Bai giang 1_Dai cuong dong dien xoay chieu

CHNG 3: DNG IN XOAY CHIU

I. KHI NIM DNG IN XOAY CHIU1) nh nghaDng in xoay chiu l dng in c cng bin thin tun hon theo thi gian (theo hm cos hay sin ca thi gian).

2) Biu thc: i = I0cos(t + i) Atrong :

i: gi tr cng dng in xoay chiu tc thi, n v l (A)

I0 > 0: gi tr cng dng in cc i ca dng in xoay chiu

, i : l cc hng s.

> 0 l tn s gc.

(t + i): pha ti thi im t.

i : Pha ban u ca dng in.

3) Chu k, tn s ca dng in

Chu k, tn s ca dng in:

V d 1: Cho dng in xoay chiu c biu thc i = 2cos(100t + /3) A.

a) Tnh cng dng in trong mch khi t = 0,5 (s); t = 0,125 (s).

b) Tm nhng thi im m cng dng in trong mch c gi tr 1 A.

c) Ti thi im t cng dng in trong mch c gi tr 1 A v ang gim. Hi sau 1/200 (s) th cng dng in c gi tr l bao nhiu?

V d 2: Cho dng in xoay chiu c biu thc i = 4cos(100t + /6) A.

a) Tnh cng dng in trong mch khi t = 0,5 (s); t = 0,125 (s).

b) Tm nhng thi im m cng dng in trong mch c gi tr 2eq \l(\r(,3)) A.

c) Ti thi im t cng dng in trong mch c gi tr 2eq \l(\r(,2)) A v ang tng. Tm cng dng in sau * (t = eq \s\don1(\f(1,120)) s* (t = eq \s\don1(\f(1,200)) s* (t = eq \s\don1(\f(1,300)) s* (t = eq \s\don1(\f(1,600)) II. IN P XOAY CHIU

Cho khung dy dn c din tch S gm c N vng dy quay u vi vn tc gc xung quanh trc i xng xx trong t trng u c xx '. Ti t = 0 gi s

Sau khong thi t, n quay c mt gc t. T thng gi qua khung l ( = NBScos(t) Wb.

t o = NBS = ocos(t), o c gi l t thng cc i. Theo hin tng cm ng in t trong khung hnh thnh sut in ng cm ng c biu thc e = = NBSsin(t).

t E0 = NBS = 0 ( e = E0sin(t) = E0cos(t - eq \l(\f((,2)))

Vy sut in ng trong khung dy bin thin tun hon vi tn s gc v chm pha hn t thng gc /2. Nu mch ngoi kn th trong mch s c dng in, in p gy ra mch ngoi cng bin thin iu ha: u = U0cos(t + u) V.

n v : S (m2), (Wb) Webe, B (T) Testla, N (vng), (rad/s), e (V) Ch : 1 vng/pht = ())eq \s\don1(\f(,60)) = ())eq \s\don1(\f(,30)) ( rad/s ); 1 cm2 = 10- 4 m2V d 1: Mt khung dy dn c din tch S = 50 cm2 gm 150 vng dy quay u vi vn tc 3000 vng/pht trong mt t trng u c cm ng t vung gc trc quay ca khung v c ln B = 0,002 T. Tnha) t thng cc i gi qua khung.b) sut in ng cc i.Hng dn giiTm tt bi:S = 50 cm2 = 50.104 m2N = 150 vng

B = 0,002 T

= 3000 vng/pht = 100 (rad/s)

a) T thng qua khung l = NBScos(t) (

t thng cc i l 0 = NBS = 150.0, 002.50.10-4 = 1, 5.10-3 Wb.

b) Sut in ng qua khung l e = ' = NBSsin(t) ( E0 = NBS = 0 = 100.1,5.10-3 = 0,47 V.

Vy sut in ng cc i qua khung l E0 = 0,47 V.

V d 2: Mt khung dy dt hnh ch nht gm 500 vng dy, din tch mi vng dy l 53,5 cm2, quay u vi tc gc l 3000 vng/pht quanh trc xx trong mt t trng u c B = 0,02 T v ng cm ng t vung gc vi trc quay xx. Tnh sut in ng cc i ca sut in ng xut hin trong khung.Hng dn gii:Tm tt bi:S = 53,5 cm2 = 53,5.104 m2N = 500 vng, B = 0,02 (T).

= 3000 vng/pht = 100 (rad/s).

Sut in ng cc i l E0 = NBS = 100.500.0,02.53,5.104 = 16,8 V.V d 3: Mt khung dy hnh ch nht, kch thc (40 cm x 60 cm), gm 200 vng dy, c t trong mt t trng u c cm ng t 0,2 (T). Trc i xng ca khung dy vung gc vi t trng. Khung dy quay quanh trc i xng vi vn tc 120 vng/pht.a) Tnh tn s ca sut in ng.b) Chn thi im t = 0 l lc mt phng khung dy vung gc vi ng cm ng t. Vit biu thc sut in ng cm ng trong khung dy.c) Sut in ng ti t = 5 (s) k t thi im ban u c gi tr no ?Hng dn gii:Tm tt bi:S = 40.60 = 2400 cm2 = 0,24 m2N = 200 vng, B = 0,2 (T).

= 120 vng/pht = 4 (rad/s).a) Tn s ca sut in ng l f = 2())eq \s\don1(\f(,)) = 2 Hz.

b) Sut in ng cc i: E0 = NBS = 4.200.0,2.0,24 = 120,64 V.

Do ti t = 0, mt phng khung vung gc vi cm ng t nn = 0 (hay )T ta c biu thc ca sut in ng l e = E0sin(t) = 120,64sin(4t) V.

c) Ti t = 5 (s) thay vo biu thc ca sut in ng vit c trn ta c e = E0 = 120,64 V.

V d 4: Mt khung dy dn phng c din tch S = 50 cm2, c N = 100 vng dy, quay u vi tc 50 vng/giy quanh mt trc vung gc vi cc ng sc ca mt t trng u c cm ng t B = 0,1 (T). Chn t = 0 l lc vect php tuyn ca khung dy cng chiu vi vect cm ng t v chiu dng l chiu quay ca khung dy.a) Vit biu thc xc nh t thng qua khung dy.b) Vit biu thc xc nh sut in ng e xut hin trong khung dy.Hng dn gii:Tm tt bi:S = 50 cm2 = 50.104 m2N = 100 vng, B = 0,1 (T).

= 50 vng/giy = 100 (rad/s).

a) Theo bi ti t = 0 ta c = 0.

T thng cc i 0 = N.B.S = 100.0,1.50.104 = 0,05 Wb.

T , biu thc ca t thng l = 0,05cos(100t) Wb.

b) Sut in ng cm ng e = - = 0,05.100 sin100t = 5sin100t V.III. LCH PHA CA IN P V DNG IN

t = u i, c gi l lch pha ca in p v dng in trong mch.

Nu > 0 thi khi in p nhanh pha hn dng in hay dng in chm pha hn in p.

Nu > 0 thi khi in p chm pha hn dng in hay dng in nhanh pha hn in p.

Ch :

- Khi lch pha ca in p v dng in l /2 th ta c phng trnh ca dng in v in p tha mn (

- Nu in p vung pha vi dng in, ng thi ti hai thi im t1, t2 in p v dng in c cc cp gi tr tng ng l u1; i1 v u2; i2 th ta c: = (

IV. CC GI TR HIU DNG

Cho dng in xoay chiu i = I0cos(t + ) A chy qua R, cng sut tc thi tiu th trn R:

p = Ri2 = RIcos2((t +() = RI

EMBED Equation.3 =

Gi tr trung bnh ca p trong 1 chu k:

=

Kt qu tnh ton, gi tr trung bnh ca cng sut trong 1 chu k (cng sut trung bnh): P = =

Nhit lng ta ra khi l Q = P.t =

Cng trong cng khong thi gian t cho dng in khng i (dng in mt chiu) qua in tr R ni trn th nhit lng ta ra l Q = I2Rt.

Cho Q = Q ( = I2Rt ( I =

I c gi l gi tr hiu dng ca cng dng in xoay chiu hay cng hiu dng.

Tng t, ta cng c in p hiu dng v sut in ng hiu dng l U = ; E =

Ngoi ra, i vi dng in xoay chiu, cc i lng nh in p, sut in ng, cng in trng, cng l hm s sin hay cosin ca thi gian, vi cc i lng ny.

Ch :Trong mch in xoay chiu cc i lng c s dng gi tr tc thi l:

v cc i lng s dng gi tr hiu dng l cng dng in I, in p U, sut in ng E.V d 1: Dng in chy qua on mch xoay chiu c dng i = 200cos(100t) A, in p gia hai u on mch c gi tr hiu dng l 12 V, v sm pha /3 so vi dng in.a) Tnh chu k, tn s ca dng in.b) Tnh gi tr hiu dng ca dng in trong mch.c) Tnh gi tr tc thi ca dng in thi im t = 0,5 (s). d) Trong mt giy dng in i chiu bao nhiu ln.e) Vit biu thc ca in p gia hai u on mch.Hng dn gii:a) T biu thc ca dng in i = 200cos(100t) A; ta c = 100 (rad/s).

T ta c chu k v tn s ca dng in l:

b) Gi tr hiu dng ca dng in trong mch l I = = eq \l(\r(,2)) Ac) Ti thi im t = 0,5 (s) th i = 2cos(10.0,5) = 0. Vy ti t = 0,5 (s) th i = 0.

d) T cu b ta c f = 50 Hz, tc l trong mt giy th dng in thc hin c 50 dao ng. Do mi dao ng dng in i chiu hai ln nn trong mt giy dng in i chiu 100 ln.

e) Do in p sm pha /3 so vi dng in nn c /3 = u i ( u = /3 (do i = 0)

in p cc i l U0 = Ueq \l(\r(,2)) = 12eq \l(\r(,2)) V Biu thc ca in p hai u mch in l u = 12eq \l(\r(,2))cos(100t + eq \l(\f((,3))) VV d 2: Mt mch in xoay chiu ch c in tr R = 50 , dng in qua mch c biu thc i = 2cos(100t + /3) A. a) Vit biu thc in p hai u mch in bit rng in p hiu dng l 50eq \l(\r(,2)) V v in p nhanh pha hn dng in gc /6.b) Tnh nhit lng ta trn in tr R trong 15 pht.Hng dn gii:a) Ta c

Biu thc ca in p l u = 100cos(100t + /2) V.

b) Cng hiu dng ca dng in: I = = eq \l(\r(,2)) AT , nhit lng ta ra trong 15 pht (15.60 = 900 (s)) l Q = I2Rt = 2.50.15.60 = 90000 J = 90 kJ.V d 3: Mt mch in xoay chiu c lch pha gia in p v cng dng in chy trong mch l /2. Ti mt thi im t, cng dng in trong mch c gi tr 2eq \l(\r(,3)) A th in p gia hai u mch l 50eq \l(\r(,2)) V. Bit in p hiu dng ca mch l 100 V. Tnh gi tr hiu dng cng dng in trong mch.Hng dn gii:Do in p v dng in lch pha nhau gc /2 nn gi s biu thc ca dng in v in p c dng nh sau: (

Thay cc gi tr bi cho ( ( I = 2eq \l(\r(,2)) AV d 4: Cho mt mch in xoay chiu c in p hai u mch l u = 50cos(100t + /6) V. Bit dng in qua mch chm pha hn in p gc /2. Ti mt thi im t, cng dng in trong mch c gi tr eq \l(\r(,3)) A th in p gia hai u mch l 25 V. Biu thc ca cng dng in trong mch l A. i = 2cos(100t + eq \l(\f((,3))) A

B. i = 2cos(100t - eq \l(\f((,3))) A C. i = eq \l(\r(,3))cos(100t - eq \l(\f((,3))) A

D. i = eq \l(\r(,3))cos(100t + eq \l(\f((,3))) A Hng dn gii: Do in p v dng in lch pha nhau gc /2 nn (

( I0 = 2A Mt khc, dng in chm pha hn in p gc /2 nn i = u - eq \l(\f((,2)) = eq \l(\f((,6)) - eq \l(\f((,2)) = - eq \l(\f((,3)) ( i = 2cos(100(t - eq \l(\f((,3))) AV. MT S BI TP TRC NGHIM IN HNHCu 1. Dng in chy qua on mch xoay chiu c dng i = 2cos(100t + /6) A, in p gia hai u on mch c gi tr hiu dng l 12 V, v sm pha /6 so vi dng in. Biu thc ca in p gia hai u on mch l

A. u = 12cos(100t + eq \l(\f((,6))) V

B. u = 12eq \l(\r(,2))cos 100t V. C. u = 12eq \l(\r(,2))cos(100t - eq \l(\f((,3))) VD. u = 12eq \l(\r(,2))cos(100t + eq \l(\f((,3))) V Hng dn gii: T gi thit ta c : ( ( u = 12eq \l(\r(,2))cos(100(t + eq \l(\f((,3)) ) VCu 2. Mt mch in xoay chiu c in p gia hai u mch l u = 200cos(100t + /6) V. Cng hiu dng ca dng in chy trong mch l 2eq \l(\r(,2)) A. Bit rng, dng in nhanh pha hn in p hai u mch gc /3, biu thc ca cng dng in trong mch l

A. i = 4cos(100t + /3) A.

B. i = 4cos(100t + /2) A. C. i = 2eq \l(\r(,2))cos(100t - /6) A. D. i = 2eq \l(\r(,2))cos(100t + /2) A.

Hng dn gii: T gi thit ta c : ( ( i = 4cos(100(t + eq \l(\f((,2))) VCu 3. Mt mch in xoay chiu c lch pha gia in p v cng dng in chy trong mch l /2. Ti mt thi im t, cng dng in trong mch c gi tr 2 A th in p gia hai u mch l 100eq \l(\r(,6)) V. Bit cng dng in cc i l 4 A. in p hiu dng gia hai u mch in c gi tr l

A. U = 100 V. B. U = 200 V. C. U = 300 V. D. U = 220 V.

Hng dn gii:

Do in p v dng in lch pha nhau gc /2 nn

Thay s ta c: ( U0= 200eq \l(\r(,2)) V ( U = 200 VCu 4. Mt khung dy quay u quanh trc xx trong mt t trng u c ng cm ng t vung gc vi trc quay xx. Mun tng bin sut in ng cm ng trong khung ln 4 ln th chu k quay ca khung phi

A. tng 4 ln. B. tng 2 ln. C. gim 4 ln.D. gim 2 ln.

Hng dn gii:T biu thc t thng ta c ( = NBScos(t + ) ( e = ( = NBSsin(t+ )Bin ca sut in ng l E0 = NBS, khi E0 tng ln 4 ln th tng 4 ln, tc l chu k T gim 4 ln.

Cu 5. Mt khung dy dn c din tch S = 50 cm2 gm 250 vng dy quay u vi tc 3000 vng/pht trong mt t trng u c vc t cm ng t vung gc vi trc quay ca khung, v c ln B = 0,02 (T). T thng cc i gi qua khung l

A. 0,025 Wb. B. 0,15 Wb. C. 1,5 Wb. D. 15 Wb.

Hng dn gii:T biu thc tnh ca t thng = NBScos(t + ) ( t thng cc i l 0 = NBS.

Thay s vi: ( (0 = 250.0,02.50.10-4 = 0,025 WbCu 6. Mt vng dy phng c ng knh 10 cm t trong t trng u c ln cm ng t B = 1/ (T). T thng gi qua vng dy khi vct cm ng t hp vi mt phng vng dy mt gc = 300 bng

A. 1,25.103 Wb. B. 0,005 Wb. C. 12,5 Wb. D. 50 Wb.

Hng dn gii:Biu thc tnh ca t thng = NBScos, vi = (), t gi thit ta c = 600. Mt khc khung dy l hnh trn c ng knh 10 cm, nn bn knh l R = 5 cm ( S = R2 = .0,052. T ta c ( = ())eq \s\don1(\f(1,)).(.0,052.cos600 = 1,25.10-3 WbI CNG V DNG IN XOAY CHIU

Cu 1: Dng in xoay chiu l dng in

A. c chiu bin thin tun hon theo thi gian.

B. c cng bin i tun hon theo thi gian.

C. c chiu bin i theo thi gian.

D. c chu k thay i theo thi gian.

Cu 2: Chn cu sai trong cc pht biu sau ?

