NHNG BI TP KH MAHC DAO NG BI 1: cho ngun in c sut in ng E=12V in
tr trong r=1 ,t c in dung C=100 F,cun dy c h s t cm L=0.2H c in tr
R0=5 ,V c in tr R=18 .Ban u ng kha K np in cho t.Tnh nhit lng ta ra
trn ien tr R trong thi gian t khi ngt K n khi dao ng trong mch tt
hon ton A;25 mJ B:28.45mJ C:24.74mJ D;31.61Mj Bi gii Ban duong k va
khi dong in trong mach n inh ta co :
in ap hai u tu in Nng lng t trng trong cun dy Nng lng in trng
trong tu in Nng lng toan phn cua mach luc ban u T khi ngt K n khi
dao ng trong mch tt hon ton thi toan b nng lng nay bin thanh nhit
lng toa ra trn cun dy va trn in tr R Goi va ln lt la nhit lng toa
ra trn cun dy va trn in tr R ta co : (1) va (2) T (1) va (2) ta
tinh c Bi2: Mch dao ng in t gm cun cm v 1 b 2 t in c cng in dung C
= 2,5 F mc song song .Trong mch c dao ng in t t do ,hiu in th cc i
gia 2 bn t in l Uo = 12V .Ti thi im hiu in th 2 u cun cm uL = 6V th
1 t in b bong ra v t dy ni .Tnh nng lng cc i trong cun cm sau :
A.0,315mJ B.0,27mJ C.0,135mJ D.0,54mJ Hai t ging nhau mc song song,
in dung tng ng ca b t: Nng lng in t trng khi c 2 t:
thi im
th : (ch rng
)
(Mch vn cn 2 t) Khi 1 t b t dy, th by gi trong mch ch cn 1 t v 1
cun dy thi. Nng lng ca mch lc ny l:
Nng lng cc i trong cun cm lc sau: Cu1: Mch dao ng c C= 10F, L=1H
. Ly pi bnh phng =10 . Khong time ngn nht t lc nng lung in trng max
n lc nng lng t bng 1 na nng lng in trng cc i l: a.1/400s b.1/300s
c.1/200s D.1/100s Bai3: Trong mch dao ng l tng LC c dao ng in t t
do (dao ng ring) vi t in c in dung C = 2nF .Ti thi im t1 cng dng in
trong mch i = 5mA ,sau T/4 hiu in th gia 2 bn t u = 10V. t cm ca
cun dy? A.0,04mH B.8mH C.2,5mH D.p n khc +Ti thi im t1 gi tr i: gi
tr u : + Sau T/4 th (1) vo Bai4: Mch dao ng gm cun dy c t cm L=20
micro H, in tr thun R=4om, t in C=2nF. Hdt cc i 2 u t l 5V. duy tr
dao ng in t trong mch ngi ta dng 1 pin c sut in ng E=5V, in lng ban
u l 30C. Cc pin c th duy tr dao ng in t ti a l? A. 3000 pht B. 50
pht C. 500 pht D. 300 pht Cu ny mnh khng chc lm xin lm th nh
sau:
Cng sut ta nhit trn in tr R l: Thi gian m cc pin duy tr dao ng
in l: (pht) Bai5: trong mch dao ng in t gm cun dy thun cm L v b 2 t
in C1 song song vi C2 vi C1=C2=6 .ti thi im dng in qua cun dy bng 1
na dng in cc i th in tch ca t C2 l q2=9 . in p cc i trn t C1? Khi
i=I0/2 th nn u=U0 /2 suy ra U0 Bai6: :cho mch dao ng LC gm t in c
in dung C=10pF,mt cun dy c t cm L = 100 mH v in tr thun R=1 m.Ni
mch vi ngun in c sut in ng E=2V.in tr trong r = 9 m thng qua mt kha
K.Ban u ng kha K t tch y in v dng in trong mch n nh v n thi im no
ngi ta ngt kha K.Xc nh nng lng dao ng in t trong mch ngay sau khi
ngt kha k l A: 2mJ B: 1mJ C 3mJ D 4mJ
tnh
W= Bai7: Cho mch dao ng l tng vi C = 1nF, L = 1mH, in p hiu dng
ca t in l UC = 4V. Lc t = 0, uC = 2cn 2V v t in ang c np in. Vit
biu thc ca: a) in p trn t in. b) Cng dng in chy trong mch dao ng.
c) Nng lng in trng. Nng lng t trng. a/ Vi uC = 2 V th bn c chn c in
th dng so vi bn kia. T ang c np in chng t dng in c chiu i n bn ny.
