th in tr ca on mch

    l R =nr. Cng t tnh cht 3 in tr tng ng ca on mch mc ni tip

    lun ln hn mi in tr thnh phn.

    C.Mch cu :

    Mch cu cn bng c cc tnh cht sau:

    - v in tr: 4

    3

    2

    1

    R

    R

    R

    R ( R5 l ng cho ca cu)

    -V dng: I5 = 0 -v HT : U5 = 0

    42314

    4

    3

    1

    2

    2

    1 ;;; IIIIR

    R

    I

    I

    R

    R

    I

    I

    Mch cu khng cn bng:4

    3

    2

    1

    R

    R

    R

    R I5 0; U5 0

    * Trng hp mch cu c 1 s in tr c gi tr bng 0; gii bi ton cn p

    dng cc quy tc bin i mch in tng ng ( phn di )

    *Trng hp c 5 in tr u khc 0 s xt sau.

    3/. Mt s quy tc chuyn mch:

    a/. chp cc im cng in th: "Ta c th chp 2 hay nhiu im c cng in

    th thnh mt im khi bin i mch in tng ng."

    (Do VA - Vb = UAB = I RAB Khi RAB = 0;I 0 hoc RAB 0,I = 0 Va = Vb

    Tc A v B cng in th)

  • TI LIU N THI HC SINH GII

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

    Su tm bi https://blogvatly.com

    Cc trng hp c th: Cc im 2 u dy ni, kha K ng, Am pe k c in

    tr khng ng k...c coi l c cng in th. Hai im nt 2 u R5 trong

    mch cu cn bng...

    b/. B in tr: ta c th b cc in tr khc 0 ra khi s khi bin i mch

    in tng ng khi cng dng in qua cc in tr ny bng 0.

    Cc trng hp c th: cc vt dn nm trong mch h; mt in tr khc 0 mc

    song song vi mt vt dn c in tr bng 0( in tr b ni tt) ; vn k c

    in tr rt ln (l tng).

    4/. Vai tr ca am pe k trong s :

    * Nu am pe k l tng ( Ra=0) , ngoi chc nng l dng c o n cn c vai

    tr nh dy ni do :

    C th chp cc im 2 u am pe k thnh mt im khi bin i mch in

    tng ng( khi am pe k ch l mt im trn s )

    Nu am pe k mc ni tip vi vt no th n o cng d/ qua vt.

    Khi am pe k mc song song vi vt no th in tr b ni tt ( ni trn).

    Khi am pe k nm ring mt mch th dng in qua n c tnh thng qua cc

    dng 2 nt m ta mc am pe k ( d theo nh l nt).

    * Nu am pe k c in tr ng k, th trong s ngoi chc nng l dng c o

    ra am pe k cn c chc nng nh mt in tr bnh thng. Do s ch ca n

    cn c tnh bng cng thc: Ia=Ua/Ra .

    5/. Vai tr ca vn k trong s :

    a/. trng hp vn k c in tr rt ln ( l tng):

    *Vn k mc song song vi on mch no th s ch ca vn k cho bit HT

    gia 2 u on mch :

    UV = UAB = IAB.RAB

    *TRong trng hp mch phc tp, Hiu in th gia 2 im mc vn k phi

    c tnh bng cng thc cng th: UAB = VA - VB = VA - VC + VC - VB = UAC +

    UCB....

    *c th b vn k khi v s mch in tng ng .

    *Nhng in tr bt k mc ni tip vi vn k c coi nh l dy ni ca vn

    k ( trong s tng ng ta c th thay in tr y bng mt im trn dy

  • TI LIU N THI HC SINH GII

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

    Su tm bi https://blogvatly.com

    RRRRR

    x

    321

    31.

    RRRRR

    z

    321

    32.

    RRRRRy

    321

    21.

    ni), theo cng thc ca nh lut m th cng qua cc in tr ny coi nh

    bng 0 ,( IR = IV = U/ = 0).

    b/. Trng hp vn k c in tr hu hn ,th trong s ngoi chc nng l

    dng c o vn k cn c chc nng nh mi in tr khc. Do s ch ca vn

    k cn c tnh bng cng thc UV=Iv.Rv...

