71
mot dion ap xoay chieu u^^g = 200\/2 cos 1007tt + - 4 Gia trj cua C va cong suat tieu thy ciia
mach khi dien ap giua hai dau R cung A ^ ^
pha voi dion np hai dau doan mach nhan cap gia tri nao sau day:
A. C = 10 ,-4 ( F ) ; P „ „ , = 3 0 0 ( W ) B. C = 10 1-3 271 -(F);Pmax=400(W) 271 -(F);Pmax=400(W) 10" ^•^^ = ^ ( F ) ; P „ . , x = 4 0 0 ( W ) D. C = 10 ,-4 271 -(F);Pmax=500(W) 365
'Phdn tich v>d hu&ng dan gidi
Ta tha'y k h i UR cung pha v a i UAB nghla la UAB cung pha v o i cuang dp dong di#n i . Vay trong mach xay ra cpng h u o n g d i # n :
Z, = Z p » c o L = =>C = — - = - = (F)
200^
Luc nay cong suat: P = P^^^ " " • ^^^^ ^ " C
C a u 2: Cho mqich d i ^ n xoay chieu n h u h i n h ve. ' 2 10"^
BietR = 200Q, L = - H , C = F.
71 71
Dat vao hai dau mach d i f n mot d i f n ap xoay chieu u = lOOcoslOOTtt (V), K h i R , L, C khong doi de so chi cua ampe ke Ion nhat, t h i tan so dong di^n phai bang bao nhieu? T i n h so' chi ampe ke luc do. (Biet rang day noi va d u n g cu do k h o n g lam anh h u o n g den mach di?n).
A. f = 5 0 H z ; I „ 3 , = 0 , 3 5 ( A ) R L C
B. f = 6 0 H z ; I „ , , = a 5 ( A ) F ^ " ^
C. f = 35,35Hz;I^^,=:0,35(A)
D. f = 4 5 H z ; I ^ ^ , = 0 , 3 ( A ) "
'Phdn tich m huong dan gidi
C a m k h a n g : Z L = coL = I O O T : . - = 200Q;
7t
D u n g khang: Z , - = — = = lOOn
10071.-—-7t 7t
Ta C O cuong dp d o n g di#n chay trong mach (so chi A m p e ke):
D o R , Z L , U la hang so'nen I = Ij^^,^ khi trong mach xay ra cpng h u o n g dien ( Z L - Z ^ = O) . Liic do: 27:f.L = — ^ => f = — ^ — = , ^ = 35,35Hz 27tf.C ^ ' 27rVLC 2n 2 1 0 ^ 7 l " 71 I =-iL = U=-i01- = a 3 5 ( A ) l Zmin R ^/2.200 ,j C h p n dap an C
Cau 3 : Cho mach d i ^ n n h u h i n h ve: u =12oV2 cos(1007it) ( V ) ; cupn day c6
r = 15Q; L = — ( H ) . C la t u di?n bien doi. 257t
Dien t r d von ke'vo cung Ion. ^ • Dieu chinh C de so chi von ke'Ion nhat. Tim C va so'chi von ke'luc nay?
A. C = — ( F ) ; U v = 1 3 6 ( v ) c. c = 10 -2 - ( F ) ; U v = 1 3 6 ( v ) B. C = D. C = 10" 471 10" 371 — ' ^ ' - - 5n
^hdn tich vd huong ddn gidi
Do von ke mac vao hai dau cupn day nen so chi von ke'la: (adsbygoogle = window.adsbygoogle || []).push({});
- ( F ) ; U v = 1 6 3 ( V ) >
- ( F ) ; U v = 1 8 6 ( v )
l J v = U d = I . Z , = ^ . Z , = U
r ' ^ + ( Z L - Z c ) '
B t Do Z J khong p h u thupc C nen no khong doi. Vay bieu thuc tren iit so khong
doi. V i the so'chi V o n ke'Ion nhat khi mau so'be nhat: ^ r ^ + ( Z L - ZQ)^ ^ ^ H ) i e u nay xay ra k h i cpng h u o n g di?n: ZQ=Z^= 8 ( Q ) .
Suy ra: C = ^ ^ (F) , luc do Z = r
H r Va so'chi von k e :
• U v = U , = - ^ . V r 2 + ( c o L ) 2 =H0.Jl52+(8)2 =1^.17 = 136V C h p n dap an A .
