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1.1. R3 1
, (.1) 1 , R3. iR3(t)
iR3(t) = iR3(t) + iR3(t), (1.1)
iR3(t)- , ;
iR3(t)- , .
iR3(t) (. 1.1) (t = ∞)
iR3(t)= E/(R2 + R3) = 150/(165 + 9) = 0,862 . (1.2)
R2 |
E |
R3 |
iR3 |
. 1.1.
(t = ∞)
iR3(t) 1 (.1.2). Z(p), Z(p) = 0.
R2 |
R3 |
Lp |
R1 |
. 1.2.
Z(p) =R1 + 1/Cp + R2(R3 + Lp)/(R2 + R3 + Lp) = 0;
CL(R1 + R2) p2 + [C(R1R2 + R1R3 + R2R3) + L] p + R2 +R3 = 0;
600 . 10-6∙100 . 10-3∙(0,01 + 165)p2 +[600 . 10-6 ∙(0,01∙165 + 0,01∙9 +165∙9)+ 100 . 10-3]p + 165 + 9 = 0;
0,0099p2 + 0,992p + 174 = 0;
p1 = -50,101 + j122,742 -1,
p2 = -50,101 j122,742 -1.
p1 p2, :
iR3(t) = Ae-αt sin(ω0t + φ), (1.3)
α = 50,101 c-1 ;
ω0 = 122,742 / .
(1.2) (1.3) (1.1),
iR3(t)= 0,862 + Ae-50,101 t sin(122,742 t + φ). (1.4)
A φ iR3(t) t = 0,
iR3(t)= 0,862 + Ae-50,101 t sin(122,742 t + φ); (1.5)
, iL(t) = iR3(t) - ,
.
iR3(0) uL(0) t = 0+.
(1.5) t = 0
iR3(0) = 0,862 + A sin φ; (1.6)
.
. 1 (.1) , iR3(0-) = iL(0-) = 0. iL(0-) = iL(0+) R3 iR3(0+) = 0.
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|
uL(0+) , t = 0+, L C JKL = iL(0-) EC = uC(0-) (.1.3)
R2 |
R3 |
E |
EC = uC(0-) |
R1 |
JKL = iL(0-) |
uL(0+) |
. 1.3. uL(0+)
R2 |
R3 |
E |
uC(0-) |
R1 |
K1 |
. 1.4. t = 0-
(. 1.4.) iL(0-) = 0 uC(0-) = 0, JKL = 0 EC = 0. (.1.3) : uL(0+) = E = 150 , (1.6)
0=0,862 + A sin φ,
.
, . ,
1500 = -50,101∙(-0,862) ∙ + 122,742∙(-0,862)∙ ;
ctg φ = -13,849, φ = -0,072 ;
= =11,982.
(1.4), R3 1
iR3(t) = 0,862 + 11,982e-50,101 t sin(122,742 t 0,072). (1.7)
1.2. R3 2
[1] , 2
t1 = 1,5:α = 1,5: 50,101 = 0,029939522 ,
α = 50,101 -1 .
2 (1.1). t 2.
(t = ∞), .1.5
iR3(t) = = =16,667 . (1.8)
E |
R3 |
iR3 |
.1.5.
(t = ∞)
Z(p) , .1.6.
R3 |
Lp |
.1.6.
Z(p) = R3 + Lp.
Z(p) = 0
R3 + Lp = 0,
p = - = - = - 90 c-1.
, iR3(t) p = - 90 c-1
iR3(t) = D , (1.9)
D .
(1.8) (1.9) (1.1),
iR3(t) = 16,667 + D ,. (1.10)
D . t=0
iR3(0) = 16,667 + D, (1.11)
iR3(0) = iL(0-) = iL(0+) = iR3(t1) (.1.7).
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E |
R3 |
iR3 |
JKL = iL(0-) |
. 1.7. t=0+
t1 = 0,029939522 . t (1.7),
iR1(0) = 0,862 + 11,982-50,101∙ 0,02993952 sin(122,742∙0,02993952 0,072) = 0,851 .
iR1(0) = 0,851 (1.11),
0,851 = 16,667 + D,
D = -15,816.
(1.10). R3 2 :
iR3(t) = 16,667 15,816-90t .
1 2 :
iR3(t) = 1(t)[0,862 + 11,982e-50,101 t sin(122,742 t 0,072)]
-1(t - 0,02993952) [0,862 + 11,982e-50,101 t sin(122,742 t 0,072)] +
+1(t - 0,02993952)[ 16,667 15,816e-90(t - 0,02993952)],
1(t) .
R3 .1.8.
1.3. R3 1
1 iL(0) = 0 uC(0) = 0, .. , L Lp, C . , (.1) 1 (.1.9).
Lp |
R2 |
R1 |
R3 |
a |
I3(p) |
b |
.1.9.
IR3(p) = = ,
Z(p) a b.
(.1.9) Z(p) = R3 + Lp + .
R3:
IR3(p) = = = =
= 1500 .
IR3(p) F(p) = :
IR3(p) = 1500 = 1500∙F(p). (1.12)
8.4 2 [2] :
f(t) = Aeat sin(ω1t + α) + K, (1.13)
ω1= = 122,742;
K = = = 0,000575; a=-50,101; d=10,101;
A = = = 0,00793;
α =arctg -arctg =arctg -arctg = -0,072 .
(1.13):
f(t) = 0,00793e-50.101t sin(122,742t 0,072) + 0,000575.
(1.12), R3 1:
iR3(t) = 1500∙f (t) = 1500∙[0,00793e-50,101t sin(122,742t 0,072) + 0,000575] = 0,862 + 11,895e-50,101t sin(122,742t 0,072).