A. Nguyn tc to ra dng in xoay chiu da trn hin tng cm ng in t.

B. Khi o cng dng in xoay chiu, ngi ta c th dng ampe k nhit.

C. S ch ca ampe k xoay chiu cho bit gi tr hiu dng ca dng in xoay chiu.

D. Gi tr hiu dng ca dng in xoay chiu bng gi tr trung bnh ca dng in xoay chiu.

Cu 3: Dng in xoay chiu hnh sin l

A. dng in c cng bin thin t l thun vi thi gian.

B. dng in c cng bin thin tun hon theo thi gian.

C. dng in c cng bin thin iu ha theo thi gian.

D. dng in c cng v chiu thay i theo thi gian. Cu 4: Cc gi tr hiu dng ca dng in xoay chiu

A. c xy dng da trn tc dng nhit ca dng in

B. ch c o bng ampe k nhit.

C. bng gi tr trung bnh chia cho 2.

D. bng gi tr cc i chia cho 2.

Cu 5: i vi dng in xoay chiu cch pht biu no sau y l ng?

A. Trong cng nghip, c th dng dng in xoay chiu m in.

B. in lng chuyn qua mt tit din thng dy dn trong mt chu k bng khng.

C. in lng chuyn qua mt tit din thng ca dy dn trong khong thi gian bt k u bng khng.

D. Cng sut to nhit tc thi c gi tr cc i bng 2 ln cng sut to nhit trung bnh.

Cu 6: Trong cc cu sau, cu no ng ?

A. Dng in c cng bin i tun hon theo thi gian l dng in xoay chiu.

B. Dng in v in p hai u mch xoay chiu lun lch pha nhau.

C. Khng th dng dng in xoay chiu m in.

D. Cng hiu dng ca dng in xoay chiu bng mt na gi tr cc i ca n.

Cu 7: Cng dng in trong mch khng phn nhnh c dng i = 2eq \l(\r(,2))cos100t A. Cng dng in hiu dng trong mch l

A. I = 4AB. I = 2,83AC. I = 2AD. I = 1,41 A.

Cu 8: in p tc thi gia hai u on mch c dng u = 141cos(100t) V. in p hiu dng gia hai u on mch l

A. U = 141 V. B. U = 50 V. C. U = 100 V. D. U = 200 V.

Cu 9: Trong cc i lng c trng cho dng in xoay chiu sau y, i lng no c dng gi tr hiu dng?

A. in p. B. chu k. C. tn s. D. cng sut.

Cu 10: Trong cc i lng c trng cho dng in xoay chiu sau y, i lng no khng dng gi tr hiu dng?

A. in p. B. Cng dng in. C. Sut in ng. D. Cng sut.

Cu 11: Pht biu no sau y l khng ng?

A. in p bin i iu ho theo thi gian gi l in p xoay chiu.

B. dng in c cng bin i iu ho theo thi gian gi l dng in xoay chiu.

C. sut in ng bin i iu ho theo thi gian gi l sut in ng xoay chiu.

D. cho dng in mt chiu v dng in xoay chiu ln lt i qua cng mt in tr th chng to ra nhit lng nh nhau.

Cu 12: Mt dng in xoay chiu chy qua in tr R = 10 , nhit lng ta ra trong 30 pht l 900 kJ. Cng dng in cc i trong mch l

A. I0 = 0,22AB. I0 = 0,32AC. I0 = 7,07AD. I0 = 10,0 A.

Cu 13: Pht biu no sau y l ng?

A. Khi nim cng dng in hiu dng c xy dng da vo tc dng ha hc ca dng in.

B. Khi nim cng dng in hiu dng c xy dng da vo tc dng nhit ca dng in.

C. Khi nim cng dng in hiu dng c xy dng da vo tc dng t ca dng in.

D. Khi nim cng dng in hiu dng c xy dng da vo tc dng pht quang ca dng in.

Cu 14: Pht biu no sau y l khng ng?

A. in p bin i theo thi gian gi l in p xoay chiu.

B. Dng in c cng bin i iu ha theo thi gian gi l dng in xoay chiu.

C. Sut in ng bin i iu ha theo thi gian gi l sut in ng xoay chiu.

D. Cho dng in mt chiu v dng in xoay chiu ln lt i qua cng mt in tr th chng ta ra nhit lng nh nhau.

Cu 15: i vi sut in ng xoay chiu hnh sin, i lng no sau y lun thay i theo thi gian?

A. Gi tr tc thi. B. Bin . C. Tn s gcD. Pha ban u.

Cu 16: Ti thi im t = 0,5 (s), cng dng in xoay chiu qua mch bng 4 A, l

A. cng hiu dng.

B. cng cc i.

C. cng tc thi.

D. cng trung bnh.

Cu 17: Cng dng in trong mt on mch c biu thc i = eq \l(\r(,2))sin(100t + eq \l(\f((,6))) A . thi im t = eq \s\don1(\f(1,100)) s cng trong mch c gi tr

A. 2A. B. - 2))eq \s\don1(\f(,2)) A. C. bng 0. D. 2 A.

Cu 18: Mt mng in xoay chiu 220 V 50 Hz, khi chn pha ban u ca in p bng khng th biu thc ca in p c dng

A. u = 220cos(50t) V.

B. u = 220cos(50t) V.

C. u = 220eq \l(\r(,2))cos(100t) V. D. u = 220eq \l(\r(,2))cos 100t V.

Cu 19: Dng in chy qua on mch xoay chiu c dng i = 2cos(100t) A, in p gia hai u on mch c gi tr hiu dng l 12 V v sm pha /3 so vi dng in. Biu thc ca in p gia hai u on mch l

A. u = 12cos(100t) V.

B. u = 12eq \l(\r(,2))sin 100t V.

C. u = 12eq \l(\r(,2))cos(100t -/3) V. D. u = 12eq \l(\r(,2))cos(100t + /3) V.

Cu 20: Dng in chy qua on mch xoay chiu c dng i = 2cos(100t + /6) A, in p gia hai u on mch c gi tr hiu dng l 12 V, v sm pha /6 so vi dng in. Biu thc ca in p gia hai u on mch l

A. u = 12cos(100t + eq \l(\f((,6))) VB. u = 12cos(100t + eq \l(\f((,3))) V

C. u = 12eq \l(\r(,2))cos(100t - eq \l(\f((,3))) VD. u = 12eq \l(\r(,2))cos(100t + eq \l(\f((,3))) V

Cu 21: Mt mch in xoay chiu c in p gia hai u mch l u = 200cos(100t + /6) V. Cng hiu dng ca dng in chy trong mch l 2eq \l(\r(,2)) A. Bit rng, dng in nhanh pha hn in p hai u mch gc /3, biu thc ca cng dng in trong mch l

A. i = 4cos(100t + /3) AB. i = 4cos(100t + /2) A.

C. i = 2eq \l(\r(,2))cos(100t - eq \l(\f((,6))) AD. i = 2eq \l(\r(,2))cos(100t + eq \l(\f((,2))) ACu 22: Mt mch in xoay chiu c in p gia hai u mch l u = 120eq \l(\r(,2))cos(100t - /4) V. Cng hiu dng ca dng in chy trong mch l 5A. Bit rng, dng in chm pha hn in p gc /4, biu thc ca cng dng in trong mch l

A. i = 5eq \l(\r(,2))sin(100t - eq \l(\f((,2))) AB. i = 5cos(100t - eq \l(\f((,2))) A

C. i = 5eq \l(\r(,2))cos(100t - eq \l(\f((,2))) AD. i = 5eq \l(\r(,2))cos(100t) A

Cu 23: Mt mch in xoay chiu c lch pha gia in p v cng dng in chy trong mch l /2. Ti mt thi im t, cng dng in trong mch c gi tr 2 A th in p gia hai u mch l 100eq \l(\r(,6)) V. Bit cng dng in cc i l 4A. in p hiu dng gia hai u mch in c gi tr l

A. U = 100 V. B. U = 200 V. C. U = 300 V. D. U = 220 V.

Cu 24: Mt mch in xoay chiu c lch pha gia in p v cng dng in chy trong mch l /2. Ti mt thi im t, cng dng in trong mch c gi tr 2eq \l(\r(,2)) A th in p gia hai u mch l 100eq \l(\r(,2)) V. Bit in p hiu dng ca mch l eq \s\don1(\f(200,3)) V. Gi tr hiu dng ca cng dng in trong mch l

A. 2A

B. 2eq \l(\r(,2))AC. 2eq \l(\r(,3)) AD. 4 A.

Cu 25: Cho mt mch in xoay chiu c in p hai u mch l u = 50cos(100t + /6) V. Bit rng dng in qua mch chm pha hn in p gc /2. Ti mt thi im t, cng dng in trong mch c gi tr eq \l(\r(,3))A th in p gia hai u mch l 25 V. Biu thc ca cng dng in trong mch l

A. i = 2cos(100t + eq \l(\f((,3))) A B. i = 2cos(100t - eq \l(\f((,3))) A

C. i = eq \l(\r(,3))cos(100t - eq \l(\f((,3))) AD. i = eq \l(\r(,3))cos(100t + eq \l(\f((,3))) ACu 26: Cho mt on mch in xoay chiu c in p cc i v dng in cc i l U0; I0. Bit rng in p v dng in vung pha vi nhau. Ti thi im t1 in p v dng in c gi tr ln lt l u1; i1. Ti thi im t2 in p v dng in c gi tr ln lt l u2; i2. in p cc i gia hai u on mch c xc nh bi h thc no di y?

A.

B.

C.

D.

Cu 27: Cho mt on mch in xoay chiu c in p cc i v dng in cc i l U0; I0. Bit rng in p v dng in vung pha vi nhau. Ti thi im t1 in p v dng in c gi tr ln lt l u1; i1. Ti thi im t2 in p v dng in c gi tr ln lt l u2; i2. Cng dng in hiu dng ca mch c xc nh bi h thc no di y?

A.

B.

C.

D.

Cu 28: Mt dng in xoay chiu c biu thc cng tc thi l i = 10cos(100t + /3) A. Pht biu no sau y khng chnh xc ?

A. Bin dng in bng 10AB. Tn s dng in bng 50 Hz.

C. Cng dng in hiu dng bng 5AD. Chu k ca dng in bng 0,02 (s).

Cu 29: Mt dng in xoay chiu c biu thc in p tc thi l u = 100cos(100t + /3) A. Pht biu no sau y khng chnh xc ?

A. in p hiu dng l 50eq \l(\r(,2)) V. B. Chu k in p l 0,02 (s.)

C. Bin in p l 100 V. D. Tn s in p l 100 Hz

Cu 30: Nhit lng Q do dng in c biu thc i = 2cos(120t) A to ra khi i qua in tr R = 10 trong thi gian t = 0,5 pht l

A. 1000 J. B. 600 J. C. 400 J. D. 200 J.

Cu 31: Mt dng in xoay chiu i qua in tr R = 25 trong thi gian 2 pht th nhit lng to ra l Q = 6000 J. Cng hiu dng ca dng in xoay chiu l

A. 3A

B. 2AC. 3AD. 2 A.

Cu 32: Chn pht biu sai ?

A. T thng qua mt mch bin thin trong mch xut hin sut in ng cm ng.

B. Sut in ng cm ng trong mt mch in t l thun vi tc bin thin ca t thng qua mch .

C. Sut in ng cm ng trong mt khung dy quay trong mt t trng u c tn s bng vi s vng quay trong 1 (s).

D. Sut in ng cm ng trong mt khung dy quay trong mt t trng u c bin t l vi chu k quay ca khung.

Cu 33: Mt khung dy phng quay u quanh mt trc vung gc vi ng sc t ca mt cm ng t trng u B. Sut in ng trong khung dy c tn s ph thuc vo

A. s vng dy N ca khung dy. B. tc gc ca khung dy.

C. din tch ca khung dy. D. ln ca cm ng t B ca t trng.

Cu 34: Mt khung dy quay u quanh trc xx trong mt t trng u c ng cm ng t vung gc vi trc quay xx. Mun tng bin sut in ng cm ng trong khung ln 4 ln th chu k quay ca khung phi

A. tng 4 ln. B. tng 2 ln. C. gim 4 ln. D. gim 2 ln.

Cu 35: Mt khung dy dn c din tch S = 50 cm2 gm 250 vng dy quay u vi tc 3000 vng/pht trong mt t trng u c vc t cm ng t vung gc vi trc quay ca khung, v c ln B = 0,02 (T). T thng cc i gi qua khung l

A. 0,025 Wb.B. 0,15 Wb.C. 1,5 Wb.D. 15 Wb.

Cu 36: Mt vng dy phng c ng knh 10 cm t trong t trng u c ln cm ng t B = 1/ (T). T thng gi qua vng dy khi vct cm ng t hp vi mt phng vng dy mt gc = 300 bng

A. 1,25.103 Wb. B. 0,005 Wb.C. 12,5 Wb. D. 50 Wb.

Cu 37: Mt khung dy quay u quanh trc ( trong mt t trng u c vc t cm ng t vung gc vi trc quay. Bit tc quay ca khung l 150 vng/pht. T thng cc i gi qua khung l (0 = ())eq \s\don1(\f(10,)) (Wb). Sut in ng hiu dng trong khung c gi tr l

A. 25 V. B. 25eq \l(\r(,2)) V. C. 50 V. D. 50eq \l(\r(,2)) V.

Cu 38: Khung dy kim loi phng c din tch S, c N vng dy, quay u vi tc gc quanh trc vung gc vi ng sc ca mt t trng u c cm ng t B. Chn gc thi gian t = 0 l lc php tuyn ca khung dy c chiu trng vi chiu ca vect cm ng t B. Biu thc xc nh t thng qua khung dy l

A. = NBSsin(t) Wb.

B. = NBScos(t) Wb.

C. = NBSsin(t) Wb.D. = NBScos(t) Wb.

Cu 39: Khung dy kim loi phng c din tch S = 50 cm2, c N = 100 vng dy, quay u vi tc 50 vng/giy quanh trc vung gc vi ng sc ca mt t trng u B = 0,1 (T). Chn gc thi gian t = 0 l lc php tuyn ca khung dy c chiu trng vi chiu ca vect cm ng t. Biu thc xc nh t thng qua khung dy l

A. = 0,05sin(100t) Wb.B. = 500sin(100t) Wb.

C. = 0,05cos(100t) Wb.D. = 500cos(100t) Wb.

Cu 40: Khung dy kim loi phng c din tch S, c N vng dy, quay u vi tc gc quanh trc vung gc vi ng sc ca mt t trng u B. Chn gc thi gian t = 0 l lc php tuyn n ca khung dy c chiu trng vi chiu ca vect cm ng t B. Biu thc xc nh sut in ng cm ng xut hin trong khung dy l

A. e = NBSsin(t) V.

B. e = NBScos(t) V.

C. e = NBSsin(t) V.

D. e = NBScos(t) V.

Cu 41: Khung dy kim loi phng c din tch S = 100 cm2, c N = 500 vng dy, quay u vi tc 3000 vng/pht quanh quanh trc vung gc vi ng sc ca mt t trng u B = 0,1 (T). Chn gc thi gian t = 0 l lc php tuyn ca khung dy c chiu trng vi chiu ca vect cm ng t B. Biu thc xc nh sut in ng cm ng xut hin trong khung dy l

A. e = 15,7sin(314t) V.

B. e = 157sin(314t) V.

C. e = 15,7cos(314t) V.

D. e = 157cos(314t) V.

Cu 42: Khung dy kim loi phng c din tch S = 40 cm2, c N = 1000 vng dy, quay `u vi tc 3000 vng/pht quanh quanh trc vung gc vi ng sc ca mt t trng u B = 0,01 (T). Sut in ng cm ng xut hin trong khung dy c tr hiu dng bng

A. 6,28 V. B. 8,88 V. C. 12,56 V. D. 88,8 V.

Cu 43: Mt khung dy quay iu quanh trc trong mt t trng u vung gc vi trc quay vi tc gc . T thng cc i gi qua khung v sut in ng cc i trong khung lin h vi nhau bi cng thc

A.

