ONTAP_VATLY12_Daodongdientu

8/3/2019 ONTAP_VATLY12_Daodongdientu

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C L

CHNG: DAO NG V SNG IN TA. TM TT L THUYT

1. Dao ng in t.a. S bin thin in tch v dng in trong mch dao ng+ Mch dao ng l mt mch in kn gm mt t in c in dung C v

mt cun dy c t cm L, c in tr thun khng ng k ni vi nhau.+ in tch trn t in trong mch dao ng: q = q0 cos( t + ).

+ in p gia hai bn t in: u = C

q

= U0 cos( t + ). Vi Uo = C

q0

Nhn xt: in p gia hai bn t in CNG PHA vi in tch trn t in

+ Cng dng in trong cun dy: i = q' = - q0sin( t + ) = I0cos( t + +2

); vi I0 = q0 .

Nhn xt : Cng dng in NHANH PHA hn in tch trn t in gc2

+ H thc lin h : 1)()( 20

2

0

=+I

i

q

qHay: 1)()( 2

0

2

0

=+I

i

I

qHay: 1).()(

2

0

2

0

=+q

i

q

q

+ Tn s gc : =LC

1

+ Chu k v tn s ring ca mch dao ng: T = 2 LC v f =LC2

1

b. Nng lng in t trong mch dao ng

+ Nng lng in trng tp trung trong t in: WC =2

1

Cq2

=2

1 20

q

Ccos2( t + ).

+ Nng lng t trng tp trung trong cun cm: WL =2

1Li2 =

2

1L 2 q 20 sin

2( t + ) =2

1 20

q

Csin2( t +

).Nng lng in trng v nng lng t trng bin thin tun hon vi tn s gc:

= 2 ; f=2f v chu k T =2T .

+ Nng lng in t trong mch: W = WC + WL =2

1 20

q

Ccos2( t + ) +

2

1 20

q

Csin2( t + )

Hay : W=2

1 20

q

C=

2

1LI 20 = 2

1CU 20 = hng s.

+ Lin h gia q0, I0 v U0 trong mch dao ng: q0 = CU0 = 0I

= I0 LC .

L: t cm, n v henry(H) C:in dung n v l Fara (F) f:tn s n v l Hc (Hz)1mH = 10-3 H [mili (m) = 310 ] 1mF = 10-3 F [mili (m) = 310 ] 1KHz = 103 Hz [ kil = 310 ]

1 H = 10-6 H [micr( )= 610 ] 1 F = 10-6 F [micr( )= 610 ] 1MHz = 106 Hz [Mga(M) = 610 ]1nH = 10-9 H [nan (n) = 910 ] 1nF = 10-9 F [nan (n) = 910 ] 1GHz = 109 Hz [Giga(G) = 910 ]

1pF = 10-12 F [pic (p) = 1210 ]

2. in t trng.* Lin h gia in trng bin thin v t trng bin thin+Nu ti mt ni c mt t trng bin thin theo thi gian th ti ni xut hin mt in trng xoy.in trng xoy l in trng c cc ng sc l ng cong kn.

+ Nu ti mt ni c in trng bin thin theo thi gian th ti ni xut hin mt t trng. ng scca t trng lun khp kn.* in t trng :Mi bin thin theo thi gian ca t trng sinh ra trong khng gian xung quanh mt intrng xoy bin thin theo thi gian, ngc li mi bin thin theo thi gian ca in trng cng sinh ramt t trng bin thin theo thi gian trong khng gian xung quanh.

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in trng bin thin v t trng bin thin cng tn ti trong khng gian. Chng c th chuyn ha lnnhau trong mt trng thng nht c gi l in t trng.3. Sng in t - Thng tin lin lc bng v tuyn.

Sng in t l in t trng lan truyn trong khng gian.* c im ca sng in t+ Sng in t lan truyn c trong chn khng vi vn tc bng vn tc nh sng (c 3.108m/s). Sngin t lan truyn c trong cc in mi. Tc lan truyn ca sng in t trong cc in mi nh hntrong chn khng v ph thuc vo hng s in mi.

+ Sng in t l sng ngang. Trong qu trnh lan truyn

E v

B lun lun vung gc vi nhau v vung gcvi phng truyn sng. Ti mi im dao ng ca in trng v t trng lun cng pha vi nhau.+ Khi sng in t gp mt phn cch gia hai mi trng th n cng b phn x v khc x nh nh sng.

Ngoi ra cng c hin tng giao thoa, nhiu x... sng in t.+ Sng in t mang nng lng. Khi sng in t truyn n mt anten, lm cho cc electron t do tronganten dao ng .+Ngun pht sng in t rt a dng, nh tia la in, cu dao ng, ngt mch in, tri sm st ... .

* Thng tin lin lc bng sng v tuyn+ Sng v tuyn l cc sng in t dng trong v tuyn, c bc sng t vi m n vi km. Theo bc sng,ngi ta chia sng v tuyn thnh cc loi: sng cc ngn, sng ngn, sng trung v sng di.+ Tng in li l lp kh quyn b ion ha mnh bi nh sng Mt Tri v nm trong khong cao t 80 kmm 800 km, c nh hng rt ln n s truyn sng v tuyn in.+ Cc phn t khng kh trong kh quyn hp th rt mnh cc sng di, sng trung v sng cc ngn nhngt hp th cc vng sng ngn. Cc sng ngn phn x tt trn tng in li v mt t.+ Nguyn tc chung ca thng tin lin lc bng sng v tuyn in:

- Bin iu sng mang:*Binm thanh (hoc hnh nh) mun truyn i thnh cc dao ng in t c tn s thp gi l tn hiu

m tn (hoc tn hiu th tn).*Trn sng: Dng sng in t tn s cao (cao tn) mang (sng mang) cc tn hiu m tn hoc th

tn i xa . Mun vy phi trn sng in t m tn hoc th tn vi sng in t cao tn (bin iu). Qua antenpht, sng in t cao tn bin iu c truyn i trong khng gian.

-Thu sng : Dng my thu vi anten thu chn v thu ly sng in t cao tn mun thu.-Tch sng: Tch tn hiu ra khi sng cao tn (tch sng) ri dng loa nghe m thanh truyn ti hocdng mn hnh xem hnh nh.

-Khuch i: tng cng ca sng truyn i v tng cng ca tn hiu thu c ngi ta dngcc mch khuch i.*. S khi ca mt my pht thanh v tuyn n gin

ng ten pht: l khung dao ng h (cc vng dy ca cun L hoc 2 bn t C xa nhau), c cun dy mc xen gncun dy ca my pht. Nh cm ng, bc x sng in t cng tn s my pht s pht ra ngoi khng gian.

*. S khi ca mt my thu thanh n gin

ng ten thu: l 1 khung dao ng h, n thu c nhiu sng, c t C thay i. Nh s cng hng vi tn s sngcn thu ta thu c sng in t c f = f0

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2

1

3 4 5

1 2 3 4

5

1.Micr2.Mch pht sng in t cao tn.3.Mch bin iu.4.Mch khuch i.5.Anten pht

1.Anten thu2.Mch khuch i dao ng in t cao tn.3.Mch tch sng.4.Mch khuch i dao ng in t m tn .5.Loa


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4.S tng t gia dao ng c v dao ng in

i lng c i lng in Dao ng c Dao ng inTa x in tch q x + 2x = 0 q + 2q = 0Vn tc v cng dng in i

k

m=

1

LC=

Khi lng m t cm L x = Acos( t + ) q = q0cos( t + ) cng k nghch o in dung

1

C

v = x = - Asin( t + )v = Acos( t + + /2)

i = q = - q0sin( t + )i = q0sos( t + + /2 )

Lc F hiu in th u 2 2 2( )v

A x

= + 2 2 20 ( )i

q q

= +

H s ma st intrR F = -kx = -m 2x 2qu L qC

= =

ng nng W NL t trng (WL) W =1

2mv2 WL =

1

2Li2

Th nng Wt NL in trng (WC) Wt =1

2kx2 WC =

2

2

q

C

B. CC DNG BI TPDNG 1: Xc nh cc i lng :T, f, , bc sng m my thu sng thu c.a. Cc cng thc:

-Chu k, tn s, tn s gc ca mch dao ng: 2T LC= ;1

2f

LC= ; =

LC

1.

-Bc sng in t: trong chn khng: =c

= cT = c2 LC (c = 3.108 m/s)

trong mi trng: = f

v

= nf

c

. (c 3.108

m/s).My pht hoc my thu sng in t s dng mch dao ng LC th tn s sng in t pht hoc thu c bng tn s

ring ca mch.Mch chn sng ca my thu v tuyn thu c sng in t c bc sng: =f

c= 2 c LC .

-Nu mch chn sng c c L v C bin i th bc sng m my thu v tuyn thu c s thay i trong gii hn t:

min = 2 c minminCL n max = 2 c maxmaxCL .

b. Bi tp trc nghim:Cu 1:Mch dao ng in t gm t C = 16nF v cun cm L = 25mH. Tn s gc dao ng l:A. = 200Hz. B. = 200rad/s. C. = 5.10-5Hz. D. = 5.104rad/s.

Gii: ChnD.Hng dn: T Cng thcLC

1= , vi C = 16nF = 16.10-9F v L = 25mH = 25.10-3H. Suy ra .

Cu 2:Mch dao ng LC gm cun cm c t cm L = 2mH v t in c in dung C = 2pF, (ly 2 =10). Tn s dao ng ca mch lA. f = 2,5Hz. B. f = 2,5MHz. C. f = 1Hz. D. f = 1MHz.

Gii:ChnB. p dng cng thc tnh tn s dao ng ca mchLC2

1f

= , thay L = 2.10-3H, C = 2.10-12F

v 2 = 10, ta c f = 2,5.106H = 2,5MHz.Cu 3: Mt mch dao ng gm mt t in c in dung C v mt cun cm c t cm L . Mch daong c tn s ring 100kHz v t in c c= 5.nF. t cm L ca mch l :A. 5.10-5H. B. 5.10-4H. C. 5.10-3H. D. 2.10-4H.

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Gii:ChnB.Hng dn: 2 2 21 1

4L

C f C = = hoc dng lnh SOLVE ca my tnh Fx 570ES, vi n s

L l bin X : Dng biu thc1

2f

LC= Nhp cc s liu vo my tnh :

5

9

110

2 5.10 =

Xx.

Sau nhn SHIFT CALC ( Lnh SOLVE) v nhn du = hin th kt qu ca L: X = 5.066059.10-4 (H)Ch :Nhp bin X l phm: ALPHA) : mn hnh xut hin X

Nhp du= lphm : ALPHA CALC:mn hnh xut hin =Chc nng SOLVE: SHIFT CALC v sau nhn phm = hin th kt quX = .....

Cu 4:Mt mch dao ng LC c t C=10 4/ F, tn s ca mch l 500Hz th cun cm phi c tcm l:A. L = 102/ H B. L = 10 2/ H C. L = 10 4/ H D. L = 10 4/ H

Cu 5:Mt mch dao ng LC vi cun cm L = 1/ mH, mch c tn s dao ng l 5kHz th t inphi c in dung l:

A. C = 10 5/ F B. C = 10 5/ F. C = 10 5/ 2 F D. C = 10 5/ FCu 6:Trong mch dao ng LC, khi hot ng th in tch cc i ca t l Q 0=1 C v cng dngin cc i cun dy l I0=10A. Tn s dao ng ca mch l:A. 1,6 MHz B. 16 MHz C. 1,6 kHz D. 16 kHz

Cu 7:Mch chn sng ca my thu v tuyn in gm t C = 880pF v cun L = 20 H. Bc sng int m mch thu c lA. = 100m. B. = 150m. C. = 250m. D. = 500m.ChnC.Hng dn: Bc sng in t m mch chn sng thu c l LC.10.3.2 8= = 250m.Cu 8:Mch chn sng ca my thu v tuyn in gm t C = 1nF v cun L = 100 H (ly ).102 =Bc sng m mch thu c.A. 300= m. B. 600= m. C. 300= km. D. 1000= mCu 9:Sng in t trong chn khng c tn s f = 150kHz, bc sng ca sng in t:A. =2000m. B. =2000km. C. =1000m. D. =1000km.ChnA.Hng dn: p dng cng thc tnh bc sng m2000

.c

4===

Cu 10:Mch dao ng ca my thu v tuyn c cun L=25 H. thu c sng v tuyn c bc sng100m th in dung C c gi tr

A. 112,6pF. B. 1,126nF. C. 1,126.10-10F D. 1,126pF.

ChnA.Hng dn: 0 2cT c LC = = . Suy ra:2

2 24C

c L

=

Cu 11:Sng FM ca i H Ni c bc sng =10

3m. Tm tn s f.

A . 90 MHz ; B. 100 MHz ; C. 80 MHz ; D. 60 MHz .

ChnA.Hng dn:c

f= .Suy ra

cf

=

c. Bi tp t lun :Cu 12:Mt mch dao ng in t LC gm cun dy thun cm c t cm L = 2 mH v t in c indung C = 0,2 F. Bit dy dn c in tr thun khng ng k v trong mch c dao ng in t ring.Xc nh chu k, tn s ring ca mch.

Gii: Ta c: T = 2 LC= 4 .10-5 = 12,57.10-5 s; f =T

1= 8.103 Hz.

Cu 13: Mch dao ng ca mt my thu thanh vi cun dy c t cm L = 5.10 -6 H, t in c in dung

2.10-8 F; in tr thun R = 0. Hy cho bit my thu c sng in t c bc sng bng bao nhiu?Gii: Ta c: = 2 c LC= 600 m.Cu 14:Mch chn sng ca mt my thu v tuyn in gm mt cun dy c t cm L = 4 H v mtt in C = 40 nF.

a) Tnh bc sng in t m mch thu c.Trang 4


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b) mch bt c sng c bc sng trong khong t 60 m n 600 m th cn phi thay t in C bngt xoay CV c in dung bin thin trong khong no? Ly 2 = 10; c = 3.108 m/s.Gii: a) Ta c: = 2 c LC= 754 m.

b) Ta c: C1 =Lc22

21

4

= 0,25.10-9 F; C2 =

Lc22

22

4

= 25.10-9 F;

vy phi s dng t xoay CV c in dung bin thin t 0,25 pF n 25 pF.Cu 15: Cho mt mch dao ng in t LC ang dao ng t do, t cm L = 1 mH. Ngi ta o cin p cc i gia hai bn t l 10 V, cng dng in cc i trong mch l 1 mA. Tm bc sng in

t m mch ny cng hng.

Gii: . Ta c:2

1CU 20 = 2

1LI 20 C = 2

0

20

U

LI; = 2 c LC= 2 c

0

0

U

LI= 60 = 188,5m.

