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:
- ( 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^^) (adsbygoogle = window.adsbygoogle || []).push({});
^ 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
^Mn tick vd huang dan giAi
D. 4N/2 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