B.

C.

D.

Cu 44: Mt khung dy t trong t trng u c trc quay ( ca khung vung gc vi cc ng cm ng t. Cho khung quay u quanh trc (, th sut in ng cm ng xut hin trong khung c phng trnh e = 200eq \l(\r(,2))cos(100t - eq \l(\f((,6))) V. Sut in ng cm ng xut hin trong khung ti thi im t = eq \s\don1(\f(1,100)) s l

A. 100eq \l(\r(,2)) V. B. 100eq \l(\r(,2)) V. C. 100eq \l(\r(,6)) V. D. 100eq \l(\r(,6)) V.

Cu 45: Mt khung dy t trong t trng u c trc quay ( ca khung vung gc vi cc ng cm ng t. Cho khung quay u quanh trc (, th t thng gi qua khung c biu thc ( = ())eq \s\don1(\f(1,)) cos(100t + eq \l(\f((,3))) Wb. Biu thc sut in ng cm ng xut hin trong khung l

A. e = 50cos(100t + ())eq \s\don1(\f(,6))) VB. e = 50cos(100t + eq \l(\f((,6))) V

C. e = 50cos(100t - eq \l(\f((,6))) VD. e = 50cos(100t - ())eq \s\don1(\f(,6))) V

1B6C11D16C21B26B31D36A41B46

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BI GING MCH IN XOAY CHIU C 2 PHN T

I. MCH IN XOAY CHIU GM R, Lc im:in p v tng tr ca mch:

nh lut Ohm cho on mch:

in p nhanh pha hn dng in gc , xc nh t biu thc

Gin vc t:

Khi : (u = (i + (Ch : vit biu thc ca u, uL, uR trong mch RL th ta cn phi xc nh c pha ca i, ri tnh ton cc pha theo quy tc

V d 1. Tnh lch pha ca u v i, tng tr trong on mch in xoay chiu RL bit tn s dng in l 50 Hz va) R = 50 , L = 3))eq \s\don1(\f(,eq \l(\l(2()))) (H).b) R = 100eq \l(\r(,2)) , L = eq \s\don1(\f(,eq \l(\l(()))) HHng dn gii:p dng cc cng thcta c

a) ZL = 50eq \l(\r(,3)) ( ( b) Z = 100eq \l(\r(,2)) ( ( V d 2. Cho mch in xoay chiu gm hai phn t R, L vi R = 50eq \l(\r(,3)) , L = ())eq \s\don1(\f(1,)) H. t vo hai u on mch mt in p xoay chiu c biu thc u = 120cos(100t + /4) V. a) Tnh tng tr ca mch.b) Vit biu thc cng dng in chy qua on mch.c) Vit biu thc in p hai u cun cm thun, hai u in tr.Hng dn gii:a) T gi thit ta tnh c Z = 50 (

b) Ta c I0 = = 1,2 A lch pha ca in p v dng in l tha mn tan = eq \s\don1(\f(ZL,R)) = eq \s\don1(\f(50,50)) = eq \s\don1(\f(1,)) ( ( = eq \l(\f((,6)) rad M in p hai u mch nhanh pha hn dng in nn u = i + ( i = u - = eq \l(\f((,4)) - eq \l(\f((,6)) = ())eq \s\don1(\f(,12))

Vy biu thc cng dng in qua mch l i = 1,2cos(100t + ())eq \s\don1(\f(,12))) Ac) Vit biu thc uL v uR.

Ta c

Do uL nhanh pha hn i gc /2 nn = i + eq \l(\f((,2)) = ())eq \s\don1(\f(,12)) + eq \l(\f((,2)) = ())eq \s\don1(\f(,12)) ( uL =60cos(100t + ())eq \s\don1(\f(,12))) Do uR cng pha vi i nn = i = ())eq \s\don1(\f(,12)) ( uR = 60eq \l(\r(,3))cos(100t + ())eq \s\don1(\f(,12))) VV d 3. Cho mt on mch in xoay chiu ch c cun cm thun L v in tr R. Nu t vo hai u on mch in p u = 100cos(100t + /4) V th cng dng in trong mch l i = eq \l(\r(,2))cos(100t) A. Tnh gi tr ca R v L.Hng dn gii: T gi thit ta c ( ( V d 4. Cho mt on mch in xoay chiu gm in tr R = 50 v cun cm thun c h s t cm L = 3))eq \s\don1(\f(,eq \l(\l(2()))) (H). Cng dng in chy qua on mch c biu thc i = 2eq \l(\r(,2))cos(100t - (/6) A. Vit biu thc in p hai u mch, hai u in tr, hai u cun cm.Hng dn gii: Cm khng ca mch ZL = (.L = 50 eq \l(\r(,3)) ( ( ZRL = 100(Vit biu thc ca u: - in p cc i hai u on mch U0L = I0.ZL = 200eq \l(\r(,2)) V - lch pha ca u v i: tan = eq \l(\r(,3)) ( = eq \s\don1(\f(,3)) = u - (i ( (u = (i + eq \l(\f((,3)) = eq \l(\f((,6)) T ta c u = 200eq \l(\r(,2))cos(100t + eq \l(\f((,6))) V Vit biu thc ca uR: - in p cc i hai u in tr U0R = I0.ZL = 100eq \l(\r(,2)) V - uR v i cng pha nn: = i = - eq \l(\f((,6)) ( uR = 100eq \l(\r(,2))cos(100t - eq \l(\f((,6))) V.

Vit biu thc ca uL: - in p cc i hai u cun cm thun U0L =I0ZL= 100eq \l(\r(,6)) V

- u nhanh pha hn i gc /6 nn: = i + eq \l(\f((,6)) = 0 ( uL = 100eq \l(\r(,6))cos(100t) V. II. MCH IN XOAY CHIU GM R, Cc im:in p v tng tr ca mch:

nh lut Ohm cho on mch:

in p chm pha hn dng in gc , xc nh t biu thc: ; ( = (u - (i Gin vc t:

Ch : vit biu thc ca u, uC, uR trong mch RC th ta cn phi xc nh c pha ca i, ri tnh ton cc pha theo quy tc

V d 1. Cho mch in xoay chiu gm hai phn t R, C vi R 100 , C = (F). t vo hai u on mch mt in p xoay chiu c biu thc u = 200cos(100t + /3) V. a) Tnh tng tr ca mch.b) Vit biu thc cng dng in chy qua on mch.c) Vit biu thc in p hai u t in, hai u in tr thun.Hng dn gii:a) Ta c ZL = 100 tng tr ca mch l ZRC = 100eq \l(\r(,2)) (b) Ta c = eq \l(\r(,2)) A

lch pha ca in p v dng in l tha mn tan = - eq \s\don1(\f(ZC,R)) = -1 ( ( = - (/4 M u - i = ( i = u - = eq \s\don1(\f(7,12)) rad.

Vy biu thc cng dng in qua mch l i = eq \l(\r(,2))cos(100t + ())eq \s\don1(\f(,12))) Ac) Vit biu thc uC v uR.

* Ta c U0C = I0.ZC = 100eq \l(\r(,2)) V v uC chm pha hn i gc /2 nn = i - eq \l(\f((,2)) = ())eq \s\don1(\f(,12)) ( uC = 100eq \l(\r(,2))cos(100t + ())eq \s\don1(\f(,12))) V * Ta c U0R = I0.R = 100eq \l(\r(,2)) V v uR cng pha vi i nn = i = ())eq \s\don1(\f(,12)) ( uR = 100eq \l(\r(,2))cos(100t + ())eq \s\don1(\f(,12))) VV d 2. Cho mt on mch in xoay chiu ch c t in C v in tr R. Nu t vo hai u on mch in p u = 100eq \l(\r(,2))cos(100(t) V th cng dng in trong mch l i = eq \l(\r(,2))cos(100t + eq \l(\f((,4))) A. Tnh gi tr ca R v C.Hng dn gii: T gi thit ta c ( (

V d 3. on mch in xoay chiu ni tip gm in tr R = 50 v t in C = (eq \s\don1(\f(200,)))) F. Vit biu thc in p tc thi gia hai bn ca t in v hai u on mch. Cho bit biu thc cng dng in i = eq \l(\r(,2))sin(100t + eq \l(\f((,3))) AHng dn gii:

Ta c = 100 rad ( ZC = (C))eq \s\don1(\f(1,)) = 50eq \l(\r(,3)) (

Tng tr ca mch ZRC = 100 ( T gi thit ta c I0 = eq \l(\r(,2)) A (

Vit biu thc in p gia hai u t C: Do uC chm pha hn i gc /2 nn - i = - eq \l(\f((,2)) ( = i - eq \l(\f((,2)) = - eq \l(\f((,6)) rad.

Biu thc hai u C l uC = 50eq \l(\r(,6))cos(100(t - eq \l(\f((,6))) VVit biu thc in p gia hai u on mch RC:

lch pha ca u v i l tan = - eq \s\don1(\f(ZC,R)) = - eq \l(\r(,3)) ( ( = - eq \l(\f((,3)) rad M ( = - (i ( = ( +(i = 0 ( uRC = 100eq \l(\r(,2))cos100t V.

III. MCH IN XOAY CHIU GM L, Cc im:

in p v tng tr ca mch: (

* Gin vc t:

- Khi UL > UC hay ZL > ZC th uLC nhanh pha hn i gc /2. (Hnh 1). Khi ta ni mch c tnh cm khng.

- Khi UL < UC hay ZL < ZC th uLC chm pha hn i gc /2. (Hnh 2). Khi ta ni mch c tnh dung khng.

V du 1. Mt on mch gm mt t in C c dung khng 100 v mt cm thun c cm khng 200 mc ni tip nhau. in p hai u cun cm c biu thc uL = 100cos(100t + /6) V. Vit biu thc in p hai u t in.Hng dn gii:Ta c I0 = ( U0C = eq \s\don1(\f(ZC,2)) = 50 VMt khc ( rad Vy biu thc hai u in p qua t C l uC = 50cos(100t - ())eq \s\don1(\f(,6))) VV du 2. on mch gm cun dy thun cm c t cm L = 2/ (H) mc ni tip vi t in ri mc vo mt in p xoay chiu tn s 50 Hz. Khi thay t C1 bng mt t C2 khc th thy cng dng in qua mch khng thay i. in dung ca t C2 c gi tr bng:

A.

B.

C.

D.

Hng dn gii:

Ta c I =

Do I khng i nn ( ( T gi thit ta tnh c ( = 300( ( C2 = FTRC NGHIM MCH IN XOAY CHIU C 2 PHN T

Cu 1: on mch in xoay chiu gm hai phn t R v L. Tng tr ca mch c cho bi cng thc

A.

B.

C. ZRL= R + ZLD. ZRL=R2+

Cu 2: on mch in xoay chiu gm hai phn t R v L. in p hiu dng gia hai u on mch c cho bi cng thc

A.

B.

C.

D.

Cu 3: on mch in xoay chiu gm hai phn t R v L. lch pha ca in p v dng in trong mch c cho bi cng thc

A. tan = -

B. tan = -

C. tan = -

D. tan =

Cu 4: Chn pht biu ng khi ni v mch in xoay chiu ch c cun cm thun v in tr thun?

A. Dng in trong mch lun nhanh pha hn in p.

B. Khi R = ZL th dng in cng pha vi in p.

C. Khi R = eq \l(\r(,3))ZL th in p nhanh pha hn so vi dng in gc /6.

D. Khi R = eq \l(\r(,3))ZLth in p nhanh pha hn so vi dng in gc /3.

Cu 5: Chn pht biu ng khi ni v mch in xoay chiu ch c cun cm thun v in tr thun? A. Khi ZL = Req \l(\r(,3)) th in p nhanh pha hn so vi dng in gc /6.

B. Khi ZL = Req \l(\r(,3)) th dng in chm pha hn so vi in p gc /3.

C. Khi R = ZL th in p cng pha hn vi dng in.

D. Khi R = ZL th dng in nhanh pha hn so vi in p gc /4.

Cu 6: Cho on mch in xoay chiu ch c in tr R v cun cm thun L. Pht biu no di y l khng ng?

A. in p nhanh pha hn dng in gc /4 khi R = ZL.

B. in p nhanh pha hn dng in gc /3 khi ZL = eq \l(\r(,3))R.

C. in p chm pha hn dng in gc /6 khi R = eq \l(\r(,3))ZL.

D. in p lun nhanh pha hn dng in.

Cu 7: on mch in xoay chiu gm hai phn t R = 50 v cun thun cm c t cm L. t vo hai u on mch mt in p xoay chiu c biu thc u = U0cos(100t) V. Bit rng in p v dng in trong mch lch pha nhau gc /3. Gi tr ca L l

A. HB. HC. HD. HCu 8: on mch in xoay chiu gm in tr R v cun cm thun c t cm (H). t vo hai u on mch mt in p xoay chiu c biu thc u = U0cos(100t) V. Tm gi tr ca R dng in chm pha so vi in p gc /6 ?

A. R = 50 . B. R = 100 . C. R = 150 D. R = 100 eq \l(\r(,3)) .

Cu 9: Mt on mch in gm mt cun dy thun cm mc ni tip vi mt in tr thun. Nu t vo hai u mch mt in p c biu thc u = 15eq \l(\r(,2))cos(100t - ())eq \s\don1(\f(,4)) ) V th in p hiu dng gia hai u cun cm l 5 V. Khi in p hiu dng gia hai u in tr c gi tr l

A. 15eq \l(\r(,2)) V. B. 5eq \l(\r(,3)) V. C. 5eq \l(\r(,2)) V. D. 10eq \l(\r(,2)) V.

Cu 10: Cho mt on mch in xoay chiu gm in tr thun R v mt cun cm thun c h s t cm L. in p hai u on mch c biu thc u = 100eq \l(\r(,2))cos(100t - eq \l(\f((,3))) V. Bit dng in chm pha hn in p gc /6. in p hai u cun cm c gi tr l

A. 50 V. B. 50eq \l(\r(,3)) V. C. 100 V. D. 50eq \l(\r(,2)) V.Cu 11: Mt cun dy c li thp, t cm L = 318 (mH) v in tr thun 100 . Ngi ta mc cun dy vo mng in khng i c in p 20 V th cng dng in qua cun dy l

A. 0,2AB. 0,14AC. 0,1AD. 1,4 A.

Cu 12: Mt cun dy c t cm L = 318 (mH) v in tr thun 100 . Ngi ta mc cun dy vo mng in xoay chiu 20 V, 50 Hz th cng dng in qua cun dy l

A. 0,2AB. 0,14AC. 0,1AD. 1,4 A.

Cu 13: Mt on mch in gm mt cun dy thun cm c t cm H v in tr thun R = 50 . t vo hai u mch mt in p c biu thc u = 100eq \l(\r(,2))cos(100t - /6) V th biu thc ca cng dng in chy qua on mch l

A. i = eq \l(\r(,2))cos(100t - /3) A B. i = eq \l(\r(,2))cos(100t - /2) A

C. i = cos(100t - /2) A D. i = cos(100t - /2) A

Cu 14: Mt on mch in gm in tr R = 50 mc ni tip vi cun thun cm c L = 0,5/ (H). t vo hai u on mch mt in p xoay chiu u = 100eq \l(\r(,2))sin(100t - /4) V. Biu thc ca cng dng in qua on mch l

A. i = 2sin(100t - /2) A B. i = 2eq \l(\r(,2))sin(100t - /4) A

C. i = 2eq \l(\r(,2))sin(100t) A

D. i = 2sin(100t) A

Cu 15: Mt on mch in gm mt cun dy thun cm c t cm L = 0,5/ (H) mc ni tip vi in tr thun R = 50eq \l(\r(,3)) (. t vo hai u mch mt in p xoay chiu th dng in trong mch c biu thc l i = 2cos(100t + /3) A. Biu thc no sau y l ca in p hai u on mch?

