. A 0,6 B 0,866 C 0,8 D 0,
^hdn tick vd huang dan gid
C a m k h a n g : Z L = (oL = I O O T I. - = 300Q; D u n g khang: ZQ = wC 10"* IOOTI. n • = lOOQ
L bien thien de UL^ax
Z , = R2 + Z2 ^ ^ R = 7 Z L . Z C - Z 2 =7300.100-100^ =iooV2n u Tj / p 2 , 72 100V2, ioo72 +(100) ^ u y R + Z c ^ VI / ^ ' =iooV3(V) Lmax ^PQ^2 U R C = KmJ-^' - J(l00Vif-(l00V^f = lOO(V) Chpn dap an D
Cty TNHH MTV DWII Khang Vi?t
Vl d u 5 : Dat vao hai dau doan mach m o t di?n ap xoay chieu u = 10oV2cos(cot) (V) g o m m o t dien tro thuan R, cuon day thuan cam c6 do t u cam L thay d o i dugc va mot tu di^n c6 d i ^ n d u n g C. K h i thay doi L ta thay U L dat cue dai va hieu dien the hai dau t u d i ^ n bSng hi^u d i ^ n the hai dau d i ^ n tro thuan. Viet bieu thuc d i ^ n ap gi&a hai dau ciia U R ^
A. U R L = 400\/5cos 1007:t + — 10 V B. U R L =400cos(1007tt)V C. u R L = 40072( cos lOOTtt- — 10 V D. U R L =100COS 1007lt + — l O j V
^hdn tick vd huang dan gidi
Nh^n xet: Thong thuang khi viel bieu thuc dien ap cua hai dau doqn mqch ndo do ta phdi tinh dugc dien dp cue dqi giua hai dau doqn mqch do vd do lech pha cua no so vai cuang dp ddng dien trong mqch. Dot vai bai nay thi deldm nhu vqy rat ddi so vai thai gian cua mot bdi trac nghiem.
Tuy nhien ta c6 theldm vice do tuang ddi dan gian ne'u chii y den nhung gift kien bdi todn cho.
Theo gia thiet U c = U R => Z c = R ma Z L = R ^ + Z ^
Z L = 2R Do l^ch pha giiia URL va i tan (p„ =.5L = 2->(P„ Do l^ch pha giiia URL va i tan (p„ =.5L = 2->(P„
Khi U L cue da U JR2 + zl r- r-
i thi ULmax = ^ = UV2 = 100V2 ( V )
Ket hap v o i bieu thuc U ^ = U R + (U L - U c ) ^
U R = 2U L = 200N/2V ^ U R L = ^ U R + U ^ = V S U L = lOOyfwV
Vi U L max nen U R C 1 U - »( p - (PRC = -
Mat khac tan (PRC = ^ = - 1 -> cpRc = — ^ ^ ~ 4"
Vay dp l^ch pha giira URL va u mach Aw - - — = — (rad)
^ ^ ^ 180 4 10 ^ ^ Bieu thuc d i f n ap giua hai dau cua d i ^ n tro va cupn day:
U R L = 400\/5cos
Chpn dap an A
1007tt + — 10
Vl dll 6: Dat difn ap xoay chieu c6 gia tri hi#u dung U = 30 V vao hai dau dogn mach RLC noi tie'p. Biet cupn day thuan cam, c6 dp cam L thay doi duoc. Khi di^n ap hieu dung hai dau cupn day dat cue dai thi hieu
dien thehieu dung hai dau tu di^n la 30V. Gia tri hieu di^n thehieu dun;.,
cue dai hai dau cugn day la:
A . 6 0 V B. 1 2 0 V C. 30N/2V D . 6 0 > / 2 V
^hdn tich v>d huomg dan gidi
i
Khi L thay doi Ui.mav khi Zi. = — (l)va Ul.max =
R , U Uc
— - —_ (adsbygoogle = window.adsbygoogle || []).push({});
Ta c6: — = 30N/2 . 2 Z ^ = R ^ + ( Z L - Z c r (2)
2 JR2 + ( Z L - Z ^ ) ^
The (1) vao (2) ta duoc pt: R " + Z^R^ - 2Z^ = O R^ = Z ^ R = Z ^ URV2
Do do Ul.max = Chpn dap an A
R = UV2 = 6 0 V
L O A I 2: C B I E N T H I E N D E Ucmax:
Lap bleu thuc duoi dang:
R
A / L
Uc = i z UZr u u
Cach 1: Phuong phap dao ham:
Dat y - ( R 2+ Z ^ J ^ - 2 Z L — + 1 = (R + Z^JX2-2X.ZL+1 (vol x = - ^ )
Z(- ZQ Q
Uc max yn„n
Khao sat ham so: y = (R^ + Z [ ) x ^ - 2X.ZL +1 ^ y ' = 2(R2 + Z ^ ) x - 2 Z L y' = 0 « 2 f R ^ + Z ? ) x - 2 Z [ =0=>x = - — t
R^+Z
Bang bien thien:
X o 00 o 00 y ' o + y ymin ' l u - ZL . 1 • ymin k h i X = • u Cmax _ U^R^ + Zl R hay (V) • r^ Z ^ = R ' + z ^
Cach 2: Phuotig phap dung tarn thuc bac hai.
Ta c6: VZr- Uc = IZc = U U R ' ^ ( Z L - Z C ) ' ( R2 +Z ^ ) 4 - 2 Z L - ' - ^ + I Vy Dat at y = ( R ' + z 2 ) - ^ - 2 Z L ^ + l = ax2 + bx + l (voi x = 1 Z ^ ' 3 - R + Z ^ ; b = -2ZL) ^ )2 , -72,
Ucm.ix khi ymin. Vi ham so' y c6 he so'
goc a > 0, nen y dat cvc tieu khi: b x = — Hay •u 2a 1 •=^Z^ = Cmax _ u ^ R^ + Z R - ( V )
Cach 3: Phuong phap dung gian do Fre-nen.
Ta c6: U = UL + UR + Uc
Ap dung djnh ly ham so sin, ta c6:
—^---Un =- smp
T
u,,