7/31/2019 Dao ng in t

1/22

Chuyn : DAO NG IN T

A. TNG QUAN KIN THC

I.Kin thc p dng :

- Sut in ng xut hin trong cun dy :'Li

dt

diLe ==

- Hiu in th gia hai u t :C

qU =

- nh lut m cho on mch tng qut:

AB

ABAB

R

eui

+=

Trong ec th l sut in ng(e>0) hoc sut phn in(e

7/31/2019 Dao ng in t

2/22

Chuyn : DAO NG IN T

Trong hnh v ny ta phi xc nh c quan h gia dng in i qua tin v in tch t in. Nu dng in c chiu t bn dng sang bn mxuyn qua t in th 'qi += v ngc li th 'qi =

3) Vit biu thc nh lut Kic xp I cho cc nt v nh lut Kic sp II

cho cc mt mng :V d: Ti A : 2121 'i'i'iiii +=+= (1)

Mt mng A(L1)B(C)A v A(L2)B(C)A:

=

=

22

11

'

'

iLC

q

iLC

q

(2)

4)Bng cch kh dng in qua cc cun dy a v dng phngtrnh vi phn hng hai,thng phng trnh vi phn hng hai c dng :

+Nu thi H hoc HSG quc gia theo ch I thng l:

( )+==+ tsinQq0q"q 0 (3)+ Nu thi HSG quc gia tr ln theo ch II c dng h sau :

( )

( )

=+++

=+++

0qmqn"qm"qn

0qmqn"qm"qn

2212222212

2111212111

V cho nghim( )

( )

+=+

+=+

222212

112111

tsin.B"qm"qn

tsin.A"qm"qn(4)

T gii (4) ta s c phng trnh dao ng ca 1q v 2q c

th l 1 phng trnh iu ha hoc khng iu ha .5)T iu kn ban u ca bi ton : 0t = th ta c c )0(');0( qq hoc)0(');0(');0();0( 2121 qqqq ,suy ra c ;Q0 trong phng trnh (3) c

21;;B;A trong phng trnh (4). Sau da vo yu cu bi ton , tac th lun gii c li gii cho ph hp .

B. P DNGDAO NG IN T

I.BI TON TH DTHEO CH IBi 1: (Trch thi chn HSG quc gia THPT - nm 2005)Cho mch in c s nh hnh v. Hai t in 21 C;C ging nhau c cng

in dungC. T in 1C c tch in n hiu in th 0U , cun dy c t

- 2 -

7/31/2019 Dao ng in t

3/22

Chuyn : DAO NG IN T

cm L , cc kha 21 k;k ban u u m. in tr ca cun dy, ca cc dy niv ca cc kha l rt nh,nn c th coi dao ng in t trong mch l iuha.1.ng kha 1k ti thi im 0t= . Hy tm biu thc ph thuc thi gian tca

:a) Cng dng in chy qua cun dy .b) in tch 1q trn bn t ni vi A ca t 1C .

2.Gi 0T l chu k dao ng ca mch 1LC v 2q l intch ca bn t ni vi kha 2k ca t 2C . ng kha 2k thi im 01 Tt = . tm biu thc ph thuc thi gian tca cng dng in chy qua cun dyL v ca 2q .

HD

1. Gi s dng in chay trong mch nh hnh v.Ta c: 'qi = v "Lq'LiuAB ==Xt mt mng A(L)B(C1)A:

+=

=+=

tLC

1sinQq

0LC

q"q"Lq

C

q

0

Ti 0t = :

=

=

=

=

=

=

20cos

1

sin

0)0()0( 00

0

000

CUQ

LCQ

CUQ

iCUq

Vy:

+==

2t

LC

1sinCUqq 01 (1)

=

+== t

LC

1sin

L

CU

2t

LC

1cos

LC

1CU'qi 00 (2)

2.Theo cu 1: LC2

2T0 =

= (3)

- Ti 0Tt = th 00 CUQq == v 0i = ; ng kha 2k . Sau mt khong

7/31/2019 Dao ng in t

4/22

Chuyn : DAO NG IN T

+ Mt mng A(L)B(C1)A : 11 'Li

C

q= (1)

+ Mt mng A(C2)B(L)A : 22 'Li

C

q= (2)

+ Ti A : 21l21l 'i'i'iiii +=+= (3)- Thay (3) vo (1),(2) ta c :

+==

=

=+

=

=++

=++

=++

'