Cu 16: Mch chn sng ca mt my thu thanh gm mt cun dy c t cm L = 2.10 -6 H, t in cin dung C thay i c, in tr thun R = 0. my thu thanh thu c cc sng in t c bc sng t57 m (coi bng 18 m) n 753 m (coi bng 240 m) th t in phi c in dung thay i trong khongno? Cho c = 3.108 m/s.

Gii: Ta c: C1 =Lc22

21

4

= 4,5.10-10 F; C2 =

Lc22

22

4

= 800.10-10 F.

Vy C bin thin t 4,5.10-10

F n 800.10-10

F.

DNG 2: Vit biu thc in tch q , n p u, dng in ia. Kin thc cn nh:

* in tch tc thi q = q0cos( t + q) Vi :LC

1= :l tn s gc ring

Khi t = 0 nu q ang tng (t in ang tch in) th q < 0;nu q ang gim (t in ang phng in) th q > 0.

* Hiu in th (in p) tc thi 0 0os( ) os( )= = + = +q uqq

u c t U c t C C

Ta thy u = q.

Khi t = 0 nu u ang tng th u < 0; nu u ang gim th u > 0.

* Dng in tc thi i = q = - q0sin( t + ) = I0cos( t + +2

) . Vi : I0 = q0

Khi t = 0 nu i ang tng th i < 0; nu i ang gim th i > 0.

* Cc h thc lin h :0

0 0

qI q

LC= = ; 0 0

0 0 0

q I LU LI I

C C C

= = = =

+ Khi t phng in th q v u gim v ngc li+ Quy c: q > 0 ng vi bn t ta xt tch in dng th i > 0 ng vi dng in n bn t ta xt.

b Phng php gii :

+ tm cc i lng c trng trn mch dao ng in t LC ta vit biu thc lin quan n cc i lng bit v i lng cn tm t suy ra v tnh i lng cn tm.+ vit biu thc ca q, i hoc u ta tm tn s gc , gi tr cc i v pha ban u ca i lng cn vit

biu thc ri thay vo biu thc tng ng ca chng.c. Bi tp t lun:

Bi1: Mt mch dao ng gm t in c in dung C = 25 nF v cun dy thun cm c t cm L = 4mH. Gi s thi im ban u cng dng in t gi tr cc i v bng 40 mA. Tm biu thc cng dng in, biu thc in tch trn cc bn t in v biu thc in p gia hai bn t.

Gii: Ta c: =LC

1= 105 rad/s; i = I0cos( t + ); khi t = 0 th i = I0 cos = 1 = 0.

Vy i = 4.10-2cos105t (A). q0 =

0I = 4.10-7 C; q = 4.10-7cos(105t - 2 )(C). u = Cq = 16.cos(10

5t - 2 )(V).

Bi2:.Cho mch dao ngl tng vi C = 1 nF, L = 1 mH, in p hiu dng ca t in l UC = 4 V. Lc t= 0, uC = 2 2 V v t in ang c np in. Vit biu thc in p trn t in v cng dng inchy trong mch dao ng.

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Gii: Ta c: =LC

1= 106 rad/s; U0 = U 2 = 4 2 V; cos =

0U

u=

2

1= cos(

3

); v t ang np in

nn = -3

rad. Vy: u = 4 2 cos(106t -

3

)(V).

I0 =L

CU0 = 4 2 .10-3 A; i = I0cos(106t -

3

+

2

) = 4 2 .10-3 cos(106t +

6

)(A).

Bi3: Mch dao ng kn, l tng c L = 1 mH, C = 10 F. Khi dao ng cng dng in hiu dngI = 1 mA. Chn gc thi gian lc nng lng in trng bng 3 ln nng lng t trng v t in angphng in. Vit biu thc in tch trn t in, in p gia hai bn t v cng dng in trn mch daong.

Gii: . Ta c: =LC

1= 104 rad/s; I0 = I 2 = 2 .10-3 A; q0 =

0I = 2 .10-7 C.

Khi t = 0 th WC = 3WtW =3

4WC q =

2

3q0 cos

0q

q= cos(

6

).V t ang phng in nn =

6

.

Vy: q = 2 .10-7cos(104t +6 )(C); u =

Cq = 2 .10-2cos(104t +

6 )(V); i = 2 .10-3cos(104t +

23 )(A).

Bi 4: Cng dng in tc thi trong mt mch dao ng LC l tng l i = 0,08cos(2000t)A. Cun dyc t cm l L = 50mH. Hy tnh in dung ca t in. Xc nh hiu in th gia hai bn t in tithi im cng dng in tc thi trong mch bng gi tr cng dng in hiu dng.

Gii: in dung ca t in

T cng thc tnh tn s goc:LC

1= , suy ra: F10.5

2000.10.50

1

L

1C 6

232

==

= hay C = 5 F.

Hiu in th tc thi:

T cng thc nng lng in t: 2022 LI21Cu

21Li

21 =+ , vi

2IIi 0== , suy ra

.66,52410.5.2

10.5008,0

2 6

3

0 VVC

LIu ====

DNG 3: Bi ton v nng lng in t trong mch dao ng LC.a. Cc cng thc:

Nng lng in trng: WC =21

Cu2 =21

C

q2. Nng lng t trng: WL =

21

Li2 .

Nng lng in t: W= WC + WL= 21

C

q 20 = 2

1CU

2

0 = 21

LI2

0

Nng lng in trng v nng lng t trng bin thin tun hon vi tn s gc :

= 2 =LC

2, vi chu k T =

2

T= LC .

Nu mch c in tr thun R 0 th dao ng s tt dn. duy tr dao ng cn cung cp cho mch mt

nng lng c cng sut: P = I2R =L

RCURUC

22

20

20

22

=

.

Lin h gia q0, U0, I0: q0 = CU0 = 0I

= I0 LC .

b. Phng php gii : tm cc i lng lin quan n nng lng in t trn mch dao ng in t LC ta vit biu thc linquan n cc i lng bit v i lng cn tm t suy ra v tnh i lng cn tm.c. Bi tp t lun:

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Bi1. Cho mt mch dao ng in t gm mt t in c in dung C = 5 F v mt cun thun cm c t cm L = 50 mH. Bit in p cc i trn t l 6 V. Tm nng lng in trng v nng lng t trngtrong mch khi in p trn t in l 4 V v cng dng in i khi .

Gii :Ta c:W =2

1CU

2

0 = 9.10-5 J; WC = 2

1Cu2 = 4.10-5 J; Wt = W WC = 5.10-5 J; i =

L

Wt

2= 0,045A.

Bi2. Trong mt mch LC, L = 25 mH v C = 1,6 F thi im t = 0, cng dng in trong mchbng 6,93 mA, in tch trn t in bng 0,8 C. Tnh nng lng ca mch dao ng.

Gii Bi2. Ta c: W = 2

1

C

q2

+ 2

1

Li2

= 0,8.10-6

J.Bi3. Mt mch dao ng in t gm mt t in c in dung 0,125 F v mt cun cm c t cm 50 H. in tr thun ca mch khng ng k. in p cc i gia hai bn t in l 3 V. Tnh cng dng in cc i, cng dng in, nng lng in trng, nng lng t trng trong mch lc in pgia hai bn t l 2 V.

Gii Bi3. Ta c: I0 =L

CU0 = 0,15 A; W =

2

1CU 20 = 0,5625.10

-6 J; WC =2

1Cu2 = 0,25.10-6 J;

Wt = W WC = 0,3125.10-6 J; i = L

Wt2 = 0,11 A.

Bi4. Mt mch dao ng gm cun cm c t cm 27 H, v t in c in dung 3000 pF; in trthun ca cun dy v dy ni l 1 ; in p cc i gia hai bn t in l 5 V. Tnh cng sut cn cungcp duy tr dao ng ca mch trong mt thi gian di.

Gii Bi4. Ta c: I0 = q0 = CU0 = U0L

C= 57,7.10-3 A ; P =

2

20RI = 1,39.10-6 W.

Bi5. Mt mch dao ng in t LC l tng gm cun cm thun c t cm 5 H v t in c indung 5 F. Trong mch c dao ng in t t do. Tnh khong thi gian gia hai ln lin tip m in tchtrn mt bn t in c ln cc i v khong thi gian gia hai ln lin tip m nng lng in trng

bng nng lng t trng.

Gii Bi5. Chu k dao ng: T = 2 LC= 10 .10-6 = 31,4.10-6 s.Trong mt chu k c 2 ln in tch trn bn t t gi tr cc i nn khong thi gian gia hai ln lin tip

m in tch trn bn t t cc i l t =2

T= 5 .10-6 = 15,7.10-6s.

Trong mt chu k c 4 ln nng lng in trng bng nng lng t trng nn khong thi gian gia

hai ln lin tip m nng lng in trng bng nng lng t trng l t =4

T= 2,5 .10-6 = 7,85.10-6 s.

Bi6. Cng dng in tc thi trong mt mch dao ng LC l tng l i = 0,08cos2000t (A). Cun dyc t cm L = 50 mH. Hy tnh in dung ca t in. Xc nh in p gia hai bn t in ti thi imcng dng in tc thi trong mch bng gi tr cng dng in hiu dng.

Gii Bi6. Ta c: C =L2

1

= 5.10-6 F; W =21 LI 20 = 1,6.10

-4 J; Wt =21 LI2 =

21 L

2

20I = 0,8.10-4 J;

WC = W Wt = 0,8.10-4 J; u =C

WC2 = 4 2 V.

Bi7. Mt mch dao ng LC l tng ang c dao ng in t t do. Bit in tch cc i ca mt bn tin c ln l 10-8 C v cng dng in cc i qua cun cm thun l 62,8 mA. Tnh tn s dao ngin t t do ca mch.

Gii Bi7. Ta c: I0 = q0 =0

0

q

I= 6,28.106 rad/s f =

2= 106 Hz.

Bi8. Khung dao ng in t gm mt cun dy thun cm c t cm L = 0,1 H v t in c in dungC = 10 F. Dao ng in t trong khung l dao ng iu ho vi cng dng in cc i I0 = 0,05 A.Tnh in p gia hai bn t thi im i = 0,03 A v cng dng in trong mch lc in tch trn t cgi tr q = 30 C.

Trang 7


8/34

Gii Bi8. Ta c: W =2

1LI 20 = 1,25.10

-4 J; Wt =2

1Li2= 0,45.10-4J; WC = W - Wt = 0,8.10-4J; u =

C

WC2 = 4V.

WC =2

1

C

q2= 0,45.10-4J; Wt = W - Wt = 0,8.10-4J; i =

L

Wt2 = 0,04 A.

Bi9. Mch dao ng l tng gm cun dy c t cm L = 0,2H v t in c in dung C = 20F.Ngi ta tch in cho t in n hiu in th cc i U0 = 4V. Chn thi im ban u (t = 0) l lc t inbt u phng in. Vit biu thc tc thi ca in tch q trn bn t in m thi im ban u n tchin dng. Tnh nng lng in trng ti thi im t=T/8, T l chu k dao ng.

*Hng dn gii: in tch tc thi

Trong : ;

Khi t = 0:Vy biu thc tc thi ca in tch q cn tm: q = 8.10-5cos500t (C)

Nng lng in trng

Vo thi im , in tch ca t in bng , thay vo ta tnh c nng lng

in trng

Bi10. Mch dao ng LC l tng thc hin dao ng in t.Hy xc nh khong thi gian, gia hai ln lin tip, nng lng

in trng trn t in bng nng lng t trng trong cun dy.*Hng dn gii: Khi nng lng in trng trn t bng nng

lng t trng trong cun dy, ta c: hay

Vi hai v tr gi tr ca q: trn trc Oq, tng ng vi 4 v tr trn ng trn, cc v tr nycch u nhau bi cc cung /2. C ngha l, sau hai ln lin tip WC = WL, pha dao ng bin thin c

mt lng l: (Pha dao ng bin thin c 2 sau thi gian mt chu k T)Tm li, c sau thi gian T/4 nng lng in li bng nnglng t.

Nhn xt: Ngoi cch trn ta cng c th gii phng trnhlng gic tm t.

Bi11. Biu thc in tch ca t trong mt mch dao ng cdng q = Q0sin(2.106t)(C). Xc nh thi im nng lng t bngnng lng in u tin.

*Hng dn gii: Phng trnh in tch

Trang 8


9/34

v coi q nh li ca mt vt dao ng iu ha. Ban u, pha dao ng bng , vt qua v tr cn bng

theo chiu dng. WC = WL ln u tin khi , vect quay ch v tr cung , tc l n qut

c mt gc tng ng vi thi gian . Vy thi im cn xc nh l t = =

Bi12.( thi i hc 2003): Trong mch dao ng (h.v) b t in gm 2 t C1 ging nhau c cp nnglng W0 = 10-6J t ngun in mt chiu c sut in ng E = 4V. Chuyn Kt (1) sang (2). C sau nhng khong thi gian nh nhau: T1= 10-6s th nnglng in trng trong t in v nng lng t trng trong cun cm bngnhau.a. Xc nh cng dng in cc i trong cun dy.

b. ng K1 vo lc cng dng in cun dy t cc i. Tnh li hiu inth cc i trn cun dy.

*Hng dn gii:

Theo bi 11 trn ta c thi gian nng lng in trng v nng lng t trng bng nhau l

;Do C1 ni tip C2 v C1 = C2 nn C1 = C2 = 2C = 0,25.10-6F

a. T cng thc nng lng:b. Khi ng k1, nng lng trn cc t in bng khng, t C1 b loi khi h dao ng nhng nng lngkhng b C1 mang theo, tc l nng lng in t khng i v bng W0.

DNG 4:Sng in t - Lin lc bng thng tin v tuyn Mch chn sng vi b t inc cc t in ghp.1. Kin thc lin quan: Sng in t l qu trnh lan truyn trong khng gian ca in t trng bin thin

theo thi gian.Sng in t l sng ngang, lan truyn trong chn khng vi vn tc bng vn tc nh sng (c =3.108 m/s). Cc loi sng v tuyn:

Tn sng Bc sng Tn s fSng di Trn 3000 m Di 0,1 MHzSng trung 3000 m 200 m 0,1 MHz 1,5 MHzSng ngn 200 m 10 m 1,5 MHz 30 MHzSng cc ngn 10 m 0,01 m 30 MHz 30000 MHz

Trong thng tin lin lc bng v tuyn pht sng in t i xa ngi ta phi trn sng m tn hoc th tnvi sng mang cao tn (gi l bin iu). C th bin iu bin , tn s hoc pha ca dao ng cao tn: lm

cho bin , tn s hoc pha ca dao ng cao tn bin thin theo tn s ca dao ng m tn hoc th tn.-Bc sng in t: trong chn khng: =

c= cT = c2 LC (c = 3.108 m/s)

trong mi trng: =f

v=

nf

c. (c 3.108m/s).

Trang 9


10/34

-Nu mch chn sng c c L v C bin i th bc sng m my thu v tuyn thu c s thay i trong gii hn t:

min = 2 c minminCL n max = 2 c maxmaxCL .