»Su 4 : Mach d i ^ n xoay chieu gom cupn day c6 L = (H) mac n o i tie'p v o i
71
ty di^n C. Dat vao hai dau doan mach dien ap u = U \/2 cos(ot(V). 2 10 "
K h i C = Ci = — t h i Ucmax = 100 S (V). K h i C = 2,5 Ci t h i cuang dp
dong dien tre pha ^ so v o i dien ap hai dau doan mach. Gia trj cua U la
,-4
A. 50V
Theo bai cho, ta c6:
B. lOOV C. 1 0 0 N / 2 V D . 5 0 ^ / 5 V
^hdn tich vd huong ddn gidi
Khi C = Ci = 2.10" - F thi U c = Ucmax liic do • ^1 Z, U Ctnax U ^ R 2 + zl
KhiC = 2,5Cithicp=^ => tan cp = - ^ - ^ ^ - ^ = tan ^ = 1
^ R = Z L - = Z L - 0,4Zcj ( vi C2 = 2,5Ci nen Z^^ = OAZ^^)
^ R = Z L - 0, 4 ^ RZL = ZL2-0,4R2-0,4ZL2 Z L => 0.4R2 + ZLR - 0.6ZL2 = 0 => R = 0,5ZL suy ra ZL = 2R U ^ R ^ + Z ^ _ U V R^ + 4R ^ Do do Ucmax •• R R ^ u = - i £ ^ = i o o (V) Chpn dap an B
Cau 5: Cho mach di?n xoay chieu gom R,L,C mac noi tiep theo thii tu nhu
tren. Von ke Vj dat vao hai dau di?n tro R, von ke dat vao hai dau cugn day va von ke V 3 dat vao hai dau tu. Di^n dung C c6 gia trj thay doi dugc va cupn day thuan cam. Dieu chinh gia tri cua C va ghi lai so chi Ian nhat tren tung von ke thi thay Ucmax = SUtmax. Khi do Ucmax gap bao nhieu Ian
URmax?
A.
C bien thien de:
U c = U c„ , a x khi'lo:
B. 3 3 c. 4N/2 (adsbygoogle = window.adsbygoogle || []).push({});
^Mn tick vd huang dan giAi
D. 4N/2 Z c = R' + z^L Cmax U ^ R ^ + Z ^ R U L = U L ^ a x khi do: (cpng huong di^n) Z , = Z ^ : ^ Z „ : „ =R U U .z, =—.z. U (2) •i£.a
U R = U B „ „ khi do: . 1 z, = => Zn,in=R
Rmax = u (3) (cpng huong di§n) Lay Lay (2) W . ^.zl u u (3) Tu (4) va (5): Lmax Cmax _ Rmax •R = ZL78 (4) (5) U Cmax _ Rmax Chpn dap an A
Cau 6: Cho mach di?n xoay chieu gom R, L, C mac noi tiep theo thu ty nhu
tren. Von ke Vj dat vao hai dau di^n tro R va von ke Vj dat vao hai dau t\i. Di?n dung C c6 gia tri thay doi dugc va cugn day thuan cam. Dieu chinh gia tri ciia C thi thay: 6 ciing thai diem so chi ciia Vi cue dai thi so'chi cua Vi gap doi so chi cua V2. Hoi khi so chi cua V2 cue dai thi so chi cua V2 gap bao nhieu Ian so chi Vi?
A. 2 Ian. B. 1,5 Ian. C. 2,5 Ian. D. 31an
^hdn tick \m huong dan gidi
Khi Vi cue dai thi mach cgng huang: UR = U = 2Uc = 2UL hay R = 2ZL (vi L va R khong doi nen UR = 2UL => R = 2ZL )
Khi V2 cue dai ta eo: Cmax
_ U^R2 + zl _ u^4zl+zl
R
Zc = ^ ^ ^ i ^ = 5 Z L=2, 5 R
Z L Z L Z = ^ R 2 + ( Z L - Z C ) ^ =
Chi so cua Vi liic nay la Uj^ = I R =
V R ^ + ( 0 , 5 R - 2 , 5 R ) 2 = R V 5 U R U u, Cmax ^1^2,5 t Vaytaco: U R ^ Chpn dap an C
C3u 7: Cho mgt doan mach RLC khong phan nhanh, cugn day thuan cam, dg
tu cam ciia cugn day c6 the thay doi dugc. Khi thay doi gia trj ciia L thi thay 6 thoi diem di?n ap hi?u dyng giiia hai dau di$n tra eye dai thi di^n ap nay gap bon di^n ap hi^u dyng giiia hai dau cugn day. Khi di$n ap hi^u dung
Bi quyei on \uyen iht dat hoc dit Aim TOl iia vaT I I , Uip 1 - LV Vim vinn -
giu-a hai dau cuon day cue dai thi dien ap nay so voi dien ap hieu dung
giira hai dau dien tro i<hi do gap:
A. 4,25ian. B. 2,5 Ian. C. 4 Jan. D. 4V2 1an.
'Phdn tich vd huong dan gidi
Khi URm.>x (mach c6 cong huong), ta c6: U i . = Uc va U R m. i x = U = 4 U L = 4UC
=> R = 4ZC (hSng so)
Khi Ui m.ix ta c6: Ui m.ix = — ^
( 1 ) (2) (2) (3) T u ( l ) s u y ra U R = 4LIc T u (2) va (3) suy ra U i m.,x = 4,25 U R Chpn dap an A
C a u 8: Cho doan mach R, L, C noi tiep voi L c6 the thay doi duoc. Trong do R va C xac dinh. Mach dien dugc dat duoi hieu dien the u = u V 2 COSM t
Voi U khong doi va m cho truoc. Khi hieu dien the hieu dung g\ua hai dan
cuon cam cue dai. Thi he thiic nao sau day dung: (adsbygoogle = window.adsbygoogle || []).push({});
A. LCa)2 = 2Cho' R^+l B. LCio^ = C^co^ R^ + 1
C. LC2ai2 = C^a)^ R^+l D. 2LCa>2 = 2Ck.^^ R^ + 1
<Phdn tich vd hu&ng ddn gidi
L bien thien de hieu dien thehieu dung giiJa hai dau cuon cam cue dai tho.i
man
=j> LC(o^ = R^(o^C^ + 1 (quy dong mau thue voi mau thuc chung la co^C^ )
Vay chpn dap an B
C a u 9: Cho mach dien xoay chieu AB gom hai doan A M va M B mac noi tiep doan A M gom bien tro R va tu dien c6 dien dung C, doan M B chi c6 cuon cam thuan c6 do t u cam thay doi duoc. Dat vao hai dau doan mach AB moi dien ap xoay chieu on djnh u = uV2eosrot(V). Ban dau, gii> L = L , , thay doi gia trj cua bien tro R ta thay dien ap hieu dung giiia hai dau doan mach A M luon khong doi voi moi gia tri cua bien tro. Sau do, gii>R = Z L , thny doi L de dien ap hi^u dung hai dnu cuon earn cue dai, gia trj dien ap hieu dung cue dai tren cuon ram bang
A. V3U/2(V). B. 72U/2(V). C. V5U/2(V). D . U / 2 ( V ) .
(phdn tich vd hu&ng ddn gidi
Hi^u di^n the hieu dung hai dau A M :
, R^ + Z
Theo bai ra: U_^^^ g R o ZL^ - 2Z(- = 0 => Z^^ = 2Z^ = R
L bien thien de U , U , = W^^^^- U V 4 Z ^ Z ^ _
Lmnx Lmax —
A?
Chpn dap an C '
Cau 10: Dat mot dien ap xoay chieu c6 dien ap hieu dung 120V va tan so 50Hz vao hai dau doan mach gom mot dien tro thuan R=30Q, mot cuon cam thuan co L = 0,4/TT ( H ) va mot tu dien eo dien dung thay doi duoc. Dieu chinh dien dung cua tu dien sao cho dien ap hieu dung giira hai dau cuon cam dat cue dai. Di^n dung eiia tu dien luc do va di^n ap ci^c dai noi tren tuong u n g bang
A. M i ^ ( F ) v a l 2 8 V B. — (F) va 120^/2 V
C. l i H ^ ( F ) v a 2 0 0 V D. l E ! ( F ) v a l 6 0 V '^"^
'Phdn tich m hu&ng ddn gidi C bien thien de <=> mach epng huong
=>Z^=Z. =coL = 1007r.—= 40Q=>C = ^ — = - = 1 ^ F '
n coZ(~ 10071.40 471
. - > U t n . a x = 7 ^ . Z L = ^ Z L ~ - 4 0 = 160V V
Chpn dap an D
Cau 11: Cho mach dien xoay chieu gom ba doan mac noi tiep. Doan A M gom dien tro thuan R, doan M N gom cuon day thuan earn, doan NB gom tu xoay CO the thay doi dien dung. Mac von ke t h u nhat vao A M , von ke thu hai vao NB. Dieu ehinh gia trj ciia C thi thay a cung thoi diem so chi ciia Vi cue dai thi so chi cua Vi gap doi so chi ciia V2. H o i khi so chi eiia V 2 cue dai va CO gia tri V2M.1v = 200V thi so chi ciia von k e t h u nhat la