LC2

TsinQqq

qq

0LC2

q"q

qq

0LC

q"q"q

0LC

q"q"q

0LC

q"q"q

0221

21

11

21

121

221

121

vi 0TtT =

Lc ( )0Tt0T == th :( )

( )2

'

0i

2

CUQq

01

00101 =

=

==

- Vy

==

== 2

2sin

22

22

2

2sin

201

012

LC

t

L

CUii

LC

tCUqq L

Bi2: ( chuyn bi dng . . .V Thanh Khit)Cho mch dao ng nh hnh v. Ti thi im ban u kho K m v t inc in tch Q0, cn t kia khng tch in. Hi sau khi ng kho K th intch cc t in v cng dng in trong mch bin i theo thi gian nhth no? Hy gi nh mt c h tng ng nh mch dao ng trn. Coi C1

= C2 = C v L bit; B qua in tr thun ca mch.HD:

- Xt ti thi im t, gi s dng in c chiu v cct tch in nh hnh v.i = - q1/ = q2/ (1)

e = - Ldtdi = - Li/ (2)

+ q1 + q2 = Q0 (3)- Ap dng nh lut m :

C

q

C

q 21 - Li/ = 0

C

q12 + Lq1// -C

Q0 = 0

q1// + LC

Q

LC

q 01

2

= 0 (4)

- 4 -

7/31/2019 Dao ng in t

5/22

7/31/2019 Dao ng in t

6/22

Chuyn : DAO NG IN T

- Ap dng nh lut KicSp cho cc mt mng v nt:

+=

+=+=

)6(

)5(.

21

/

2

/

11

iii

LiLC

q

b

T (6) ta suy ra: i/ = i1/ + i2/ - q// = +bb CL

qCLq 21+

q// + )11

(1

21 LLCb+ q = 0

Hay q// +2121

21

)(

)(

LLCC

LL

++

q = 0 q = Q0.sin[2121

21

)(

)(

LLCC

LL

++

. t + ]

Ti t = 0

==

==

cos0

sin

0)0(

)0( 00101 QUC

i

UCq

==

2

010

UCQ

Vy q = C1U0.sin [2121

21

)( )( LLCC LL+ + .t + 2 ] (7)

i = - C1U02121

21

)(

)(

LLCC

LL

++

cos[2121

21

)(

)(

LLCC

LL

++

.t + 2 ]

= C1U02121

21

)(

)(

LLCC

LL

++

.sin(2121

21

)(

)(

LLCC

LL

++

.t) (8)

T (5) L1i1/ = L2i2/ L1i1 = L2i2 v i2 =2

1

L

L.i1 (9)

Thay vo (6) ta c:

i1 = 212

LL

L

+ i = C1U0).

)((sin.

))(( 2121

21

12121

2 tLLCC

LL

LCCLL

L

++

++

i2 = 211

LL

L

+ i = C1U0).

)((sin.

))(( 2121

21

22121

1 tLLCC

LL

LCCLL

L

++

++

Thay s ta c: i1 = 32

.10-3.sin105t (A) =3

2.sin105t (mA)

=

3

2sin(100000t) (mA)

i2 = 31

.sin(100000t) (mA)

Bi4 : (Trch : thi Olympic Vt l ti Lin

bang Nga nm 1987)Cho mch in nh hnh bn. Cc phn ttrong mch u l l tng .

a) ng kha K , tm Imax trong cun dy vU1max trn t in C1 .

b) Kho st s bin thin in tch ca t in khi ng kha K .

- 6 -

7/31/2019 Dao ng in t

7/22

Chuyn : DAO NG IN T

HD:+ Khi K m : cc t C1 v C2 c in tch :

1 201 021 2

C CQ Q E

C C= =

+

- Khi K ng :Gi s chiu ca cc dng in trong mch v

in tch ca cc bn t (hnh v)Ta c : 1 2Li i i= + (1)

'21

2

qLi

C= (2)

'1 1i q= (3)

'2 2i q= (4)

'1 2 1

1 2 1

Lq q q Li EC C C+ = + = (5)

T (5)

1 2

' '

1 2 22 1

1 2 1 2 1

0 0q q i i C

i iC C C C C

+ = + = = (6)

T (5)'

" "1 1

1 1

0 0L Lq i

Li LiC C

+ = + = (7)

T (6) v (1) suy ra : 2 11 1 11 1 2

L L

C Ci i i i i

C C C

= =

+Thay vo (7) c :

