chuyen de ve lai mach dien

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I. T VN 1. L do chn ti

Do thc t trnh nhn thc ca hc sinh THCS cha cao, c bit l i vi vng nng thn, thi gian tip thu trn lp cn t so vi lng kin thc v kh nng t duy, nhn dng, phn loi bi ton xc nh c yu cu ca bi ton l ht sc kh khn i vi phn ln hc sinh. Bn cnh , do nhu cu ham hc, ham hiu bit ca s hc sinh c trin vng, do mc quan trng ca vt l 9 i vi vic thi hc sinh gi i cc c p v tip tc hc ban khoa hc t nhin cc lp trn nn yu cu t ra l phi chn la, sng lc v phn loi bi tp hng dn cho hc sinh l cng vic v cng quan trng i vi mi gio vin dy bi dng.

Thc t cho th y: kin thc l v hn, cc loi, cc dng bi tp ni chung, bi tp v mch in ni ring l r t phong ph v a dng: - Mch in mc ni tip, mc song song. - Mch in hn hp tng minh.

- Mch in hn hp khng tng minh. - Mch cu, mch i xng, mch tun hon, mch bc thang... Trong qu trnh bi dng vt l THCS cho hc sinh, nu ta ch phn ra cc phn c, nhit, in, quang; mi phn lm mt vi bi hc sinh quan st, ghi chp v ghi nh my mc theo kiu ti hin th r t kh c th ghi nh bn vng v p dng khi cn thit.

Vic bi dng hc sinh c trin vng i h i gio vin phi nh hng c v phn loi tng dng bi tp cho hc sinh, vi mi dng trc ht cung c p cho hc sinh h thng kin thc c bn, nhng im cn lu , cung c p cch gii c th, chn la bi tp cho hc sinh luyn gii nm vng phng php vi mc t n gin n phc tp. Trong cc dng bi tp th vic hc sinh bit phn tch cch mc cc b phn trong mch in phc tp th mi c th bt tay vo vic gii cc bi tp khc.

Trong qu trnh bi dng cho hc sinh mi nhn, hc sinh thuc i tuyn d thi hc sinh gi i, iu m ti nhn th y hu ht hc sinh l i vi nhng mch in phc tp, cc em u b lng tng, b tc khng tm ra hng phn tch mch in.

Song do iu kin c hn v thi gian, iu kin v phng tin, dng, vt ch t.. nn khng th nghin cu k trnh by cc cho cc dng bi tp v cc loi mch in m y ti ch a ra mt vi kinh nghim nh gip hc sinh bin i t mch in hn hp khng tng minh tr v mch in hn hp tng minh c th thc hin gii mt cch n gin v nh vy, khi hc sinh bit cch v li mch in th khi hc sinh s c s hng th bt tay vo vic khai thc nhiu dng ton, bi ton v mch in. Vy gip hc sinh c kh nng gii ton vt l phn nh lut m, bi dng hc sinh c trin vng chn i tuyn hc sinh gi i... t kt qu cao, ti la chn chuyn N N N V L N T N T N T TRONG B T VT L L 9 trong chuyn ny ti ch nu ra cc bi ton v mch in hn hp khng tng minh cung c p cho hc sinh c thm gii php gii bi ton loi mch in ny. 2. c ch nghin cu Tm ra mt phng php tng qut a nhng mch in phc tp v cc dng n gii nh t . C th khi gp mt bi v mch in, phn nh mch in thuc dng no ri dng phng php ko mch, a chng v dng n gin hn.

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3. i tng v phm vi nghin cu Cc bi ton v g mch in mt chiu khng tng minh trong chng trnh c bn cng nh nng cao ca lp 9. 4. Nhim v nghin cu Tm nhng mch in t n gin n phc tp, phn loi ri a v nhng dng chung nh t. a ra cc v d c th v tng loi mch in, tng dng mt cho hc sinh d hnh dung. 5. hng php nghin cu Lit k nhng dng mch in Phn loi nhng dng mch in Thu thp thng tin t cc ti liu bi dng v tham kho.

. T VN 1. C u n A. Mt kin thc c bn Mt mch in c th gm nhiu on mch in. Mi on mch in gia hai im ca on mch in c th gm mt hay nhiu b phn, cc b phn c th mc ni tip hoc mc song song vi nhau.

