7.89
Dane:
Szukane:
R = 110 Ω
Y=
L = 0,35 H
IR =
U = 220 V
IL =
f = 50 Hz
I=
ϕ=
Wzory:
1
1
=
ωC 2π fC
X L = ω L = 2π fL
XC =
Z 2 = R2 + X 2
1
fr =
2π LC
1
G=
R
1
1
=
= 0, 00909 S
R 110
1
1
1
1
1
BL =
=
=
=
=
= 0, 00909 S
X L ω L 2π fL 2 ⋅ 3,14 ⋅ 50 ⋅ 0,35 110
G=
Y = G − jBL = (0, 00909 − j 0, 00909) S
Y = G 2 + BL 2 = 0, 009092 + 0, 00909 2 = 0, 00909 2 = 0, 01285
I R = UG = 220 ⋅ 0, 00909 = 2 A
I L = U (− jBL ) = 220 ⋅ (− j 0, 00909) = − j 2 A
I = I R + I L = ( 2 − j2) A
I = I R 2 + I L 2 = 2 2 + 22 = 2 2 = 2,83 A
tgϕ =
− BL −0, 00909
=
= −1
G
0, 00909
ϕ = −450
______________________________________________________________________
7.90
Dane:
Szukane:
R = 70 Ω
Y=
I R = 1,5 A
BC =
ϕ = 600
IC =
I=
Wzory:
1
1
=
ωC 2π fC
X L = ω L = 2π fL
XC =
Z 2 = R2 + X 2
1
fr =
2π LC
1
G=
R
1 1
=
≈ 0, 01429 S
R 70
B
tgϕ = C
G
2
Y = G 2 + BC 2
G=
Y 2 = G 2 + ( Gtgϕ )
2
2
Y = G 2 + G 2 tg 2ϕ = 0, 01429 2 + 0, 014292 ⋅ 3 = 0, 01429 1 + 3 ≈ 0, 02858 S
BC = Gtgϕ = 0, 01429 3 ≈ 0, 02475 S
U = IR = 1, 5 ⋅ 70 = 105 V
I C = UBC = 105 ⋅ 0, 02475 ≈ 2, 6 A
I = UY = 105 ⋅ 0, 02858 ≈ 3 A
I
IC
U
IR
______________________________________________________________________
7.91
Dane:
Szukane:
I = 2,5 A
IR =
I C = 1, 2 A
Wzory:
1
1
=
ωC 2π fC
X L = ω L = 2π fL
XC =
Z 2 = R2 + X 2
1
fr =
2π LC
1
G=
R
I 2 = I R 2 + IC 2
I R = I 2 − I C 2 = 2,52 − 1, 22 ≈ 2,193 A
I
IC
U
IR
______________________________________________________________________
7.92
Dane:
Szukane:
X L = 2, 5 Ω
BL =
I
=2
IL
R=
G=
Y=
Wzory:
1
1
=
ωC 2π fC
X L = ω L = 2π fL
XC =
Z 2 = R2 + X 2
1
fr =
2π LC
1
1
1
G= ; B= ; Y =
R
X
Z
BL =
1
1
=
= 0, 4 S
X L 2,5
I
=2
IL
I = 2I L
U = I ⋅ Z = 2I L ⋅
1
U 1
=2
⋅
Y
XL Y
1
Y
Y = 2 ⋅ BL = 2 ⋅ 0, 4 = 0,8 S
U = 2UBL ⋅
Y = G 2 + BL 2
Y 2 = G 2 + BL 2
G = Y 2 − BL 2 = 0,82 − 0, 42 = 0, 48 = 0, 692 S
R=
1
1
=
≈ 1, 44 Ω
G 0, 692
______________________________________________________________________
7.93
Dane:
Szukane:
U = = 500 V
tgδ =
I = = 100 µ A = 100 ⋅10
−6
A
C = 0,1 µ F = 0,1 ⋅10−6 F
f = 50 Hz
ϕ=
Wzory:
1
1
=