"

1 2

0( )

LL

ii

L C C+ =

+ (*) t2

1 2

1

( )L C C =

+Nghim phng trnh (*) l : 0 ( )L Li I Sin t = +

- Ti t=0 th 0 0Li = = '

0L Li I Cos t =

T (5) suy ra : EtcosLIC

qL0

1

1 =+

- Ti t=0 th 1 01q Q

= nn01 2

0 0

1 1 2

L L

Q ECLI E L I E

C C C + = + =

+

1 1

0

1 2 1 2( )L

E C ECI

L C C L C C = =

+ +

Ta c :1

max 0

1 2( )L

ECI I

L C C= =

+

Suy ra : 111 2( )

LECu E Cos t

L C C

=

+

- 7 -

7/31/2019 Dao ng in t

8/22

Chuyn : DAO NG IN T

11

1 2

1 1 21max

1 2 1 2

( )

(2 )

( )

ECu E Cos t

C C

EC E C CU E

C C C C

= +

+= + =

+ +

1

1 1 1 1

1 2

' 1

2 2 2

1 2

1 2

2

1 2

(1 )( )

( )L

Cq C u C E Cos t

C C

ECq LC i LC Cos t

L C C

C Cq E Cos t

C C

= = +

= =+

=+

Bi5: (Trch thi chn HSG quc gia THPT - nm2003)Trong mch in nh hnh v, t in c in dung l

C, hai cun dy L1 v L2 c t cm ln lt l L1=L,L2=2L; in tr ca cc cun dy v dy ni khngng k. thi im t=0 khng c dng qua cun dyL 2 , t in khng tch in cn dng qua cun dy L1l I1.

a) Tnh chu k ca dao ng in t trong mch.b) Lp biu thc ca cng dng in qua mi cun dy theo thigian.

HD:- Chn chiu dng in nh hnh vGi q l in tch bn t ni vi BTa c: 1 2Ci i i= + (1)

' '22 0CLi Li = (2)

'1

qLi

C= (3)

'Ci q= (4)o hm hai v ca (1) (2) v (3):

" " "C 1 2

" "1 2

" C1

i =i +i (1)

Li -2Li =0 (2)

iqLi =+ =- (3)

C C

"C C

3i - i2LC=

- 8 -

7/31/2019 Dao ng in t

9/22

Chuyn : DAO NG IN T

Chng t iC dao ng iu ha vi3

=2LC

3

LC22

2T =

=

+ 0 ( ) (5)Ci I Sin t = +T (2) cons)i2i(const)'Li2Li( 2121 ==

Ti t=0 th : 1 1 2 1 2 1i =I , i =0 i -2i =I (6)+ 1 2 C 0Ci +i =i =I Sin(t+ )Gii h c :

01

1

0 1

2

' 0

1

2Sin(t+ )

3 3

Ii Sin(t+ )-

3 3

2L C.Cos(t+ )

3

C

C

C

AB

IIi

I

Iqu Li

C

= +

=

= = =

Ti thi im t=0 : 1 1 2; 0; 0ABi I i u= = = .

Gii h c : 0 1; 2CI I= =

Vy : 1 112 3

3 3 2

I Ii Cos t

LC= +

1 123

3 2 3

I Ii Cos t

LC=

.II.BI TON LUYN TPTHEO CH I

Bi 6: Cho mch in c s nh hnh v. Hai t C1, C2 c in dung bngnhau: C1 = C2 = C ; cun dy thun cm c t cm L; ngun c sut inng E, b qua in tr dy ni v kho K. Ban u kho K cht a, sau ng sang cht b.1) Vit biu thc in tch ph thuc thi gian trn cct C1,C2 khi kho K ng sng cht b. Ly mc thigian l lckho K ng vo cht b.2) Tnh in lng chy qua tit din thng ca dydn sau mt chu k bin i ca in tch trn t C1.

Ap dng s: C = 0,5F ; L = 5mH ; E = 6V.

S: 1) q1 = ]1)10.22[cos(5,1]1).2

[cos(2

4 +=+ ttLC

CE

c

q2 = ]1)10.22[cos(5,1]1).2

[cos(2

4 = ttLC

CE c

2) i = q1/ = - tLCLC

CE.