1.1 nh u t m:

1.2 nh u t m i vi cc oi on mch a/ on mch ni tip:

Tnh ch t: ta c : I = I1 = I2. (1a)

U = U1 + U2. (2a)

R = R1 + R2. (3a)

2

1

2

1

R

R

U

U . (4a)

Ch : R

U

21

1

RR

R

R

U

21

2

RR

R

Chia U thnh U1 v U2 t l thun vi R1 v R2. 2

1

2

1

R

R

U

U .

- Nu R2 = 0 th theo (5a) ta th y : U2 = 0 v U1 = U. Do trn s (H.1). Hai im C v B: UCB = I.R2 = 0. Khi im C coi nh trng vi im B (hay im C v B c cng in th). - Nu R2 = (r t ln) U1 = 0 v U2 = U

R1 R2

A C B

H.1

(5a)

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b/ on mch mc song song:

Tnh ch t: ta c: U = U1 = U2 . (1b)

I = I1 + I2. (2b)

1

2

2

1

R

R

I

I . (3b)

21

111

RRRtd . (4b)

Ch : 21

2

211

21

11

11 .

)(

..

R RR

RI

RRR

RRIU

R

UI

21

1

212

21

22

22 .

)(

..

R RR

RI

RRR

RRIU

R

UI

(5b)

Chia I thnh I1 v I2 t l nghch vi R1 v R2 : 1

2

2

1

R

R

I

I

- Nu R2 = 0 th theo (5b) ta c: I1 = 0 v I2 = I. Do trn s (H.2). Hai im A v B c: UAB = 0. Khi hai im A v B c th coi l trng nhau (hay hai im A v B c cng in th). - Nu R2 = (r t ln) th ta c : I2 = 0 v I1 = I. (Khi R2 c in tr r t ln so vi R1 th kh nng cn tr dng in ca vt dn l r t ln. Do ta c th coi dng in khng qua R2.)

1.3 t im u - Trong mt mch in, cc im ni vi nhau bng dy ni (hoc ampe k) c in tr khng ng k c coi l trng nhau. Khi ta chp cc im li v v li mch tnh ton. - Trong cc bi ton, nu khng c ghi ch g c bit th ta c th coi:

RA 0 v RV . - Khi gii bi ton vi nhng s mch in mc hn hp tng i phc tp, nn tm cch a v mt s tng ng n gin hn. Trn s tng ng, nhng im c in th nh nhau (bng nhau) c gp li (chp li) lm r nhng b phn phc tp ca on mch c ghp li to thnh on mch n gin hn. B. ch in hn hp khng tng minh Mch in hn hp khng tng minh cng l mt loi mch in mc hn hp, song cch mc kh phc tp, khng n gin m phn tch cch mc cc b phn trong mch in c ngay. V vy, thc hin c k hoch gii, bt buc phi tm cch mc li a v mch in tng ng n gin hn. Nh rng, gia cc im ni vi nhau bng dy dn, ampe k... c in tr khng ng k l nhng im c cng in th, ta gp li (chp li). Khi v li mch in, ta s c mch in tng ng dng n gin hn.

B

R1

R2 I2

I1

I A

H.2

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- Phn tch cch mc cc b phn trong mch in l bc kh quan trng, n gip ta thc hin yu cu ca bi ton trnh c nhng sai st. Cui cng, ta p dng cc tnh ch t v h qu ca nh lut m i vi tng loi on mch ni tip v song song.

2. Th c t ng t c khi th c hin ti Trc khi thc hin ti, qua ging dy trng, qua tm hiu v trao i vi

ng nghip ti nhn th y a s hc sinh ham m hc b mn Vt l, nhng khi lm cc bi tp vt l cc em thng lng tng trong vic nh hng gii, c th ni hu nh cc em cha bit cch gii cng nh trnh by li gii. Theo ti, thc trng nu trn c th do mt s nguyn nhn sau: + Hc sinh cha c phng php tng quan gii mt bi tp Vt l, bi tp in ni ring. + Hc sinh cha bit vn dng lin kt cc kin thc + Ni dung c u trc chng trnh sch gio khoa mi hu nh khng dnh thi lng cho vic hng dn hc sinh gii bi tp hay luyn tp, dn n hc sinh khng c iu kin b sung, m rng v nng cao kin thc cng nh rn k nng gii bi tp in. 3. Cc bin php tin hnh h n 1: V i mch in

Cc em thng th y s nhng mch in phc tp v n kh rc ri v kh nhn ra tnh ch t song song ni tip ca cc in tr. Sau y l mt s bc chuyn i mch c bn cng nh phc tp.