A. u = 200cos(100t+ /3) V. B. u = 200cos(100t+ /6) V. C. u = 100eq \l(\r(,2))cos(100t+ /2) V.D. u = 200cos(100t+ /2) V.Cu 16: Cho mt on mch in xoay chiu gm cun cm thun L v in tr R. Nu t vo hai u on mch in p u = 100cos(100t +/4) V th cng dng in trong mch l i = eq \l(\r(,2))cos(100t) A. Gi tr ca R v L l

A. R = 50( , L = H

B. R = 50( , L = H

C. R = 50( , L = H

D. R = 50eq \l(\r(,3)) ( , L = HCu 17: Mt on mch in gm mt cun dy thun cm c t cm L = 1/ (H) v in tr thun R = 100 . t vo hai u mch mt in p xoay chiu u = 200cos(100t + /4) V th biu thc no sau y l ca in p hai u cun cm thun ?

A. uL = 100eq \l(\r(,2))cos(100t + /4) V. B. uL = 100cos(100t + /2) V.

C. uL = 100eq \l(\r(,2))cos(100t - /2) V. D. uL = 100eq \l(\r(,2))cos(100t + /2) V.

Tr li cc cu hi 18, 19, 20, 21 vi cng d kin sau:Cho on mch in xoay chiu gm in tr thun R = 50 , cun dy thun cm c h s t cm L = eq \s\don1(\f(,2eq \l(\l(()))) (H). t in p u = 100cos(100t + /6) V vo hai u on mch. Cu 18: Biu thc cng dng in chy qua on mch l

A. i = cos(100t - eq \l(\f((,6))) A

B. i = eq \l(\r(,2))cos(100t - eq \l(\f((,6))) A C. i = cos(100t - eq \l(\f((,2)) ) A

D. i = eq \l(\r(,2))cos(100t + eq \l(\f((,2))) ACu 19: in p hiu dng gia hai u L, R c gi tr ln lt l

A. 25eq \l(\r(,6)) V, 25eq \l(\r(,3)) V. B. 25eq \l(\r(,2)) V, 25eq \l(\r(,6)) V.C. 25eq \l(\r(,6)) V, 25eq \l(\r(,2)) V. D. 25 V, 25eq \l(\r(,2)) V.

Cu 20: Biu thc in p hai u cun cm thun l

A. uL =50eq \l(\r(,3))cos(100t+ /3) V. B. uL =50cos(100t+ /2) V. C. uL =50eq \l(\r(,3))cos(100t+ /2) V. D. uL =50cos(100t+ /3) V. Cu 21: Biu thc in p hai u in tr R l

A. uR = 50cos(100t + (/6) VB. uR = 25eq \l(\r(,2))cos(100t + (/6) V C. uR = 25eq \l(\r(,2))cos(100t - (/6) VD. uR = 50cos(100t - (/6) VDCu 22: Cho on mch in xoay chiu gm in tr R = 50 v cun cm thun c h s t cm L = eq \s\don1(\f(,eq \l(\l(2()))) H. in p v dng in lch pha nhau gc /6 th tn s ca dng in c gi tr no sau y?

A. f = 50eq \l(\r(,3)) Hz. B. f = 25eq \l(\r(,3)) Hz.C. f = eq \s\don1(\f(50,3)) Hz.D. f = eq \s\don1(\f(100,3)) Hz.

Cu 23: Cho on mch RL ni tip, in p hai u on mch c dng u =100eq \l(\r(,2))sin(100t) V th biu thc dng in qua mch l i = 2eq \l(\r(,2))sin(100t - /6) A . Tm gi tr ca R, L.

A. R = 25eq \l(\r(,3)) , L = ())eq \s\don1(\f(1,)) H.B. R = 25 , L = 3))eq \s\don1(\f(,eq \l(\l(4()))) H.

C. R = 20 , L = ())eq \s\don1(\f(1,)) H

D. R = 30 , L = eq \s\don1(\f(,eq \l(\l(()))) H.

Cu 24: Cho mt on mch in xoay chiu AB gm in tr thun R ni tip cun dy thun cm L. Khi tn s dng in bng 100 Hz th in p hiu dng UR = 10 V, UAB = 20 V v cng dng in hiu dng qua mch l 0,1 A. Gi tr ca R v L l

A. R = 100 , L = eq \s\don1(\f(,eq \l(\l(2()))) H

B. R = 100 , L = eq \s\don1(\f(,eq \l(\l(()))) H C. R = 200 , L = eq \s\don1(\f(2,eq \l(\l(()))) H

D. R = 200 , L = eq \s\don1(\f(,eq \l(\l(()))) HCu 25: on mch in xoay chiu gm hai phn t R v C. in p hiu dng gia hai u on mch c cho bi cng thc

A.

B.

C.

D.

Cu 26: on mch in xoay chiu gm in tr thun R v t in c in dung C th tng tr ca mch l

A.

B.

C. ZRC=

D. ZRC = Cu 27: on mch in xoay chiu gm hai phn t R v C. lch pha ca in p v dng in trong mch c cho bi cng thc

A. tan = -

B. tan = -

C. tan =

D. tan = -

Cu 28: on mch in xoay chiu AB ch cha mt trong cc phn t: in tr thun, cun dy hoc t in. Khi t in p u = U0cos(t /6) V ln hai u A v B th dng in trong mch c biu thc i = Iocos(t + /3)A. on mch AB cha

A. in tr thun.

B. cun dy c in tr thun.

C. cun dy thun cm.

D. t in.

Cu 29: Chn pht biu ng khi ni v mch in xoay chiu ch c t in v in tr thun?

A. Dng in trong mch lun chm pha hn in p.

B. Khi R = ZC th dng in cng pha vi in p.

C. Khi R = eq \l(\r(,3))ZC th in p chm pha hn so vi dng in gc /3.

D. Dng in lun nhanh pha hn in p.

Cu 30: Mt on mch in xoay chiu gm in tr thun R mc ni tip vi t in C. in p hai u mch l u. Nu dung khng ZC = R th cng dng in chy qua in tr lun

A. nhanh pha /2 so vi u. B. nhanh pha /4 so vi u.

C. chm pha /2 so vi u. D. chm pha /4 so vi u.

Cu 31: Mt on mch gm t c in dung C = (F) ghp ni tip vi in tr R = 100 , mc on mch vo in p xoay chiu c tn s f. dng in lch pha /3 so vi in p th gi tr ca f l

A. f = 25 Hz. B. f = 50 Hz. C. f = 50eq \l(\r(,3)) Hz. D. f = 60 Hz.

Cu 32: Mt on mch in gm t in c in dung C = 104/ (F) v in tr thun R = 100 . t vo hai u mch mt in p c biu thc u = 200eq \l(\r(,2))cos(100t - /4) V th biu thc ca cng dng in trong mch l

A. i = eq \l(\r(,2))cos(100t - /3) A. B. i = eq \l(\r(,2))cos100t A.

C. i = 2cos 100t A

D. i = 2cos(100t - /2) A.

Cu 33: Mt on mch in xoay chiu RC c C = (F), R = 50(. t vo hai u mch mt in p xoay chiu th dng in trong mch c biu thc l i = cos(100t + /6) A. Biu thc no sau y l ca in p hai u on mch?

A. u = 100cos(100t - /6) V. B. u = 100cos(100t +/2) V

C. u = 100eq \l(\r(,2))cos(100t - /6) V. D. u = 100cos(100t + /6) V.

Cu 34: Cho mt on mch in xoay chiu gm in tr thun v t in c in dung C, f = 50 Hz. Bit rng tng tr ca on mch l 100 v cng dng in lch pha gc /3 so vi in p. Gi tr ca in dung C l

A. C = (F). B. C = (F)C. C = (F)D. C = (F)

Cu 35: Cho mt on mch in xoay chiu RC. t vo hai u on mch in p u = 100eq \l(\r(,2))cos 100t V th cng dng in trong mch l i = eq \l(\r(,2))cos(100t + /4) A. Gi tr ca R v C l

A. R = 50eq \l(\r(,2)) , C = (F).B. R = 50eq \l(\r(,2)) , C = (F).

C. R = 50 , C = (F).D. R = 50eq \l(\r(,2)) , C = (F).

Cu 36: Mt on mch in xoay chiu RC c R = 100 , C = (F). t vo hai u mch mt in p xoay chiu u = 200cos(100t + /4) V th biu thc no sau y l ca in p hai u t in?

A. uC = 100eq \l(\r(,2))cos100t V. B. uC = 100cos(100t + (/4) V

C. uC = 100eq \l(\r(,2))cos(100t - (/2) V.D. uC = 100cos(100t + (/2) V.

Cu 37: on mch in xoay chiu ch c cun cm thun c t cm L v t in c in dung C th cng dng in trong mch

A. lun nhanh pha hn in p gc /2.

B. lun tr pha hn in p gc /2.

C. lun nhanh pha hn in p gc /2 khi ZL > ZC

D. lun nhanh pha hn in p gc /2 khi ZL < ZCCu 38: Chn pht biu khng ng. on mch in xoay chiu ch c cun cm thun c t cm L v t in c in dung C th cng dng in trong mch

A. lun nhanh pha hn in p gc /2 khi ZL < ZC

B. lun tr pha hn in p gc /2.

C. lun tr pha hn in p gc /2 khi ZL > ZC

D. lun nhanh pha hn in p gc /2 khi ZL < ZC.

Cu 39: on mch in xoay chiu ch c cun cm thun c t cm L v t in c in dung C. t vo hai u on mch in p xoay chiu u = U0cos(t) V th cng dng in hiu dng trong mch l

A. I0 = B. I0 = C. I0 =

D. I0 =

Cu 40: on mch gm cun dy c t cm L = 2/ (H) mc ni tip vi t in C1 = 104/ (F) ri mc vo mt in p xoay chiu tn s 50 Hz. Khi thay t C1 bng mt t C2 khc th thy cng dng in qua mch khng thay i. in dung ca t C2 bng

A. C2 = FB. C2 = FC. C2 = FA. C2 = FCu 41: Mt on mch gm mt t in c dung khng ZC = 100 v cun dy c cm khng ZL = 200 mc ni tip nhau. in p ti hai u cun cm c dng uL = 100cos(100t + /6) V. Biu thc in p hai u t in c dng nh th no?

A. uC = 100cos(100t + /6) V. B. uC = 50cos(100t /3) V.

C. uC = 100cos(100t /2) V. D. uC = 50cos(100t 5/6) V.

Cu 42: Mt on mch gm mt t in c dung khng ZC = 200 v cun dy c cm khng ZL = 120 mc ni tip nhau. in p ti hai u t in c dng uC = 100cos(100t /3) V. Biu thc in p hai u cun cm c dng nh th no?

A. uL = 60cos(100t + /3) V. B. uL = 60cos(100t + 2/3) V.

C. uL = 60cos(100t /3) V. D. uL = 60cos(100t + /6) V.

Cu 43: t mt in p xoay chiu u = 60sin(100t) V vo hai u on mch gm cun thun cm L = 1/ (H) v t C = 50/ (F) mc ni tip. Biu thc ca cng dng in chy trong mch l

A. i = 0,2sin(100t + /2) A. B. i = 0,2sin(100t /2) A.

C. i = 0,6sin(100t + /2) A. D. i = 0,6sin(100t /2) A.

Cu 44: Mt on mch xoay chiu ch cha 2 trong 3 phn t R, L, C mc ni tip. Bit rng in p hai u on mch sm pha /3 so vi cng dng in. on mch cha

A. R, C vi ZC < R. B. R, C vi ZC > R.C. R, L vi ZL < R. D. R, L vi ZL > R.

Cu 45: Mt on mch xoay chiu ch cha 2 trong 3 phn t R, L, C mc ni tip. Bit rng in p hai u on mch chm pha /4 so vi cng dng in. on mch cha

A. R, C vi ZC < R. B. R, C vi ZC = R.C. R, L vi ZL = R. D. R, C vi ZC > R.

Cu 46: Cho mt on mch in xoay chiu gm hai phn t mc ni tip. in p gia hai u on mch v cng dng in trong mch c biu thc u = 100eq \l(\r(,2))cos(100t - eq \l(\f((,2))) V, i = 10eq \l(\r(,2))cos(100t - eq \l(\f((,4))) A. Chn kt lun ng ?

A. Hai phn t l R, L. B. Hai phn t l R, C.

C. Hai phn t l L, C. D. Tng tr ca mch l 10eq \l(\r(,2)) (Cu 47: Cho mt on mch in xoay chiu gm in tr thun R v mt cun cm thun c h s t cm L. in p hai u on mch c biu thc u = 100eq \l(\r(,2))cos(100t + ) V. Cng dng in trong mch c gi tr hiu dng l 2 A v chm pha hn in p gc /3. Gi tr ca in tr thun R l

A. R = 25 . B. R = 25eq \l(\r(,3)) . C. R = 50 . D. R = 50eq \l(\r(,3)) .P N TRC NGHIM IN XOAY CHIU C 2 PHN T

1B6C11A16A21D26D31D36A41D46B

2C7C12B17D22C27B32C37D42B47C

3D8B13B18A23A28D33A38B43C48

4C9D14A19C24A29D34C39B44D49

5B10A15D20A25B30B35D40C45B50

BI GING MCH IN XOAY CHIU RLC - PHN 1

I. MCH IN XOAY CHIU RLC NI TIPc im:in p v tng tr ca mch:

nh lut Ohm cho mch:

lch pha ca in p v cng dng in trong mch l , c cho bi ;( = (u- (i- Khi UL > UC hay ZL > ZC th u nhanh pha hn i gc . (Hnh 1). Khi ta ni mch c tnh cm khng. - Khi UL < UC hay ZL < ZC th u chm pha hn i gc . (Hnh 2). Khi ta ni mch c tnh dung ghng.

Gin vc t: V d 1: Cho mch in RLC c R = 10eq \l(\r(,3)) (, L = ())eq \s\don1(\f(3,))(H), C=(F). t vo hai u mch in p xoay chiu c gi tr hiu dng 120 V, tn s 50 Hz. a) Tnh tng tr ca mch.b) Tnh cng hiu dng ca dng in qua mch. c) in p hiu dng trn tng phn t R, L, C.Hng dn gii:a) Tnh tng tr ca mch

Ta c ZL = (.L = 30 ; ZC = 20

Tng tr ca mch = 20 (b) Cng hiu dng qua mch I = eq \s\don1(\f(U,R)) = 6 Ac) in p hiu dng trn tng phn t l UL

V d 2: Cho mch in RLC c R = 10 (, L = eq \s\don1(\f(,eq \l(\l(()))) (H), C = (F). in p hai u mch l u = 60eq \l(\r(,2))cos(100t + eq \l(\f((,3))) V. Vit biu thc ca i, uR; uL; uC; uRLHng dn gii:

V d 3: Cho mch in RLC c R = 10 (, L = eq \s\don1(\f(,eq \l(\l(()))) (H), C = (F). in p hai u mch l uC = 50eq \l(\r(,2))cos(100t + ())eq \s\don1(\f(,4))) V. Vit biu thc ca i, uR; uLHng dn gii:

V d 4: Cho on mch RLC gm R = 80 , L = 318 (mH), C = 79,5 (F). in p gia hai u on mch c biu thc u = 120eq \l(\r(,2))cos100t V.a) Vit biu thc cng dng in chy trong mch v tnh in p hiu dng gia hai u mi dng c. b) Tnh in p hiu dng gia hai u R, hai u L v hai u C.c) Vit biu thc in p hai u R, hai u L, hai u C.Hng dn gii: a) Ta c = 100n rad ( ZL = (L ( 100 ( v ZC = (C))eq \s\don1(\f(1,)) ( 40 (

Tng tr ca mch l = 100 (Cng dng in ca mch : I = eq \s\don1(\f(U,Z)) = 1 A ( I0 = eq \l(\r(,2)) A

Gi ( l lch pha ca u v i, ta c = eq \s\don1(\f(3,4)) ( ( ( 0.64 rad M = u - i ( i = u - = -0,64 rad.

Vy biu thc cng dng in trong mch l i = eq \l(\r(,2))cos(100t - 0,64) A.

b) Theo a ta c I = 1 A, in p hiu dng gia hai u mi phn t l

c) Vit biu thc hai u mi phn t R, L v C.

Biu thc in p gia hai u R UR = 80 V ( U0R = 80eq \l(\r(,2)) V.