    6/ mt s quay tc i mch

    *Quy tc bin i mch hnh

    sao thnh mch hnh tam

    gic:

    R1=z

    zxyzxy ,

    R1=x

    zxyzxy ,

    R1=y

    zxyzxy

    *Quy tc chuyn mch hnh tam gic thnh hnh sao:

    III/ BI TP P DNG:

    Bi 1: Cho mch in MN nh hnh v di y, hiu in th hai u mch

    in khng i UMN = 7V; cc in tr R1 = 3 v R2 = 6 . AB l mt dy dn

    in c chiu di 1,5m tit din khng i S = 0,1mm2, in tr sut = 4.10

    -7

    m ; in tr ca ampe k A v cc dy ni khng ng k :

    M UMN N a/ Tnh in tr ca dy dn AB

    ?

    R1 D R2 b/ Dch chuyn con chy c sao cho AC

    = 1/2 BC. Tnh

    cng dng in qua ampe

    k ?

    A c/ Xc nh v tr con chy C Ia

    = 1/3A ?

  • TI LIU N THI HC SINH GII

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

    Su tm bi https://blogvatly.com

    Hng dn gii:

    a/ i 0,1mm2 = 1. 10

    -7 m

    2 . p dng cng thc tnh in tr

    S

    lR . ; thay s

    v tnh RAB = 6

    b/ Khi 2

    BCAC RAC =

    3

    1.RAB RAC = 2 v c RCB = RAB - RAC = 4

    Xt mch cu MN ta c 2

    321 CBAC R

    R

    R

    R nn mch cu l cn bng. Vy IA

    = 0

    c/ t RAC = x ( K : 0 x 6 ) ta c RCB = ( 6 - x )

    * in tr mch ngoi gm ( R1 // RAC ) ni tip ( R2 // RCB ) l

    )6(6

    )6.(6

    3

    .3

    x

    x

    x

    xR

    = ?

    * Cng dng in trong mch chnh : R

    UI ?

    * p dng cng thc tnh HT ca mch // c : UAD = RAD . I = Ix

    x.

    3

    .3

    = ?

    V UDB = RDB . I = Ix

    x.

    12

    )6.(6

    = ?

    * Ta c cng dng in qua R1 ; R2 ln lt l : I1 = 1R

    U AD = ? v I2 =

    2R

    U DB = ?

    + Nu cc dng ca ampe k gn vo D th : I1 = Ia + I2 Ia = I1 - I2 = ?

    (1)

    Thay Ia = 1/3A vo (1) Phng trnh bc 2 theo x, gii PT ny c x = 3 (

    loi gi tr -18)

    + Nu cc dng ca ampe k gn vo C th : Ia = I2 - I1 = ? (2)

    Thay Ia = 1/3A vo (2) Phng trnh bc 2 khc theo x, gii PT ny c x =

    1,2 ( loi 25,8 v > 6 )

  • TI LIU N THI HC SINH GII

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

    Su tm bi https://blogvatly.com

    * nh v tr im C ta lp t s CB

    AC

    R

    R

    CB

    AC = ? AC = 0,3m

    Bi 2: Cho 3 in tr c gi tr nh nhau bng R0, c mc vi nhau theo nhng

    cch khc nhau v ln lt ni vo mt ngun in khng i xc nh lun mc

    ni tip vi mt in tr r . Khi 3 in tr trn mc ni tip th cng dng

    in qua mi in tr bng 0,2A, khi 3 in tr trn mc song song th cng

    dng in qua mi in tr cng bng 0,2A.

    a/ Xc nh cng dng in qua mi in tr R0 trong nhng trng hp cn

    li ?

    b/ Trong cc cch mc trn, cch mc no tiu th in nng t nht ? Nhiu nht ?

    c/ Cn t nht bao nhiu in tr R0 v mc chng nh th no vo ngun in

    khng i c in tr r ni trn cng dng in qua mi in tr R0 u

    bng 0,1A ?