-B t mc song song:C = C1 + C2 + + Cn.Xt 2 t mc song song::

+Chu k: )(2 21 CCLTSS += + Lin h gia cc chu k: 2 2 21 2=+SST T T

+Tn s:1 2

1

2 ( )=

+SSf

L C C ; + Lin h gia cc tn s: 2 2 2

1 2

1 1 1= +

SS f f f

+Tn s gc: )(1

21 CCLSS

+

=

-B t mc ni tip : ...111

21

++=CCC

+nC

1. Xt 2 t mc ni tip :

+Chu k:)(

.221

21

CC

CCLTNT

+= Hay )11(1

2

21 CCL

TNT

+

=

; + Lin h gia cc chu k: 2 2 21 2

1 1 1= +

NTT T T

+Tn s:1 2

1 1 1 1( )

2= +NTf

L C C Hay 1 2

1 2

( )1

2 . .

+=NTC C

fLC C

+ Lin h gia cc tn s: 2 2 21 2= +NT f f f

+Tn s gc :21

21

..)(

CCLCCNT +=

2. Phng phpa. Mi gi tr ca L hc C, cho ta mt gi tr tn s, chu k tng ng, vit tt c cc biu thc tn s hoc chuk ri gn nhng gi tr bi cho tng ng (nu c).VD: -Khi t cm cun dy l L1, in dung t in l C1 th chu k dao ng l T1

-Khi t cm cun dy l L2, in dung t in l C2 th chu k dao ng l T2...........

-Ta c cc biu thc chu k tng ng:

211CL2T =

222 CL2T=

..........-Lp mi lin h ton hc gia cc biu thc . Thng l lp t s; bnh phng hai v ri cng, tr cc

biu thc; phng php th.....b. T cng thc tnh bc sng ta thy, bc sng bin thin theo L v C. L hay C cng ln, bc sng cngln. Nu iu chnh mch sao cho C v L bin thin t C m, Lm n CM, LM th bc sng cng bin thintng ng trong di t mmm CLc2= n MMM CLc2=

3. Mt s bi tp minh haBi 1: Nu iu chnh in dung ca mt mch dao ng tng ln 4 ln th chu k dao ng ring camch thay i nh th no ( t cm ca cun dy khng i)?

Gii:C hai gi tr ca in dung: C v C = 4C, tng ng vi hai gi tr chu kLC2T = v

T2C.L22C4.L2'LC2'T ==== Vy chu k tng 2 ln.Khi lm bi trc nghim, tit kim thi gian, ta c nhn nh sau:T biu thc tnh chu k ta thy T t l vi cn bc hai ca in dung C v t cm L.Tc l, nu C tng (hay gim) n ln th T tng (hay gim) n ln,

nu L tng (hay gim) m ln th T tng (hay gim) m ln. Ngc li vi tn s f.Nh bi tp trn, do C tng 4 ln, suy ra ngay chu k tng 24 = ln.

Bi 2:Nu tng in dung ca mt mch dao ng ln 8 ln, ng thi gim t cm ca cun dy i 2 ln

th tn s dao ng ring ca mch tng hay gim bao nhiu ln?

Trang 10


11/34

Gii: .f2

1'fH a y

2

1

f

'f

C8.L2

1

2

1

'C'L2

1'f

L C2

1f

==

=

=

=

=> Tn s gim i hai ln.

C th suy lun: C tng 8 ln, L gim 2 ln suy ra nu s tng: 22

1.8 = ln.=> Tn s gim hai ln.

Bi 3: Mt mch dao ng gm c mt cun cm c t cm L = 10 -3H v mt t in c in dung iuchnh c trong khong t 4pF n 400pF (1pF = 10-12F).Mch ny c th c nhng tn s ring nh th no?

Gii: T cng thcLC2

1f

= suy ra 22 Lf4

1C

= Theo bi ra: F10.400CF10.4 1212 ta c

F10.400

Lf4

1F10.4 12

22

12

, vi tn s f lun dng, ta suy ra:

Hz10.52,2fHz10.52,2 65

Vi cch suy lun nh trn th rt cht ch nhng s bin i qua li kh rc ri, mt nhiu thi gian v haynhm ln.

Nh ni phn phng php, tn s lun nghch bin theo C v L,nn fmax ng vi Cmin, Lmin v fmin ng vi Cmax v Lmax.

=>

===

=

=

=

H z1 0.5 2,21 0.4.1 02

1

L C2

1f

H z1 0.5 2,21 0.4 0 0.1 02

1

L C2

1f

61 23

m i n

m a x

5

1 23m a x

m i n

tc l tn s bin i t 2,52.105Hz n 2,52.106HzBi 4: Mt cun dy c in tr khng ng k mc vi mt t in c in dung 0,5F thnh mt mch daong. H s t cm ca cun dy phi bng bao nhiu tn s ring ca mch dao ng c gi tr sau y:

a) 440Hz (m).b) 90Mhz (sng v tuyn).

Gii: T cng thcLC2

1f

= suy ra cng thc tnh t cm:

22Cf4

1L

=

a) f = 440Hz; .H26,0440.10.5,0.41

Cf4

1L 26222 ===

b) f = 90MHz = 90.106Hz

.pH3,6H10.3,6)10.90.(10.5,0.4

1

Cf4

1L 12

266222==

=

=

Bi 5: Mt mch dao ng gm cun dy L v t in C. Nu dng t C1 th tn s dao ng ring ca mchl 60kHz, nu dng t C2 th tn s dao ng ring l 80kHz. Hi tn s dao ng ring ca mch l baonhiu nu:

a) Hai t C1 v C2 mc song song.b)Hai t C1 v C2 mc ni tip.

Gii: Bi ton cp n mch dao ng vi 3 b t khc nhau, ta lp 3 biu thc tn s tng ng:

Trang 11


12/34

+ Khi dng C1:

=

=

=

1

2

2

1

1

2

2

1

1

1

L4

1f

L4f

1

L C2

1f

+ Khi dng C2:

=

=

=

2

2

2

2

2

2

2

2

2

2

L4

1f

L4f

1

L C2

1f

a) Khi dng hai t C1 v C2 mc song song, in dung ca b t C = C1 + C2

)CC(L4f

1

)CC(L2

1f 21

2

2

21

+=+

=

Suy ra: .kHz488060

80.60

ff

fff

f

1

f

1

f

1

222

2

2

1

21

2

2

2

1

2=

+=

+=+=

b) Khi dng hai t C1 v C2 mc ni tip, in dung ca b t c xc nh bi21 C

1

C

1

C

1+=

+

=

+

=

212

2

21 C

1

C

1

L4

1f

C

1

C

1

L

1

2

1f

Suy ra: .kHz1008060ffffff222

2

2

1

2

2

2

1

2

=+=+=+=

Bi 6: Mch dao ng ca mt my thu v tuyn gm cun cm L = 1H v t in bin i C, dng thusng v tuyn c bc sng t 13m n 75m. Hi in dung C ca t in bin thin trong khong no?

Gii: Cch 1: T cng thc tnh bc sng: LCc2= suy raLc4

C22

2

=

Do > 0 nn C ng bin theo ,

FLc

C 126282

2

22

2

minmin 10.6,47

10.)10.3.(.4

13

4

===

= 47,6 pF

FLcC

12

6282

2

22

2max

max 10.158310.)10.3.(.4

75

4

===

=1583 pFVy in dung bin thin t 47.10-12C n 1583.10-12C.Cch 2: Dng lnh SOLVE trong My Tnh cm tay 570ES: ( Ch dng trong COMP: MODE 1 )

Ch : Phm ALPHA) :gn bin X; phm:SHIFT CALC : SOLVE;phmALPHA CALC l du=-Cng thc : LCc2= : Vi =13m ;L = 10-6H ; C l bin X

-Bm: 13ALPHA CALC =2 SHIFTX10X X 3X10X 8 X10X -6 X ALPHA) X

Mn hnh hin th: 8 613 2 .3 1 0 10

= x x

-Tip tc bm: SHIFT CALC SOLVE = ( ch khong 6s )

Mn hnh hin th: X l i lng C

Vy :C = 47,6 10-12 ( F) = 47,6 ( pF)

Trang 12

8 6

13 2 .3 10 10

=

x xXX= 4.756466x 10-11

L--R = 0


13/34

-Tng t: Vi =75m ;L = 10-6H ; C l bin X :Ch : xem hoc sa cng thc va nhp ta ch nhn phm DEL v khng nhn phm AC-Bm: 75ALPHA CALC =2 SHIFTX10X X 3X10X 8 X10X -6 X ALPHA) X

Mn hnh hin th: 8 675 2 .3 10 10

= x x

-Tip tc bm: SHIFT CALC SOLVE = ( ch khong 6s )

Mn hnh hin th: X l i lng C

Vy :C = 1,5831434. 10-9 (F)= 1583,1434. 10-12 (F)=1583 (pF)Bi 7:Mch dao ng chn sng ca mt my thu thanh gm mt cun dy c t cm L = 11,3H vt in c in dung C = 1000pF.

a) Mch in ni trn c th thu c sng c bc sng 0 bng bao nhiu?b) thu c di sng t 20m n 50m, ngi ta phi ghp thm mt t xoay CV vi t C ni trn. Hi

phi ghp nh th no v gi tr ca CV thuc khong no?c) thu c sng 25m, CV phi c gi tr bao nhiu? Cc bn t di ng phi xoay mt gc bng bao

nhiu k t v tr in dung cc i thu c bc sng trn, bit cc bn t di ng c th xoay t0 n 1800?

Gii:a) Bc sng mch thu c: m20010.1000.10.3,1110.3.2LCc2 12680 ===

b) Nhn xt:Di sng cn thu c bc sng nh hn bc sng 0 nn in dung ca b t phi nh hn C. Do

phi ghp CV ni tip vi C.Khi :

222

2

V

V

V

LCc4

CC

CC

C.CLc2

=+

=

Vi > 0, CV bin thin nghch bin theo .

F10.7,662010.10.3,11.)10.3(4

10.1000.20

LCc4

CC

F10.1,105010.10.3,11.)10.3(4

10.1000.50

LCc4

C

C

12

296282

122

2

min

22

2

minmaxV

12

296282

122

2

max

22

2

max

minV

=

=

=

==

=

Vy pF7,66CpF1,10 V

c) thu c sng 1 = 25m,

F10.9,152510.10.3,11.)10.3.(.4

10.25

LCc4

CC 12

296282

92

2

1

22

2

1V

=

=

V CV t l vi gc xoay nn ta c

0

minVmaxV

1VmaxV

minVmaxV

1VmaxV 1621,107,669,157,66180

CCCC180

180CCCC =

=

==

4. Bi tp t lun:Bi1. Trong thng tin lin lc bng sng v tuyn, ngi ta s dng cch bin iu bin , tc l lm cho bin ca sng in t cao tn (sng mang) bin thin theo thi gian vi tn s bng tn s ca dao ng m tn.Cho tn s sng mang l 800 kHz, tn s ca dao ng m tn l 1000 Hz. Xc nh s dao ng ton phnca dao ng cao tn khi dao ng m tn thc hin c mt dao ng ton phn.

Bi2. Mt mch thu sng in t gm cun dy thun cm c h s t cm khng i v t in c in dungbin i. thu c sng c bc sng 30m, ngi ta phi iu chnh in dung ca t l 300 pF. thu

c sng 90m th phi iu chnh in dung ca t in n gi tr no?Bi3. Mch chn sng ca mt my thu v tuyn in gm t in c in dung C0 v cun cm thun c tcm L, thu c sng in t c bc sng 20 m. thu c sng in t c bc sng 60 m th phi mcvi C0 mt t in c in dung CX. Hi phi mc CX th no vi C0? Tnh CX theo C0.

Trang 13

8 675 2 .3 10 10= x xXX= 1.5831434 x10-9

L--R =0


14/34

Bi4. Mch chn sng ca mt my thu v tuyn l mt mch dao ng c mt cun thun cm m tcm c th thay i trong khong t 10 H n 160 H v mt t in m in dung c th thay i 40 pFn 250 pF. Tnh bng sng v tuyn (theo bc sng) m my ny bt c.

Bi5. Mch chn sng ca mt my thu v tuyn l mt mch dao ng c mt cun thun cm c t cm10 H v mt t in c in dung bin thin trong mt gii hn nht nh. My ny thu c bng sng vtuyn c bc sng nm trong khong t 10 m n 50 m. Hi khi thay cun thun cm trn bng cun thuncm khc c t cm 90 H th my ny thu c bng sng v tuyn c bc sng nm trong khongno?

Bi6. Mt mch dao ng c cu to t mt cun thun cm L v hai t in C1 v C2. Khi dng L vi C1th mch dao ng bt c sng in t c bc sng 1 = 75 m. Khi dng L vi C2 th mch dao ng btc sng in t c bc sng 2 = 100 m. Tnh bc sng in t m mch dao ng bt c khi:

a) Dng L vi C1 v C2 mc ni tip.b) Dng L vi C1 v C2 mc song song.

Bi7. Mt mch dao ng LC l tng gm cun cm thun c t cm khng i. Khi mc cun cm vit in c in dung C1 th tn s dao ng ring ca mch l 7,5 MHz v khi mc cun cm vi t in cin dung C2 th tn s dao ng ring ca mch l 10 MHz. Tnh tn s dao ng ring ca mch khi mccun cm vi:

a) Hai t C1 v C2 mc ni tip.b) Hai t C1 v C2 mc song song.

Bi8. Xt hai mch dao ng in t l tng. Chu k dao ng ring ca mch th nht l T 1, ca mch thhai l T2= 2T1. Ban u in tch trn mi bn t in c ln cc i Q 0. Sau mi t in phng in quacun cm ca mch. Khi in tch trn mi bn t ca hai mch u c ln bng q (0 < q < Q0) th t s ln cng dng in trong mch th nht v ln cng dng in trong mch th hai l bao nhiu?

5. Hng dn gii v p s:

Bi1. Thi gian dao ng m tn thc hin c mt dao ng ton phn: TA =1

Af= 10-3 s. Thi gian

dao ng cao tn thc hin c mt dao ng ton phn TC =1

Cf= 0,125.10-5 s. S dao ng ton phn ca

dao ng cao tn khi dao ng m tn thc hin c mt dao ng ton phn: N =A

C

TT

= 800.

Bi2. Ta c:2

1

2

1

C

C=

C2 = 2

1

2

21

C= 2,7 nF.

Bi3. Ta c: 0 = 2 c 0LC ; X = f

c= 2 c bLC

0 0

bX C

C

= = 3

Cb = 9C0. V Cb > C0 nn phi mc CX song song vi C0 v CX = Cb C0 = 8C0.

Bi4. Ta c: min = 2 c min minL C = 37,7 m; max = 2 c ax axm mL C = 377 m.

Bi5.Ta c: min = 2 c minLC ; 'min = 2 c min'L C

'min = 'L

L. min = 30 m;

'max = 'L

L. max

=150 m.