Do uR cng pha vi i nn = 1 = 0,64 rad ( uR = 80eq \l(\r(,2))cos(100t - 0,64) V.

Biu thc in p gia hai u L UL = 100 V ( U0L = 100eq \l(\r(,2)) V

Do uL nhanh pha hn i gc /2 nn - i = eq \l(\f((,2)) ( = eq \l(\f((,2)) + i = eq \l(\f((,2)) - 0.64 rad Biu thc in p hai u L l uL = 100eq \l(\r(,2))cos(100(t + eq \l(\f((,2)) - 0,64 V.

Biu thc in p gia hai u C UC = 40 V ( U0C = 40eq \l(\r(,2)) V.

Do uC chm pha hn i gc /2 nn - i = - eq \l(\f((,2)) ( = i - eq \l(\f((,2)) = - eq \l(\f((,2)) - 0.64 rad

Biu thc in p hai u t C l uC = 40eq \l(\r(,2))cos(100(t - eq \l(\f((,2)) - 0,64 V.V d 5: Cho on mch RLC gm R = 10(, L = ())eq \s\don1(\f(1,)) (H), C = (F). in p hai u cun cm c biu thc u = 20eq \l(\r(,2))cos(100t + eq \l(\f((,2))) V.a) Vit biu thc cng dng in chy trong mch.b) Vit biu thc in p hai u on mch u, hai u in tr uR, hai u t in uC, uRL, uRC.Hng dn gii:a) T gi thit ta c

T ta c I0 = = 2eq \l(\r(,2)) A Do uL nhanh pha hn i gc /2 nn - i = eq \l(\f((,2)) (i = - eq \l(\f((,2)) = 0 ( i = 2eq \l(\r(,2))cos100t A.

b) Vit biu thc u, uR, uC, uRL, uRC Vit biu th ca u: + Ta c U0 = I0.Z = 2eq \l(\r(,2)).10eq \l(\r(,2)) = 40 V.

+ lch pha ca u v i: = - 1 ( ( = - ())eq \s\don1(\f(,4)) = u - i ( u = i - eq \s\don1(\f(,4)) T ta c biu thc ca in p hai u mch u = 40cos(100t - eq \l(\f((,4))) V Vit biu thc ca uR: + Ta c U0R = I0.R = 2eq \l(\r(,2)).10 = 20eq \l(\r(,2)) V.

+ lch pha ca uR v i: u = i = 0 ( uR = 20eq \l(\r(,2))cos(100t - eq \l(\f((,4))) V.

Vit biu thc ca uC: + Ta c U0C = I0.ZC = 2eq \l(\r(,2)).20 = 40eq \l(\r(,2)) V.

+ lch pha ca uC v i:- i = - eq \l(\f((,2)) ( uC = 40eq \l(\r(,2))cos(100t - eq \l(\f((,2))) V. Vit biu thc ca uRL: + Ta c U0RL = I0.ZRL = 2eq \l(\r(,2)).= 40 V + lch pha ca uRL v i: = 1 ( (RL = eq \l(\f((,4)) = - (i ( = eq \l(\f((,4)) T ta c: uRL = 40cos(100(t + eq \l(\f((,4))) V Vit biu thc ca uRC: + Ta c U0RC = I0.ZRC = 2eq \l(\r(,2)).= 20eq \l(\r(,10)) V + lch pha ca uRC v i: = - 2

( (RC ( - ())eq \s\don1(\f(,180)) = - (i ( = - ())eq \s\don1(\f(,180)) + (i = - ())eq \s\don1(\f(,180)) T ta c: uRC = 20eq \l(\r(,10))cos(100(t - ())eq \s\don1(\f(,180)) ) VV d 6: Cho on mch RLC gm R = 40(, L = ())eq \s\don1(\f(3,)) (H), C = (F). in p hai u on mch RL c biu thc uRL = 120cos100t V.a) Vit biu thc cng dng in chy trong mch. b) Vit biu thc in p hai u on mch.Hng dn gii:a) T gi thit ta c

T ta c I0 = = 2,4 A Mt khc ( (RL = ())eq \s\don1(\f(,180)) = - (i = - ())eq \s\don1(\f(,180)) ( i = 2,4cos(100t - ())eq \s\don1(\f(,180)) ) Ab) lch pha ca u v i: ( = - eq \l(\f((,4)) = u - i ( u = (i - eq \l(\f((,4)) = - ())eq \s\don1(\f(,90)) ng thi U0 = I0.Z = 96 eq \l(\r(,2)) V ( u = 96eq \l(\r(,2))cos(100t - ())eq \s\don1(\f(,90)) ) VV d 7: Cho mch in xoay chiu RLC vi R = 30 (; C = F; u = 120cos(100t)V; UL = 120 V. Tm gi tr ca L v vit biu thc cng dng in.Hng dn gii:

II. CNG HNG IN TRONG MCH RLC NI TIP* Khi nim v cng hng inKhi ZL = ZC ( (L = (C))eq \s\don1(\f(1,)) ( (2 = eq \s\don1(\f(1,LC)) ( ( = eq \s\don1(\f(1,)) th trong mch c xy ra hin tng cng hng in.* c im ca hin tng cng hng in

+ Khi xy ra hin tng cng hng in th tng tr ca mch t gi tr nh nht, Zmin = R cng hiu dng ca dng in t gi tr cc i vi Imax= eq \s\don1(\f(U,R)) .

+ in p gia hai u in tr R bng vi in p hai u mch, UR = U.

+ Cng dng in trong mch cng pha vi in p hai u mch

+ Cc in p gia hai u tu in v hai u cun cm c cng ln (tc UL = UC) nhng ngc pha nn trit tiu nhau.

+ iu kin cng hng in = eq \s\don1(\f(1,)) ( ( (2LC = 1Ch : Khi ang xy ra cng hng th tng tr ca mch t cc tiu, cng dng in t cc i. Nu ta tng hay gim tn s dng in th tng tr ca mch s tng, ng thi cng dng in s gim.V d 1. Mt on mch ni tip gm mt in tr R = 10 , cun dy thun L = 5 mH v t in C = 5.104 F. Hiu in th hai u on mch U = 220 V.a) Xc nh tn s ca dng in c cng hng.b) Tnh cng qua mch v cc hiu in th UL, UC khi c cng hng.Hng dn gii:a) Khi cng hng th = ( 100 Hzb) Vi f = 100 Hz th = 200 ( ZL = L = 200.5.10-3 ( 3,14 ( = ZC Khi c cng hng th I = Imax = eq \s\don1(\f(U,R)) = 22 A ( UL = UC = I.ZL = 69 VV d 2. t vo hai u on mch RLC ni tip mt in p xoay chiu c gi tr hiu dng khng i th in p hiu dng trn cc phn t R, L v C ln lt l 30 V, 50 V v 90 V. Khi thay t C bng t C mch c cng hng in th in p hiu dng gia hai u in tr R bng A. 50 V. B. 70eq \l(\r(,2)) V. C. 100 V. D. 100eq \l(\r(,2)) V.

Hng dn gii:

T gi thit ta tnh c in p hai u mch l U = = 50 V

Khi thay t C bng t C c cng hng in, theo c im cng hng ta c UR = U = 50 V. Vy A ng.

TRC NGHIM MCH IN XOAY CHIU RLC - PHN 1Cu 1: Cho mch in xoay chiu RLC ni tip. t vo hai u on mch mt in p u = U0cos(t) V. Cng thc tnh tng tr ca mch l

A.

B.

C.

D.

Cu 2: Cng tc tnh tng tr ca on mch RLC mc ni tip l

A.

B.

C.

D. Z = R + ZL + ZCCu 3: Mch in xoay chiu gm RLC mc ni tip, c R = 30 , ZC = 20 , ZL = 60 . Tng tr ca mch l

A. Z = 50 . B. Z = 70 . C. Z = 110 . D. Z = 2500 .

Cu 4: Cho mch in xoay chiu RLC ni tip. t vo hai u on mch mt in p u = U0cos(t) V. Cng dng in hiu dng ca mch l

A.

B.

C.

D.

Cu 5: Cho mch in xoay chiu RLC ni tip. Cng dng in chy trong mch c biu thc i = I0cos(t) A. in p hiu dng hai u on mch c cho bi

A.

B.

C.

D.

Cu 6: Cho on mch RLC ni tip c R = 60 , L = 0,2/ (H), C = 104/ (F). t vo hai u on mch mt in p xoay chiu u = 50eq \l(\r(,2))cos 100t V. Cng dng in hiu dng trong mch l

A. 0,25A.B. 0,50 A.C. 0,71 A.D. 1,00 A.

Cu 7: Cho on mch gm in tr R = 100 , t in C = 104/ (F) v cun cm L = 2/ (H) mc ni tip. t vo hai u on mch mt in p xoay chiu u = 200cos(100t) V. Cng dng in hiu dng trong mch l

A. 2AB. 1,4AC. 1AD. 0,5 A.

Cu 8: Cho on mch RLC ni tip, in p hai u on mch c gi tr hiu dng l 100 V. Tm UR bit ZL = eq \s\don1(\f(8,3))R = 2ZC .

A. 60 V . B. 120 V. C. 40 V . D. 80 V.

Cu 9: Khi t mt in p u = U0cos(120t + ) V vo hai u on mch RLC khng phn nhnh th in p hiu dng gia hai u in tr, hai u cun dy v gia hao bn t in c gi tr ln lt l 30 V, 120 V v 80 V. Gi tr ca U0 bng

A. 50 V. B. 60 V. C. 50eq \l(\r(,2)) V. D. 30eq \l(\r(,2)) V. Cu 10: on mch in xoay chiu RLC mc ni tip. lch pha ca in p v cng dng in trong mch c cho bi cng thc

A.

B.

C.

D.

Cu 11: Trong mch in xoay chiu khng phnh nhnh RLC th

A. lch pha ca uR v u l /2. B. pha ca uL nhanh hn pha ca i mt gc /2.

C. pha ca uC nhanh hn pha ca i mt gc /2. D. pha ca uR nhanh hn pha ca i mt gc /2. Cu 12: Trong mch RLC mc ni tip, lch pha gia dng in v in p ph thuc vo

A. cng dng in hiu dng trong mch. B. in p hiu dng gia hai u on mch.

C. cch chn gc tnh thi gian. D. tnh cht ca mch in.

Cu 13: Mt in tr thun R mc vo mch in xoay chiu tn s 50 Hz, mun dng in trong mch sm pha hn in p gia hai u on mch mt gc /2 ngi ta phi

A. mc thm vo mch mt t in ni tip vi in tr.

B. thay in tr ni trn bng mt t in.

C. mc thm vo mch mt cun cm ni tip vi in tr.

D. thay in tr ni trn bng mt cun cm.

Cu 14: iu kin xy ra hin tng cng hng in trong mch RLC ni tip l

A.

B.

C.

D.

Cu 15: on mch RLC ni tip ang xy ra cng hng. Tng dn tn s ca dng in mt lng nh v gi nguyn cc thng s khc ca mch, kt lun no di y khng ng?

A. Cng dng in gim, cm khng ca cun dy tng, in p hai u cun dy khng i.

B. Cm khng ca cun dy tng, in p hai u cun dy thay i.

C. in p hai u t gim.

D. in p hai u in tr gim.

Cu 16: Pht biu no sau y l khng ng. Trong mch in xoay chiu khng phn nhnh khi in dung ca t in thay i v tho mn iu kin th

A. cng dng in cng pha vi in p gia hai u on mch.

B. cng dng in hiu dng trong mch t cc i.

C. cng sut tiu th trung bnh trong mch t cc i.

D. in p hiu dng gia hai u t in t cc i.

Cu 17: Chn pht biu khng ng. Trong mch in xoay chiu khng phn nhnh khi in dung ca t in thay i v tho mn iu kin L = (C))eq \s\don1(\f(1,)) th

A. in p hiu dng gia hai u cun cm t cc i.

B. in p hiu dng gia hai u t in v cun cm bng nhau.

C. tng tr ca mch t gi tr ln nht.

D. in p hiu dng gia hai u in tr t cc i.

Cu 18: Trong on mch RLC, mc ni tip ang xy ra hin tng cng hng. Tng dn tn s dng in v gi nguyn cc thng s ca mch, kt lun no sau y l khng ng?

A. h s cng sut ca on mch gim. B. cng hiu dng ca dng in gim.

C. in p hiu dng trn t in tng. D. in p hiu dng trn in tr gim.

Cu 19: Dung khng ca mt on mch RLC ni tip c gi tr nh hn cm khng. Ta lm thay i ch mt trong cc thng s ca on mch bng cch nu sau y. Cch no c th lm cho hin tng cng hng in xy ra?

A. Tng in dung ca t in. B. Tng h s t cm ca cun dy.

C. Gim in tr ca on mch. D. Gim tn s dng in.

Cu 20: Mch in xoay chiu RLC ni tip. Kt lun no sau y l ng ng vi lc u L > eq \s\don1(\f(1,C))?

A. Mch c tnh dung khng.

B. Nu tng C n mt gi tr C0 no th trong mch c cng hng in.

C. Cng dng in sm pha hn in p gia hai u mch.

D. Nu gim C n mt gi tr C0 no th trong mch c cng hng in.

Cu 21: t mt in p xoay chiu c tn s thay i c vo hai u mt on mch RLC khng phn nhnh. Khi tn s trong mch ln hn gi tr th

A. in p hiu dng gia hai u cun dy nh hn in p hiu dng gia hai bn t.

B. in p hiu dng gia hai u in tr bng in p hiu dng gia hai u mch.

C. dng in trong sm pha so vi in p gia hai u mch.

D. dng in trong tr pha so vi in p gia hai u mch.

Cu 22: Dng in xoay chiu qua in tr thun bin thin iu ho cng pha vi in p gia hai u in tr trong trng hp no?

A. Mch RLC xy ra cng hng in. B. Mch ch cha in tr thun R.

C. Mch RLC khng xy ra cng hng in. D. Trong mi trng hp.

Cu 23: Chn phng n ng nht. Trong mch xoay chiu RLC ni tip, dng in v in p cng pha khi

A. on mch ch c in tr thun.

B. trong on mch xy ra hin tng cng hng in.

C. on mch ch c in tr thun hoc trong mch xy ra cng hng.

D. trong on mch dung khng ln hn cm khng.

Cu 24: Pht biu no sau y l khng ng. Trong mch in xoay chiu khng phn nhnh ta c th to ra in p hiu dng gia hai u

A. cun cm ln hn in p hiu dng gia hai u on mch.

B. t in ln hn in p hiu dng gia hai u on mch.

C. in tr ln hn in p hiu dng gia hai u on mch.

D. t in bng in p hiu dng gia hai u cun cm.

Cu 25: t vo mt on mch RLC khng phn nhnh mt in p u = U0cos(t) V th cng dng in trong mch c biu thc i = I0cos(t /3) A. Quan h gia cc tr khng trong on mch ny tha mn h thc

A.

B.

C.

D.

Cu 26: t vo mt on mch RLC khng phn nhnh mt in p u = U0cos(t /3) V th cng dng in trong mch c biu thc i = I0cos(t /6) A. Quan h gia cc tr khng trong on mch ny tha mn

A.

B.

C.

D.

Cu 27: t vo hai u on mch RLC khng phn nhnh mt in p xoay chiu u = U0cos(t) V. K hiu UR, UL, UC tng ng l in p hiu dng hai u in tr thun R, cun dy thun cm (cm thun) L v t in C. Nu UR = 0,5UL = UC th dng in qua on mch

A. tr pha /2 so vi in p hai u on mch.

B. tr pha /4 so vi in p hai u on mch.

C. tr pha /3 so vi in p hai u on mch.

D. sm pha /4 so vi in p hai u on mch. Cu 28: t vo hai u on mch RLC khng phn nhnh mt in p xoay chiu u = U0cos(t) V. K hiu UR, UL, UC tng ng l in p hiu dng hai u in tr thun R, cun dy thun cm (cm thun) L v t in C. Khi 3))eq \s\don1(\f(2,3))UR = 2UL = UC th pha ca dng in so vi in p l

A. tr pha /3. B. tr pha /6. C. sm pha /3. D. sm pha /6.

Cu 29: Cho on mch RLC ni tip, gi tr ca R bit, L c nh. t mt in p xoay chiu n nh vo hai u on mch, ta thy cng dng in qua mch chm pha /3 so vi in p trn on RL. trong mch c cng hng th dung khng ZC ca t phi c gi tr bng

A. eq \s\don1(\f(R,)) B. R. C. Req \l(\r(,3)) . D. 3R.

Cu 30: Cn ghp mt t in ni tip vi cc linh kin khc theo cch no di y, c c on mch xoay chiu m dng in tr pha /4 i vi in p gia hai u on mch. Bit t in trong mch ny c dung khng bng 20 .