    Hng dn gii:

    a/ Xc nh cc cch mc cn li gm :

    cch mc 1 : (( R0 // R0 ) nt R0 ) nt r cch mc 2 : (( R0 nt R0 )

    // R0 ) nt r

    Theo bi ta ln lt c cng dng in trong mch chnh khi mc ni tip :

    Int = 03Rr

    U

    = 0,2A (1)

    Cng dng in trong mch chnh khi mc song song :

    AR

    r

    UI 6,02,0.3

    3

    0

    SS

    (2)

    Ly (2) chia cho (1), ta c : 3

    3

    3

    0

    0

    Rr

    Rr r = R0 . em gi tr ny ca r thay vo

    (1) U = 0,8.R0

    + Cch mc 1 : Ta c (( R0 // R0 ) nt R0 ) nt r (( R1 // R2 ) nt R3 ) nt r t R1

    = R2 = R3 = R0

  • TI LIU N THI HC SINH GII

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    Su tm bi https://blogvatly.com

    Dng in qua R3 : I3 = AR

    R

    RRr

    U32,0

    .5,2

    .8,0

    20

    0

    0

    0

    . Do R1 = R2 nn I1 = I2 =

    AI

    16,02

    3

    + Cch mc 2 : Cng dng in trong mch chnh I =

    AR

    R

    R

    RRr

    U48,0

    3

    .5

    .8,0

    .3

    ..2 0

    0

    0

    00

    .

    Hiu in th gia hai u mch ni tip gm 2 in tr R0 : U1 = I. 0

    00

    .3

    ..2

    R

    RR =

    0,32.R0 cng dng in qua mch ni tip ny l I1 =

    AR

    R

    R

    U16,0

    .2

    .32,0

    .2 0

    0

    0

    1 CD qua in tr cn li l

    I2 = 0,32A.

    b/ Ta nhn thy U khng i cng sut tiu th mch ngoi P = U.I s nh

    nht khi I trong mch chnh nh nht cch mc 1 s tiu th cng sut nh

    nht v cch mc 2 s tiu th cng sut ln nht.

    c/ Gi s mch in gm n dy song song, mi dy c m in tr ging nhau v

    bng R0 ( vi m ; n N)

    Cng dng in trong mch chnh ( Hv ) I +

    -

    n

    mR

    n

    mr

    UI

    1

    8,0

    . 0

    ( B sung vo hv cho y )

    cng dng in qua mi in tr R0 l 0,1A ta phi c :

    n

    n

    mI .1,0

    1

    8,0

    m + n = 8 . Ta c cc trng hp sau

    m 1 2 3 4 5 6 7

    n 7 6 5 4 3 2 1

    S in tr R0 7 12 15 16 15 12 7

    Theo bng trn ta cn t nht 7 in tr R0 v c 2 cch mc chng :

  • TI LIU N THI HC SINH GII

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

    Su tm bi https://blogvatly.com

    a/ 7 dy //, mi dy 1 in tr. b / 1 dy gm 7

    in tr mc ni tip.

    Bi 3 Cho mch in sau

    Cho U = 6V , r = 1 = R1 ; R2 = R3 = 3 U r

    bit s ch trn A khi K ng bng 9/5 s ch R1 R3

    ca A khi K m. Tnh :

    a/ in tr R4 ? R2 K R4 A

    b/ Khi K ng, tnh IK ?

    Hng dn gii:

    * Khi K m, cch mc l ( R1 nt R3 ) // ( R2 nt R4 ) in tr tng ng ca

    mch ngoi l

    4

    4

    7

    )3(4

    R

    RrR

    Cng dng in trong mch chnh : I =

    4

    4

    7

    )3(41

    R

    R

    U

    . Hiu

    in th gia hai im A v B l UAB = IRRRR

    RRRR.

    ))((

    4321

    4231

    I4 =

    4321

    31

    42

    ).(

    RRRR

    IRR

    RR

    U AB

    Thay s ta c I = 4519

    4

    R

    U

    * Khi K ng, cch mc l (R1 // R2 ) nt ( R3 // R4 ) in tr tng ng ca

    mch ngoi l

    4

    4

    412

    159'

    R

    RrR

    Cng dng in trong mch chnh lc ny l : I =

    4

    4

    412

    1591

    R

    R

    U

    . Hiu in th gia hai im A v B l UAB = '..

    43

    43 IRR

    RR

    I4 =

    43

    3

    4

    '.

