bai tap mach dien
TRANSCRIPT
-
7/31/2019 Bai Tap Mach Dien
1/22
1U
1I
O x
3U
3I
ll/2
Bi t p: Mch in2Bi 1:Cho ngdy v i cc thng s sau:
0
-30
-90
60
1
l = 100 km
R = 6 /m
L = 1,6.10 H/kmC = 6,4.10 F/km
G = 10 S/km
f = 100 Hz
U = 1 kV0
1I = 500 -15 A
1/ Xc nhcc h s: C, Z , V,
2/ Xc nhU v h sphn x sng gi a ngdyGii:
a/Ta c:2 f 2. .100 200 (rad/s)
o o o
-6 -6o o o
o o
Z R j L = 6 + j ( )
Y G j C 10 +4,0212.10 j (S)
Z .Y 0,0037 + 0,0034j (1/km)
= 0,0037 (neper/km)= 0,0034
(rad/km)
oC
o
5
3
ZZ 1013,2 - 664,56j ( )
Y
V= 1,8428.10 (km/s)
V= 1,8428.10 (km)
f
b/ Ta c:
5 41 1 C 1
5 42 1 C 1
1A U Z .I 2,8816.10 - 9,4922.10 j
21
A U Z .I -2,8716.10 9.4922.10 j2
-
7/31/2019 Bai Tap Mach Dien
2/22
Mc khc:x x
x 1 2 x x
x x1 2x x x
C C
U A .e A .e U U
A AI .e .e I I
Z Z
Ti im: x = l/2 th.l / 2 .l / 2
3 l / 2 1 2 l / 2 l / 2
5 4
U U A .e A .e U U
-1,3683.10 - 6,4567.10 j (V)
-136,83 - 64,567j (kV)
.l / 2 .l / 21 23 l / 2 l / 2 l / 2
C C
A AI I .e .e I I
Z Z
479,95 144,69j (A)
-
7/31/2019 Bai Tap Mach Dien
3/22
1U
1I
Ox
2U
2I
l
cZ
Bi 2:
c
-3
2
l 30 km
Z 500
=3.10 Neper/km
Z 500
GTHD c a inp du ngdy l U 1 = 120Va/ Xc nhGTHD c aU2;I2 cui ngdyb/ Xc nh hiusut truyn ti ca ngdy.
Gii:
a/ Do: c cZ R ngdy g n nh vn hnh ch ho h p ti nn: 0 . Tcl mi im trn ngdy ch c ST m khng c SPXXy d ngcng th c theo h trc nh hnh v :
x x xx 1 2 x x x x 1
x x x1 2 1x x x x x
C C C
U A .e A .e U U U U A .e(1)A A A
I .e .e I I I I .eZ Z Z
Ta c
.01 (x 0) (x 0) 1 1
1 1
U U U A .e A
U A
U1
1
j.1 1 U 1 1U U U .e A (2)
Th (2) vo (1) ta c:
-
7/31/2019 Bai Tap Mach Dien
4/22
U1
U1
U1
U1
U1
j. ( j )xx 1
j.( j )x1
xC
j( x )xx 1
j( x )x1x
C
j( l)l2 (x l) 1
j( l)l12 (x l)
C
l (30.2 1
U U .e .e U .e
I .eZ
U U .e e
UI .e eZ
U U U .e e U
I I .e eZ
U U .e 120.e
3
3
3.10 )
l (30.3.10 )12C
109.6717 (V)
U 120I .e .e 0.2193 (A)Z 500
b/ Hi usut truy n ti :Ta c
2
1
P.100%
P
Vi: 2 21 1
2 2 2 U I
1 1 1 U I
P U .I .cos( )
P U .I .cos( )
Ch ho h p ti nn:
1 1
2 2
1 1 1 1
2 2 2 2
1 1c c
1 1
2 2