Bi6. a) Ta c: nt = 2 c21

21

CC

CLC

+ nt = 2

221

21

+= 60 m.

b) Ta c: // = 2 c )( 21 CCL + => // =22

21 + = 125 m.

Bi7. a) Ta c: fnt =

21

212

1

CC

CLC

+

fnt = 22

21 ff + = 12,5 Hz.

b) Ta c: f// = )(2

1

21 CCL + f// = 2

22

1

21

ff

ff

+= 6 Hz.

Trang 14


15/34

Bi8. Ta c: 1 =1

2

T

; 2 =

2

2

T

=

12

2

T

=

2

1 1 = 2 2; I01 = 1Q0; I02 = 2Q0 I01 = 2I02.

V:2

01

1

Q

q+

2

01

1

I

i= 1;

2

02

2

Q

q+

2

02

2

I

i= 1; Q01 = Q02 = Q0 v |q1| = |q2| = q > 0

2

01

1

I

i=

2

02

2

I

i

||

||

2

1

i

i=

02

01

I

I= 2.

C. TRC NGHIM N TP:Cu 1. Tn s gc ca dao ng in t t do trong mch LC c in tr thun khng ng k c xc nh

bi biu thc

A. = LC2 . B. = LC1 . C.

= LC21 . D. = LC 1 .

Cu 2. Pht biu no sau y l sai khi ni v nng lng ca mch dao ng in t LC c in tr thunkhng ng k ?A. Nng lng in t ca mch dao ng bin i tun hon theo thi gian.B. Nng lng in t ca mch dao ng bng nng lng t trng cc i cun cm.C. Nng lng in t ca mch dao ng bng nng lng in trng cc i t in.D. Nng lng in trng v nng lng t trng cng bin thin tun hon theo mt tn s chung.Cu 3. Mt mch dao ng in t c tn s f = 0,5.10 6Hz, vn tc nh sng trong chn khng l c =3.108m/s. Sng in t do mch pht ra c bc sngA. 6m. B. 600m. C. 60m. D. 0,6m.

Cu 4. Cng thc tnh nng lng in t ca mt mch dao ng LC l

A. W =C

Qo2

. B. W =L

Qo2

. C. W =C

Qo2

2

. D. W =L

Qo2

2

.

Cu 5. Mt mch dao ng c t in C =

2.10-3F v cun dy thun cm L. tn s in t trong mch

bng 500Hz th L phi c gi tr l

A. 5.10-4H. B.500

H. C.

310H. D.

2

10 3H.

Cu 6. Trong dng c no di y c c my pht v my thu sng v tuyn ?A. My thu thanh. B. Chic in thoi di ng.C. My thu hnh (Ti vi). D. Ci iu khin ti vi.Cu 7. Trong mch dao ng in t LC, nu in tch cc i trn t in l Q o v cng dng in cci trong mch l Io th chu k dao ng in t trong mch l

A. T = 2 QoIo. B. T = 2 .o

o

Q

I. C. T = 2 LC. D. T = 2

o

o

I

Q.

Cu 8. Trong mch dao ng in t LC, in tch t in bin thin vi chu k T. Nng lng in trng t in

A. bin thin iu ho vi chu k T. B. bin thin iu ho vi chu k2

T.

C. bin thin iu ho vi chu k 2T. D. khng bin thin theo thi gian.Cu 9*. Trong mch dao ng in t LC, khi dng t in c in dung C1 th tn s dao ng l f1 =30kHz, khi dng t in c in dung C2 th tn s dao ng l f2 = 40kHz. Khi dng hai t in c cc indung C1 v C2 ghp song song th tn s dao ng in t lA. 38kHz. B. 35kHz. C. 50kHz. D. 24kHz.Cu 10. Chu k dao ng in t t do trong mch dao ng LC c tnh theo cng thc

Trang 15


16/34

A. T = 2C

L. B. T =

LC

2. C. T = 2

L

C. D. T = 2 LC .

Cu 11. Trong mt mch dao ng in t LC, in tch ca mt bn t bin thin theo hm s q = Qocos t.Khi nng lng in trng bng nng lng t trng th in tch ca cc bn t c ln l

A.4

oQ . B.22

oQ . C.2

oQ . D.2oQ .

Cu 12.Chn cu tr li sai. Khi mt t trng bin thin khng u v khng tt theo thi gian s sinhra:A. mt in trng xoy. B. mt t trng xoy. C. mt dng in dch. D. Mt dng in dn.Cu 13*.Mt mch dao ng in t c L = 5mH; C = 31,8F, hiu in th cc i trn t l 8V. Cng dng in trong mch khi hiu in th trn t l 4V c gi tr:A. 5mA B. 0,25mA C. 0,55A D. 0,25ACu 14*.Mt mch dao ng LC c cun thun cm L = 0,5H v t in C = 50F. Hiu in th cc igia hai bn t l 5V. Nng lng dao ng ca mch v chu k dao ng ca mch l:

A. 2,5.10-4J ;100

s. B. 0,625mJ;

100

s. C. 6,25.10-4J ;

10

s. D. 0,25mJ ;

10

s.

Cu 15.Mch dao ng gm cun dy c t cm L = 30 H mt t in c C = 3000pF. in tr thunca mch dao ng l 1 . duy tr dao ng in t trong mch vi hiu in th cc i trn t in l 6V

phi cung cp cho mch mt nng lng in c cng sut:A. 1,8 W B. 1,8 mW C. 0,18 W D. 5,5 mWCu 16.Mt mch dao ng gm t c C = 125nF v cun cm c L = 50 H. in tr thun ca mchkhng ng k. Hiu in th cc i gia hai bn t in U 0 = 1,2V. Cng dng in cc i trong mchlA. 6.10-2A B. 3 2 A C. 3 2 mA D. 6mACu 17. Mch dao ng in t LC c L = 0,1mH v C = 10 -8F. Bit vn tc ca sng in t l 3.108m/s th

bc sng ca sng in t m mch c th pht ra lA. 60 m. B. .103m. C. 600 m. D. 6 .103m.Cu 18. Mch dao ng ca mt my thu v tuyn in gm cun dy c t cm L = 1mH v mt t inc in dung thay i c. my thu bt c sng v tuyn c tn s t 3MHz n 4MHz th in dungca t phi thay i trong khong:A. 1,6pF C 2,8pF. B.2 F C 2,8 F. C. 0,16pF C 0,28 pF. D.0,2 F C 0,28 F.Cu 19. Trong thng tin lin lin lc di nc ngi ta thng s dngA. sng di. B. sng trung. C. sng ngn. D. sng cc ngn.Cu 20.Mch dao ng gm t in c in dung 4500pF v cun dy thun cm c t cm 5H. Hiuin th cc i hai u t in l 2V. Cng dng in cc i chy trong mch lA. 0,03A. B. 0,06A. C. 6.10-4A. D. 3.10-4A.Cu 21. Pht biu no sau y l sai v sng in t ?A. Sng in t mang nng lng t l vi lu tha bc 4 ca tn s.B. Sng in t l sng ngang. C. Sng in t c y cc tnh cht ging sng c.

D. Ging nh sng c, sng in t cn mi trng vt cht n hi lan truyn.Cu 22. Mt mch chn sng ca my thu v tuyn gm cun cm L = 5 H v mt t xoay c in dungbin thin t 10pF n 240pF. Di sng my thu c lA. 10,5m 92,5m. B. 11m 75m. C. 15,6m 41,2m. D. 13,3 m 65,3m.Cu 23. Mt mch dao ng in t c in dung ca t l C = 4 F. Trong qu trnh dao ng hiu in thcc i gia hai bn t l 12V. Khi hiu in th gia hai bn t l 9V th nng lng t trng ca mch lA. 2,88.10-4J. B. 1,62.10-4J. C. 1,26.10-4J. D. 4.50.10-4J.Cu 24. Mch chn sng u vo ca my thu v tuyn in gm t in C = 1nF v cun cm L = 100 H

(ly ).102 = Bc sng in t m mch thu c l.A. 300= m. B. 600= m. C. 300= km. D. 1000= m.

Cu 25. Mt mch dao ng gm mt cun cm c t cm L = 1 mH v mt t in c in dung C =

1,0 F . Mch thu c sng in t c tn s no sau y?

A. 50Hz. B. 50kHz. C. 50MHz. D. 5000Hz.

Trang 16


17/34

HNG DN TR LI :Cu 1. B;Cu 2. A;

Cu 3. Bf

c= = 6

8

10.5,0

10.3= 600m

Cu 4.C W =2

1

C

Qo2

;

Cu 5. D f =LC2

1 L =

Cf2241

=

210 3 H.

Cu 6. B;

Cu 7. D Wm =Wtm = W 2

1LI2o =

2

1

C

Qo2

LC = 20

2

I

Qo T = 2 LC = 20I

Qo

Cu 8. B

Cu 9*. D f1=12

1

LC; f2 =

22

1

LC; f =

LC2

1; C1 //C2 C = C1 + C2 f

1=

1

1

f+

2

1

f

f = 2212

2

2

2

1

ff

ff

+ = 2222

4030

40.30

+ = 24kHzCu 10. D;

Cu 11. D Khi W = Wt th W = W + Wt = 2W 2

1

C

Qo2

= 22

1

C

Q2

Q =2oQ

Cu 12. D

Cu 13*. C W = W + Wt Wt = W - W 2

1LI2 =

2

1CU2o-

2

1CU2 I =

L

UUC )( 220 = 0,55A

Cu 14*. B W = Wm=2

1CU2o=

2

150.10-6.52= 0,625mJ; T = 2 LC= 2 610.50.5,0 =

100

s

Cu 15. B Wm =Wtm 21 CU2o=

21 LI2o Io = Uo

L

C I = UoL

C

2= 6

5

9

10.3.2

10.3

= 4,25.10-2A

P = RI2 = 1,8 mW

Cu 16. A Gii nh cu 15 : Io = UoL

C= 1,2

5

9

10.50

10.125

= 6.10-2

Cu 17. C = cT = c 2 LC = 3.108. 2 84 10.10 = 600m

Cu 18. A f1=12

1

LC C1 =

Lf2124

1

= 2,8pF

Cu 19. A. sng di;

Cu 20. B Gii nh cu 15 : Io = UoLC = 0,06A

Cu 21. D Sng in t truyn c trong chn khngCu 22. D 1 = c 2 1LC = 3.10

8. 2 126 10.10.10.5 =13,3 m

Cu 23.C W = W + Wt Wt = W - W = =2

1CU2o-

2

1CU2 =

2

14.10-6(122-92) = 1,26.10-4J

Cu 24. B = cT = c 2 LC = 3.108.2 49 10.10.10 = 600 m

Cu 25. B. Sng thu phi c tn s bng tn s ring: f =LC2

1=

731010

2

1

= 5.104Hz= 50kHz

D. BI TP T LUN:Cu 26: Cho mch dao ng gm mt cun dy thun cm c t cm L = 2.10-4H, C = 8pF. Nng lngca mch l E = 2,5.10-7J. Vit biu thc ca cng dng in trong mch v biu thc hiu in th gia2 bn t. Bit O rng ti thi im ban u cng dng in trong mch c ga tr cc i.

Trang 17


18/34

Li gii: Tn s gc ca mch dao ng l: =124 10.8.10.2

1

LC

1

= = 25.106 Rad/s

Biu thc ca in tch trn t in c dng:

q = Q0sin ( t + ) = Q0sin (25.106+ ) (1)

i = I0cos(25.106t + ) (2)

Theo b khi t = 0 ; i = I0 cos = 1 = 0

Nng lng ca mch E =C2

Q

2

LT 2020 = . I0=

410.2

710.25.2

L

E2

= = 5.10-2 A

Q0= 127 10.8.10.5,2.2EC2 = = 2.10-9C.

i = 5.10-2cos (25.106t) (A)

u = C

Q0sin(25.10

6

t) = 250.sin (25.106

t) (V)Cu 27: Mt mch dao ng in t l tng ang c dao ng in t

t do. Ti thi im t = 0, t in bt u phng in. Sau khong thi

gian ngn nht t = 10-6s th in tch trn mt bn t in bng mt

na gi tr cc i. Tnh chu k dao ng ring ca mch.

Li gii: thi im u (t = 0), in tch trn mt bn t l: q1 = qo

Sau khong thi gian ngn nht t, in tch trn mt bn t in l:

q2 =2

oq ;Ta c: 21 MOM= Hay: = 3

rad => t =

62.

3

TT==

Vy, chu k dao ng ring ca mch l: T = 6t = 6.10-6s

Cu 28: Mt mch dao ng LC l tng ang c dao ng in t t do. in tch trn mt bn t in c

biu thc: q = qocos(106 t - )2

(C). K t thi im ban u( t = 0), sau mt khong thi gian ngn nht l

bao lu th nng lng in trng trn t in bng ba ln nng lng t trng cun cm?

Li gii: thi im ban u t = 0, in tch trn mt bn t l q 1 = 0.

Sau mt khong thi gian ngn nht t, th WL =3

1WC

=> W =3

1WC + WC =

3

4WC

C

q

C

qo

23

4

2

2

2

2

= => q2 =2

3 qo

hoc q2 = -2

3 qo Ta c:

=t .vi =

2; m:cos =

2

32 =oq

q=> =

6

=> =

3

. Vy: st

3

10

10.3

6

6

==

=

Trang 18

q-qo qo q

2

q1

M1

O

M2

q-qo qo

O

M2

M1

q1

q2


19/34

Cu 29: Mt mch dao dng LC l tng c chu k dao ng l T. Ti mt thi im in tch trn t in

bng 6.10-7C, sau mt khong thi gian t = 3T/4 cng dng in trong mch bng 1,2 .10-3A. Tm

chu k T.

Li gii: Gi s thi im ban u t1, in tch trn t in c gi tr q1.

thi im t2, sau mt khong thi gian t = T4

3ta c

2

3

4

3.

2 ===

T

Tt rad

Theo gin vc t: 1 + 2 =2

=> sin 2 = cos 1= q1/q0 (29.1)

T cng thc:2

222

iqqo += =>

oq

i

22sin =

Do , (29.1) oo q

q

q

i 12

.

=

=>

2000

10.6

10.2,17

3

1

2 ===

q

irad/s .

Vy: T = 10-3s

D.TRC NGHIM LUYN TP THEO DNG:Dng 1: I CNG V DAO NG IN TCu 1: Trong mch dao ng in tLC, nu in tch cc i trn t in l Q0 v cng dng in cc i trongmch l I0 th chu k dao ng in t trong mch l

A. T = 2 q0I0 B. T = 2 q0/I0 C. T = 2 I0/q0 D. T = 2 LCCu 2: Mt mch dao ng in t LC, c in tr thun khng ng k. Hiu in th gia hai bn t in bin thiniu ha theo thi gian vi tn s f. Pht biu no sau y l sai?