    RR

    IR

    R

    U AB

    Thay s ta c I = 41921

    12

    R

    U

    * Theo bi th I4 = 4.5

    9I ; t tnh c R4 = 1

  • TI LIU N THI HC SINH GII

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    Su tm bi https://blogvatly.com

    b/ Trong khi K ng, thay R4 vo ta tnh c I4 = 1,8A v I = 2,4A UAC =

    RAC . I = 1,8V

    I2 = AR

    U AC 6,02

    . Ta c I2 + IK = I4 IK = 1,2A

    Bi 4: Mt hp kn cha mt ngun in c hiu in th khng i U = 150V

    v mt in tr r = 2. Ngi ta mc vo hai im ly in A v B ca hp mt

    bng n c cng sut nh mc P = 180W ni tip vi mt bin tr c in tr

    Rb ( Hv )

    A U

    B

    1/ n sng bnh thng th phi iu chnh Rb = 18. Tnh r

    hiu in th nh mc ca n ?

    2/ Mc song song vi n mt bng n na ging ht n. Hi

    Rb

    c hai n sng bnh thng th phi tng hay gim Rb ? Tnh

    tng ( gim ) ny ?

    3/ Vi hp in kn trn, c th thp sng ti a bao nhiu bng n nh n ?

    Hiu sut s dng in khi l bao nhiu phn trm ?

    Hng dn gii:

    1/ Gi I l cng dng in trong mch chnh th U.I = P + ( Rb + r ).I2 ; thay

    s ta c mt phng trnh bc 2 theo I : 2I2 - 15I + 18 = 0 . Gii PT ny ta

    c 2 gi tr ca I l I1 = 1,5A v I2 = 6A.

    + Vi I = I1 = 1,5A Ud = dI

    P = 120V ; + Lm tt vi I = I2 = 6A Hiu sut

    s dng in trong trng hp ny l : H = 206.150

    180

    .

    IU

    p nn qu thp

    loi b nghim I2 = 6A

    2/ Khi mc 2 n // th I = 2.Id = 3A, 2 n sng bnh thng nn: Ud = U - (

    r + Rb ).I Rb ? gim ca Rb ? ( S : 10 )

    3/ Ta nhn thy U = 150V v Ud = 120V nn cc n sng bnh thng, ta

    khng th mc ni tip t 2 bng n tr ln c m phi mc chng song song.

    Gi s ta mc // c ti a n n vo 2 im A & B

  • TI LIU N THI HC SINH GII

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

    Su tm bi https://blogvatly.com

    cng dng in trong mch chnh I = n . Id .

    Ta c U.I = ( r + Rb ).I2 + n . P U. n . Id = ( r + Rb ).n

    2 .I

    2d + n . P U.Id

    = ( r + Rb ).n.Id + P

    Rb = 0.

    .2

    rIn

    PIU

    d

    d 10

    )5,1.(2

    1805,1.150

    .

    .22

    d

    d

    Ir

    PIUn n max = 10

    khi Rb = 0

    + Hiu sut s dng in khi bng : H = U

    U d = 80

    Bi 5: Mt m in c 2 in tr R1 v R2 . Nu R1 v R2 mc ni tip vi nhau th

    thi gian un si nc ng trong m l 50 pht. Nu R1 v R2 mc song song vi

    nhau th thi gian un si nc trong m lc ny l 12 pht. B qua s mt nhit

    vi mi trng v cc iu kin un nc l nh nhau, hi nu dng ring tng

    in tr th thi gian un si nc tng ng l bao nhiu ? Cho hiu in th U l

    khng i .

    Hng dn gii:

    * Gi Q (J) l nhit lng m bp cn cung cp cho m un si nc th Q

    lun khng i trong cc trng hp trn. Nu ta gi t1 ; t2 ; t3 v t4 theo th t l

    thi gian bp un si nc tng ng vi khi dng R1, R2 ni tip; R1, R2 song

    song ; ch dng R1 v ch dng R2 th theo nh lut Jun-lenx ta c :

    2

    4

    2

    1

    3

    2

    21

    21

    2

    2

    21

    1

    22 ..

    .

    ...

    R

    tU

    R

    tU

    RR

    RR

    tU

    RR

    tU

    R

    tUQ

    (1)

    * Ta tnh R1 v R2 theo Q; U ; t1 v t2 :

    + T (1) R1 + R2 = Q

    tU 12 .