2sin)

2(

2)

==

4

0

2)(4

T

CEdtiq = 6 c

- 9 -

7/31/2019 Dao ng in t

10/22

Chuyn : DAO NG IN T

Bi 7: Mt mch dao ng LC gm mt t in 1,0nF v mt cun cm 3,0mHc in p chnh bng 3,0V.

a) Hi in tch cc i trn t in.

b) Hi dng in cc i chy qua mch? Hi nng lng cc i c dtr trong t trng ca cun dy.p s:a)Qmax=3.10-9C

b)Imax= 3 10-3A;W= 4,5.10-9JBi8: Trong mch in nh HV:U=34V; R=14 ; C=6,2F ;L=54mH, o in v tr a trong mt thi gian

di. By gi n c gt sang v tr b.a) Hy tnh tn s ca dng dao ng.

b) Tnh bin ca dao ng dng in.

p s a) f=0,275kHzb)Ima x=0,364ABi9: Bn c a cho mt cun cm L=10mH v hai tC1= 5,0F vC2= 2,0F. Hy k ra cc tn s dao ng c th c bng cch ni cc yu t theocc t hp khc nhau.p s: (LC1) 712 Hz; (LC2) 1125Hz; (L,C1ntC2) 1331Hz; (L,C1song songC2)602HzBi 10:Mt cun cm c ni vo mt t in c in dung thay i c nhxoay mt nm. Ta mun lm cho tn s ca cc dao ng LC thay i tuyntnh vi gc quay ca nm, i t 2x105 n 4x105Hz khi nm quay 1 gc 1800.

Nu L = 1,0mH hy biu din bng th C nh mt hm s ca gc quay.

p s:f=.6,3662.104 2910.25,6

= C

( l gc quay ca nm xoay)Bi 11:Trong mt mch LC, L = 25,0mH v C = 7,80F thi im t = 0,dng bng 9,20mA, in tch trn t in bng 3,80F v t ang c np.

a) Hi nng lng tng cng trong mch bng bao nhiu?b) Hi in tch cc i trn t in?c) Hi dng cc i?d)Nu in tch trn t in c cho bi q = Qcos( +t ) th gc pha

bng bao nhiu?e) Gi s cc d kin vn nh vy, tr thi im t = 0 , t ang phng

in. Khi gc pha bng bao nhiu?p s:a)W=1,98 J

b)Q=5,56 Cc)I=12,6mA.d) 09,46=e) 09,46=

Bi12: Mt mch ni tip gm cun cm L1 v t in C1 dao ng vi tn sgc . Mt mch ni tip th hai , cha cun cm L2 v t C2, cng dao ng

- 10 -

7/31/2019 Dao ng in t

11/22

Chuyn : DAO NG IN T

vi cng tn s gc nh vy. Hi tn s gc ca dao ng(tnh theo ) camch ni tip cha c bn yu t ? B qua in tr c trong mch.(gi : dng cc cng thc cho in dung tngng v t cm tng ng).

p s:2211

1 11 CLCL ===

Bi 13: Trn HV t C1 =900F mi u c npn 100V v t in C2=100F khng c in tch.Hy m t chi tit lm th no np t in C2 n300V nh cc kho S1 v S2.Bit L=10H.Bi14: (Trch thi chn HSG QG nm 1992 1993)Mt mch dao ng gm 1 t in v 1 cun dy thun cm. Mch c ni

qua kho K vi mt b pin c sut in ng (E,r)(HV). K ng v dng in

n nh th ngi ta m kho K, trong mch LC c dao ng in vi chu kT. Bit rng hiu in th cc i gia hai bn t ln gp n ln sut in ngb pin. Hy tnh theo T v n in dung C ca t v t cm L ca cundy.HD:i vi bi ny mch LC dao ng iuho nn ch cn p dng nh lut bo ton nng

lng: C =rn

T

2v L =

2

Trn

Bi15: Cho mch in nh hnh v. Cc t inc cng gi tr in dung C,cc cun dy c cngh s t cm Lphn t trong mch u l tng.1) ng kho K, tm max)( Li trong cun dyv max)( 1cu trn t C12) Kho st s bin thin in tch ca cct in khi kho K ng.

S: 1) max)( Li = 06U

L

C.

max1 )(u = 03

4U .

2) q1 = CU0 - tLC

UC 1

cos3 0

q2 =q3= tLC

UC 1

cos.3 0

.