A. Nng lng in t bng nng lng t trng cc i.B.Nng lng in t bin thin tun hon vi tn s f.C. Nng lng in trng bin thin tun hon vi tn s 2f .D. Nng lng in t bng nng lng in trng cc i.

Cu 3: Tn s gc ca dao ng in t t do trong mch LC c in tr thun khng ng k c xc nh bi biu

thc A. LC

1

= B.LC

1= C.

LC

2

1= D.

LC

2=

Cu 4: Mt mch dao ng in t LC gm t in c in dung C v cun dy thun cm c t cm L. Bit dydn c in tr thun khng ng k v trong mch c dao ng in t ring. Gi Q0, U0 ln lt l in tch cc i vhiu in th cc i ca t in, Io l cng dng in cc i trong mch. Biu thc no sau y khng phi l

biu thc tnh nng lng in t trong mch ?

A.

2

0

2LIW = B. LqW2

2

0= C.2

0

2CUW = D. CqW2

2

0=Cu 5: Pht biu no sau y l sai khi ni v nng lng ca mch dao ng in t LC c in tr thun khng ngk? A. Nng lng in t ca mch dao ng bng nng lng in trng cc i t in.

B. Nng lng in trng v nng lng t trng cng bin thin tun hon theo mt tn s chung.C.Nng lng in t ca mch dao ng bin i tun hon theo thi gian.D. Nng lng in t ca mch dao ng bng nng lng t trng cc i cun cm.

Cu 6: Trong mch dao ng in t LC, in tch ca t in bin thin iu ho vi chu k T. Nng lng intrng t in

A. bin thin iu ho vi chu k 2T B. khng bin thin iu ho theo thi gianC.bin thin iu ho vi chu k T/2 D. bin thin iu ho vi chu k T

Cu 7: Mt mch dao ng gm c cun dy L thun in cm v t in C thun dung khng. Nu gi I 0 dng in

cc i trong mch, hiu in th cc i U0 gia hai u t in lin h vi I 0 nh th no ? Hy chn kt qu ngtrong nhng kt qu sau y:

A.C

LIU

00 = B.

L

CIU 00 = C.

C

LIU 00 = D.

C

LIU 00 =

Cu 8: Cng thc tnh nng lng in t ca mt mch dao ng LC l

Trang 19

q-qo

q2 q1q

o

O

M2

1

2

M1


20/34

A.C

IW

2

2

0= B.C

qW

2

2

0= C.C

qW

2

0= D. LIW /20=

Cu 9: Trong mch dao ng, dng in trong mch c c im no sau y ?A. Tn s rt ln. B. Cng rt ln. C. Nng lng rt ln. D. Chu k rt ln.

Cu 10: Trong mch dao ng LC c in tr thun bng khng thA. Nng lng t tp trung t in v bin thin vi chu k bng na chu k dao ng ring ca mch.B. Nng lng t tp trung cun cm v bin thin vi chu k bng chu k dao ng ring ca mch.C. Nng lng tt tp trung t in v bin thin vi chu k bng na chu k dao ng ring ca mch.D. Nng lng tt tp trung cun cm v bin thin vi chu k bng chu k dao ng ring ca mch.

Cu 11: S hnh thnh dao ng in t t do trong mch dao ng l do hin tng no sau y ?A. Hin tng cng hng in. B. Hin tng t ho.C. Hin tng cm ng in t. D. Hin tng t cm.Cu 12: Mt mch dao ng in t LC gm cun dy thun cm c t cm L khng i v t in c in dung Cthay i c. Bit in tr ca dy dn l khng ng k v trong mch c dao ng in t ring. Khi in dung cgi tr C1 th tn s dao ng ring ca mch l f1. Khi in dung c gi tr C2 = 4C1 th tn s dao ng in t ringtrong mch l

A. f2 = 4f1 B. f2 = f1/2 C. f2 = 2f1 D. f2 = f1/4Cu 13: Mt mch LC ang dao ng t do, ngi ta o c in tch cc i trn 2 bn t in l q 0 v dng incc i trong mch l I0. Nu dng mch ny lm mch chn sng cho my thu thanh, th bc sng m n bt ctnh bng cng thc:

A. = 2 c00Iq

. B. = 2 cq0/I0. C. = 2 cI0/q0. D. = 2 cq0I0.Cu 14: Trong mch dao ng LC c dao ng in t vi tn s 1MHz, ti thi im t = 0, nng lng t trngtrong mch c gi tr cc i. Thi gian ngn nht k t thi im ban u nng lng t trng bng mt na gi trcc i ca n l:

A. 0,5.10-6s. B. 10-6s. C. 2.10-6s. D. 0,125.10-6s

Cu 15: Trong mt mch dao ng LC, in tch trn mt bn t bin thin theo phng trnh ).2

cos(0

= tqq

Nh vy: A. Ti cc thi im T/4 v 3T/4, dng in trong mch c ln cc i, chiu ngc nhauB. Ti cc thi im T/2 v T, dng in trong mch c ln cc i, chiu ngc nhau.C. Ti cc thi im T/4 v 3T/4, dng in trong mch c ln cc i, chiu nh nhau.

D. Ti cc thi im T/2 v T, dng in trong mch c ln cc i, chiu nh nhau

Cu 16: in tch ca t in trong mch dao ng LC bin thin theo phng trnh q = q ocos(2

T

t + ). Ti thi

im t = T/4 , ta c: A. Hiu in th gia hai bn t bng 0. B. Dng in qua cun dy bng 0.C. in tch ca t cc i. D. Nng lng in trng cc i.

Cu 17: Trong mch dao ng LC l tng, gi i v u l cng dng in trong mch v hiu in th gia hai ucun dy ti mt thi im no , I0 l cng dng in cc i trong mch. H thc biu din mi lin h gia i, uv I0 l :

A. ( ) 2220 uC

LiI =+ B. ( ) 2220 u

L

CiI = C. ( ) 2220 u

C

LiI = D. ( ) 2220 u

L

CiI =+

Cu 18: Trong mch LC in tch ca t in bin thin iu ho vi gi tr cc i bng q 0. in tch ca t in khinng lng t trng gp 3 ln nng lng in trng l

A. q = 0Q3 B. q =0Q

4 C. q = 0Q 2

2 D. q = 0Q2

Cu 19: Mt mch dao ng LC c L = 2mH, C=8pF, ly 2 =10. Thi gian t lc t bt u phng in n lc cnng lng in trng bng ba ln nng lng t trng l:

A. 2.10-7s B. 10-7s C.510

75s

D.610

15s

Cu 20: Trong mch dao ng LC c in tr thun khng ng k, chu k dao ng ca mch l T = 10-6s, khongthi gian ngn nht nng lng in trng li bng nng lng t trng

A. 2,5.10-5s B. 10-6s C. 5.10-7s D. 2,5.10-7s

Dng 2: XC NH CHU K, TN S V BC SNG

Cu 1: Tn s dao ng ca mch LC tng gp i khi:A. in dung t tng gp i B. t cm ca cun dy tng gp iC. in dung gim cn 1 na D. Chu k gim mt na

Cu 2: Trong mch thu sng v tuyn ngi ta iu chnh in dung ca t C = 1/4000 (F) v t cm ca cun dyL = 1,6/ (H). Khi sng thu c c tn s bao nhiu ? Ly 2 = 10.

Trang 20


21/34

A. 100Hz. B. 25Hz. C. 50Hz. D. 200Hz.Cu 3: Mch dao ng bt tn hiu ca mt my thu v tuyn in gm mt cun cm L = 2 H v mt t in

1800C 0 = pF. N c th thu c sng v tuyn in vi bc sng l:A. 11,3m B. 6,28m C. 13,1m D. 113m

Cu 4: Khung dao ng vi t in C v cun dy c t cm L ang dao ng t do. Ngi ta o c in tch cci trn mt bn t l q0 = 106C v dng in cc i trong khung I0 = 10A. Bc sng in t cng hng vi khungc gi tr:

A. 188m B. 188,4m C. 160m D. 18mCu 5: Mun tng tn s dao ng ring mch LC ln gp 4 ln th:

A. Ta tng in dung C ln gp 4 ln B. Ta gim t cm L cn16L

C. Ta gim t cm L cn4

LD. Ta gim t cm L cn

2

L

Cu 6: Mt t in mFC 2,0= . mch c tn s dao ng ring 500Hz th h s t cm L phi c gi tr bng baonhiu ? Ly 102 = .A. 1mH. B. 0,5mH. C. 0,4mH. D. 0,3mH.

Cu 7: Mt mch dao ng LC gm mt cun cm c t cm HL

1= v mt t in c in dung C. Tn s dao

ng ring ca mch l 1MHz. Gi tr ca C bng:

A. pFC 4

1

= B. FC 41

= C. mFC 41

= D. FC 41

=Cu 8: Mt mch dao ng LC l tng ang c dao ng in t t do. Bit in tch cc i ca mt bn t in c ln l 10-8 C v cng dng in cc i qua cun cm thun l 62,8 mA. Tn s dao ng in t t do camch l

A. 2,5.103 kHz. B. 3.103 kHz. C. 2.103 kHz. D. 103 kHz.Cu 9: Mt mch dao ng in t LC gm cun dy thun cm c t cm khng i v t in c in dung thayi c. in tr ca dy dn khng ng k v trong mch c dao ng in t ring. Khi in dung c gi tr C 1 thtn s dao ng ring ca mch l f1. Khi in dung c gi tr C2 = 4C1 th tn s dao ng in t ring trong mch l

A. f2 = 0,25f1. B. f2 = 2f1. C. f2 = 0,5f1. D.f2 = 4f1.Cu 10: Mt mch dao ng in t LC gm cun dy thun cm c t cm L = 2mH v t in c in dung C =0,2 F. Bit dy dn c in tr thun khng ng k v trong mch c dao ng in t ring. Chu k dao ng in t

ring trong mch lA. 6,28.10-4s. B. 12,57.10-4s. C. 6,28.10-5s. D.12,57.10-5s.

Cu 11: Mt mch dao ng LC l tng gm cun cm thun c t cm khng i, t in c in dung C thayi. Khi C = C1 th tn s dao ng ring ca mch l 7,5 MHz v khi C = C 2 th tn s dao ng ring ca mch l 10MHz. Nu C = C1 + C2 th tn s dao ng ring ca mch l

A. 12,5 MHz. B. 2,5 MHz. C. 17,5 MHz. D. 6,0 MHz.Cu 12: Mt mch dao ng in t LC l tng gm cun cm thun t cm L v t in c in dung thay ic t C1 n C2. Mch dao ng ny c chu k dao ng ring thay i c.

A. t 14 LC n 24 LC . B. t 12 LC n 22 LC .

C. t 12 LC n 22 LC . D. t 14 LC n 24 LC .

Cu 13: Mt mch dao ng gm mt cun cm c t cm L = 1mH v mt t in c in dung C = 0,1 F. Tns ring ca mch c gi tr no sau y?A. 1,6.104Hz. B. 3,2.104Hz. C. 1,6.103Hz. D. 3,2.103Hz.

Cu 14 : Mch dao ng in t LC l tng gm cun cm thun c t cm 1 mH v t in c iin dung 0,1 F.Dao ng in t rin ca mch c tn s gc

A. 3.105 rad/s. B. 2.105 rad/s. C. 105 rad/s. D. 4.105 rad/s.Cu 15: Trong mch dao ng LC c in tr thun khng ng k, c sau nhng khong thi gian bng 0,25.10 -4s thnng lng in trng li bng nng lng t trng. Chu k dao ng ca mch l

A. 10-4s. B. 0,25.10-4s. C. 0,5.10-4s D. 2.10-4sCu 16: Mch dao ng LC gm cun cm thun c t cm L = 2 H v t in c in dung 8 F. Tn s daong ring ca mch bng

A. 8106

Hz. B. 4106

Hz C. 8108

Hz D. 4108

HzCu 17: . Mch dao ng c to thnh t cun cm L v hai t in C1 v C2. Khi dng L v C1 th mch c tn sring l f1 = 3MHz. Khi dng L v C2 th mch c tn s ring l f2 = 4MHz. Khi dng L v C1, C2 mc ni tip th tns ring ca mch lA. 7MHz. B. 5MHz. C. 3,5MHz. D. 2,4MHz.

Trang 21


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Cu 18: Mch dao ng c to thnh t cun cm L v hai t in C1 v C2. Khi dng L v C1 th mch c tn sring l f1 = 3MHz. Khi dng L v C2 th mch c tn s ring l f2 = 4MHz. Khi dng L v C1, C2 mc song song thtn s ring ca mch l

A. 7MHz. B. 5MHz. C. 3,5MHz. D. 2,4MHz

Dng 3: XC NH CNG DNG IN V HIU IN THCu 1: Mt mch dao ng gm mt t 20nF v mt cun cm 8 H, in tr khng ng k. Hiu in th cc i hai u t in l U0 = 1,5V. Tnh cng dng in hiu dng chy qua trong mch.

A. 43 mA B. 73mA C. 53 mA D. 63 mA

Cu 2: Trong mt mch dao ng LC khng c in tr thun, c dao ng in t t do (dao ng ring). Hiu inth cc i gia hai bn t v cng dng in cc i qua mch ln lt l U0 v I0 . Ti thi im cng dngin trong mch c gi tr I0/2 th ln hiu in th gia hai bn t in l

A. 3U0 /4. B. 3 U0 /2 C. U0/2. D. 3 U0 /4Cu 3: Mt mch dao ng LC l tng c L = 40mH, C = 25F, in tch cc i ca t q 0 = 6.10-10C. Khi in tchca t bng 3.10-10C th dng in trong mch c ln.

A. 5. 10-7 A B. 6.10-7A C. 3.10-7 A D. 2.10-7ACu 4: Mt mch dao ng gm t in c in dung FC 50= v cun dy c t cm L = 5mH. in p cci trn t in l 6V. Cng dng in trong mch ti thi im in p trn t in bng 4V l:

A. 0,32A. B. 0,25A. C. 0,60A. D. 0,45A.Cu 5: Cng dng in tc thi trong mch dao ng LC l tng l i = 0,08cos(2000t)(A). Cun dy c t

cm L = 50mH. Hiu in th gia hai bn t ti thi im cng dng in tc thi trong mch bng cng dng in hiu dng l.:A. 22 V. B. 32V. C. 24 V. D. 8V.

Cu 6: Khi trong mch dao ng LC c dao ng t do. Hiu in th cc i gia 2 bn t l U o=2V. Ti thi imm nng lng in trng bng 2 ln nng lng t trng th hiu in th gia 2 bn t l

A. 0,5V. B.2

3V. C. 1V. D. 1,63V.

Cu 7: Mt mch dao ng gm mt t 20nF v mt cun cm 80 H , in tr khng ng k. Hiu in th cc ihai u t in l U0 = 1,5V. Tnh cng dng in hiu dng chy qua trong mch.