    + Cng t (1) R1 . R2 = 221

    4

    212

    2 ..).(

    .

    Q

    ttURR

    Q

    tU

    * Theo nh l Vi-et th R1 v R2 phi l nghim s ca phng trnh : R2 -

    Q

    tU 12 .

    .R + 2

    21

    4 ..

    Q

    ttU = 0(1)

  • TI LIU N THI HC SINH GII

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

    Su tm bi https://blogvatly.com

    Thay t1 = 50 pht ; t2 = 12 pht vo PT (1) v gii ta c = 102 .

    2

    4

    Q

    U =

    Q

    U 2.10

    R1 =

    Q

    UtQ

    U

    Q

    tU

    .2

    ).10(

    2

    .10.2

    1

    2

    1

    2

    30.Q

    U 2 v R2 = 20.

    Q

    U 2

    * Ta c t3 = 21.

    U

    RQ = 30 pht v t4 = 2

    2.

    U

    RQ = 20 pht . Vy nu dng ring tng

    in tr th thi gian un si nc trong m tng ng l 30ph v 20 ph .

    Bi 6. Cho mch in nh hnh v

    U = 60V, R1 = R3 = R4 = 2 m, R2 =

    10 m, R6 = 3,2 m. Khi dng

    in qua R5 l 2A v c chiu nh

    hnh v. Tm R5?

    Hng dn gii:

    Ti nt C. I3 +I5 = I1 => I3 = I1- 2

    Ti nt D. I2 +I5 = I4 => I4 = I2+2

    UAE = U1 + U3= U2 + U4 => 2I1+2( I1- 2) = 10 I2 + 2( I2 + 2)

    => 4I1 = 12I2 + 8 => I1 = 3I2 + 2

    dng in qua R6 : I6 = I1 + I2 = 4I2 + 2

    Ta c UAB = UAE + U6 => I2 = 2A => I1= 8A

    U5 = UCD = - UAC + UAD = - U1 + U2 = 4V

    Vy in tr R5 l 2 m

    Bi 7: Mt m un nc bng in c 3 dy l xo, mi ci c in tr R=120 ,

    c mc song song vi nhau. m c mc ni tip vi in tr r=50 v c

    A

    I1

    I2

    R1

    I3R3

    R2

    I4

    R4

    I5

    R5I6 R6

    BE

    C

    D

  • TI LIU N THI HC SINH GII

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

    Su tm bi https://blogvatly.com

    mc vo ngun in. Hi thi gian cn thit un m ng y nc n khi si

    s thay i nh th no khi mt trong ba l xo b t?

    Hng dn gii:

    *Lc 3 l xo mc song song:

    in tr tng ng ca m:

    R1 = )(403

    R

    Dng in chy trong mch:

    I1 = rR

    U

    1

    Thi gian t1 cn thit un m nc n khi si:

    Q = R1.I2.t1 2

    1

    1

    2

    1

    1

    rR

    UR

    Q

    IR

    Qt hay t1 =

    1

    2

    2

    1 )(

    RU

    rRQ (1)

    *Lc 2 l xo mc song song: (Tng t trn ta c )

    R2 = )(602

    R

    I2 = rR

    U

    2

    t2 = 2

    2

    2

    2 )(

    RU

    rRQ

    ( 2 )

    Lp t s 2

    1

    t

    tta c: 1

    242

    243

    )5060(40

    )5040(60

    )(

    )(2

    2

    2

    21

    2

    12

    2

    1

    rRR

    rRR

    t

    t *Vy t1 t2

    Bi 8: trang tr cho mt quy hng, ngi ta dng cc bng n 6V-9W mc

    ni tip vo mch in c hiu in th U=240V chng sng bnh thng. Nu

    c mt bng b chy, ngi ta ni tt on mch c bng li th cng sut tiu

    th ca mi bng tng hay gim i bao nhiu phn trm?