A. mt cun thun cm c cm khng bng 20 .

B. mt in tr thun c ln bng 20 .

C. mt in tr thun c ln bng 40 v mt cun thun cm c cm khng 20 .

D. mt in tr thun c ln bng 20 v mt cun thun cm c cm khng 40 .

Cu 31: Cho mch in xoay chiu R, L, C. Khi ch ni R, C vo ngun in th thy i sm pha /4 so vi in p trong mch. Khi mc c R, L, C ni tip vo mch th thy i chm pha /4 so vi in p hai u on mch. Xc nh lin h ZL theo ZC.

A. ZL = 2ZC

B. ZC = 2ZL.

C. ZL = ZC

D. khng th xc nh c mi lin h.

Cu 32: Mch RLC ni tip c R = 100 , L = 2/ (H), f = 50 Hz. Bit i nhanh pha hn u mt gc /4 rad. in dung C c gi tr l

A. FB. FC. FD. F

Cu 33: Cho on mch xoay chiu khng phn nhnh gm cun dy thun cm c h s t cm L = 2/ (H), t in F v mt in tr thun R. in p t vo hai u on mch v cng dng in qua on mch c biu thc l u = U0cos(100(t) V v i = I0cos(100(t /4) A. in tr R c gi tr l

A. 400 . B. 200 . C. 100 . D. 50 .

Cu 34: t mt in p xoay chiu c gi tr hiu dung lun khng i v hai u on mch RLC khng phn nhnh. in p gia hai u

A. cun dy lun vung pha vi in p gia hai bn t in.

B. cun dy lun ngc pha vi in p gia hai bn t in.

C. t in lun sm pha /2 so vi cng dng in.

D. on mch lun cng pha vi cng dng in trong mch

Cu 35: Khi in p gia hai u on mch RLC mc ni tip sm pha /4 i vi dng in trong mch th

A. cm khng bng in tr thun.

B. dung khng bng in tr thun.

C. hiu ca cm khng v dung khng bng in tr thun.

D. tng ca cm khng v dung khng bng in tr thun.

Cu 36: in p gia hai u on mch RLC mc ni tip sm pha 3/4 so vi in p hai u t in. Pht biu no sau y l ng vi on mch ny?

A. Tng tr ca mch bng hai ln in tr thun ca mch.

B. Dung khng ca mch bng vi in tr thun.

C. Hiu s gia cm khng v dung khng bng in tr thun ca mch.

D. Cm khng ca mch bng vi in tr thun.

Cu 37: Cng hiu dng ca dng in chy trn on mch RLC ni tip khng c tnh cht no di y?

A. Khng ph thuc vo chu k dng in. B. T l thun vi in p hai u on mch.

C. Ph thuc vo tn s dng in. D. T l nghch vi tng tr ca on mch. Cu 38: Mt on mch khng phn nhnh RLC c dng in sm pha hn in p hai u on mch.

A. Trong on mch khng th c cun cm, nhng c t in.

B. H s cng sut ca on mch c gi tr khc khng.

C. Nu tng tn s dng in ln mt lng nh th lch pha gia dng in v in p gim.

D. Nu gim tn s ca dng in mt lng nh th cng hiu dng gim.

Cu 39: Mt on mch in xoay chiu gm mt cun dy thun cm c cm khng ZL mc ni tip vi t in c dung khng ZC. Pht biu no sau y l ng?

A. Tng tr ca mch c xc nh bi biu thc Z = ZL ZC.

B. Dng in chm pha hn /2 so vi in p gia hai u mch.

C. Dng in nhanh pha hn /2 so vi in p gia hai u mch.

D. in p gia hai bn t v hai u cun dy ngc pha nhau.

Cu 40: Mt on mch in xoay chiu gm in tr thun, cun cm thun v t in mc ni tip. Bit cm khng gp i dung khng. Dng vn k xoay chiu (in tr rt ln) o in p gia hai u t in v in p gia hai u in tr th s ch ca vn k l nh nhau. lch pha ca in p gia hai u on mch so vi cng dng in trong on mch l

A. /4 B. /6. C. /3. D. /3.

Cu 41: Cng dng in lun lun tr pha so vi in p hai u on mch khi

A. on mch ch c t in C. B. on mch c R v C mc ni tip.

C. on mch c R v L mc ni tip. D. on mch c L v C mc ni tip.

Cu 42: Trong mch in xoay chiu gm R, L, C mc ni tip th dng in nhanh pha hay chm pha so vi in p ca on mch l tu thuc vo

A. R v C. B. L v C. C. L, C v . D. R, L, C v .

Cu 43: Trong mch in xoay chiu gm R, L, C mc ni tip th tng tr Z ph thuc vo

A. L, C v . B. R, L, C. C. R, L, C v . D. .

Cu 44: Trong on mch xoay chiu RLC ni tip. Gi U, UR, UL, UC ln lt l in p hiu dng gia Cai u on mch, hai u in tr R, hai u cun dy L v hai bn t in C. iu no sau y khng th xy ra?

A. UR > UC B. UL > U C. U = UR = UL = UC D. UR > U

Cu 45: Mch in c i = 2cos(100t) A, v C = 250/ (F), R = 40 , L = 0,4/ (H) ni tip nhau th c

A. cng hng in.

B. uRL = 80cos(100t /4) V.

C. u = 80cos(100t + /6) V. D. uRC = 80cos(100t + /4) V.

Cu 46: Trong mch in xoay chiu khng phn nhnh c f = 50 Hz v ln lt C = 1000/ (F), R = 40 , L = 0,1/ (H). Chn kt lun ng ?

A. ZC = 40 , Z = 50 .

B. tanu/i = 0,75.

C. Khi R = 30 th cng sut cc i. D. in p cng pha so vi dng in.1B6A11B16D21D26D31A36C41C46D

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BI GING MCH IN XOAY CHIU RLC - PHN 2

II. MCH IN RLC NI TIP KHI CUN DY C THM IN TR rCho mch in xoay chiu RLC trong cun dy khng thun cm m c thm mt in tr r

Khi R v r c gi l tng tr thun ca mch v do R, r ni

tip nn tng tr thun k hiu l R0 = R + r ( UR0 = UR + Urc im ca mch in: in p v tng tr ca mch

nh lut m

lch pha ca in p v cng dng in trong mch l , c cho bi h thc

Nhn xt :

Cun dy c thm in tr hot ng r nn c th coi nh mt mch in (r, L) thu nh. Cc cng thc tnh ton vi cun dy cng nh tnh ton vi on mch RL kho st trn:

- in p hai u cun dy Ud = ULr =

- Tng tr ca cun dy Zd = ZLr =

- lch pha ca ud v i c cho bi tand = eq \s\don1(\f(ZL,r)) ( in p ud nhanh pha hn i gc d hay d = ud iCh : Trong mt s bi ton m khi bi cho nhp nhng khng bit c cun dy c thun cm hay khng hoc i khi yu cu chng minh rng cun dy c thm in tr hot ng r th ta lm theo cch sau:

- Gi s rng cun dy khng c in tr hot ng, r = 0.

- Thit lp cc biu thc vi r = 0 th s mu thun vi gi thit cho.

- Kt lun l cun dy phi c in tr hot ng r 0.V d 1: Cho mch in xoay chiu RLC vi R = 30 ; L = (H);r = 20 (();C = (F). Cng dng in chy qua on mch c biu thc i = 2eq \l(\r(,2))cos(100t + eq \l(\f((,6)) ) A.a) Tnh tng tr v in p hai u mch.b) Tnh tng tr v in p hai u cun dy.c) Vit biu thc in p hai u mch, in p hai u cun dy. d) Vit biu thc uR; uL; uC; ur.Hng dn gii:

V d 2: Cho mch in xoay chiu AB gm hai on mch AM v MB. on mch AM gm in tr R = 70 (. on mch MB l mt cun dy khng thun cm c L = 1,2))eq \s\don1(\f(,eq \l(\l(()))) (H); r = 90 ( v in p hai u on mch AB l uAB = 200eq \l(\r(,2))cos100t V.a) Vit biu thc cng dng in i. b) Vit biu thc udHng dn gii:

V d 3: Cho mch in xoay chiu AB gm hai on mch AM v MB. on mch AM gm cun dy khng thun cm, on MB gm mt t in. Bit uAM = 100eq \l(\r(,2))cos100t V; uMB = 100eq \l(\r(,2))cos( 100t - eq \s\don1(\f(2,3)) ) Va) Tnh r, C.b) Vit biu thc uABHng dn gii:

V d 4: Cho mch in xoay chiu RLC mc ni tip, cun dy c in tr r. Cc thng s ca mch in R = 60(; r = 20( ;C= ; i = 2eq \l(\r(,2))cos(100t - eq \l(\f((,6))) A; U = 160 V. Tnh h s t cm ca cun dy. /s: L = ())eq \s\don1(\f(2,)) Hng dn gii:

V d 5: Cho mch in xoay chiu gm in tr R v cun dy khng thun cm. Bit R = 40(; r = 20eq \l(\r(,3)) (; u= 120eq \l(\r(,2))cos(100t + eq \l(\f((,4))) V; ZL = 60 (. Hy vit biu thc in p hai u cun dy? /s: u = 120eq \l(\r(,6))cos(100t + ())eq \s\don1(\f(,12))) VHng dn gii:

V d 6: Cho mch in xoay chiu gm in tr R, t in C v cun dy khng thun cm. in p hai u mch l u = 50eq \l(\r(,2))cost V. Bit R = 30 (; r =Z= 10 (; ZC = 40 (. Hy vit biu thc in p hai u cun dy? /s: ud = 20cos(t + ())eq \s\don1(\f(,90))) VHng dn gii:

V d 7. Cho mch in xoay chiu nh hnh v, bit R = 50 (, C = 2.104/ (F), u = 80cos(100t) V, u = 200eq \l(\r(,2)) cos(100t + eq \l(\f((,2))) V.a) Tnh gi tr ca r v L.b) Vit biu thc ca cng dng in v in p hai u mch.Hng dn gii:a) Ta c = 100 rad ZC = (C))eq \s\don1(\f(1,)) = 50 .

Tng tr ca on mch AM l ZAM = ZRC = = 50eq \l(\r(,2)) .

Cng dng in = 0,8 A ( ZAM = ZLr = eq \s\don1(\f(UMB,I)) = 250 ( ( r2 + Z = 2502 (1) lch pha ca uAM vi i tha mn tanAM = - eq \s\don1(\f(ZC,R)) = -1 ( (AM = - eq \l(\f((,4)) , hay uAM chm pha hn i gc /4.

M uMB nhanh pha hn uAM gc / ( uMB nhanh pha hn i gc /4.

T taneq \s\don1(\f(,4)) = eq \s\don1(\f(ZL,r)) = 1 ( r = ZL (2) Gii h (1) v (2) ta c r =ZL = 125eq \l(\r(,2))( v L = eq \s\don1(\f(5,eq \l(\l(4()))) Hb) Vit biu thc ca u v i.Vit biu thc ca i : T cu a ta c AM = - ())eq \s\don1(\f(,4)) = - i ( i = eq \s\don1(\f(,4)) M I = 0,8 A ( i = 0,8eq \l(\r(,2))cos(100t + eq \l(\f((,4))) AVit biu thc ca in p hai u mch:Tng tr ca mch = 150eq \l(\r(,3)) ( in p hai u mch U =I.Z = 0,8.150eq \l(\r(,3)) V ( U0 = 120eq \l(\r(,6)) V lch pha ca u v i l ( 0,56 V ( ( ( 0,51 rad

M ( = (u - (i ( (u = (i + ( = ())eq \s\don1(\f(,4)) + 0,51 ( u = 120eq \l(\r(,6))cos(100(t + ())eq \s\don1(\f(,4)) + 0,51) VV d 8. Cho mch in xoay chiu nh hnh v, bit u = 120eq \l(\r(,2)) cos(100t) V, L = ())eq \s\don1(\f(3,)) (H). Tm R v C bit uAN tr pha /3 so vi uAB v uMB sm pha /3 so vi uAB.Hng dn gii:

Ta c gin vc t nh hnh v.

T gi thit ta c ZL = 300 .

on mch MB cha L v C, do uMB nhanh pha hn uAB nn ZL > ZC v uAB nhanh pha hon i gc /6.

Mt khc, uAN chm pha hn uAB gc /3, m uAB nhanh pha hn i gc /6 nn uAN chm pha hn i gc /6.

T cc lp lun ta c:

( UR = eq \l(\r(,3))UC v UL = 2UC M UAB = 120 V = (

Li c, I = = eq \s\don1(\f(120,300)) = 0,4 A ( (

Cch 2: (S dng gin vc t) T gin ta tnh c: (

Vi UR tnh c, ta li c UC = UR.taneq \l(\f((,6)) = 60 V ( UL = 120 V T ta gii tip nh trn thu c kt qu.

TRC NGHIM MCH IN XOAY CHIU RLC - PHN 2

Cu 1: Mt mch in xoay chiu gm R, L, C mc ni tip. Bit L, C khng i v tn s dng in thay i c. Bit rng ng vi tn s f1 th ZL = 50 v ZC = 100 . Tn s f ca dng in ng vi lc xy ra cng hng in phi tho mn

A. f > f1. B. f < f1. C. f = f1. D. f = 0,5f1.

Cu 2: Cho mt on mch RLC ni tip. Bit L = 1/ (H), C = 2.10-4/ (F), R thay i c. t vo hai u on mch mt in p c biu thc u = U0cos(100t) V. uC chm pha 3/4 so vi u th R phi c gi tr

A. R = 50 . B. R = 50eq \l(\r(,2)) (C. R = 100 . D. R = 100eq \l(\r(,2)) (Cu 3: Cho mt on mch RLC ni tip. Bit L = ())eq \s\don1(\f(1,)) (H), C = (F), R thay i c. t vo hai u on mch mt in p c biu thc u = U0cos(100t) V. uL nhanh pha 2/3 so vi u th R phi c gi tr

A. R = 50 . B. R = 50eq \l(\r(,3)) (C. R = 100 . D. R = 100eq \l(\r(,3)) (Cu 4: Khi mc ln lt R, L, C vo mt in p xoay chiu n nh th cng dng in hiu dng qua ca chng ln lt l 2A, 1A, 3A. Khi mc mch gm R, L, C ni tip vo in p trn th cng dng in hiu dng qua mch bng

A. 1,25 A B. 1,2 A. C. 3eq \l(\r(,2)) A. D. 6 A.

Cu 5: t mt in p xoay chiu u = U0sin(t) V vo hai u on mch ch c cun dy thun cm L. Gi U l in p hiu dng hai u on mch; i, I0, I ln lt l gi tr tc thi, gi tr cc i v gi tr hiu dng ca cng dng in trong mch. H thc no sau y khng ng?

A.

B.

C.

D.

Cu 6: Khi ta mc R, C vo mt in p c biu thc khng i, gi tr hiu dng U = 100 V, th thy i sm pha so vi u l /4, khi ta mc R, L vo in p ny th thy in p sm pha so vi dng in l /4. Hi khi ta mc c ba phn t trn vo in p th in p hai u L v C c gi tr l

A. 100eq \l(\r(,2)) V. B. 50eq \l(\r(,2)) V. C. 0 V. D. 200 V.

Cu 7: Khi ta mc R, C vo mt in p c biu thc khng i th thy i sm pha so vi u l /4, khi ta mc R, L vo in p ny th thy in p sm pha so vi dng in l /4. Hi khi ta mc c ba phn t trn vo in p th u v i lch pha nhau l

A. .

B. 0. C. /2. D. /4.

Cu 8: Cho mch R, L, C vi cc gi tr ban u th cng trong mch ang c gi tr I, v dng in sm pha /3 so vi in p. Nu ta tng L v R ln hai ln, gim C i hai ln th I v lch pha ca u v i s bin i th no?