Bi 16:Mt t in c in dung C v hai cun dy thun cm c cc h s tcm L1 v L2 ( in tr khng ng k ) c mc thnhmt mach in c s nh hnh bn .

thi im ban u t in cha tch in v khngc dng in trong cun dy L2 nhng c dng in I0 trong

- 11 -

7/31/2019 Dao ng in t

12/22

Chuyn : DAO NG IN T

cun dy L1 . Hy tnh in tch cc i ca t in v cng cc i cadng in trong cun dy L2 .Bi 26: Cho mch dao ng gm t C v cun dy thun cm L 1= L .Ti thiim khi in tch ca t l Q v cng dng in qua cun dy l I th

ngi ta mc thm cun dy thun cm L 2= 2L song song vi cun L1.a) Tm qui lut bin thin in tch ca t.b) Khi q ma th dng in qua hai cun cm c chiu nh th no v c gi

tr bng bao nhiu ?

DAO NG IN T LIN KTI.BI TON TH DTHEO CH II

- 12 -

7/31/2019 Dao ng in t

13/22

Chuyn : DAO NG IN T

Bi 1:Hai t in c in dung CC;C2C 21 == , ban u mi ci c tch in n

hiu in th 0U , sau ghp ni tip vi nhau , bn m t 1C c ni vi bndng t 2C . Cng mt lc ngi ta ng c hai

kha 1k v 2k . Bit hai cun dy thun cm c t cm L2L;LL 21 == mc nh hnh v.

a)Tm dng in cc i qua mi cun cm .b)Hi sau bao nhiu lu t lc ng 2 kha ,

dng in qua cun cm t cc i .HD

a)Xt ti thi im t no ( 0t> ), gi s dngin trong mch c chiu nh hnh v . Khi ta c:

=

===

====

C

qu

C2

q'Lieu

'Li2eu

'qi;'qi

2MB

111AM

22AB

2213

- Xt mt mng :

A(L1)M(C1)A : 0'LiC2q 11 = (1)

A(L2)B(C2)M(C1)A : 0C2

q

C

q'Li2 122 = (2)

Ti M : 21231213 "q"q'i'i'iiii +==+= (3)Thay (3) vo (1),(2) ta c h theo q1 v q2 :

( )

( ) ( )

+=

+=+

=+

=+++

=++

=++

=++

=+

212

121

1212

2121

212

211

212

121

2sin.2

sin.

024

1'"2

01

""

024

"

024

3"

024

"

02

""

LC

tBqq

LC

tAqq

qqLC

qq

qqLC

qq

LC

q

LC

qq

LC

q

LC

qq

LC

q

LC

qq

LC

qqq

- Gi thit cho : 0t = th 0)0(';0)0(';)0(;2)0( 210201 ==== qqCUqCUq .Thay ttc iu kin ban u vo (4) ta c:

10 sin.3 ACU = (a)

- 13 -

(4)

7/31/2019 Dao ng in t

14/22

Chuyn : DAO NG IN T

2sin.0 B= (b)

1cos0 LC

A= (c)

2cos

2

0

LC

B= (d)

Gii h (a),(b),(c),(d) ta c : 0B;CU3A;2

01 ==

= thay vo (4) ta

c :

+=

+=

2sin

2sin2

02

01

LC

tCUq

LC

tCUq

- VyLCtsin

LCUqi 022 ==

( )LC

t

L

CUi

qqqiii sin

2

''' 01

121231 ====

b)Vy khi LC24

Tt

== th dng 21 i;i cc i.

Bi 2: (Trch d b thi Olympic VL Chu 2004)Cho mt mch in gm 2 t in, mi t c in

dung C, ni vi 3 cun cm, mt cun c t

cm L0, cn hai cun kia mi cun c t cm L(Hnh v bn ).Ban u trong cc on mch u khng c dng

in v cc t tch in nh sau: bn A1 mang intch Q1 = Q, bn B2 mang in tch Q2.ng kho K1 v K2 cng mt lc .

1. Hy vit biu thc cho cc cng dngin i1, i2 v i3 theo thi gian trong iu kin : Q1 = Q2 = Q.

2. Vi gi tr no ca Q2 i3 = 0 qua cun L0 mi thi im. Vit biu

thc i1, i2 khi .3. Vi gi tr ca Q2 nh th no ta lun c i1 = i2 = i3/2 .

Bi gii:- Gi q1, q2 l in tch ln lt trn cc bn A1 v B2 v dng in c chiu

nh hnh v ti thi im t:i1 = - q1/ (1)i2 = - q2/ (2)i1 + i2 = i3 (3)

1. Ap dng nh lut Kic Sp cho cc mt

mng.