Tc l i0 > 0. in tch t in tng nn uC tng. Tn s gc ca dao ng: omg =
1/cn LC = 10^6 rad/s. Thi im u: u0 = U0cos(phi); suy ra cos(phi) =
2cn2/4cn2 = 1/2; uC ang tng nn (phi) = -60 = -pi/3 rad. Vy: uC =
4cn2cos(10^6.t - pi/3) (V). Bi 8: Mch dao ng kn, l tng c L = 1 mH,
C = 10 F. Khi dao ng cng dng in hiu dng I = 1 mA. Chn gc thi gian
lc nng lng in trng bng 3 ln nng lng t trng v t in ang phng in. Vit
biu thc in tch trn t in, in p gia hai bn t v cng dng in trn mch dao
ng. 1 I0 = LC = 104 rad/s; I0 = I 2 = 2 .10-3 A; q0 = = 2 .10-7 C;
q 4 3 q0 = cos( 6 ); t ang phng in: = 6 khi t = 0 th WC = 3Wt =>
W = 3 WC => q = 2 q0 cos
; q = 2 .10-7cos(104t + 6 )(C); q 3 u = C = 2 .10-2cos(104t + 6
)(V); i = 2 .10-3cos(104t + 2 )(A). Bi 9: Trong mt mch LC, L = 25
mH v C = 1,6 F thi im t = 0, cng dng in trong mch bng 6,93 mA, in
tch trn t in bng 0,8 C v t ang c np in. Tnh nng lng ca mch dao ng,
vit biu thc in tch trn t in v cng dng in trong mch dao ng 1 q2 1 2
W= + Li = 0,8.10-6J; q0 = 2CW = 1,6.10-6C; 2 C 2 q 1 = = 5000
rad/s; cos = = cos( ); t ang np in nn = - ; q0 LC 3 3 q =
1,6.10-6cos(5000t - )(C); I0 = q0 = 8.10-3 A; 3 i = 8.10-3cos(5000t
+ )(A). 6Bi 10:
Mt mch dao ng LC ang thc hin dao ng in t t do vi nng lng dao ng
l 2J. Khong thi gian ngn nht nng lng in trng ca t tng thm 1J l 1ms.
Hi trong 2 pht, mch thc hin c bao nhiu dao ng ton phn? Bi 11: Mch
dao ng LC, cun dy c t cm L=0,05H, in tr ton mch R=0,5. Thi gian mch
giao ng tt dn l: A.0,5 s B.0,1 s C.0,05 s D.0,055 s Bi 12: Mch dao
ng LC l tng gm L v 2 t C ging nhau mc ni tip. Mch ang hot ng th
ngay ti thi im nng lng in trng v t trng = nhau, mt t b nh thng hon
ton. Dng in cc i trong mch sau s = bao nhiu ln so vi lc ban u:
A.Khng i B. 1/4 C.0,5 nhn Cn bc 2 ca 3 D. Bi 13: cho mch in c (ngun
E nt in tr trong r)//(L nt R)//C.E=12V,r=1m,C=100.10^-6F,cun dy c h
s t cm l L=0.2H v in tr R0=5m; in tr R=18m. ban u K ng,khi trng thi
trong mch n nh ngi ta ngt kha K.Tnh nhit lng ta ra trn R trong thi
gian t khi ngt K n khi dao ng trong mch tt hon ton. LIo^2/2+
cUo^2/2 = QR +Qr ( 1 ) QR/Qr =R/r ( 2 ) Giai ( 1 ) Va (2 ) Ra dap
an1. Hai t in C1=3C, C2=6C mc ni tip. Ni hai u b t vi pin c E= 3V
np in cho cc t ri ngt ra v ni vi cun dy thun cm L to thnh mch dao
ng in t t do. Ti thi im dng in qua cun dy c ln bng 1/2 gi tr I max,
nguoi ta ni tt C1. HDT cc i trn t C2 ca mch sau bng bao nhiu.?
do mc ni tip nn mnh quy v dien luong (vi mac noi tiep dien luong
qua cc t bng nhau) i=Io/2 ==> Wt=1/4W ==> Wd=3/4W m C2=6C
trong khi Cb=2C ==> Wd2=1/3Wd=1/4W - khi ngt C1 ==> mch dao
ng vi W'=1/4W+1/4W=1/2W ==> ==> ---------------------------Mt
mch dao ng gm cun thun cm L v hai t C ging nhau mc ni tip, kha K mc
hai u mt t C (nh hnh v). Mch ang hot ng vi dng in cc i trong mch l
I0 th ta ng kha K ngay ti thi im nng lng in trng v
nng lng t trng trong mch ang bng nhau. Gi tr cc i ca dng in
trong mch ca mch sau s thay i nh th no? ngay khi ng kha k ta c:
1/2.L.Io^2 = 1/2.Cb.u^2 + 1/2.L.i^2 v Cb =C/2 v nng lng t trng bng
nng lng in trng nn(1/2.Cb.u^2 = 1/2.L.i^2) 1/2.L.Io^2 =
2.1/2.C/2.u^2 =>u = Io.cn(L/C) hiu in th hai u mt t l: u1=u/2 =
Io.cn(L/C)/2 nng lng ca mt t khi l: w1 = 1/2.C.u1^2 = 1/8.L.Io^2
nng lng ca t trng khi l: 1/2.L.i^2 =1/2.Cb.u^2 =1/4.L.Io^2 vy nng
lng ca mch khi ng kha k l ( mt t b loi) w' = w1 +1/2.L.i^2
=3/8.L.Io^2 ( khi ta c mt mch LC vi nng lng ban u l 3/8.L.Io^2) m
nng lng: w' =1/2.L.Io'^2 =>Io'^2 =Io.cn3/2 -----------Mt mch dao
ng l tng LC gm cun dy thun cm L v 2 t in C ging nhau mc ni tip. Mch
ang hot ng th ngay ti thi im nng lng t trng bng nng lng in trng, mt
t b nh thng hon ton. Dng in cc i trong mch sau s bng bao nhiu ln so
vi lc u: A. khng i. B. 2 C. 0,53. D. . Cc bn gii gip nh. goj W la
nang luong tu cuc daj khi chua danh thung luc nang luong tu bang
nang luong dien truong thi Wt=Wd=W/2 ma hai tu giong nhau ma mac
noj tiep nua nen Wd/2=Wd1=Wd2, do vay khi dang thung mot tu thi
nang luong tong cong luc do la W'=3/4 W vay luc nang luong tu cuc
dai luc sai chi bang 3/4 nang luong cuc daj luc dau