V d: ho mch in Bc 1: t tn im.

t tn cho t t c nhng im c t 3 dy ni tr ln. Bc 2 : Chp mch v loi b mch. Xc nh t t c nhng im ni 2 u Ampe k, dy in, chp thnh 1, xc nh im u v im cui ca mch in. Theo mch in ta c: A, B l im u v im cui. A trng C ; D ; E trng F trng B. B dy ni nhng im ni Vn k n c ). Bc 3: Vit t t c cc im ln mt ng thng, ch nhng im trng nhau ch v thnh mt im, hai im mt hai bn l hai cc ca ngun in.

Bc 4: V in tr ln ng thng va v. Xt xem tng in tr hay nhiu in tr c ni gia 2 im gn nh t

no, v in tr v dy ni vo gia cc im. T im C n im D c mt in tr R2

A

R1

R3 R4 R2

A B

E

F D

A

R1

R3 R4 R2

A B

C

D E = F= B A = C

5

T im im D n im F c mt in tr R4

T im D n im E c mt in tr R3

T im C n im E c mt in tr R1

Bc 5: c mch in. Nh vy mch in AB gm c cc in tr c mc nh sau: [(R3//R4)ntR2]//R1 y l cch c bn v chnh xc nh t, cch ny kh lu nhng nu cc em cha nm vng cch chuyn i mch th nn s dng cch ny, khi quen ri th c th khng cn dng na, khng nn t u tp cch nhn mch bng mt s d dn n nhng sai lm khng ng c.

* Bi t p p dng Cho s mch in nh hnh v Hy v s tng ng tnh: a. RAB

b. RAC c. RBC

ii a. Tnh AB

- V mch in tnh RAB th ta xem im A, B l hai u mt. - Ta xc nh c cc im A; im D trng vi im C; im B.

+ T im A n im B c mt in tr R1 + T im A n im D c mt in tr R3

R2 D E = F= B A=C

R4 R2 D E = F= B A=C

R3

R4 R2 D E = F= B A=C

R3

R4 R2

R1

D E = F= B A=C

R1

R3 R4

R2 A B C

D

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+ T im D n im B c mt in tr R4 + T im C n im B c mt in tr R2

Nh vy mch in AB gm cc in tr: [(R2//R4)ntR3]//R1 b. Tnh AC

Tng t, tnh RAC ta xc nh c A v C l hai u mt.

Nh vy mch in AC gm cc in tr: [(R2//R4)ntR3]//R1

c. Tnh BC Ta xc nh c hai im B, C l hai u mt. S dng phng php tng t ta c mch in

Nh vy mch in BC gm cc in tr: [(R1ntR4)//R2]//R3

h n 2: Tnh gi t in t th m n u c u bi ton gii mt bi tp vt l ni chung v bi tp in hc ni ring v nh t l nhng bi ton ngc tm gi tr in tr th a mn yu cu bi ton nh th ny, hc sinh thng cm th y lng tng khng bit nn bt u t u, vn dng nhng cng thc v h thc no gii quyt. Do trong cc dng bi tp nh th ny ti a ra cc bc gip hc sinh c th d dng tip cn n cc yu cu ca bi ton.

R2

R4

R1

R3 C = D B A

R2

R4

R3

R1 B D=C A

R2

R4

R3

R1 A D=C B

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V d: Cho mch in nh hnh v:

Bit R1 = 30, R2 = 24, R3 = 90 UAB = 120V, I4 = 1,2A

in tr Ampe k v cc dy ni khng ng k. Tnh gi tr R4

ng dn gii: Bc 1: Hc sinh c k t 3 n 5 ln cho n khi nm c cc yu cu ca bi ton: cho g? Yu cu g? tm tt li bi ton. Bc 2: V li mch in a v mch in tng minh

(nu c).

Bc 3: Vit cc cng thc nh lut m ca on mch ni tip v song song. Cc cng thc ny c th gi l cng thc lin h gia cc i lng gtrong mch in. Lu vit t mch tng qut bn ngoi sau vit dn vo cc mch nhnh bn trong.