- 14 -

7/31/2019 Dao ng in t

15/22

Chuyn : DAO NG IN T

+ Mt mng: (MA1NM) :C

q1 - Li1/ - L0i3/ = 0 (4)

(MB2NM) :C

q2 - Li2/ - L0i3/ = 0 (5)

+ Ly (4) tr (5) : (q1 q2 ) C1

+ L (i2/ - i1/) = 0

(q1// -q2//) + LC1

(q1 q2) = 0

q1 q2 = A.sin( 1.1

+tLC

) (6)

+ Ly (4) cng (5) : (q1 + q2)C

1- L(i1/ + i2/) 2L0i3/ = 0

Thay (1), (2) v (3) vo ta c: (q1 + q2) C1

+ L(q1// + q2//) + 2L0(q1// + q2//) =

0

(q1// + q2//) + )2(1

0LLC +.(q1 + q2) = 0

q1 + q2 = B.Sin( 20

.)2(

1+

+t

CLL ) (7)

T (6) v (7) - i1 + i2 = 1.1

cos(. +tLCLC

A) (8)

- i1 i2 = 200 )2(

cos(.)2(

+++ LLCt

LL

B) (9)

T (6) v (7) ta c:

q1 =2

ASin(

LC

t+ 1 ) + 2

BSin( )2( 0LLC

t

+ + 2 ) (10)

q2 = -2

A.Sin(

LC

t+ 1 ) +

2

BSin( )2( 0LLC

t

+ + 2 ) (11)

T (8) v (9) ta c:

i1 = -LC

A

2cos(

LC

t+ 1 ) - )2(2 0LLC

B

+ cos( )2( 0LLCt

+ + 2 ) (12)

i2 =LCA

2cos(

LCt + 1 ) - )2(2 0LLC

B+ cos( )2( 0LLC

t+ + 2

) (13)

Ap dng iu kin ban u: lc t = 0 th:

====

0)0(

0)0(

)0(

)0(

2

1

2

1

i

i

Qq

Qq

Thay vo (10), (11), (12), (13) ta c:

- 15 -

7/31/2019 Dao ng in t

16/22

Chuyn : DAO NG IN T

+=

+=

+=

+=

)(cos)2(2

_cos2

0

)(cos)2(2cos20

)(22

)(22

2

0

1

20

1

21

21

dLLC

B

LC

AcLLC

B

LC

A

bSinB

SinA

Q

aSinB

SinA

Q

T (a), (b) v (c), (d) ta c h:

+=

=+

=

=

)(cos)2(0

)(sin0

)(cos.)2(

0

)(sin.2

/

10

/1

/2

0

/2

dLLC

AcA

bLLC

BaBQ

T (a/) v (b/) ta c 22 = v B = 2Q

T (c/) v (d/) ta c A = 0Thay kt qu trn vo (12) v (13):

i1 = i2 = - 2)2(cos(.

)2( 00

+

++ LLCt

LLC

Q)

i3 = - 2)2(cos(.

)2(

2

00

+

++ LLCt

LLC

Q

2.a) Mun i3 = 0 vi mi t th:

i3 = i1 + i2 = - 200 )2(

cos(.)2(

+++ LLCt

LLC

B) = 0

Mun vy B = 0

+=

+=

+=

+=

)sin(2

)sin(2

)cos(.2

)cos(2

12

11

12

11

LC

tAq

LCtAq

LC

t

LC

Ai

LC

t

LC

Ai

Kt hp iu kin ban u:

- 16 -

7/31/2019 Dao ng in t

17/22

Chuyn : DAO NG IN T

====

0)0(

0)0(

)0(

)0(

2

1

2

1

i

i

Qq

Qq

==

=

=

1

1

12

11

cos2

0

cos20

2

2

LC

ALC

A

SinA

Q

SinA

Q

Q1 = - Q2 21

=

Vi Q1 = Q Q2 = - Q A = 2Q1

+=

+=

)2cos(

)2cos(

2

1

LC

t

LC

Qi

LC

t

LC

Qi

b) i1 = i2 =23i th :

-LC

A

2cos(

LC

t+ 1 )-

2

Bcos( )2( 0LLC

t

+ + 2 )=

LC

A

2cos(

LC

t+ 1 )-

2

Bcos(

)2( 0LLC

t

+ + 2 )

