Download - Bai Tap Mach Dien

7/31/2019 Bai Tap Mach Dien

1/22

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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

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2/22

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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)

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3/22

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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:

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4/22

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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

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5/22

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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%

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6/22

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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)

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7/22

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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)

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8/22

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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


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

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9/22

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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


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

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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

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11/22

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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

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12/22

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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

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13/22

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(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)

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14/22

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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)

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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

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16/22

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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]


17/22

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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

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18/22

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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

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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

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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

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21/22


22/22

[email protected] c:

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

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