    Hng dn gii:

    in tr ca mi bng:

    R = )(42

    d

    d

    P

    U

    S bng n cn dng chng sng bnh thng:

  • TI LIU N THI HC SINH GII

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

    Su tm bi https://blogvatly.com

    n = 40dU

    U(bng)

    Nu c mt bng b chy th in tr tng cng ca cc bng cn li l:

    R = 39R = 156 ( )

    Dng in qua mi n by gi:

    I = )(54,1156

    240A

    R

    U

    Cng sut tiu th mi bng by gi l:

    P = I2.R = 9,49 (W)

    Cng sut mi bng tng ln so vi trc:

    Pm - P = 9,49 - 9 = 0,49 (W)

    Ngha l tng ln so vi trcl:

    %4,5.%9

    100.49,0

    Bi 9: Mt m in bng nhm c khi lng 0,5kg cha 2kg nc 25oC. Mun

    un si lng nc trong 20 pht th m phi c cng sut l bao nhiu? Bit

    rng nhit dung ring ca nc l C = 4200J/kg.K. Nhit dung ring ca nhm l

    C1 = 880J/kg.K v 30% nhit lng to ra mi trng xung quanh.

    Hng dn gii:

    *Nhit lng cn tng nhit ca m nhm t 25oC ti 100

    oC l:

    Q1 = m1c1 ( t2 t1 ) = 0,5.880.(100 25 ) = 33000 ( J )

    *Nhit lng cn tng nhit ca nc t 25oC ti 100

    oC l:

    Q2 = mc ( t2 t1 ) = 2.4200.( 100 25 ) = 630000 ( J )

    *Nhit lng tng cng cn thit:

    Q = Q1 + Q2 = 663000 ( J ) ( 1 )

    *Mt khc nhit lng c ch un nc do m in cung cp trong thi gian 20

    pht ( 1200 giy ) l:

    Q = H.P.t ( 2 )

    ( Trong H = 100% - 30% = 70%; P l cng sut

    ca m; t = 20 pht = 1200 giy )

    *T ( 1 ) v ( 2 ) : P = W)Q 663000.100

    789,3(

    H.t 70.1200

    A

    +

    V A

    B

    C

    R1

    M N D

    -

  • TI LIU N THI HC SINH GII

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    Bi 10: Cho mch in nh hnh v. Bit UAB = 10V;

    R1 = 2 ; Ra = 0 ; RV v cng ln ; RMN = 6 .

    Con chy t v tr no th ampe k ch 1A. Lc ny

    vn k ch bao nhiu?

    Hng dn gii:

    *V in tr ca ampe k Ra = 0 nn:

    UAC = UAD = U1 = I1R1. = 2.1 = 2 ( V ) ( Ampe k ch dng qua R1 )

    *Gi in tr phn MD l x th:

    x DN 1 x

    DN

    AB AD DN

    2 2I ;I I I 1

    x x

    2U 1 6 x

    x

    2U U U 2 1 6 x 10

    x

    *Gii ra c x = 2 . Con chy phi t v tr chia MN thnh hai phn MD c

    gi tr 2 v DN c gi tr 4 . Lc ny vn k ch 8 vn ( Vn k o UDN.

    Bi 11:Cho mch in nh hnh v. Hiu in th hai du on mch

    U = 60 V, R1 = 10 ,R2 = R5 = 20 , R3 = R4 = 40

    Vn k V l l tng, b qua in tr cc dy ni.

    Cu a: Tm s ch ca vn k

    u b: Nu thay vn k V bng mt bng n c dng in nh mc

    Id = 0,4 A mc vo hai im P v Q ca mch in th bng n

    sng bnh thng.Tm in tr ca bng n

    Hng dn gii:

    a) Khi vn k mc vo hai im P v Q ta c (R2 n tR3)// (R4 nt R5)

    R23 = R45 = 60

    => RMN = 30

    - in tr tng dng ton mch:

    R = RMN + R1 = 30 + 10 = 40

    - Cng dng in trong mch chnh

    V

    R1

    R2 R3

    R4 R5

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    601,5

    40

    UI A

    R

    - Cng dng n qua R2 v R4

    I2 = I4 =1,5

    0,752 2

    IA

    => UPQ = R4.I4 R2.I2 = 40.0,75 -20. 0,75 = 15 V

    Vy s ch ca vn k l 15 V

    b) Khi thay vn k V bi n .