A. I khng i, lch pha khng i. B. I gim, lch pha khng i.

C. I gim eq \l(\r(,2)) ln, lch pha khng i. D. I v lch u gim.

Cu 9: t vo hai u on mch RLC ni tip mt in p xoay chiu c gi tr hiu dng khng i th in p hiu dng trn cc phn t R, L v C ln lt l 30 V, 50 V v 90 V. Khi thay t C bng t C mch c cng hng in th in p hiu dng gia hai u in tr R bng

A. 50 V. B. 70eq \l(\r(,2)) V. C. 100 V. D. 100eq \l(\r(,2)) V.Cu 10: Trong mch in gm r, R, L, C mc ni tip. Gi Z l tng tr ca mch. lch pha gia in p hai u mch v cng dng in trong mch c tnh bi cng thc

A.

B.

C.

D.

Cu 11: Trong mch in gm r, R, L, C mc ni tip. Gi Z l tng tr ca mch. lch pha gia in p hai u mch v cng dng in trong mch c tnh bi cng thc

A.

B.

C.

D.

Cu 12: Cho on mch in xoay chiu gm cun dy mc ni tip vi mt t in. in p hiu dng gia hai u cun dy, gia hai bn t, gia hai u on mch ln lt l: Ud, UC, U. Bit Ud = eq \l(\r(,2))UC; U = UC

A. V UL UC nn ZL ZC, vy trong mch khng xy ra cng hng.

B. Cun dy c in tr thun ng k,trong mch khng xy ra hin tng cng hng.

C. Cun dy c in tr thun ng k, trong mch xy ra hin tng cng hng.

D. Cun dy c in tr thun khng ng k.

Cu 13: Biu thc hiu in th hai u mt on mch u = 200cos(t) V. Ti thi im t, in p u = 100 V v ang tng. Hi vo thi im t = t + eq \s\don1(\f(T,4)) in p u c gi tr bng bao nhiu ?

A. 100 V. B. 100eq \l(\r(,2)) V. C. 100eq \l(\r(,3)) V. D. 100 V.

Cu 14: Ti thi im t, in p xoay chiu u = 200eq \l(\r(,2))cos(100t - /2) V c gi tr 100eq \l(\r(,2)) V v ang gim. Sau thi im eq \s\don1(\f(1,300)) (s) , in p ny c gi tr l

A. - 100eq \l(\r(,2)) V. B. 100 V. C. 100eq \l(\r(,3)) V. D. 200 V.

Cu 15: in p gia hai u mt on mch c biu thc u = 220eq \l(\r(,2))cos(100t + /2) V. Ti mt thi im t1 no in p ang gim v c gi tr tc thi l 110eq \l(\r(,2)) V. Hi vo thi im t2 = t1 + 0,005 (s) th in p c gi tr tc thi bng bao nhiu ?

A. - 110eq \l(\r(,3)) V. B. 110eq \l(\r(,3)) V. C. -110eq \l(\r(,6)) V. D. 110eq \l(\r(,6)) V.

Cu 16: Dng in chy qua mt on mch c biu thc i = I0cos(100t) A. Trong khong thi gian t 0 n 0,018 (s) cng dng in c gi tr tc thi c gi tr bng 0,5I0 vo nhng thi im no?

A. eq \s\don1(\f(1,400))s; eq \s\don1(\f(2,400)) sB. s; sC. s; sD. s; s

Cu 17: Cho mt ngun xoay chiu n nh. Nu mc vo ngun mt in tr thun R th dng in qua R c gi tr hiu dng I1 = 3A. Nu mc t C vo ngun th c dng in c cng hiu dng I2 = 4A. Nu mc R v C ni tip ri mc vo ngun trn th dng in qua mch c gi tr hiu dng l

A. 1 A .B. 2,4 A. C. 5 A. D. 7 A.

Cu 18: Mt mch in gm in tr thun R, cun dy thun cm v mt t in c in dung thay i c mc ni tip. t vo hai u on mch trn mt in p xoay chiu c biu thc u = U0cos(t) V. Khi thay i in dung ca t cho in p gia hai bn t t cc i v bng 2U. Mi quan h gia ZL v R l

A. ZL = eq \s\don1(\f(R,)) B. ZL = 2R. C. ZL = Req \l(\r(,3)) . D. ZL = 3R.

Cu 19: Nu t vo hai u cun dy mt in p mt chiu 9 V th cng dng in trong cun dy l 0,5 A. Nu t vo hai u cun dy mt in p xoay chiu tn s 50 Hz v c gi tr hiu dng l 9 V th cng dng in hiu dng qua cun dy l 0,3 A. in tr thun v cm khng ca cun dy l

A. R = 18 , ZL = 30 .

B. R = 18 , ZL = 24 .

C. R = 18 , ZL = 12 .

D. R = 30 , ZL = 18 .

Cu 20: t vo hai u mt cun dy c t cm L = 0,4/ (H) mt in p mt chiu U1 = 12 V th cng dng in qua cun dy l I1 = 0,4A. Nu t vo hai u cun dy ny mt in p xoay chiu c gi tr hiu dng U2 = 100 V, tn s f = 50 Hz th cng hiu dng ca dng in chy qua cun dy l

A. I = 2,5 A. B. I = 2 AC. I = 0,5 AD. I = 2,4 A.

Cu 21: Mt chic n nn t di mt in p xoay chiu 119 V 50 Hz. N ch sng ln khi in p tc thi gia hai u bng n ln hn 84 V. Thi gian bng n sng trong mt chu k l

A. (t = 0,0100 (s). B. (t = 0,0133 (s). C. (t = 0,0200 (s). D. (t = 0,0233(s). Cu 22: Mt n non t di in p xoay chiu c gi tr hiu dng 220 V v tn s 50 Hz. Bit n sng khi in p gia hai cc khng nh hn 155 V. Trong mt giy n sng ln hoc tt i bao nhiu ln?

A. 50 ln. B. 100 ln. C. 150 ln. D. 200 ln.

Cu 23: Mt n non t di in p xoay chiu c gi tr hiu dng 220 V v tn s 50 Hz. Bit n sng khi in p gia hai cc khng nh hn 155 V. T s gia thi gian n sng v thi gian n tt trong mt chu k l

A. 0,5 ln. B. 1 ln. C. 2 ln. D. 3 ln

Cu 24: Cho on mch gm cun dy c in tr thun R = 100 , h s t cm L = 1/ (H) mc ni tip vi t in c in dung C (F). t vo hai u on mch mt in p xoay chiu u = 200sin(100t)V. Biu thc in p tc thi gia hai u cun dy l

A. ud = 200sin(100t + /2) V. B. ud = 200sin(100t + /4) V.

C. ud = 200sin(100t - /4) V.D. ud = 200sin(100t) V.

Cu 25: Cho mt on mch xoay chiu gm cun dy c in tr r, t cm L mc ni tip vi in tr thun R = 50 . in p hai u mch v cng dng in qua mch c biu thc u = 100eq \l(\r(,2)) cos(100t + /2) V v i = eq \l(\r(,2))cos(100t + /3) A. Gi tr ca r bng

A. r = 20,6 . B. r = 36,6 . C. r = 15,7 . D. r = 25,6 . Cu 26: Trong mch in xoay chiu gm R, L, C mc ni tip, lch pha gia in p gia hai u in tr R v in p gia hai u on mch l = /3. Chn kt lun ng ?

A. Mch c tnh dung khng. B. Mch c tnh cm khng.

C. Mch c tnh tr khng. D. Mch cng hng in.

Cu 27: Cho on mch in xoay chiu RLC, cun dy khng thun cm. Bit r = 20 , R = 80 , C =F. Tn s dng in trong mch l 50 Hz. mch in p hai u mch nhanh pha hn dng in gc /4 th h s t cm ca cun dy l

A. L = HB. L = HC. L = HD. L =

Tr li cc cu hi 28, 29, 30: Mt on mch xoay chiu gm in tr thun R = 100 , mt cun dy thun cm c t cm L = 2/ (H) v mt t in c in dung C = (F) mc ni tip gia hai im c in p u = 200eq \l(\r(,2))cos(100t)V.Cu 28: Biu thc tc thi cng dng in qua mch l

A. i = 2eq \l(\r(,2))cos(100t - eq \l(\f((,4))) A B. i = 2cos(100t - eq \l(\f((,4))) A C. i = 2cos(100t + eq \l(\f((,4))) A D. i = eq \l(\r(,2))cos(100t + eq \l(\f((,4))) ACu 29: in p hai u cun cm l

A. uL = 400eq \l(\r(,2))cos(100t + ())eq \s\don1(\f(,4))) V B. uL = 200eq \l(\r(,2))cos(100t + ())eq \s\don1(\f(,4))) V C. uL = 400cos(100t + ())eq \s\don1(\f(,4))) V D. uL = 400cos(100t + ())eq \s\don1(\f(,2))) VCu 30: in p hai u t in l

A. uC = 200eq \l(\r(,2))cos(100t - ())eq \s\don1(\f(,4))) V B. uC = 200eq \l(\r(,2))cos(100t - ())eq \s\don1(\f(,4))) V C. uC = 200cos(100t - ())eq \s\don1(\f(,2))) V D. uC = 200cos(100t - ())eq \s\don1(\f(,4))) V Cu 31: Cho on mch xoay chiu gm R, L mc ni tip c R = 40 , L = 0,4/ (H). on mch c mc vo in p u = 40eq \l(\r(,2))cos(100t )V. Biu thc cng dng in qua mch l

A. i = cos(100t - ())eq \s\don1(\f(,4))) Aeq \f(,)

B. i = cos(100t + ())eq \s\don1(\f(,4))) Aeq \f(,) C. i = eq \l(\r(,2))cos(100t - ())eq \s\don1(\f(,4))) Aeq \f(,)D. i = eq \l(\r(,2))cos(100t + ())eq \s\don1(\f(,4)) ) Aeq \f(,)Cu 32: Cho on mach xoay chiu gm R, L mc ni tip. R = 20 , L = 0,2/ H. on mch c mc vo in vo in p u = 40eq \l(\r(,2))cos(100t )V. Biu thc cng dng in qua mch l

A. i = 2cos(100t - ())eq \s\don1(\f(,4))) Aeq \f(,)

B. i = 2cos(100t + ())eq \s\don1(\f(,4))) Aeq \f(,) C. i = eq \l(\r(,2))cos(100t - ())eq \s\don1(\f(,4))) Aeq \f(,)D. i = eq \l(\r(,2))cos(100t + ())eq \s\don1(\f(,4)) ) Aeq \f(,)Cu 33: Cho mch R, L, C mc ni tip c R = 20eq \l(\r(,3)) , L = eq \s\don1(\f(,eq \l(\l(()))) (H), C = (F). t vo hai u mch in mt in p u = 200eq \l(\r(,2))cos(100t )V. Biu thc cng dng in trong mch l

A. i = 5eq \l(\r(,2))cos(100t + ())eq \s\don1(\f(,3))) A B. i = 5eq \l(\r(,2))cos(100t - ())eq \s\don1(\f(,3)) ) A C. i = 5eq \l(\r(,2))cos(100t + ())eq \s\don1(\f(,6)) ) A D. i = 5eq \l(\r(,2))cos(100t - ())eq \s\don1(\f(,3)) ) ACu 34: t in p xoay chiu vo hai u on mch c R, L, C mc ni tip. Bit R = 10 , cun cm thun c L = ())eq \s\don1(\f(1,)) H, t in c C = (F) v in p gia hai u cun cm thun l uL = 20eq \l(\r(,2))cos(100t + eq \l(\f((,2))) V. Biu thc in p gia hai u on mch l

A. u = 40cos(100t + eq \l(\f((,4))) VB. u = 40cos(100t - eq \l(\f((,4))) V C. u = 40eq \l(\r(,2))cos(100t + eq \l(\f((,4))) VD. u = 40eq \l(\r(,2))cos(100t - eq \l(\f((,4))) VCu 35: t in p xoay chiu c gi tr hiu dng 60 V vo hai u on mch R, L, C mc ni tip th cng dng in qua on mch l i1 = I0cos(100t + eq \l(\f((,4))) A. Nu ngt b t in C th cng dng in qua on mch l i2 = I0cos(100t - ())eq \s\don1(\f(,12))) A. in p hai u on mch l

A. u = 60eq \l(\r(,2))cos(100t - ())eq \s\don1(\f(,12)) ) V B. u = 60eq \l(\r(,2))cos(100t - ())eq \s\don1(\f(,6))) V C. u = 60eq \l(\r(,2))cos(100t + ())eq \s\don1(\f(,12)) ) V D. u = 60eq \l(\r(,2))cos(100t + ())eq \s\don1(\f(,6)) ) VCu 36: Khi t in p khng i 30 V vo hai u on mch gm in tr thun mc ni tip vi cun cm thun c t cm L = ())eq \s\don1(\f(1,)) (H) th dng in trong on mch l dng in mt chiu c cng 1A. Nu t vo hai u on mch ny in p u = 150eq \l(\r(,2))cos120t V th biu thc ca cng dng in trong on mch l

A. i = 5eq \l(\r(,2))cos(120t - (/4) A B. i = 5cos(120t + (/4) A C. i = 5eq \l(\r(,2))cos(120t + (/4) A D. i = 5cos(120t - (/4) ACu 37: t in p u = U0cos(100t - (/3) V vo hai u mt t in c in dung C = (F) . thi im in p gia hai u t in l 150 V th cng dng in trong mch l 4A. Biu thc ca cng dng in trong mch l

A. i = 4eq \l(\r(,2))cos(100t + (/6) A B. i = 5cos(100t + (/6) A C. i = 5cos(100t - (/6) A D. i = 4eq \l(\r(,2))cos(100t - (/6) ACu 38: t in p xoay chiu u = U0cos(100t + (/3) V vo hai u mt cun cm thun c t cm L = ())eq \s\don1(\f(1,)) H. thi im in p gia hai u cun cm l 100eq \l(\r(,2)) V th cng dng in qua cun cm l 2A. Biu thc ca cng dng in qua cun cm l: A. i = 2eq \l(\r(,3))cos(100t - (/6) A B. i = 2eq \l(\r(,3))cos(100t + (/6) A C. i = 2eq \l(\r(,2))cos(100t + (/6) A D. i = 2eq \l(\r(,2))cos(100t - (/6) ACu 39: on mch xoay chiu nh hnh v, bit L = 2/ (H), C = 31,8 (F), R c gi tr xc nh. Cng dng in trong mch c biu thc i = 2cos(100t - (/3) A. Biu thc uMB c dng

A. uMB = 200cos(100t - (/3) VB. uMB = 600cos(100t + (/6) V

C. uMB = 200cos(100t + (/6) VD. uMB = 600cos(100t - (/2) VCu 40: in p hai u on mch xoay chiu ch c t C = (F) c biu thc u = 100eq \l(\r(,2))cos(100t + (/3) V, biu thc cng dng in qua mch trn l nhng dng no sau y?