A. lOOV. B . 5 0 V . C . 8 0 V . D. 120V.
f i'hdn tich vd hu&ng ddn gidi
Khi Vi cue dai thi mach cpng huong: ' '
U R = U = 2Uc = 2Ui, hay R = 2ZL H ) (adsbygoogle = window.adsbygoogle || []).push({});
U /R^ + Z^
Biquyei on luyjtt thi d^i hqc ilnl iliOm toi U, tap 1-Le Van Vinh
u Cmax 2Z,
Khi do lai c6: = — = — - theo (1) ta dupe:
ZC = 5ZL = 2,5R, Z = R S
Chi so cua Vi liic nay la U,^ = I R = U R U
(2) (3) (4) IF..'- •':••)•' Tu (3) va (4) ta c6: = ^ = 2,5 Vay V , = 80V. U R 2 Chpn dap an C
Cau 12: Dat di?n ap u = Uocos27Tft vao hai dau doan mach gom di^n tro thuan R, cupn cam thuan c6 dp tu cam L va tu di?n c6 dien dung C mac noi tiep. Gpi UR, UL, UC Ian lupt la di?n ap hi^u dung giua hai dau di#n tro, giiia hai dau cupn cam va giira hai dau tu di^n. Truong hop nao sau day trong mach xay ra hif n tupng cpng huong di^n?
A. Thay doi C de Ucmax B. Thay doi f de Utmax
C. Thay doi R de Pmax D. Thay doi C de URmax
'Phdn tkh vd huang dan gidi
Thay doi C de Uc^ax '
=> A sai
Thay doi/de U L „ , a , c
=>Bsai
Thay doi R de P^ax «
=>Csai
Z ^ =
u Cmax u,
khong phai cpng huong
R
U 2U.L
khong phai cpng huong
Lmax
R N / 4 L C - R 2 C 2 R = Z L - Z C R = Z L - Z C
U^^_
2R 2
U^ khong phai cpng huong
Thay doi C de UR n,ax <=> i x r _ T T '^^ng huong xay ra => D diing
[URmax - (adsbygoogle = window.adsbygoogle || []).push({});
V|y chpn dap an D
CUf TNHH MTV DWH Khang Viet
Cau 13: Mach noi tiep gom di#n tra R, cupn day thuan cam L , t\ di^n C c6 di?n dung thay doi dupe. Hi^u di?n the hai dau doan mach la 220V - 50Hz. Thay doi C ta thay Ucmax = 440V. Chpn phuang an dung.
A . ZL = R . B. ZL = 3R. C. = N/SR . D . ZL = R .
'Phdn tich vd hu&ng ddn gidi
C bien thien de hi^u dien the hieu dung giua hai dau tu cue dai thoa man :
Cmax /R2 + Z2 = . Cmax 440 r -
•R = ^ R = 2 R = ^ Z L = V 3 R
R ^ U
Chpn dap an C
Cau 14: Dat dien ap u = 100>/6cos(1007:t) (V) vao hai dau doan mach RLC noi tiep. Biet cupn cam thuan c6 dp tu cam thay doi dupe. Dieu chinh dp tu cam L de di?n ap hi^u dung giiia hai dau cupn cam dat gia trj evrc dai thi di#n ap hi#u dung giira hai dau t\ di^n la 200 V. Gia tri cue dai do la
A . lOOV. B.250V. C.300V. D. 150 V.
Phdn tich v>d hu&ng ddn gidi
L bien thien de hi#u dien the hi^u dung giira hai dau cupn cam cue dai nen :
Z L = — — ^ « Z L Z C = R ^ + Z 2 : ^ U L U C = U | + U^(1) ^ Zc Zc U L = U L „ , a x _ U ^ R 2 + zl _ Uyjvl + ul R (2) U^=Ul+ ( U L - U C ) ' = U 2 + U 2 + U 2 - 2 U L U C = - U ^ - U^. + U ^ (3) Tu (2) va (3) ta c6: u^fu^+uM ul = u'+ui+ul=- ^
o U^U^ + U^ + U^U^ = U"U^ + U^U^ 2TT2 T2TT2 T2TT2 < » U ^ + U ^ U R - U ^ U ^ =0 < » U ^ + U ^ U R - U ^ U ^ =0
« U ^ + 40000UR - 30000.40000 = 0 =:> U^ = 20000
U^ = -60000(1) • U L = 7u^ + UR + Uc = V30000 + 20000 + 40000 = 300(V) • U L = 7u^ + UR + Uc = V30000 + 20000 + 40000 = 300(V) Chpn dap an C
C a u 1 5 : Dat dien ap xoay chieu u = 80%/2 coslOOnt (V) vao hai dau doari mach mic noi tiep gom dien tro thuan R, tu dien c6 dien dung C va cuon mach mic noi tiep gom dien tro thuan R, tu dien c6 dien dung C va cuon cam thuan c6 dp t y cam L thay doi dupe. Dieu chinh L de di^n ap hieu dung 6 hai dau cuon cam dat gia tri cue dai thi thay gia tri cue dai do bang
100 V, dien ap hi?u dung 6 hai dau dien tro bang bao nhieu?
A. 48 V B. 64 V C. 60 V D. 36 V
'Phdn tich vd huang ddn gidi