    Do R2=R5 v R3=R4 (mch i xng)

    Ta c: I2=I5 ; I3=I4

    => I=I2+I3 v I=I2-I3=0,4A (1)

    Mt khc ta c: U = U1 + U2 + U3 = (I2+I3)R1 + R2I2 + R3I3

    60 = 10(I2 + I3) + 20I2 + 40I3

    6 = 3I2 + 5I3 (2)

    Gii 2 h phng trnh (1) v (2)

    Ta c: I2 = 1A = I5 ; I3 = 0,6A = I4

    Mt khc ta c: UMN = I2R2 + I3R3 = I2R2 + IR + I5R5

    I3R3 = IR + I5R5

    0,6.40 = 0,4R + 1.20

    => R=10

    Bi 12: Cho mch in c s nh hnh v bn.

    in tr ton phn ca bin tr l Ro , in tr ca

    vn k rt ln. B qua in tr ca ampe k, cc

    dy ni v s ph thuc ca in tr vo nhit .

    Duy tr hai u mch mt hiu in th U khng

    i. Lc u con chy C ca bin tr t gn pha M. Hi s ch ca cc dng c

    o s thay i nh th no khi dch chuyn con chy C v pha N? Hy gii thch

    ti sao?

    Hng dn gii:

    Khi dch chuyn con chy C ca bin tr v pha N th s ch ca cc dng c o

    s tng. (nu khng gii thch ng th khng cho im ny)

    V

    A R

    M

    C

    N

  • TI LIU N THI HC SINH GII

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    Gii thch:

    Gi x l phn in tr ca on MC ca bin tr; IA v UV l s ch ca ampe k

    v vn k.

    in tr tng ng ca on mch:

    Rm = (Ro x) + 1

    1

    Rx

    xR

    Rm 1

    2

    Rx

    xR

    = R

    2

    1

    x

    R

    x

    1

    1

    Khi dch con chy v pha N th x tng => (

    2

    1

    x

    R

    x

    1

    1

    ) tng => Rm gim

    => cng dng in mch chnh: I = U/Rm s tng (do U khng i).

    Mt khc, ta li c: xR

    I

    R

    II

    x

    I AA

    => IA =

    x

    R1

    I

    xR

    x.I

    Do , khi x tng th (1 + )x

    Rgim v I tng (c/m trn) nn IA tng.

    ng thi UV = IA.R cng tng (do IA tng, R khng i)

    Bi 13: Cho hai vn k V1, V2 ging ht nhau,

    hai in tr c gi tr mi ci bng R hai in

    tr kia gi tr mi ci bng 3R (hnh v ) S ch

    ca cc my o l 6 mA, 6 V v 2 V.Tnh R ?

    *Hng dn gii:

    * Hng dn hc sinh xc nh cch mc :

    * Hng dn hc sinh xc nh c s ch

    cc my o:

    V1 ch 2V , V2ch 6V , A ch 6mA

    *Tm c in tr ca vn k:

    RV= 2

    2

    V

    v

    I

    U= 1000( ).

    V1AV2C

    P

    Q

    DM N

    V 1 A V 2 C

    P

    Q

    D M N

    R

    R 3R

    3R

    V2

    A V1

    C A B

    + -

  • TI LIU N THI HC SINH GII

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    Su tm bi https://blogvatly.com

    * Xc nh IV1 = V

    V

    R

    U 1 = 0,002(A).

    * Xc nh c chiu dng in i t P n Q v do

    mch i xng nn I2 = I4 ; I1 = I3

    * I1= IV1+I2 I1 - I2 = 0,002A, I1 + I2= 0,006. Tnh I2, I1

    * Ta c UPQ=UPC + UCQ=UV1 thay vo tnh c: - I1R + I23R = 1

    R.

    *M rng:- Nu thay i s ch ca V1 l 1V th bi ton s i n mt iu v l.

    Bi 14: C mt ampek, hai vn k ging nhau v bn in

    tr gm hai loi m gi tr ca chng gp bnln nhau c

    mc vi nhau nh hnh v. Sch ca cc my o l 1V, 10V

    v 20mA.

    a) CMR cng dng in chy qua bn in tr trn ch

    c hai gi tr?

    b) Xc nh gi tr ca cc in tr mc trong mch?