A. 73mA. B. 43mA. C. 16,9mA. D. 53mA.

Cu 8: Khung dao ng (C = 10 F; L = 0,1H). Ti thi im uC = 4V th i = 0,02A. Cng cc i trong khung bng:A. 4,5.102A B. 4,47.102A C. 2.104A D. 20.104ACu 9: Mt mch dao ng in t, cun dy thun cm c h s t cm 0,5mH, t in c in dung 0,5nF. Trongmch c dao ng in t iu ha.Khi cng dng in trong mch l 1mA th in p hai u t in l 1V. Khicng dng in trong mch l 0 A th in p hai u t l:

A. 2 V B. 2 V C. 22 V D. 4 VCu 10: Ti thi im ban u, in tch trn t in ca mch dao ng LC c ga tr cc i q 0 = 10-8C. Thi gian t phng ht in tch l 2 s. Cng hiu dng trong mch l:

A. 7,85mA. B. 78,52mA. C. 5,55mA. D. 15,72mA.Cu 11: Cng dng in tc thi trong mch dao ng LC c dng i = 0,02cos2000t (A).T in trong mch cin dung 5F. t cm ca cun cm l

A. L = 50 H B. L = 5.10

6

H C. L = 5.10

8

H D. L = 50mHCu 12: Mt mch dao ng LC, gm t in c in dung C = 8nF v mt cun dy thun cm c t cm L =2mH. Bit hiu in th cc i trn t 6V. Khi cng dng in trong mch bng 6mA, th hiu in th gia 2 ucun cm gn bng.

A. 4V B. 5,2V C. 3,6V D. 3VCu 13: Trong mch dao ng LC c dao ng in t t do (dao ng ring) vi tn s gc 104rad/s. in tch cci trn t in l 10-9C. Khi cng dng in trong mch bng 6.10-6A th in tch trn t in l

A. 8.10-10 C. B. 4.10-10 C. C. 2.10-10 C. D. 6.10-10 C.Cu 14: Mt mch dao ng LC c =107rad/s, in tch cc i ca t q0 = 4.10-12C. Khi in tch ca t q = 2.10-12C th dng in trong mch c gi tr:

A. 52.10 A B. 52 3.10 A C. 52.10 A D. 52 2.10 A Cu 15: Mt t in c in dung C = 8nF c np in ti in p 6V ri mc vi mt cun cm c L = 2mH.Cng dng in cc i qua cun cm l

A. 0,12 A. B. 1,2 mA. C. 1,2 A. D. 12 mA.

Cu 16: Mt mch dao ng gm cun dy thun cm v t in th hiu in th cc i gia hai bn t in U0C linh vi cng dng in cc i I0 bi biu thc:

Trang 22


23/34

A.C

LU C

10 = B. 0C 0

LU = I

CC.

0C 0

LU = I

CD.

0C 0

LU = I

C

Cu 17: . Mt mch dao ng LC l tng, gm cun cm thun c t cm L v t in c in dung C. Trong mchc dao ng in t t do. Gi U0, I0 ln lt l hiu in th cc i gia hai u t in v cng dng in cci trong mch th

A. 00I

ULC

= . B.0 0

LU I

C= . C.

0 0

CU I

L= . D. 0 0U I LC= .

Cu 18: Mt mch dao ng in t gm mt t in c in dung 0,125 F v mt cun cm c t cm 50 H.

in tr thun ca mch khng ng k. in p cc i gia hai bn t l 3V. Cng dng in cc i trong mchlA. 7,5 2 mA. B. 15mA. C. 7,5 2 A. D. 0,15A.Cu 19: Trong mch dao ng in t LC, nu in tch cc i trn t in l Q o v cng dng in cc itrong mch l Io th chu k dao ng in t trong mch l

A. T = 2 qoIo. B. T = 2 .o

o

q

I. C. T = 2 LC. D. T = 2

o

o

I

q.

Cu 20: Mt mch dao ng in t c L = 5mH; C = 31,8F, hiu in th cc i trn t l 8V. Cng dng introng mch khi hiu in th trn t l 4V c gi tr:

A. 5,5mA. B. 0,25mA. C. 0,55A. D. 0,25A.

Cu 21: Mch dao ng gm t in c C = 125nF v mt cun cm c L = 50 H. in tr thun ca mch khngng k. in p cc i gia hai bn t in U0 = 1,2V. Cng dng in cc i trong mch lA. 6.10-2A. B. 3 2 A. C. 3 2 mA. D. 6mA

Cu 22: Mot mach dao ong gom mot cuon cam co o t cam L va mot tu ienco ien dung C thc hien dao ong t do khong tat. Gia tr cc ai cua ien apgia hai ban tu ien bang U0. Gia tr cc ai cua cng o dong ien trong machla

A. I0 = U0 LC . B. I0 = U0C

L. C. I0 = U0

L

C. D. I0 =

LC

U0.

Cu 23: Mch dao ng gm t in c in dung 4500pF v cun dy thun cm c t cm 5H. in p cc i hai u t in l 2V. Cng dng in cc i chy trong mch l

A. 0,03A. B. 0,06A. C. 6.10-4A. D. 3.10-4A.Cu 24: Mch dao ng c cun thun cm c t cm L = 0,1H, t in c in dung C = 10 F. Khi uC = 4V th i= 30mA. Tm bin I0 ca cng dng in.

A. I0 = 500mA. B. I0 = 50mA. C. I0 = 40mA. D. I0 = 20mA.Cu 25: Mch dao ng c cun thun cm L = 0,1H, t in c in dung C = 10 F. Trong mch c dao ng int. Khi in p gia hai bn t l 8V th cng dng in trong mch l 60mA. Cng dng in cc i trongmch dao ng l

A. I0 = 500mA. B. I0 = 40mA. C. I0 = 20mA. D. I0 = 0,1A.

Dng 4: NNG LNG IN TRNG V T TRNGCu 1: Trong mch dao ng l tng, t in c in dung C = 5 F, in tch ca t c gi tr cc i l 8.10 -5C.

Nng lng dao ng in t trong mch l:

A. 6.10-4J. B. 12,8.10-4J. C. 6,4.10-4J. D. 8.10-4J.Cu 2: Dao ng in t trong mch l dao ng iu ho. Khi hiu in th gia hai u cun cm bng 1,2V thcng dng in trong mch bng 1,8mA.Cn khi hiu in th gia hai u cun cm bng 0,9V th cng dngin trong mch bng 2,4mA. Bit t cm ca cun dy L = 5mH. in dung ca t v nng lng dao ng in ttrong mch bng:

A. 10nF v 25.10-10J. B. 10nF v 3.10-10J. C. 20nF v 5.10-10J. D. 20nF v 2,25.10-8J.Cu 3: Hiu in th cc i gia hai bn t trong khung dao ng bng 6V, in dung ca t bng 1 F. Bit daong in t trong khung nng lng c bo ton, nng lng t trng cc i tp trung cun cm bng:A. 18.106J B. 0,9.106J C. 9.106J D. 1,8.106J

Cu 4: Mt t in c in dung FC2

10 3

= c np mt lng in tch nht nh. Sau ni 2 bn t vo 2 u

1 cun dy thun cm c t cm HL51= . B qua in tr dy ni. Sau khong thi gian ngn nht bao nhiu

giy (k t lc ni) nng lng t trng ca cun dy bng 3 ln nng lng in trng trong t ?A. 1/300s B. 5/300s C. 1/100s D. 4/300s

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24/34

Cu 5: Mt mch dao ng LC c in tr thun khng ng k, t in c in dung 0,05F. Dao ng in t ring(t do) ca mch LC vi hiu in th cc i hai u t in bng 6V. Khi hiu in th hai u t in l 4V thnng lng t trng trong mch bng

A. 0,4 J B. 0,5 J C. 0,9 J D. 0,1 J

Cu 6: Mch dao ng LC gm t C = 6 F v cun cm thun. Bit gi tr cc i ca in p gia hai u t in lUo = 14V. Ti thi im in p gia hai bn ca t l u = 8V nng lng t trng trong mch bng:

A. 588 J B. 396 J C. 39,6 J D. 58,8 JCu 7: Trong mch dao ng LC l tng c mt dao ng in t t do vi tn s ring f0 = 1MHz. Nng lng ttrng trong mch c gi tr bng na gi tr cc i ca n sau nhng khong thi gian l

A. 1ms B. 0,5ms C. 0,25ms D. 2msCu 8: Trong mch LC l tng cho tn s gc: = 2.104rad/s, L = 0,5mH, hiu in th cc i trn hai bn t 10V.Nng lng in t ca mch dao ng l:

A. 25 J. B. 2,5 J. C. 2,5 mJ. D. 2,5.10-4 J.Cu 9: T in ca mch dao ng c in dung C = 1F, ban u c in tch n hiu in th 100V, sau chomch thc hin dao ng in t tt dn. Nng lng mt mt ca mch t khi bt u thc hin dao ng n khi daong in t tt hn l bao nhiu ?

A. W = 10 kJ B. W = 5 mJ C. W = 5 k J D. W = 10 mJCu 10: Mt mch dao ng in t LC l tng ang dao ng vi in tch cc i trn bn cc ca t in l q0.

C sau nhng khong thi gian bng nhau v bng 10 -6s th nng lng t trng li c ln bngC

q

4

2

0 . Tn s ca

mch dao ng:A. 2,5.105Hz. B. 106Hz. C. 4,5.105Hz. D. 10-6Hz.

Cu 11: Chn tnh cht khng ng khi ni v mch dao ng LC:A. Dao ng trong mch LC l dao ng t do v nng lng in trng v t trng bin thin qua li vi nhau.B. Nng lng t trng t p trung cun cm L.C. Nng lng in trng t p trung t in C.D. Nng lng in trng v nng lng t trng cng bin thin tun hon theo mt tn s chung.

Cu 12: Mt mch dao ng gm cun thun cm L v hai t C ging nhau mc ni tip, kha K mc hai u mt tC (hnh v). Mch ang hot ng th ta ng kha K ngay ti thi im nng lng in trng v nng lng ttrng trong mch ang bng nhau. Nng lng ton phn ca mch sau s:

A. gim cn 3/4B. gim cn 1/4C. khng iD. gim cn 1/2

Cu 12 b: Mch dao ng in t gm cun dy c t cm L v hai t in ging ht nhau ghp ni tip .Mch dao ng vi hiu in th cc i hai u cun dy l U0, vo lc nng lng in trng trn cc t

bng nng lng t trng trong cun dy th ngi ta ni tt mt t. Hiu in th cc i trong mch l baonhiu?.

A. 2/30U hay8

30U

Gii: Nng lng ban u ca mch: W0 =

22002 2 4

CU

CU=

Khi ni tt mt t (ng kho k): Nng lng ca mch W =3

4W0 =

2

03

4 4

CU

W ='2

' 00

W2

CU= Do o U0 =

8

30U

Cu 13: Mch dao ng LC l tng gm t in c in dung C, cun cm thun c t cm L. Trong mch c daong in t t do. Bit hiu in th cc i gia hai bn t in l U0. Nng lng in t ca mch bng

A. 21

LC2

. B.2

0U

LC2

. C. 20

1CU

2. D. 2

1CL

2.

Cu 14: Trong mch dao ng LC l tng c dao ng in t t do thA. nng lng in trng tp trung cun cm.B. nng lng in trng v nng lng t trng lun khng i.C. nng lng t trng tp trung t in. D. nng lng in t ca mch c bo ton.

Trang 24

L

C C

K

L

C C

K


25/34

Cu 15: Mt mch dao ng in t LC gm t in c in dung C v cun dy thun cm c t cm L. Bit dydn c in tr thun khng ng k v trong mch c dao ng in t ring. Gi q0, U0 ln lt l in tch cc i vin p cc i ca t in, I0 l cng dng in cc i trong mch. Biu thc no sau y khng phi l biu thctnh nng lng in t trong mch ?

A. W =2

1CU 20 . B. W =

C

q

2

20 . C. W =

2

1LI 20 . D.W =

L

q

2

20 .

Cu 16: Mt mch dao ng in t c in dung ca t l C = 4 F. Trong qu trnh dao ng in p cc i giahai bn t l 12V. Khi in p gia hai bn t l 9V th nng lng t trng ca mch l

A. 2,88.10-4J. B. 1,62.10-4J. C. 1,26.10-4J. D. 4.50.10-4J.

Cu 17: Mt mch dao ng LC c cun thun cm L = 0,5H v t in C = 50F. Hiu in th cc i gia hai bnt l 5V. Nng lng dao ng ca mch v chu k dao ng ca mch l:

A. 2,5.10-4J ;100

s. B. 0,625mJ;

100

s. C. 6,25.10-4J ;

10

s. D. 0,25mJ ;

10

s.

Dng 5: CHO BIU THC DNG IN XC NH CC I LNG CN LICu 1: Cng dng in tc thi trong mt mch dao ng LC l tng l i = 0,08cos2000t(A). Cun dy c tcm l 50Mh. Xc nh hiu in th gia hai bn t in ti thi im cng dng in tc thi bng gi tr hiudng ? A. V54 B. V24 C. V34 D. V4Cu 2: Mch dao ng l tng LC gm t in c in dung 25nF v cun dy c t cm L. Dng in trong mch

bin thin theo phng trnh i = 0,02cos8000t(A). Tnh nng lng in trng vo thi im st48000

= ?

A. 38,5 J B. 39,5 J C. 93,75 J D. 36,5 JCu 3: Mch dao ng l tng LC gm t in c in dung 25nF v cun dy c t cm L. Dng in trong mch

bin thin theo phng trnh i = 0,02cos8000t(A). Xc nh L v nng lng dao ng in t trong mch ?A. 0,6H, 385 J B. 1H, 365 J C. 0,8H, 395 J D. 0,625H, 125 J

Cu 4: Mch dao ng l tng LC c cung cp mt nng lng J4 t mt ngun in mt chiu c sut inng 8V. Xc nh in dung ca t in ?

A. 0,145 J B. 0,115 J C. 0,135 J D. 0,125 JCu 5: Mch dao ng l tng LC c cung cp mt nng lng J4 t mt ngun in mt chiu c sut inng 8V. Bit tn s gc ca mch dao ng 4000rad/s. Xc nh t cm ca cun dy ?

A. 0,145H B. 0,5H C. 0,15H D. 0,35H

Cu 6: Mch dao ng l tng LC gm t in c in dung C v cun dy c t cm L = 0,125H. Dng ngunin mt chiu c sut in ng cung cp cho mch mt nng lng 25 J th dng in tc thi trong mch l I= I0cos4000t(A). Xc nh ?

A. 12V B. 13V C. 10V D. 11VCu 7 : Mt mch dao ng gm c cun dy L thun cm v t in C thun dung khng. Khong thi gian hai ln lintip nng lng in trng trong t bng nng lng t trng trong cun dy l:

A. LC B.2

LC C.4

LC D.3

LC

Cu 8: Mch dao ng in t LC gm mt cun dy thun cm c t cm 1mH v t in c in dung F

1,0.