c c2 2
1 1
2 2
U I
U I
U U ; I I
U U ; I I
U UZ R
I I
U UZ R
I I
U ;I cu`ng pha
U ;I cu`ng pha
cos( ) 1
cos( ) 1
-
7/31/2019 Bai Tap Mach Dien
5/22
l1 1 2 1
.0 l1 1 11 (x 0) 2
C C C
l l11
C2 2
11 11
C
2 l
U A 120 U U .e;A U UI I .e I .e
Z Z Z
UU .e . .eZU .I
.100% 100%UU .I U .Z
e .100%
85%
-
7/31/2019 Bai Tap Mach Dien
6/22
1U
1I
Ox
2U
2I
l
2Z
Bi 3:Cho ngdy khng tiu tn c:
3
2(t)
l 100 km
3,4.10 rad / km
U 110 2 sin( t) kV
Xc ngU2(t) u ngdy trong cc tr nghpc:a/ Z 2 = Z c b/ Z 2 = 0,5Z c
Gii:
Ta c:
x 2 2 c
2x 2
c
U U cos x jI Z sin x
UI I cos x j sin x
Z
M2
22
UI
Z nn:
cx 2 2 (x ) (x )
2
2
2 cx
2
c c(x )
2 2
ZU U cos x j sin x U .M
Z
ZM cos x sin x
Z
Z Zsin xarctan arctan tan x
Z cos x Z
a/ c
2 C2
ZZ = Z 1
Z
2 2x
(x )
M cos x sin x 1
sin xarctan arctan tan xcos x
1 (x l)
1 (x l)
1 2 1 1
1(t)
M M 1
arctan tan l 0,34 (rad)
U U .M 110.1 0,34 110 0,34
U 110 2 sin( t 0,34) (kV)
-
7/31/2019 Bai Tap Mach Dien
7/22
c2 C
2
ZZ = 0,5.Z 2
Z
22 2 2x
(x )
M cos x 2sin x cos x 4sin x
sin xarctan 2. arctan 2. tan x
cos x
2 21 (x l)
1 (x l)
1 2 1 1
1(t)
M M cos l 4sin l 1,915
arctan tan l 0,616 (rad)
U U .M 110.1,915 0.6157 210,62 0,616
U 210,62 2 sin( t 0,616 ) (kV)
-
7/31/2019 Bai Tap Mach Dien
8/22
L
1I
Ox
2U
2I
l
1U1V
Z
Bi 4:Mt ngdy khng tiu tn. C chi udi l, Z C = R C, dng inc tns f,ticui ngdy l cu ncmL.Xc nh L h ngdy v t i tr thnh m chcnghngpGii:
1
( x )
1V
1
x 2 2 c
2x 2
c
2 L 2
x 2 L c
Lx 2c
L cV
L
c
UZ
I
U U cos x jI Z sin x
UI I cos x j sin x
Z
U j.X .I
U j.I X cos x Z sin x
XI I cos x sin xZ
X cos x Z sin xZ j
Xcos x sin x
Z
1
1
L c
L
c
L cV
L
c
V
L c
L c
c
X Z tan x j
X1 tan x
Z
X Z tan lZ j
X1 tan l
Z
Z 0
X Z tan l 0
X Z tan l
ZL tan l
2. .f
-
7/31/2019 Bai Tap Mach Dien
9/22
1I
Ox
2hU
2I
l
1U1V
Z
Bi 5:
( x )
1
1
x 2 2 c
2x 2
c
2 2h
2
x 2h
2hx
c
2hV c
2h
c
V c
V
U U cos x jI Z sin x
UI I cos x j sin x
Z
U U
I 0U U cos x
UI j sin x
Z
U cos xZ j.Z .cotan x
U j sin x
Z
Z j.Z .cotan l
Z 0 cotan l 0
l k 2
6
k=1,3,5,....,2n+1,....