    * Hng dn gii:

    a) *Tng t, hng dn hc sinh cch xc nh cch mc

    cc in tr v s o ca cc dng c o, t v hnh.

    * Khi V1 ch 10V, V2 ch 1V v A ch 20mA.

    * T xc nh c RV = 5002

    2 2V

    UI

    R (mA)

    * UAB = RI1 + 4RI3 = 4RI2 + RI4

    * T hng dn hc sinh chng minh c : I1 = I4,

    I2 = I3

    Vy cng dng in chy qua 4 in tr trn ch c hai

    gi tr.

    b) * V I1 + I2 = Ia = 20mA. T hng dn hc sinh tnh I1 v I2: I1 = 11mA v I2 =

    9mA.

    * Xt mch vng ACD:

    UAD = UAC + UCD thay s vo tnh c: R = 40 v 4R = 160

    A B

    V2

    A V1

    C

    D

    + -

    R

    R

    4R

    4R I4

    I3

    I2

    I1

  • TI LIU N THI HC SINH GII

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    Bi 15: Hai cm dn c dng chung mt trm in, in tr

    ti hai cm bng nhau v bng R (nh hnh v), cng sut

    nh mc mi cm l P0 bng 48,4 KW, hiu in th nh

    mc mi cm l Uo , hiu in th hai u trm lun c

    duy tr l U0. Khi ch cm I dng in (ch K1 ng) th cng

    sut tiu th cm I l

    P1 = 40 KW, khi ch cm II dng in (ch K2 ng) th

    cng sut tiu th cm II \l P2 = 36,6 KW.

    1) Hy tm biu thc lin h gia r1, r2 v R?

    2) Khi c hai cm dng in th tng cng sut tiu th trn

    hai cm l bao nhiu?

    Hng dn gii:

    * Khi ch cm I dng in( ch K1 ng):

    + Cng sut nh mc trn mi cm: P0= 2

    0U

    R (1)

    + Khi cng sut tiu th trn cm I: P1 = 2

    1U

    R (2)( U1l hiu in th trn cm I

    khi ch cm I dng in)

    + T (1) v (2) ta c: 1 1

    0 0

    1

    1,1

    U P

    U P

    + Theo bi ra ta c: 01 1 11 0 1

    10,1

    1,1

    UU U Rr R

    R R r U R r

    * Khi ch cm II dng in( ch K2 ng):

    + Khi cng sut tiu th trn cm II: P2 = 2

    2U

    R (3) ( U2l hiu in th trn cm II

    khi ch cm II dng in)

    + T (1) v (3) ta c: 2 2

    0 0

    1

    1,15

    U P

    U P

    + Theo bi ra ta c: 2 21 2 0

    0,05UR

    r RR r r U

    *Khi c hai cm dng in (K1 v K2 u ng) ta c in tr ton mch RM:

    r1 r2A C

    RR

    DB

    K1 K2

  • TI LIU N THI HC SINH GII

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    Su tm bi https://blogvatly.com

    + RM = r1+ 2

    2

    0,61222

    R R rR

    R r

    . in tr on mch AB: RAB =

    2

    2

    0,51222

    R R rR

    R r

    + Ta c: 0

    0,5122

    0,6122

    AB AB

    M

    U R

    U R

    * Gi cng sut tiu th trn cm I khi c hai cm dng in l PI ta c:

    + 2 2

    2 2

    0 0

    0,512233,88

    0,6122

    I ABI

    P UP

    P U (KW)

    + Ta c: 2 0

    1 0,5122 1. 0,7968

    1,05 0,6122 1,05

    CB CB

    AB

    U UR

    U R r U

    * Gi cng sut tiu th trn cm II khi c hai cm dng in l PII ta c

    + 2

    2

    2

    0 0

    0,7968 30,73CBII IIUP

    PP U

    (KW)

    * Vy khi c hai cm dng in th tng cng sut tiu th trn hai cm l:

    P = PI + PII P = 64,61(KW)

    * M rng

    Nu khng tnh c hai cm dng chung th tng cm dng in khi c hai kho u

    ng th kt qu nh th no? y l mt bi tp rt hay, s dng nhiu kin thc c

    bn v gip hc sinh t