T A = 0 i1 = i2 = -2

Bcos(

)2( 0LLC

t

++ 2 )

Ap dng iu kin ban u :

====

0)0(

0)0(

)0(

)0(

2

1

2

1

i

i

Qq

Qq

+=

=

=

2

0

22

21

cos)2(2

0

sin.2

sin.2

LLC

B

BQ

BQ

2 = 2 v Q1 = Q2 ; B = 2Q1

Vi Q1 = Q. Vy khi i1 = i2 = -

)2)2(cos(

)2( 00

+++ LLCt

LLC

Q

Bi 3:(Trch thi chn i tuyn HS d thiOlympc Vt l quc t nm 2001).Gia hai im A v B c ba onn mch in

mc song song nh HV. Mi on mch u

c mt t in in dung C; c hai on mchcha cun cm c t cm L; tt c cc cun

- 17 -

7/31/2019 Dao ng in t

18/22

Chuyn : DAO NG IN T

cm v dy ni u c in tr thun bng khng. Hai cun cm t cch nhau c th b qua nh hng ca t trng ca cun cm ny ln cun cm kia.Trong mch c dao ng in.1. K hiu q1, q2 , q3 ln lt l in tch ca bn A1, A2, A3 ca t in; i1, i2, i3

ln lt l cng dng in i t cc bn A1, A2, A3 ca t in ti A (chiudng c chn l chiu ca mi tn trn hnh v).a) Vit phng trnh cho mi quan h gia cng dng in ik . (k = 1, 2, 3. ..)

b) Vit biu thc ca hiu in th uBA = VA VB theo cc d kin ca tngon mch BA1A, BA2A, BA3A.2) Tm biu thc cho s ph thuc vo thi gian ca cng dng in i2trong on mch khng cha cun cm.3) Chng t rng , cng dng in trong mi on mch c cha cun cml tng ca hai s hng bin i iu ho theo thi gian. Hy tnh cc tn s gc.4) Xt trng hp c bit khi i1(t) = i3(t) v i1(t) = - i3(t).

HD:

1)a. Theo hnh v ta c: i1 = -dt

dq1 (1) ; i2 = -dt

dq2 (2) ; i3 = -dt

dq3 (3)

b. Ta c:

uAB = VA VB =C

q1 - Ldt

di1 (4)

uAB = Cq2

(5)uAB =

C

q3 - L.i3/ (6)

2) Theo quy tc Kicxp, ti nt A ta c:i1 + i2 + i3 = 0 i2 = - i1 i3 (7)

(4) v (5) cho ta :C

q1 - Li1/ =C

q2 (8)

(5) v (6) cho ta :C

q3 - Li3/ =C

q2 (9)

(8) v (9) cho ta:C

qq 31 + - Ldt

iid )( 31 + = 2Cq2

Ch n (7) v h qu ca (7):Q2 = - q1 q3 + K ( K l hng s )

Ta c th bin i phng trnh ni trn thnh:

L i2/ = 3C

q2 +C

K

Ly o hm theo thi gian hai v ca phng trnh ny ta c phng trnhvi phn :

Li2// = - Ci23

- 18 -

7/31/2019 Dao ng in t

19/22

Chuyn : DAO NG IN T

i2// +LC

3i2 = 0 (10)

Chng t i2 bin i iu ho theo thi gian vi tn s gcLC

32 = (11);

Ngha l ta c : i2 = B.cos( 22 +t ) (12)3. Tr (8) v (9) v vi v ta c:

C

qq 31 - Ldt

iid )( 31 = 0 (13)

t i4 = i1 i3 (14) ta c : i4 = -dt

iid )( 31

Ly o hm (13) theo thi gian ta c phng trnh (vi phn) :

Li4// +C

i4 = 0 i4// + LC1

i4 = 0 (15)

Rt ra: i4 = A.cos( 11 +t ) (16)

Vi LC11 = (17)

T (7) v (14) ta thu c:

i1 = - (i2 i4) = 2A

cos( 11 +t ) - 2B

cos( 22 +t ) (18)

i3 = - (i2 + i4) = -2

Acos( 11 +t ) -

2

Bcos( 22 +t ) (19)

viLC

11 = ; LC

32 =

4. + Xt trng hp c bit th nht: i1(t) = i3(t) i1(t) = i3(t) =2

)(2 ti : Trong h ch c dao ng in t theo mt tn s gc

LC

32 = .

in tch ca cc t in tho mn cc h thc:q2 = -2q1 = - 2q3, khi c s i xng gia hai on mch c cun cm.+ Trng hp c bit th hai: i1(t) = - i3(t).Trong trng hp ny i2(t) = 0. Nh vy on mch khng cha cun cm

khng tham gia vo dao ng in. V khi , c th coi c h nh mtmch kn AA3BA1A (mch ny gm 2 cun cm ni tip 2C v hai t nitip vi in dung tng ng bng C/2), mch ny c dao ng in vi

tn s gcLC

11 = , v lun lun c q1 = - q3.