2 f 2 f (dd tren ko: V=c)
V c2 fl c
k f k 2,5.10 k (Hz)c 2 4.l
2,5.k (MHz)
k 1 3 5 9 11 f(MHz) 2.5 7.5 12.5 17.5 22.5
-
7/31/2019 Bai Tap Mach Dien
10/22
[email protected] t p6 : Cho m ch innh hnh v :
ngkho K khi ( t ) me E sin( t ) (V) tgi tr cc imXc nh 2(t)i bit:
1 2
m
R 25 R 50
L 0.25H C 400 F
E 400V f 50Hz
Gii: Ta c: 2(t) 2td(t) 2xl(t)i i i Xc nh 2xl(t)i
Mch insau ngm ch xc l p
L
C
X 2 f .L 78,5
1X 7,96
2 f .C
Ti thi imt = 0 th c hinqu trnh ngct Nn o(t ) m me E sin E 90
o(t )e 400sin(314t 90 )
E 400 j (V)
Ta c:ab 2 1 CZ R //(R j.X ) 17 3,5j ( )
oxl
L ab
E 400 jI 5,2 167,2 (A)
j.X Z 78,5 j 17 3,5 j
o
ab LU E jX .I 90,5 178.8 (V) oab
2xl2
UI 1,8 178,8 (A)
R
LX
2R1R
CX
xlI a
E
b
1xlI
2xlI
L
2R1R
C
K
i 1i
2i
( t )e
-
7/31/2019 Bai Tap Mach Dien
11/22
2xl(t)i 1,8sin(3,14t 178.8 ) (A) Xc nh 2td(t)i
Xc nh sm ctnh p:
ab 2
v(p) ab 1 6
2
12,5pZ R // pL
50 0,25p
1 12,5p 1Z Z R 25
pC 50 0,25p p.400.10
75p 7500p 50000(200 p)p
2
v(p)
2
1
2
75p 7500p 50000Z 0
(200 p)p
75p 7500p 50000 0
p 50 64,55j
p 50 64,55j
50t2td(t)i 2.A.e .cos(64,55t+ )
Trong A v l cc h s cnxc nh. Xc nh s kin: v trong bi u thc thnh ph n t do c hai h s cnxc nh nnta cnxc nh2 s kin l i 2(0) ;i2(0)
Xc nh: i (o),u c(o) theo lu t ngm ch nh: (0) ( 0)c(0) c( 0 )
i i
u u
Xt m chtrc ngm (khi kho K ch a m )
L c
1
o
X X 78,5 7,96tg 2,8216
R 25
70,4
pL
2R1R
1/pC
a
b
L
1R
C
i
( t)e
-
7/31/2019 Bai Tap Mach Dien
12/22
mm 2 2 2 2
1 L C
E 400I 5,34(A)
R (X X ) 25 (78,5 7,96)
o(t )i 5,34.sin(314t 160,4 ) (A)
Cm m CU I .X 5,34.7,96 42,50 (V) o
C(t)u 42,50.sin(314t 250,4 ) (V) o
(0 )
oC(0)
i 5,34.sin( 160,4 ) 1,79 (A)
u 42,50.sin( 250,4 ) 40.03 (V)
Hph ng trnh m t sau ngm :
(t ) 1( t) 2(t )
2( t ) 2 (t )
2( t ) 2 1(t ) 1( t) 1
i i i 0
diL i .R e
dt1
i .R i .dt i .R 0C
(I)
Thay t = 0 vo h (I) ta c (0) 1(0) 2(0)
'(0) 2(0) 2 (0)
2(0) 2 C(0) 1(0) 1
i i i 0L.i i .R e
i .R u i .R 0
1(0) 2(0)'(0) 2(0)
2(0) 1(0)
1,79 i i 0
0,25.i 50.i 400
50.i 40,03 25.i 0
1(0) 2(0)
1(0) 2(0)'(0) 2(0)
i i 1,79
25.i 50.i 40,03
0,25.i 50.i 400
1(0)
2(0)'(0 )
i 1,7272 (A)
i 0,0628 (A)i 1578,44 (A / s)
ohm cc v cacc ph ng trnh trong h pt(I)
L
2R1R
C
K
i 1i
2i