A. i = eq \l(\r(,2))cos(100t - (/2) AB. i = eq \l(\r(,2))cos(100t - (/6) A C. i = eq \l(\r(,2))cos(100t - 5(/6) A D. i = 2cos(100t - (/6) ACu 41: Mch in xoay chiu gm in tr R = 40 ghp ni tip vi cun cm L. in p tc thi hai u on mch u = 80cos(100t) V v in p hiu dng hai u cun cm UL = 40 V. Biu thc cng dng in qua mch l

A. i = eq \s\don1(\f(,2))cos(100(t - (/4) A. B. i = eq \s\don1(\f(,2))cos(100(t + (/4) A.

C. i = eq \l(\r(,2))cos(100(t - (/4) A. D. i = eq \l(\r(,2))cos(100(t + (/4) A.

Cu 42: Mt on mch gm t C = (F) v cun dy thun cm c t cm L = 2/ (H) mc ni tip. in p gia 2 u cun cm l uL = 100eq \l(\r(,2))cos(100t + (/3) V. in p tc thi hai u t c biu thc nh th no

A. uC = 50eq \l(\r(,2))cos(100t - 2(/3) V B. uC = 50cos(100t - (/6) V C. uC = 50eq \l(\r(,2))cos(100t + (/6) V D. uC = 100eq \l(\r(,2))cos(100t + (/3) VCu 43: Mch xoay chiu gm R, L, C mc ni tip (cun dy thun cm), R = 100 , C = 31,8 F, h s cng sut mch cos = eq \s\don1(\f(,2)) , in p hai u mch u = 200cos(100t) V. t cm L v cng dng in chy trong mch l:

A. H, i = AB. H, i = A

C. H, i = AD. H, i = A

Cu 44: Mt bn l 200 V 1000 W c mc vo in p xoay chiu u = 100eq \l(\r(,2))cos100t V. Bn l c t cm nh khng ng k. Dng in chy qua bn l c biu thc no ?

A. i = 2,5eq \l(\r(,2))cos(100t) A. B. i = 2,5eq \l(\r(,2))cos(100t+ (/2) A.

C. i = 2,5cos(100t) A.

D. i = 2,5cos(100t - (/2) A.

Cu 45: Mt mch gm cun dy thun cm c cm khng bng 10 mc ni tip vi t in c in dung C = F. Dng in qua mch c biu thc i = 2eq \l(\r(,2))cos(100t + (/3) A. Biu thc in p ca hai u on mch l

A. u = 80eq \l(\r(,2))cos(100t - (/6) V B. u = 80eq \l(\r(,2))cos(100t + (/6) V C. u = 120eq \l(\r(,2))cos(100t - (/6) V D. u = 80eq \l(\r(,2))cos(100t - 2(/3) VCu 46: Cho on mch xoay chiu mc ni tip gm in tr c R = 100 , t in c dung khng 200, cun dy c cm khng 100 . in p hai u mch cho bi biu thc u = 200cos(120t + /4)V. Biu thc in p hai u t in l

A. uC = 200eq \l(\r(,2))cos(120t + (/4) VB. uC = 200eq \l(\r(,2))cos(120t) V C. uC = 200eq \l(\r(,2))cos(120t - (/4) V D. uC = 200cos(120t - (/2) VCu 47: on mch R, L, C mc ni tip c R = 40 , L = ())eq \s\don1(\f(1,)) (H), C = (F). t vo hai u mch in p xoay chiu c biu thc u = 120eq \l(\r(,2))cos100t V. Cng dng in tc thi trong mch l

A. i = 1,5cos(100t + (/4) AB. i = 1,5cos(100t - (/4) A

C. i = 3cos(100t + (/4) AD. i = 3cos(100t - (/4) ACu 48: Nu t vo hai u mt mch in cha mt in tr thun R v mt t in C mc ni tip mt in p xoay chiu c biu thc u = U0cos(t - (/2) V, khi dng in trong mch c biu thc i=I0cos(t - (/4) A. Biu thc in p gia hai bn t s l

A. uC = I0Rcos((t - 3(/4) VB. uC = eq \s\don1(\f(U0,R)) cos((t + (/4) V

C. uC = I0ZCcos((t + (/4) VD. uC = I0Rcos((t - (/2) VCu 49: Mt on mch xoay chiu gm R v C ghp ni tip. t gia hai u on mch in p xoay chiu c biu thc tc thi u = 220eq \l(\r(,2))cos(100t - (/2) V th cng dng in qua on mch c biu thc tc thi i = 4,4cos(100t - (/4) A. in p gia hai u t in c biu thc tc thi l

A. uC = 220cos(100(t - (/4) VB. uC = 220cos(100(t - 3(/4) V C. uC = 220eq \l(\r(,2))cos(100(t + (/2) VD. uC = 220eq \l(\r(,2))cos(100(t - 3(/4) VCu 50: Mt on mch gm cun dy thun cm c t cm L = ())eq \s\don1(\f(1,)) (H) mc ni tip vi t in c in dung C = (F). Dng in chy qua on mch c biu thc i = 2eq \l(\r(,2))cos(100t + (/3) A. Biu thc in p hai u on mch s l

A. u = 80eq \l(\r(,2))cos(100t + (/6) V B. u = 80eq \l(\r(,2))cos(100t - (/3) V C. u = 80eq \l(\r(,2))cos(100t - (/6) V D. u = 80eq \l(\r(,2))sin(100t - (/6) VCu 51: in p v cng dng in trong on mch ch c t in c dng u = U0cos(t + /4) V v i = I0cos(t + ) A. Hi I0 v c gi tr no sau y ?

A. I0 = (CU0; ( = 3(/4

B. I0 = (CU0; ( = - (/2 C. I0 = ; ( = 3(/4

D. I0 = ; ( = -(/2Cu 52: Dng in xoay chiu i = I0cos(t + /4) A qua cun dy thun cm L. in p gia hai u cun dy l u = U0cos(t + ) V. Hi U0 v c cc gi tr no sau y ?

A. U0 = ; ( = (/2B. U0 = I0(L; ( = 3(/4C. U0 = ; ( = 3(/4D. U0 = I0(L; ( = -(/4

P N TRC NGHIM MCH IN XOAY CHIU RLC - PHN 21A6C11D16C21B26B31A36D41C46B51A

2A7B12C17B22B27D32A37B42A47D52B

3B8B13C18C23C28B33B38A43A48A

4B9A14A19B24A29C34B39C44A49B

5B10C15C20B25B30D35C40C45A50C

CC LOI MCH IN XOAY CHIU

I. MCH IN CH C IN TR THUN Rc im:* in p v dng in trong mch cng pha vi nhau (tc u = i):

* nh lut Ohm cho mch:

* Gin vc t: * th ca uR theo i (hoc ngc li) c dng ng thng i qua gc ta .* Nhit lng ta ra trn in tr R trong thi gian t l: Q = I2Rt =

* Nu hai in tr R1 v R2 ghp ni tip th ta c cng thc R = R1 + R2, ngc li hai in tr mc song song th

V d 1. Mc in tr thun R = 55 vo mch in xoay chiu c in p u = 110cos(100t + /2) V. a) Vit biu thc cng dng in qua mch.b) Tnh nhit lng ta ra trn in tr trong 10 pht.Hng dn gii:a) Ta c U0 = 110 V, R = 55 ( ( I0 = = 2A Do mch ch c R nn u v i cng pha. Khi u = i = eq \l(\f((,2)) ( i = 2cos(100t + eq \l(\f((,2))) A b) Nhit lng ta ra trn in tr R trong 10 pht: Q = I2Rt = R.t = (eq \l(\r(,2)))255.10.60 = 66000 J = 66 kJ.

V d 2. iu no sau y l ng khi ni v on mch xoay chiu ch c in tr thun? A. Dng in qua in tr v in p hai u in tr lun cng pha.B. Pha ca dng in qua in tr lun bng khng.C. Mi lin h gia cng dng in v in p hiu dng l U = I/R.D. Nu in p hai u in tr l u = U0sin(t + ) V th biu thc dng in l i = I0sin(t) A.Hng dn gii:Phng n B sai v pha ca dng in bng vi pha ca in p ch khng phi lun bng 0. Phng n C sai v biu thc nh lut Ohm l U = I.R

Phng n D sai v dng in v in p cng pha nn u = Uosin(t + ) V ( i = I0sin(t + ) A.

II. MCH IN CH C CUN CM THUN VI T CM Lc im:* in p nhanh pha hn dng in gc /2 (tc u = i + /2):

* Cm khng ca mch: ZL = L = 2f.L ( th ca cm khng theo L l ng thng i qua gc ta (dng y = ax).

* nh lut Ohm cho mch

Gin vc t:

* Do uL nhanh pha hn i gc /2 nn ta c phng trnh lin h ca uL v i c lp vi thi gian

( T h thc trn ta thy th ca uL theo i (hoc ngc li) l ng elipH qu:Ti thi im t1 in p v dng in c gi tr l u1; i1, ti thi im t2 in p v dng in c gi tr l u2; i2 th ta c = 1 = ( ( (

V d 1. Tnh cm khng ca cun cm thun trong on mch in xoay c tn s f = 50 Hz bita) L = ())eq \s\don1(\f(1,)) H. . b) L = eq \s\don1(\f(,eq \l(\l(2()))) H. . c) L = eq \s\don1(\f(1,()))) H. . V d 2. Tnh cm khng ca cun cm thun trong on mch in xoay c tn s f = 60 Hz bit a) L = ())eq \s\don1(\f(2,)) H. . b) L = eq \s\don1(\f(,eq \l(\l(()))) H. . c) L = ())eq \s\don1(\f(1,)) H. . V d 3. Vit biu thc uL trong on mch in xoay chiu ch c cun cm thun L bita) L = ())eq \s\don1(\f(1,)) H, i = 2eq \l(\r(,3))cos(100t + eq \l(\f((,6))) A b) L = eq \s\don1(\f(,eq \l(\l(()))) H, i = eq \l(\r(,2))cos(100t - eq \l(\f((,3))) A c) L = eq \s\don1(\f(,eq \l(\l(2()))) H, i = eq \l(\r(,6))cos(100t - eq \l(\f((,4)) ) A Hng dn gii:Vi mch in ch c L th ta lun c a) L = ())eq \s\don1(\f(1,)) H ( ZL= 50(. T ta c

( uL = 100eq \l(\r(,3))cos(100t + ())eq \s\don1(\f(,3))) Vb) L = eq \s\don1(\f(,eq \l(\l(()))) H ( ZL = 100 eq \l(\r(,3)) (. .

. . c) L = eq \s\don1(\f(,eq \l(\l(2()))) H ( ZL = 50eq \l(\r(,2)) (. . . . V d 4. Cho mch in xoay chiu ch c cun cm c t cm L vi L = 2/ (H). t vo hai u mch in p xoay chiu c gi tr hiu dng 200 V, tn s 50 Hz, pha ban u bng khng.a) Tnh cm khng ca mch.b) Tnh cng hiu dng ca dng in.c) Vit biu thc cng dng in qua mch.

V d 5. ( thi i hc 2009).t in p u = U0cos(100t + eq \l(\f((,3)) ) V vo hai u mt cun cm thun c t cm L = ())eq \s\don1(\f(1,)) (H) . thi im in p gia hai u cun cm l 100eq \l(\r(,2)) V th cng dng in trong mch l 2 A. Biu thc cng dng in trong mch l A. i = 2eq \l(\r(,3))cos(100t + eq \l(\f((,6)) ) A B. i = 2eq \l(\r(,2))cos(100t - eq \l(\f((,6)) ) A

C. i = 2eq \l(\r(,2))cos(100t + eq \l(\f((,6))) A D. i = 2eq \l(\r(,3))cos(100t - eq \l(\f((,6))) A Hng dn gii:Cm khng ca mch l Z = L =100.())eq \s\don1(\f(1,)) = 50 (Do mch ch c L nn u - i = ())eq \s\don1(\f(,2)) ( i = u- eq \l(\f((,2)) = - eq \l(\f((,6)) rad T h thc lin h ( ( ( I0 = 2eq \l(\r(,3)) A Vy biu thc cng dng in qua mch l i = 2eq \l(\r(,3))cos(100( - ())eq \s\don1(\f(,6)) ) AV d 6. Cho mt on mch in xoay chiu ch c cun cm thun. t vo hai u mch in mt in p xoay chiu c tn s 50 Hz. Ti thi im t1 in p v dng in qua cun cm c gi tr ln lt l 117 V; 0,6 A. Ti thi im t2 in p v dng in qua cun cm c gi tr ln lt l 108 V; 1 A. Tnh h s t cm L.

III. MCH IN CH C T IN VI IN DUNG Cc im:* in p chm pha hn dng in gc /2 (tc u = i - /2):

* Dung khng ca mch: ZC = eq \s\don1(\f(1,C)) = eq \s\don1(\f(1,2f.C)) ( th ca dung khng theo C l ng cong hupebol (dng y = eq \s\don1(\f(1,x)) ).

* nh lut Ohm cho mch

Gin vc t: * Do uC chm pha hn i gc /2 nn ta c phng trnh lin h ca uL v i c lp vi thi gian

( T h thc trn ta thy th ca uC theo i (hoc ngc li) l ng elipH qu:Ti thi im t1 in p v dng in c gi tr l u1; i1, ti thi im t2 in p v dng in c gi tr l u2; i2 th ta c = 1 = ( ( (

V d 1. Tnh dung khng ca t in trong on mch in xoay c tn s f = 50 Hz bita) C = (F) .............

b) C = (F) .. .........c) C =(F) .. . . . . . . V d 2. Vit biu thc cng dng in tc thi trong on mch in xoay chiu ch c t C bita) C = F, uC = 100eq \l(\r(,2))cos(100t + ())eq \s\don1(\f(,12))) V b) C = F, uC = 200eq \l(\r(,3))cos(100t - eq \l(\f((,3)) ) V c) C = = F, uC = 50eq \l(\r(,3))cos(100t - eq \l(\f((,6)) ) VHng dn gii:Vi mch in ch c C th ta lun c

a) C = F ( ZC= =100eq \l(\r(,2)) (. T ta c

( i = cos(100t + 7())eq \s\don1(\f(,12)) ) Ab) C = F

........................................................................................................................c) C = F

........................................................................................................................V d 3. Cho mch in xoay chiu ch c t in vi in dung C = (F) . Dng in trong mch c biu thc l i = 2cos(100t + /3) A. a) Tnh dung khng ca mch.b) Tnh hiu in p hiu dng gia hai u t in. c) Vit biu thc in p hai u mch.

V d 4. t in p u = U0cos(100t + eq \l(\f((,6))) V vo hai u mt t in c in dung C = (F) . thi im in p gia hai u t in l 300 V th cng dng in trong mch l 2eq \l(\r(,2)) A. Vit biu thc cng dng in chy qua t in.Hng dn gii:Mch ch c t in nn in p chm pha hn dng in gc /2, khi u = i /2 ( i = 2/3 rad.Dung khng ca mch l ZC = (C))eq \s\don1(\f(1,)) = 50eq \l(\r(,3)) ( ( U0C = 50eq \l(\r(,3))I0p dng h thc lin h ta c ( ( I0=2eq \l(\r(,5)) A Vy cng dng in chy qua bn t in c biu thc i = 2eq \l(\r(,5))cos(100t + ())eq \s\don1(\f(,3)) ) AV d 5. Cho on mch in xoay chiu ch c t in vi in dung C. Ti thi im t1 in p v dng in qua t in c gi tr ln lt l 65 V; 0,15 A. Ti thi im t2 in p v dng in qua t in c gi tr ln lt l 63 V ; 0,25 A. Dung khng ca mch c gi tr l bao nhiu?Hng dn gii:Mch ch c C nn u v i vung pha. Khi

Ti thi im t1:

Ti thi im t2:

T ta c:

EMBED Equation.3 ( (

( ZC . Thay s ta c ZC = 80 (Vy dung khng ca mch l 80

IV. MT S BI TP TRC NGHIM IN HNHCu 1. Mch in xoay chiu ch c cun thun cm vi t cm L. t vo hai u cun thun cm mt in p xoay chiu u = Ueq \l(\r(,2))cos(t + ) V. Cng dng in cc i ca mch c cho bi cng thc

A.

B.

C.

D.

Hng dn gii:

Vi on mch ch c L th

Cu 2. Mch in xoay chiu ch c cun thun cm vi t cm L. t vo hai u cun thun cm mt in p xoay chiu c biu thc u = U0cos(t + ) V. Cng dng in tc thi ca mch c biu thc l

A. i = AB. i = A

C. i = AD. i = AHng dn gii:Vi on mch ch c L th ( i = ACu 3. Dng in xoay chiu chy qua mt on mch ch c cun dy thun cm c t cm L = 1/ (H) c biu thc i = 2eq \l(\r(,2))cos(100t - eq \l(\f((,6))) A. Biu thc in p xoay chiu gia hai u on mch ny l

A. u = 200cos(100t + eq \l(\f((,6))) V B. u = 200eq \l(\r(,2))cos(100t + eq \l(\f((,3)))