II.bi ton LUYN TP theo CH IIBi 4: Ba cun cm L ging nhau v hai t in C ging nhau c mc thnhmt mch c hai vng nh HV.a)Gi thit cc dng in nhHV. Hi dng trong cun dy gia? Vit cc phng trnh

- 19 -

7/31/2019 Dao ng in t

20/22

Chuyn : DAO NG IN T

mch vng v chng c tho mn nu dng in dao ng vi tn s gc

LC

1= .

b)By gi gi s cc dng nh HV. Hi dng trong cun dy gia? Vit

phng trnh cho cc mch vng v chng minh chng c tho mn nu dngin dao ng vi tn s gc

LC3

1= .

c)Do mch c th dao ng hai tn s khc nhau, chng minh rng khng ththay mch hai vng cho bng mt mch LC n vng tng ng.

Bi 5:Hai t in c in dung CC;C2C 21 == , hai cun

dy thun cm c t cm L2L;LL 21 == ,mtngun in(E,r) v hai kho K1,K2 mcphi hp nh

hnh v. Ban u kho K2 ngv K1 m. Cng mtlc ngi ta ng K1 v v m kho K2.a)Tm tn s dao ng ca mch.

b)Vit biu thc dng in qua mi cun cm v biuthc in tch trn mi t.

- 20 -

7/31/2019 Dao ng in t

21/22

Chuyn : DAO NG IN T

Phn III: KT LUNVi kinh nghim trnh by trong chuyn ny, chng ta c th p dng

kin thc v dng in mt cch thnh tho gii mt bi ton v mch dao

ng in t. i vi hc sinh luyn thi i hc hoc luyn thi HSG cp tnhch cn quan tm n cc bi tp theo ch I ( dng mch LC thng thng )trong L v C l cc gi tr tng ng cho nhiu phn t. i vi hc sinhluyn thi hc sinh gii cp quc gia bt buc phi thnh tho gii cc bi tontheo ch II ( mch dao ng lin kt ), y l bi ton c t hai mch vngtr ln nn khng th xem kiu mch LC thng thng c.

Trong nhng nm qua chng ti p dng c kinh nghim gii bi tonmch dao ng in t theo hai ch trn rt hiu qu,hc sinh tip nhn kinthc rt nhanh, to cho hc sinh k nng x l kiu mch dao ng in t

trong cc thi rt tt .c bit trong k thi HSG cp QG,i tuyn HSG vt lca chng ti gii tt bi 3 thi HSG cpQG(10/3/2005) ,gp phn t thnh tch cao trong k thi ny.

y l mt chuyn , c th lm ti liu tham kho tt cho gio vin gingdy vt l v hc sinh THPT. Da trn c s gio vin c th sng tc cc

bi tp hoc dng bi tp theo ch ca mnh .Chng ti hy vng rng, sau khi cc bn c xong ti liu ny s gip cho

cc bn c mt ci nhn thng sut v bi ton mch dao dng in, ng this khng gp kh khn khi gii mt bi ton mch dao ng LC.

- 21 -

7/31/2019 Dao ng in t

22/22

Chuyn : DAO NG IN T

TI LIU THAM KHO1. Cc bi ton vt l chn lc. Tc gi PGS-TS V Thanh Khit.2. Chuyn bi dng hc sinh gii vt l THPT. Tc gi PGS-TS

V Thanh Khit.3. Bi ton c s vt l. Tc gi Lng Duyn Bnh-Nguyn Quang

Hu.4. Bi tp vt l12. Tc gi Dng Trng Bi-V Thanh Khit.5. 3000 bi ton in. Tc gi T Quang Hng.6. Tuyn tp bi tp vt l nng cao .Tc gi PGS-TS V Thanh

Nguyn Th Khi.7. Tp ch vt l v tui tr .8. Mt s ti liu chuyn mn khc.