Theo mch in t c: [( 3//R4)ntR2]//R1

Xt on mch gm 3 in t 2, 3 v 4 t c:

V t ong on mch BD c h i in t 3 v 4 mc ong ong

Bc 4: Xy dng lp lun i t yu cu ca bi ton, s dng h thc ca nh lut m v cc h thc trong on mch ni tip v song song va phn tch bc 3 i n cc i lng cho. V d trong bi ny yu cu ca l I4 = 1,2A v cho l UAB = 120V. Ta s xy dng lp lun nh sau: Gi gi tr in tr cn tm l x

Ta c:

A

R1

R3 R4 R2

A B

Tm tt

R1 = 30

R2 = 24

R3 = 90 UAB = 120V

I4 = 1,2A

Tm R4 = ?

A

R3

R4 R2

R1

D B

8

=>

(*)

n y ta th y trong cng thc ch cn cha in tr v gi tr hiu in th UAB cho trc. Vit t t c cc in tr c trong cng thc di dng bin x, sau thay vo (*) gii v tm x

Th cc gi t vo (*) t c:

= > 114x + 2160 = 9000 => x = 60

Vy gi tr in tr R4 cn tm l: R4 = 60. Qua bi ny ta th y trong qu trnh xy dng cng thc i t yu cu ca bi ton cho n i lng cho c s an xen gia hai cng thc l nh lut m v cc cng thc ca mi lin h trong mch in. Ngha l nu bc ny chng ta dng h thc ca nh lut m th bc tip theo chng ta s s dng cng thc ca mi lin h. lm c iu ny vic vit cc cng thc trong on mch ni tip v song song l r t quan trng. N gip hc sinh c mt ci nhn tng qut v mch in v d dng chuyn i qua li gia cc i lng gii quyt v n .

h n 3: Bi t p n u n Bi 1. Cho mch in nh hnh v:

I = 0,5A

U4 = 2,4V

R1 = ?

9

Bi 2. Cho mch in nh hnh v:

R1 = 3r; R2 = 6r; R3 = 4r

R4 = 2r; R5 = r

a. Tnh RAB theo r.

b. Tnh cd qua cc in tr khi r = 3 ;

UAB = 18V

Coi rng i n tr ca cc dy ni kh ng ng

k

Bi 3. Cho mch in (nh hnh v) c:

R1 = R2 = R3 = 40 , R4 = 30 , ampe k ch

0,5A.

a.Tm cng dng in qua cc in tr, v

qua mch chnh.

b. Tnh UAB

c. Gi nguyn v tr cc in tr, hon v ampe k

v ngun in U, th ampe k ch bao nhiu? Trong bi ton ny, ampe k l tng.

Bi 4. Cho mch in nh hnh v:

R1= 40; R3=30 ; R4=40;

R5= 36 ; R6 = 90 ; hiu in th UAB lun

khng i

Khi ng kha K, dng in qua R3 c cng

I3 = 0,08A, hiu in th gia hai u in tr R4

khi l U4 =7,2V. Tnh:

a. Gi tr R2?

b. Hiu in th UAB?

c. S ch ca Ampe k khi kha K m?

3. t qu th c hin

T vic hng dn hc sinh gii mt bi tp in nu trn, trong nm hc 2013

2014 ti th y a s hc sinh vn dng mt cc linh hot vo vic gii bi tp, hc

sinh c kh nng t duy tt hn, c k nng vn dng kin thc vo gii bi tp tt

hn, linh hot hn.

C th thng qua kho st ch t lng hc sinh sau khi ng dn hc inh v

i mch in v tnh gi t in t th m n u c u bi ton v t p 9 ti

thu c kt qu nh sau:

t qu o nh i chng.

* Kt qu kho st trc khi thc hin ti .

_+U

R3

R2R1

R4

C

BA

A

R1 R2

R4

A

+ _

R3

A

B

R6

K

R5

10

ng

im t

9 10

im t

7 8

im t

5 7 im di 5

SL % SL % SL % SL %

25 0 0,0% 3 12,0% 10 40,0% 12 48,0%

* Kt qu kho st sau khi thc hin ti.

ng

im t

9 10

im t

7 8

im t

5 7 im di 5

SL % SL % SL % SL %

25 14 56,0% 8 32,0% 3 12,0% 0 0,0%

Qua so snh i chng kt qu ti th y t l im: Kh, Gi i tng, im yu

gim c th l:

C th: im gi i tng 56% ; im kh tng 20% ; im trung bnh v yu

gim 76 %.

. T L N

1. Bi hc kinh nghim:

Trong phn in hc vt l 9, kin thc v bi tp r t a dng. y, ti ch

a ra mt phm vi nh v bi tp trong ni dung bi dng hc sinh gi i c p huyn.