Tnh khong thi gian t lc hiu in th trn t cc i U0 n lc hiu in th trn t2

0U+ ?

A. 3 s B. 1 s C. 2 s D. 6 s

Cu 9: Xt mch dao ng l tng LC. Thi gian t lc nng lng in trng cc i n lc nng lng t trngcc i l:

A. LC B.4

LC C.2

LC D. LC2

Cu 10: Trong mch dao ng b t in gmg hai t in C1, C2 ging nhau c cp mt nnglng 1 J t ngun in mt chiu c sut in ng 4V. Chuyn kho K t v tr 1 sang v tr 2.C sau nhng khong thi gian nh nhau 1 s th nng lng trong t in v trong cun cm li

bng nhau. Xc nh cng dng in cc i trong cun dy ?A. 0,787A B. 0,785A C. 0,786A D. 0,784A

Cu 12: Trong mch dao ng t in c cp mt nng lng 1 J t ngun in mt chiu csut in ng 4V. C sau nhng khong thi gian nh nhau 1 s th nng lng trong t in v trong cun cm libng nhau. Xc nh t cm ca cun dy ?

A. H234

B. H235

C. H232

D. H230

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26/34

Cu 13: Mch dao ng in t LC gm mt cun dy thun cm c t cm L v t c in dung. Dng ngun inmt chiu c sut in ng 6V cung cp cho mch mt nng lng 5 J th c sau khong thi gian ngn nht 1 s dng in trong mch trit tiu. Xc nh L ?

A. H2

3B. H

2

6,2C. H

2

6,1D. H

2

6,3

Cu 14: Mch dao ng LC l tng, cng dng in tc thi trong mch bin thin theo phng trnh i = 0,04cost (A). Xc nh C ? Bit c sau nhng khong thi gian nhn nht 0,25 s th nng lng in trng v nng

lng t trng bng nhau v bng J

8,0.

A. pF

125 B. pF

100 C. pF

120 D. pF

25

Dng 6: VIT BIU THC IN TCH, CNG DNG IN V HIU IN TH

Cu 1: Mt cun dy thun cm, c t cm HL

2= , mc ni tip vi mt t in c in dung FC 18,3= . in p

tc thi trn cun dy c biu thc ))(6

100cos(100 VtuL

= . Biu thc ca cng dng in trong mch c dng l:

A. )3

100cos(

= ti (A) B. )3

100cos(

= ti (A)

C. )3100cos(51,0= ti (A) D. )3100cos(51,0

+= ti (A)

Cu 2: Mch dao ng gm t in c in dung C v cun dy c t cm L = 10-4H. in tr thun ca cun dyv cc dy ni khng ng k. Bit biu thc ca in p gia hai u cun dy l: u = 80cos(2.106t - /2)V, biu thcca dng in trong mch l:

A. i = 4sin(2.106t )A B. i = 0,4cos(2.106t - )A C. i = 0,4cos(2.106t)A D. i = 40sin(2.106t -2

)A

Cu 3: Mt mch dao ng LC gm mt cun cm HL 640= v mt t in c in dung pFC 36= . Ly

102 = . Gi s thi im ban u in tch ca t in t gi tr cc i Cq 60 10.6

= . Biu thc in tch trnbn t in v cng dng in l:

A. )(10.6,6cos10.676

Ctq

= v ))(210.1,1cos(6,67

Ati

=

B. )(10.6,6cos10.6 76 Ctq = v ))(2

10.6,6cos(6,39 7 Ati

+=

C. )(10.6,6cos10.6 66 Ctq = v ))(2

10.1,1cos(6,66 Ati

=

D. )(10.6,6cos10.6 66 Ctq = v ))(2

10.6,6cos(6,39 6 Ati

+=

Cu 4: Cng dng in tc thi trong mt mch dao ng l )(100cos05,0 Ati = . H s t cm ca cun dyl 2mH. Ly 102 = . in dung v biu thc in tch ca t in c gi tr no sau y ?

A. FC2

10.5

= v ))(2100cos(10.5 4

Ctq

=

B. FC3

10.5

= v

))(2

100cos(10.5 4

Ctq

=

C. FC 310.5 = v ))(2

100cos(10.5 4

Ctq

+=

D. FC 210.5 = v

)(100cos10.5 4

Ctq

=

Cu 5: Trong mch dao ng LC l tng th dng in trong mch

A. ngc pha vi in tch t in. B. tr pha

3

so vi in tch t in.

C. cng pha vi in in tch t in. D. sm pha2

so vi in tch t in.

CH II. MCH DAO NG C CC T GHP, C IN TR THUNDng 1: MCH GHP

Trang 26


27/34

Cu 1: Mch dao ng ca my thu sng v tuyn c t in vi in dung C v cun cm vi t cm L, thu csng in t c bc sng 20m. thu c sng in t c bc sng 40m, ngi ta phi mc song song vi t inca mch dao ng trn mt t in c in dung C bng

A. 4C. B. C. C. 3C. D. 2C.

Cu 2: Mt mch dao ng in t khi dng t C1 th tn s dao ng ring ca mch l f1= 3 MHz. Khi mc thm tC2 song song vi C1 th tn s dao ng ring ca mch l f= 2,4MHz. Nu mc thm t C2 ni tip vi C1 th tn sdao ng ring ca mch s bng

A. 0,6 MHz B. 5,0 MHz C. 5,4 MHz D. 4,0 MHzCu 3: Cho mt mch dao ng in t gm mt t in C v mt cun cm L. B qua in tr thun ca mch. Nu

thay C bi cc t in C1, C2 ( C1 > C2 ) mc ni tip th tn s dao ng ring ca mch l 12,5Hz, cn nu thay bi hait mc song song th tn s dao ng ring ca mch l 6Hz. Xc nh tn s dao ng ring ca mch khi thay C biC1 ? A. 10MHz B. 9MHz C. 8MHz D. 7,5MHzCu 4: Khi mc t C1 vo mch dao ng th mch c f1 = 30kHz khi thay t C1 bng t C2 th mch c f2 = 40kHz.Vy khi mc song song hai t C1, C2 vo mch th mch c f l:

A. 24(kHz) B. 50kHz C. 70kHz D. 10(kHz)Cu 5: Mt mch dao ng in t, t in c in dung 40nF, th mch c tn s 2.10 4 Hz. mch c tn s 104Hzth phi mc thm t in c gi tr

A. 120nF ni tip vi t in trc. B. 120nF song song vi t in trc.C. 40nF ni tip vi t in trc. D. 40nF song song vi t in trc.

Cu 6: Mt mch dao ng LC gm cun cm thun c t cm L = 640mH v t in c in dung C bin thin t36pF n 225pF. Tn s ring ca mch bin thin trong khong:

A. 0,42kHz 1,05kHz B. 0,42Hz 1,05Hz C. 0,42GHz 1,05GHz D. 0,42MHz 1,05MHzCu 7: Mch dao ng LC l tng c t cm L khng i. Khi t in c in dung C 1 th tn s dao ng ringca mch l f1 = 75MHz. Khi ta thay t C1 bng t C2 th tn s dao ng ring ca mch l f2 = 100MHz. Nu ta dngC1 ni tip C2 th tn s dao ng ring f ca mch l :

A. 175MHz B. 125MHz C. 87,5MHz D. 25MHzCu 8: Mt mch dao ng in t c cun cm khng i L. Nu thay t in C bi cc t in C1, C2, C1 ni tip C2,C1 song song C2 th chu k dao ng ring ca mch ln lt l T1, T2, Tnt = 48 s , Tss = 10 s . Hy xc nh T1, bitT1 > T2 ? A. 9 s B. 8 s C. 10 s D. 6 sCu 9: Mt cun cm L mc vi t C1 th tn s ring ca mch dao ng f1 = 7,5MHz. Khi mc L vi t C2 th tn sring ca mch dao ng l f2 = 10MHz. Tm tn s ring ca mch dao ng khi ghp C1 song song vi C2 ri mc voL. A. 2MHz. B. 4MHz. C. 6MHz. D. 8MHz.Cu 10: Trong mch dao ng in t LC l tng, khi dng cun cm L 1 th tn s dao ng in t trong mch l f1= 30 kHz, khi dng cun cm L2 th tn s dao ng in t trong mch l f2 = 40kHz. Khi dng c hai cun cm trnmc ni tip th tn s dao ng in t l

A. 24 kHz B. 50 kHz C. 35 kHz D. 38 kHz

Cu 11: Khi mc t in C1 vi cun cm L th mch thu c sng c bc sng 1 = 60m; Khi mc t in c indung C2 vi cun cm L th mch thu c sng c bc sng 2 = 80m. Khi mc C1 ni tip C2 vi cun cm L thmch thu c sng c bc sng l bao nhiu ?

A. = 140m. B. = 100m C. = 48m. D. = 70m.Cu 12: Mt my thu thanh c mch chn sng l mch dao ng LC l tng, vi t C c gi tr C1 th sng bt cc bc sng 300m, vi t C c gi tr C2 th sng bt c c bc sng 400m. Khi t C gm t C1 mc ni tip vi tC2 th bc sng bt c l

A. 500m B. 240m C. 700m D. 100m

Cu 13: Mch dao ng LC trong my thu v tuyn c in dung C0 =8,00.10-8

F v t cm L = 2.10-6

H, thu csng in t c bc sng 240m. thu c sng in t c bc sng 18 m ngi ta phi mc thm vo mchmt t in c in dung C bng bao nhiu v mc nh th no ?

A. Mc ni tip v C = 4,53.10-10F B. Mc song song v C = 4,53.10-10FC. Mc song song v C = 4,53.10-8F D. Mc ni tip v C = 4,53.10-8F

Cu 14: Mt mch dao ng in t gm mt cun dy thun cm c t cm L v hai t in C 1 v C2. Khi mccun dy ring vi tng t C1 v C2 th chu k dao ng ca mch tng ng l T 1 = 3ms v T2 = 4ms. Chu k dao ngca mch khi mc ng thi cun dy vi C1 song song C2 l

A. 11ms B. 5 ms C. 7 ms D. 10 ms

Cu 15: Mt mch dao ng l tng gm cun cm c t cm L v t in c in dung C thc hin dao ng int vi chu k T= 10-4s. Nu mc ni tip thm vo mch mt t in v mt cun cm ging ht t in v cun cmtrn th mch s dao ng in t vi chu k

A. 0,5.10-4s . B. 2.10-4s . C. 2 .10-4s . D. 10-4s .Cu 16: Mch dao ng gm cun cm v hai t in C1 v C2. Nu mc hai t C1 v C2 song song vi cun cm L thtn s dao ng ca mch l f1 = 24kHz. Nu dng hai t C1 v C2 mc ni tip th tn s ring ca mch l f2 = 50kHz.

Nu mc ring l tng t C1, C2 vi cun cm L th tn s dao ng ring ca mch lA. f1 = 40kHz v f2 = 50kHz B. f 1 = 50kHz v f2 = 60kHz

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C. f1 = 30kHz v f2 = 40kHz D. f1 = 20kHz v f2 = 30kHz

Dng 2: CNG SUT CN CUNG CP CHO MCH B VO PHN HAO PH DO TO NHIT

Cu 1: Mt mch dao ng gm mt t in 350pF, mt cun cm 30 H v mt in tr thun 1,5 . Phi cung cpcho mch mt cng sut bng bao nhiu duy tr dao ng ca n, khi in p cc i trn t in l 15V.

A. 1,69.10-3 W B. 1,79.10-3 W C. 1,97.10-3 W D. 2,17.10-3 W

Cu 2: Mt mch dao ng gm mt cun cm c in tr r = 0,5 , t cm 275 H, v mt t in c in dung4200pF. Hi phi cung cp cho mch mt cng sut bao nhiu duy tr dao ng vi in p cc i trn t l 6V.

A. 513 W B. 2,15mW C. 137mW D. 137 WCu 3: Mch dao ng gm cun dy c L = 210-4H v C = 8nF, v cun dy c in tr thun nn duy tr mt hiuin th cc i 5V gia 2 bn cc ca t phi cung cp cho mch mt cng sut P = 6mW. in tr ca cun dy cgi tr: A. 100 B. 10 C. 50 . D. 12

CH III. S PHT V THU SNG IN TCu 1: Dao ng in t trong mch chn sng ca my thu khi my thu bt c sng l:

A. Dao ng t do vi tn s bng tn s ring ca mchB. Dao ng cng bc c tn s bng tn s ring ca mchC. Dao ng tt dn c tn s bng tn s ring ca mchD. C 3 cu trn u sai

Cu 2: Sng in t dng trong thng tin lin lc di nc l

A. sng ngn B. sng di C. sng trung D. sng cc ngnCu 3: Mt mch dao ng LC ang thu c sng trung. mch c th thu c sng ngn th phi

A. mc ni tip thm vo mch mt t in c in dung thch hpB. mc ni tip thm vo mch mt in tr thun thch hpC. mc ni tip thm vo mch mt cun dy thun cm thch hpD. mc song song thm vo mch mt t in c in dung thch hp

Cu 4: Chn phng n sai khi ni v b sung nng lng cho mch:A. b sung nng lng ngi ta s dng my pht dao ng iu ho.B. Dng ngun in khng i cung cp nng lng cho mch thng qua tranzito.C. Sau mi chu k, mch c b sung ng lc mt nng lng ln hn hoc bng nng lng tiu hao.D. My pht dao ng iu ho dng tranzito l mt mch t dao ng sn ra dao ng in t cao tn.

Cu 5: Chn pht biu sai.A. Bin iu sng l lm cho bin ca sng cao tn bin thin tun hon theo m tn.B. Mch chn sng trong my thu v tuyn hot ng da vo hin tng cng hng in t.C. Trong tn hiu v tuyn c pht i, sng cao tn l sng in t, m tn l sng c.D. Mt ht mang in dao ng iu ha th n bc x ra sng in t cng tn s vi dao ng ca n.

Cu 6: iu no sau y l sai khi ni v nguyn tc pht v thu sng in t ?A. Khng th c mt thit b va thu v pht sng in t. B. thu sng in t cn dng mt ng ten.C. Nh c ng ten m ta c th chn lc c sng cn thu.D. pht sng in t phi mc phi hp mt my dao ng iu ho vi mt ng ten.

Cu 7: Gia hai mch dao ng xut hin hin tng cng hng, nu cc mch c:A. Tn s dao ng ring bng nhau. B. in dung bng nhauC. in tr bng nhau. D. cm ng t bng nhau.