( t )e
-
7/31/2019 Bai Tap Mach Dien
13/22
(t ) 1( t ) 2(t )
(t ) 2(t ) 2 ( t )
2( t ) 2 1(t ) 1( t ) 1
i i i 0
Li i .R e
1i .R i i .R 0
C
(t ) 1(t ) 2( t)
( t ) 2( t ) ( t )
2(t ) 1(t ) 1( t )
i i i 0
0,25.i i .50 e
i .50 2500.i i .25 0
(0) 1(0) 2(0)
(0) 2(0) (0)
2(0) 1(0) 1(0)
i i i 0
0,25.i i .50 e
i .50 2500.i i .25 0
1(0) 2(0)
2(0) 1(0)
(0) 2(0) (0)
1587,44 i i 0
i .50 2500.1,7272 i .25 0
0,25.i i .50 e
1(0) 2(0)
1(0) 2(0)
(0) 2(0) (0)
i i 1587,44
i .25 i .50 4318
0,25.i i .50 e
1(0)
2(0)
i 1000,72 (A / s)
i 586,72 (A / s)
Ta c:2(t ) 2td(t ) 2xl( t )i i i
o2xl(t)i 1,8sin(3,14t 178.8 ) (A)
50t2td(t)i 2.A.e .cos(64,55t+ ) (A) 2(0) 2td(0) 2xl(0)i i i
0,0628 2.A.cos( ) 0, 0377
A.cos( ) 0,01255 (1)
Ta c:2(t ) 2td(t ) 2xl( t )i i i
o2xl(t)i 314.1,8.cos(3,14t 178.8 ) (A / s) 50t2td(t)i 2A.e 50.cos(64,55t ) 64,55.sin(64,55t ) (A /s)
2(0) 2td(0) 2xl(0)
o
i i i
586,72 2A 50.cos 64,55.sin 314.1,8.cos( 178.8 )
21,72 129,1.Asin 100Acos (2)
T (1)(2) ta c
o
Acos 0,54
129,1.Asin 100Acos 21,72Acos 0,01255 tg 14,18
Asin 0,178 Asin 0,178
85,97
A 0,178
Vy:o
2xl(t)i 1,8sin(3,14t 178.8 ) (A)
-
7/31/2019 Bai Tap Mach Dien
14/22
2td(t)i 0,35e .cos(64,55t+85,97 ) (A)
o 50t o2(t)i 1,8sin(3,14t 178.8 ) 0,35e .cos(64,55t+85,97 ) (A)
-
7/31/2019 Bai Tap Mach Dien
15/22
[email protected] t p7: Cho m ch innh hnh v :
Xc nh ( t )i bit cc ngun trong mch l ngun hng v cc thng s sau:1 2 3
3 4
1 2
R 300 R R 600
C 300 F C 200 F
E 36 V E 6 V
Gii: A. Phng php tch phn kinh in: Ta c: (t) td(t) xl(t)i i i Xc nh xl(t)i
Mch insau ngm ch xc l p
V ngun E1 l ngun hng nn Ic = 0Ti thi im trc ngct
1xl( t) xl
1 3
E 36i I 0.04(A)
R R 900
Xc nh td(t)i
Xc nh sm ctnh p:Mch in sau ng m c i s ha theo p
1E 2E
1R
3R
2R
3C
4C
K
12i(t)
3U 4U
1E
1R
3R 3C
xlI
4C
xlI
cI 0
-
7/31/2019 Bai Tap Mach Dien
16/22
v(p) 1 33 4
3 4
3 4
4
1 1Z R // R
pC pC
1 1200 200
1 1 p(C C )pC pC
12005.10 p
v(p) 4
4
1Z 200 0
5.10 p
1200
5.10 p
p 10
Dng ca thnh phn t do l: 10t
td(t)i A.e Trong A h scnxc nh. Xc nh s kin: v trong bi u thc thnh ph n t do c mth s cnxc nh nnta cnxc nh1 s kin l i (0) Xc nh: u3(o) ,u4(o) theo lu t ngm khng ch nh:Xt m chtrc ngm (khi kho K ch a m )