Qua vic i mi chng trnh vt l 9, ti th y rng hc sinh cn b ng vi phng

php hc, cha c k nng gii bi tp, cho d l nhng hc sinh gi i. V vy,

ngi gio vin cn phi h thng ha kin thc a ra phng php gii bi tp

cho hc sinh l iu cn thit. Hn na, theo chng trnh th r t t tit luyn tp, cn

phi tng cng cho hc sinh lm bi tp.

Vi chuyn ny,chng ti ch cp n bi tp trong phm vi nh , vi

nhng kin thc v bi tp c bn, phn dng bi tp v cch gii. Tuy ti ny ngn

gn, n gin nhng nu p dng c trong tnh hnh thc t, n s gip cho cc

hc sinh gi i c r t nhiu kin thc b ch khi cc em lm bi tp mch in, gp phn

nng cao ch t lng hc tp v yu thch mn hc ca hc sinh.Trong qu trnh ging

dy b mn Vt l trng THCS vic hnh thnh cho hc sinh phng php, k nng

gii bi tp Vt l l ht sc cn thit, t gip cc em o su, m rng nhng

kin thc c bn ca bi ging, vn dng tt kin thc vo thc t, pht trin nng lc

t duy cho cc em, gp phn nng cao ch t lng gio dc, c th l :

+ Gip hc sinh c thi quen phn tch u bi, hnh dung c cc hin tng

Vt l xy ra trong bi ton sau khi tm hng gii.

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+ Trong mt bi tp gio vin cn hng cho hc sinh nhiu cch gii (nu c

th ). kch thch s hng th, say m hc tp cho hc sinh rn thi quen tm ti li

gii hay cho mt bi ton in.

+ Khc su cho hc sinh nm chc cc kin thc b tr khc. C nh vy vic

gii bi tp in ca hc sinh mi thun li v hiu qu.

lm c iu ny:

- Gio vin cn t bi dng nng cao nghip v chuyn mn, thng xuyn

trao i, rt kinh nghim vi ng nghip.

- Nm vng chng trnh b mn ton c p hc.

- Gio vin cn hng dn hc sinh nghin cu k cc kin thc cn nh n

tp, nh li kin thc c bn, kin thc m rng, ln lt nghin cu k cc phng

php gii bi tp sau gii cc bi tp theo h thng t d n kh, so snh cc dng

bi tp khc su ni dung kin thc v cch gii. Trn c s hc sinh t hnh

thnh cho mnh k nng gii bi tp.

Trn dy l mt s kinh nghim m bn thn ti rt ra c t thc t qua

qu trnh ging dy t chn chng in hc, lp 9 trng THCS ni chung, cng

l kinh nghim rt ra c sau khi thc hin ti ny ni ring.

2. t u n chung:

Sau thi gian nghin cu tm hiu, c s quan tm gip ca ban gim

hiu nh trng cng nh t chuyn mn ti thc hin thnh cng vic: ip hc

sinh v i mch in v tnh gi t in t th m n u c u bi ton v t p

9 vi mong mun: pht trin nng lc t duy, rn luyn k nng, k xo cho hc sinh

trong vic bi dng b mn Vt l. Nhm nng cao ch t lng b mn ni ring, gp

phn nng cao ch t lng gio dc ni chung.

Tuy nhin v iu kin thi gian, cng nh tnh hnh thc t nhn thc ca hc

sinh a phng ni ti cng tc v nng lc c nhn c hn, nn vic thc hin

ti ny chc hn khng trnh kh i thiu st. Knh mong cc ng ch v cc bn ng

nghip, trao i v gp gip ti hon thin hn trong chuyn mn.

Cam Ha, ngy 28 thng 10 nm 2014

Ngi vit

Phan Ngc Linh

12

L

. T VN Trang 1

1. L do chn ti Trang 1

2. c ch nghin cu Trang 1

3. i tng v phm vi nghin cu Trang 2

4. Nhim v nghin cu Trang 2

5. hng php nghin cu Trang 2

. T VN Trang 2

1. u n Trang 2

2. Th c t ng t c khi th c hin ti Trang 4

3. Cc bin php tin hnh Trang 4

3. t qu th c hin Trang 9

. T L N Trang 10

1. Bi hc kinh nghim Trang 11

2. t u n chung Trang 11