Cu 8: Nguyn tc thu sng in t da voA. hin tng hp th sng in t ca mi trng B. hin tng giao thoa sng in tC. hin tng bc x sng in t ca mch dao ng h D. hin tng cng hng in trong mch LC

Cu 9: Mch chn sng ca mt my thu thanh gm mt cun dy thun cm v mt t in c in dung bin ic. Khi t in dung ca t in c gi tr 20pF th bt c sng c bc sng 30m. Khi in dung ca t in gitr 180pF th s bt c sng c bc sng l

A. 150 m. B. 270 m. C. 90 m. D. 10 m.

Cu 10: Mch chn sng ca mt my thu v tuyn in gm mt t in c in dung 0,1nF v cun cm c tcm 30 H. Mch dao ng trn c th bt c sng v tuyn thuc di

A. sng trung B. sng di C. sng ngn D. sng cc ngn

Cu 11: Mch chn sng ca mt my thu v tuyn in gm mt t in c in dung thay i t pF

10n

pF

160v cun dy c t cm F

5,2. Mch trn c th bt c sng in t c bc sng nm trong khong

no ?A. mm 122 B. mm 123 C. mm 153 D. mm 152

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Cu 12: Mt mch chn sng gm cun cm c t cm 4 H v mt t in c in dung bin i t 10pF n360pF. Ly 2 = 10. Di sng v tuyn thu c vi mch trn c bc sng trong khong:

A. T 120m n 720m B. T 12m n 72m C. T 48m n 192m D. T 4,8m n 19,2m

Cu 13: Mch chn sng ca mt my thu v tuyn in gm mt t in c in dung 1F v cun cm c t cm25mH. Mch dao ng trn c th bt c sng v tuyn thuc di

A. sng trung B. sng di C. sng cc ngn D. sng ngnCu 14: Mt mch chn sng gm cun dy c h s t cm khng i v mt t in c in dung bin thin. Khiin dung ca t l 20nF th mch thu c bc sng 40m. Nu mun thu c bc sng 60m th phi iu chnhin dung ca t

A. Gim 4nF B. Gim 6nF C. Tng thm 25nF D. Tng thm 45nF

Dng 2: IU CHNH MCH THU SNGCu 1: Mch dao ng LC ca mt my thu v tuyn c L bin thin t 4mH n 25mH, C=16pF, ly 102 = .My ny c th bt c cc sng v tuyn c bc sng t:

A. 24m n 60m B. 480m n 1200m C. 48m n 120m D. 240m n 600m

Cu 2: Mch chn sng ca mt my thu thanh gm cun dy c t cm L = 2.10 -6 H, in tr thun R = 0. mythu thanh ch c th thu c cc sng in t c bc sng t 57m n 753m, ngi ta mc t in trong mch trn

bng mt t in c in dung bin thin. Hi t in ny phi c in dung trong khong no?A. 2,05.10-7F C 14,36.10-7F B. 0,45.10-9F C 79,7.109FC. 3,91.10-10F C 60,3.10-10F D. 0,12.10-8F C 26,4.10-8F

Cu 3: Mt my thu thanh ang thu sng ngn. chuyn sang thu sng trung, c th thc hin gii php no sau ytrong mch dao ng antenA. Gim C v gim L. B. Gi nguyn C v gim L. C. Tng L v tng C. D. Gi nguyn L v gim

Cu 4: Mch chn sng ca mt my thu gm mt t in c in dung pF29

4

v cun cm c t cm bin

thin. c th bt c sng in t c bc sng 100m th t cm cun dy bng bao nhiu ?A. 0,0645H B. 0,0625H C. 0,0615H D. 0,0635H

Cu 5: in dung ca t in phi thay i trong khong no mch c th thu c sng v tuyn c tn s nmtrong khong t f1 n f2 ( f1 < f2 ). Chn kt qu ng:

A. 22

22

1

2 2

1

2

1

LfC

Lf >> B. 2

2

22

1

2 2

1

2

1

LfC

Lf > C

Lf

Cu 6: Mt mch chn sng thu c sng c bc sng 20 m th cn chnh in dung ca t l 200 pF. thuc bc sng 21 m th chnh in dung ca t l

A. 220,5 pF. B. 190,47 pF. C. 210 pF. D. 181,4 mF.Cu 7: Mt sng in t c tn s 100 MHz truyn vi tc 3.108 m/s c bc sng l

A. 300 m. B. 0,3 m. C. 30 m. D. 3 m.

Cu 8: Mt mch thu sng in t gm cun dy thun cm c h s t cm khng i v t in c in dung bini. thu c sng c bc sng 90 m, ngi ta phi iu chnh in dung ca t l 300 pF. thu c sng 91 mth phi

A. tng in dung ca t thm 303,3 pF. B. tng in dung ca t thm 306,7 pF.C. tng in dung ca t thm 3,3 pF. D. tng in dung ca t thm 6,7 pF.Dng 3: T XOAY

Cu 1: Mch chn sng ca mt my thu v tuyn gm mt cun dy c t cm mF2108

1

v mt t xoay. Tnh

in dung ca t thu c sng in t c bc sng 20m ?A. 120pF B. 65,5pF C. 64,5pF D. 150pF

Cu 2: Mch chn sng ca mt my thu v tuyn gm mt cun dy c t cm mF2108

1

v mt t xoay. T

xoay c in dung bin thin theo gc xoay C = + 30(pF). thu c sng in t c bc sng 15m th gc xoaybng bao nhiu ?

A. 35,50

B. 37,50

C. 36,50

D. 38,50

Cu 3: Mch chn sng ca my thu v tuyn in gm cun dy thun cm c L = 2.10 -5H v mt t xoay c indung bin thin t C1 = 10pF n C2 = 500pF khi gc xoay bin thin t 00 n 1800. Khi gc xoay ca t bng 900 thmch thu sng in t c bc sng l:

A. 26,64m. B. 188,40m. C. 134,54m. D. 107,52m.

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Cu 4: Mch chn sng ca mt my thu gm mt t in c in dung 100pF v cun cm c t cm H2

1.

c th bt c sng in t c bc sng t 12m n 18m th cn phi ghp thm mt t in c in dung bin thin.in dung bin thin trong khong no ?

A. nFCnF 9,03,0 B. nFCnF 8,03,0 C. nFCnF 9,04,0 D. nFCnF 8,04,0 Cu 5: Mch chn sng ca mt my thu v tuyn gm mt t in c in dung 2000pF v cun cm c t cm 8,8

H . c th bt c di sng ngn c bc sng t 10m n 50m th cn phi ghp thm mt t in c in dungbin thin. in dung bin thin trong khong no ?A. nFCnF 3,92,4 B. nFCnF 9,03,0 C. nFCnF 8,04,0 D.

nFCnF 3,82,3

Dng 4: XC NH C TRNG L0C0Cu 1: Mch chn sng ca mt my thu v tuyn gm mt cun dy c t cm L v mt b t in gm mt t

in c nh C0 mc song song vi mt t xoay C. T C c in dung thay i t pF23

1n pF5,0 . Nh vy mch

c th thu c cc sng c bc sng t 0,12m n 0,3m. Xc nh t cm L ?

A.2

5,1

H B.

2

2

H C.

2

1

H D.

1H

Cu 2: Mch chn sng ca mt my thu v tuyn gm mt cun dy c t cm L v mt b t in gm mt tin c nh C0 mc song song vi mt t C. T C c in dung thay i t 10nF n 170nF. Nh vy mch c th thu

c cc sng c bc sng t n 3 . Xc nh C0 ?A. 45nF B. 25nF C. 30nF D. 10nFCu 3: Mch chn sng ca mt my thu v tuyn gm mt cun dy c t cm L v mt b t in gm mt tin c nh C0 mc song song vi mt t xoay C. T C c in dung thay i t 10pF n 250pF. Nh vy mch c ththu c cc sng c bc sng t 10m n 30m. Xc nh t cm L ?

A. 0,93 H B. 0,84 H C. 0,94 H D. 0,74 HDng 5: T XOAY V MCH C IN TR THUNCu 1: Mch chn sng ca mt my thu v tuyn gm mt cun dy v mt t xoay. Gi s khi thu c sng in tc bc sng 15m m sut in ng hiu dng trong cun dy l 1 V th tn s gc v dng in cc i chy trongmch l bao nhiu ? Bit in tr thun trong mch l m01,0 .

A. A

s

rad2,0;10 7 B. A

s

rad1,0;10.4 7 C. A

s

rad3,0;10.4 7 D. A

s

rad1,0;10.2 7

Cu 2: Mch chn sng ca mt my thu v tuyn gm mt cun dy c t cm H5,2 v mt t xoay. in trthun ca mch l m3,1 . Sau khi bt c sng in t c bc sng 21,5m th xoay nhanh t sut in ngkhng i nhng cng hiu dng dng in th gim xung 1000ln. Hi in dung t thay i bao nhiu ?A. 0,33pF B. 0,32pF C. 0,31pF D. 0,3pFCu 3: Mch chn sng ca mt my thu v tuyn gm mt cun dy c t cm H5,2 v mt t xoay. Sau khi

bt c sng in t c bc sng 21,5m th tn s gc v in dung t in bng bao nhiu ?

A. pFs

rad2,5;10

7 B. pFs

rad42;10.4 7 C. pF

s

rad2,4;10.2

7 D. pFs

rad52;10.8,8 7

Cu 4: Mt mch chn sng ca my thu v tuyn gm cun cm L = 5 H v mt t xoay c in dung bin thin t10pF n 240pF. Di sng my thu c l

A. 10,5m 92,5m. B. 11m 75m. C. 15,6m 41,2m. D. 13,3 65,3m.

Cu 5: Chn cu ng. Mt mch dao ng gm cun cm c t cm 27H, mt in tr thun 1 v mt t in3000pF. in p cc i gia hai bn t in l 5V. duy tr dao ng cn cung cp cho mch mt cng sut:

A. 0,037W. B. 112,5 kW. C. 1,39mW. D. 335,4 W.

E. MT S CU TRC NGHIM THI H-C CC NM TRC* H C nm 2009:Cu 1. Mt mch dao ng in t LC l tng gm cun cm thun c t cm 5 H v t in c indung 5 F. Trong mch c dao ng in t t do. Khong thi gian gia hai ln lin tip m in tch trnmt bn t in c ln cc i l

A. 5 .10-6 s. B. 2,5 .10-6 s. C.10 .10-6 s. D. 10-6 s.

Cu 2. Trong mch dao ng LC l tng ang c dao ng in t t do, in tch ca mt bn t in vcng dng in qua cun cm bin thin iu ha theo thi gian

A. lun ngc pha nhau. B. vi cng bin .C. lun cng pha nhau. D. vi cng tn s.

Cu 3. Khi ni v dao ng in t trong mch dao ng LC l tng, pht biu no sau y sai?

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A. Cng dng in qua cun cm v hiu in th gia hai bn t in bin thin iu ha theo thigian vi cng tn s.

B. Nng lng in t ca mch gm nng lng t trng v nng lng in trng.C. in tch ca mt bn t in v cng dng in trong mch bin thin iu ha theo thi gian lch

pha nhau2

.

D. Nng lng t trng v nng lng in trng ca mch lun cng tng hoc lun cng gim.Cu 4. Mt mch dao ng LC l tng gm cun cm thun c t cm khng i, t in c in dung C

thay i. Khi C = C1 th tn s dao ng ring ca mch l 7,5 MHz v khi C = C2 th tn s dao ng ringca mch l 10 MHz. Nu C = C1 + C2 th tn s dao ng ring ca mch lA. 12,5 MHz. B. 2,5 MHz. C. 17,5 MHz. D. 6,0 MHz.

Cu 5. Mt sng in t c tn s 100 MHz truyn vi tc 3.108 m/s c bc sng lA. 300 m. B. 0,3 m. C. 30 m. D. 3 m.

Cu 6. Mt mch dao ng in t LC l tng gm cun cm thun t cm L v t in c in dungthay i c t C1 n C2. Mch dao ng ny c chu k dao ng ring thay i c.

A. t 14 LC n 24 LC . B. t 12 LC n 22 LC .

C. t 12 LC n 22 LC . D. t 14 LC n 24 LC .

Cu 7. Mch thu sng in t gm cun dy thun cm c t cm khng i v t in c in dung bin

i. thu c sng c bc sng 90 m, ngi ta phi iu chnh in dung ca t l 300 pF. thu csng 91 m th phi

A. tng in dung ca t thm 303,3 pF. B. tng in dung ca t thm 306,7 pF.C. tng in dung ca t thm 3,3 pF. D. tng in dung ca t thm 6,7 pF.

Cu 8. Mt mch chn sng thu c sng c bc sng 20 m th cn chnh in dung ca t l 200 pF. thu c bc sng 21 m th chnh in dung ca t l

A. 220,5 pF. B. 190,47 pF. C. 210 pF. D. 181,4 mF.Cu 9. Trong mch dao ng LC l tng c dao ng in t t do th

A. nng lng in trng tp trung cun cm.B. nng lng in trng v nng lng t trng lun khng i.C. nng lng t trng tp trung t in.D. nng lng in t ca mch c bo ton.

Cu 10. Mt mch dao ng LC l tng ang c dao ng in t t do. Bit in tch cc i ca mt bnt in c ln l 10-8 C v cng dng in cc i qua cun cm thun l 62,8 mA. Tn s dao ngin t t do ca mch l

A. 2,5.103 kHz. B. 3.103 kHz. C. 2.103 kHz. D. 103 kHz.Cu 11. Mch dao ng LC l tng gm t in c in dung C, cun cm thun c t cm L. Trongmch c dao ng in t t do. Bit hiu in th cc i gia hai bn t in l U0. Nng lng in t camch bng

A. 21

LC2

. B.2

0U LC2

. C. 201

CU2

. D. 21

CL2

.

Cu 12. Mt mch dao ng LC l tng, gm cun cm thun c t cm L v t in c in dung C.Trong mch c dao ng in t t do. Gi U 0, I0 ln lt l hiu in th cc i gia hai u t in vcng dng in cc i trong mch th

A. 00I

ULC

= . B.0 0

LU I

C= . C.

0 0

CU I

L= . D. 0 0U I LC= .

H C nm 2010Cu 13. Mt mch dao ng l tng gm cun cm thun c t cm 4 H v mt t in c in dung

bin i t 10 pF n 640 pF. Ly 2 = 10. Chu k dao ng ring ca mch ny c gi trA. t 2.10-8 s n 3,6.10-7 s. B. t 4.10-8 s n 2,4.10-7 s.

C. t 4.10-8 s n 3,2.10-7 s. D. t 2.10-8 s n 3.10-7 s.HNG DN: Theo cng thc tnh chu k dao ng ca mch:

min aX2 2 2 MT c LC c LC T c LC = , thay s vo ta c 4.10-8s n 3,2.10-7s

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Cu 14. Mt mch dao ng l tng gm cun cm thun c t cm L khng i v t in c in dungC thay i c. iu chnh in dung ca t in n gi tr C1 th tn s dao ng ring ca mch l f1. tn s dao ng ring ca mch l 5 f1 th phi iu chnh in dung ca t in n gi tr

A. 5C1.

ONTAP_VATLY12_Daodongdientu

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