1E 2E
1R
3R
2R
3C
i(t)
3( 0)U 4( 0)U
4C
1R
3R
3
1pC
4
1pC
H mch [email protected]
-
7/31/2019 Bai Tap Mach Dien
17/22
13(t ) 3
1 3
3( 0)
Eu U 0.04 600 24V
R R
u 24
4(t ) 4 2
4( 0)
u u E 6V
u 6(V)
Theo lut ng m khng chnh: 3 4 3(0) 3 3( 0) 4 4( 0)
3(0)
3(0) 4(0)
(C C )U C U C U
500 U 300 24 200 6
U U 12 V
Hph ng trnh m t sau ngm :
1 3(t)( t )
1
1 3(0)
(0 )1
E Ui
R
E U 36 12i 0.08(A)R 300
Xc nh A: (0) xl(0)A i i 0.04
Vy:10t
2(t)i 0.04 (1 e )
B/ Gii bng phng php ton t LAPLACE
Cc s kin c lp c tnh nh trn (phng php tch phn kinh in): 3(0) 4(0)U U 12 V
S ton t ha:
1E
1R
3R 3C
i(t)
3(t)U
4C
-
7/31/2019 Bai Tap Mach Dien
18/22
Chn 2(p) 0 Ta c in th nh to im 1:
3(0) 4(0)1
3 4 1(p)1 2 1
3 4
3 4 11(p) 3 3(0) 4 4(0)
1
3 41(p)
1(p) 4 3
1(p)
U UE1 1 p p ppC pC
1 1R R RpC pC
E5 10 5 10 p C .U C .U
pR
0.125 10 5 10 p 0.006
p
0.12 0.006p 240 12pp(5 10 p 5 10 ) p(p 10)
24 12
p p 10
(V)
Do : 10t
1(t)
10t1 1( t )
( t )
1
24 12e
E 36 24 12ei
R 300
10t(t )i 0.04(1 e )
1Ep
1(p)I
1R
3R
3
1pC 4
1pC
3(0)U
p4(0)U
p
3(p)I
C(p)I
C3(p)I
C4(p)I1
2
-
7/31/2019 Bai Tap Mach Dien
19/22
[email protected] tp 8:
Hy xc nh dng in i(t) khi dch chuyn K sang v tr 3. Bit khi kha K cn tr 1 th mch ch xc lp
Gii: A. Khi t < 25 ms Tnh s kin c lp: i(0)
Trc khi kha K chuyn t v tr 1 sang 2:
( 0)1
E 6i 3 (A)
R 2
Theo LDM chnh ta c:L(0) L( 0)i i 3 (A) S phc ha:
Ta c:
(0 )
(p )2
E 6Li 0.1 3
60 3p 40 pp pI 1.5 3pL R p 0.1 4 p p 40 p p 40 p 40
40t 40t 40t(t )i 1.5(1 e ) 3e 1.5(1 e )
1
2
3
E 6 (E)L 100 mH
R 2
R 4
R 6
1R 2R 3R
1 2 3E
( t )i
L
Cho mch in vi cc thng ssau:
Ep
pL (0 )Li
2R
t = 0 chuyn K t 1 sang 2 t = 25 ms chuyn K t 2 sang 3
-
7/31/2019 Bai Tap Mach Dien
20/22
[email protected]. Khi t > 25ms Tnh s kin c lp: i1(0)
Trc khi kha K chuyn t v tr 2 sang 3:3
340 25 10
1( 0) (t 2.5 10 )i i 1.5(1 e ) 2.05 (A) S phc ha:
Ta c:
1(0)
(p)13
E 6Li 0.1 2.05
60 2.05p 60 pp pI 2.05pL R p 0.1 6 p p 60 p p 60 p 60
60t 60t 60t
(t)1i (1 e ) 2.05e 1 1.05e
Ep
pL (0 )Li
3R
-
7/31/2019 Bai Tap Mach Dien
21/22
-
7/31/2019 Bai Tap Mach Dien
22/22
100tC(t ) 1( t) 2(t ) 1( t)
100tC(t) 100t
2(t)2
u 50(1 e )
u 50(1 e )i 0.25(1 e )
R 200