ωC 2π fC
X L = ω L = 2π fL
XC =
Z 2 = R2 + X 2
1
fr =
2π LC
1
1
1
G= ; B= ; Y =
R
X
Z
IR
tgδ =
IX
R=
U=
500
=
= 5 ⋅106 Ω
I = 100 ⋅10 −6
1
1
106
=
=
Ω
2π fC 2 ⋅ 3,14 ⋅ 50 ⋅ 0,1 ⋅10 −6 31, 4
1
1
= 0, 2 ⋅10 −6 S
G= =
R 5 ⋅106
1
1
BC =
=
= 31, 4 ⋅10−6 S
X C 106
31, 4
XC =
tgϕ =
BC 31, 4 ⋅10−6
=
= 157
0, 2 ⋅10−6
G
ϕ = 89038'
1
X
I R UG
tgδ =
=
= R = C
1
I X UBC
R
XC
106
31, 4
tgδ =
≈ 0, 006369
5 ⋅106
______________________________________________________________________
7.94
Dane:
Szukane:
U = 380 V
C=
I1 = 5, 2 A
I1
2
cos ϕ = 0,3
I2 =
f = 50 Hz
Odbiornik o charakterze
indukcyjnym
Wzory:
1
1
=
ωC 2π fC
X L = ω L = 2π fL
XC =
Z 2 = R2 + X 2
1
fr =
2π LC
1
1
1
G= ; B= ; Y =
R
X
Z
I1 5, 2
=
= 2, 6 A
2
2
I R = I1 cos ϕ = 5, 2 ⋅ 0,3 = 1,56 A
I2 =
I L1 = I12 − I R 2 = 5, 2 2 − 1,56 2 = 4,96 A
I L 2 = I 2 2 − I R 2 = 2, 62 − 1, 562 = 2, 08 A
I C = I L1 − I L 2 = 4,96 − 2, 08 = 2,88 A
I C 2,88
=
= 0, 00758 S
U
380
B
B
0, 00758
0, 00758
C= C = C =
=
= 24,1 ⋅10 −6 F = 24,1 µ F
ω 2π f 2 ⋅ 3,14 ⋅ 50
314
BC =
______________________________________________________________________
7.95
Dane:
Szukane:
U = 380 V
I1 = 5, 2 A
C=
Wzory:
1
1
=
ωC 2π fC
X L = ω L = 2π fL
XC =
I 2 = minimum
cos ϕ = 0,3
f = 50 Hz
Odbiornik o charakterze
indukcyjnym
Z 2 = R2 + X 2
1
fr =
2π LC
1
1
1
G= ; B= ; Y =
R
X
Z
I R = I1 cos ϕ
I12 = I R 2 + I L12
I L12 = I12 − I R 2 = I12 − ( I1 cos ϕ ) = I12 − I12 cos 2 ϕ = I12 (1 − cos 2 ϕ )
2
Jeżeli składowa bierna indukcyjna zostanie zrównoważona równolegle dołączoną składową
bierną pojemnościową to natężenie prądu pobierane przez układ będzie najmniejsze (co
wynika z trójkąta prądów)
I L12 = I12 − I R 2 = I12 − ( I1 cos ϕ ) = I12 − I12 cos 2 ϕ = I12 (1 − cos 2 ϕ )
2
I L1 = I C
I12 (1 − cos 2 ϕ ) = I C
1
= 2π fC
1
2π fC
I C = BCU = 2π fCU
BC =
I1
1
=
XC
(1 − cos ϕ ) = 2π fCU
I (1 − cos ϕ ) 5, 2 (1 − 0, 3 )
2
2
C=
1
2π fU
2
=
2 ⋅ 3,14 ⋅ 50 ⋅ 380
=
5, 2 ⋅ 0,9539
≈ 41, 6 ⋅106 F = 41, 6 µ F
119320
______________________________________________________________________
7.96
Dane:
Szukane:
U = 50 V
Y=
G = 0, 03 S
I R ;C ;L =
BL = 0,16 S
I=
BC = 0,12 S
ϕ=
Wzory:
1
1
=
ωC 2π fC
X L = ω L = 2π fL
XC =
Z 2 = R2 + X 2
1
fr =
2π LC
1
1
1
G= ; B= ; Y =
R
X
Z
Y 2 = G2 + B2
Y 2 = G 2 + ( BL − BC )
2
Y = G 2 + ( BL − BC ) = 0, 032 + ( 0,16 − 0,12 ) = 9 ⋅10−4 + 16 ⋅10−4 = 0, 05 S
2
I R = UG = 50 ⋅ 0, 03 = 1,5 A
I L = UBL = 50 ⋅ 0,16 = 8 A
I C = UBC = 50 ⋅ 0,12 = 6 A
tgϕ =
I L − I C 0, 04
=
= 1, 3333
IR
0, 03
ϕ = 530 07 '
2
IL
I
IC
U
IR
______________________________________________________________________
7.97
Dane:
Szukane:
Wzory:
U = 50 V
Y=
G = 0, 03 S
I R ;C ;L =
BL = 0,16 S
I=
BC = 0,12 S
ϕ=
1
1
=
ωC 2π fC
X L = ω L = 2π fL
1
1
1
G= ; B= ; Y =
R
X
Z
XC =
Y 2 = G2 + B2
Y 2 = G 2 + ( BL − BC )
2
Y = G 2 + ( BL − BC ) = 0, 032 + ( 0,16 − 0,12 ) = 9 ⋅10−4 + 16 ⋅10−4 = 0, 05 S
2
I R = UG = 50 ⋅ 0, 03 = 1,5 A
I L = UBL = 50 ⋅ 0,16 = 8 A
I C = UBC = 50 ⋅ 0,12 = 6 A
tgϕ =
I L − I C 0, 04
=
= 1, 3333
IR
0, 03
ϕ = 530 07 '
2
IC
IL
I
U
IR
Ponieważ prądy płynące w cewce i kondensatorze są przesunięte względem siebie o 180
stopni. . Dlatego suma tych prądów (są przecież o przeciwnych znakach) daje wartość mniejszą
niż oddzielnie prąd cewki czy kondensatora.
______________________________________________________________________
7.98
Dane:
Szukane:
L = 20 mH = 20 ⋅10−3 H
fr =
C = 0,12 µ F = 0,12 ⋅10−6 F
Z=
Y=
RC = ∞
fr =
=
1
2π LC
=
1
2 ⋅ 3,14 20 ⋅10−3 ⋅ 0,12 ⋅10 −6
1
6, 28 20000 ⋅10−6 ⋅ 0,12 ⋅10−6
=
Wzory:
1
1
=
ωC 2π fC
X L = ω L = 2π fL
1
fr =
2π LC
1
1
1
G= ; B= ; Y =
R
X
Z
XC =
=
1
6, 28 20000 ⋅10−6 ⋅ 0,12 ⋅10 −6
106
≈ 3250 Hz = 3, 25 kHz
6, 28 ⋅ 48, 99
=
1 1
= =0
R ∞
1
1
1
103
BL =
=
=
=
= 2, 449 ⋅10−3 S
−3
X L 2π fL 2 ⋅ 3,14 ⋅ 3250 ⋅ 20 ⋅10
408200
G=
BC =
1
= 2π fC = 2 ⋅ 3,14 ⋅ 3250 ⋅ 0,12 ⋅10 −6 ≈ 2, 449 ⋅10−3 S
XC
Y = G 2 + ( BL − BC ) 2 = 02 + (2, 449 ⋅10 −3 − 2, 449 ⋅10 −3 )2 ≈ 0 = 0
Z=
1
1
=
≈∞
Y ց0
______________________________________________________________________
7.99
Dane:
Szukane:
Wzory:
C = 30 pF do Cmax
L=
f1 = 0,52 MHz
Cmax =
1
1
=
ωC 2π fC
X L = ω L = 2π fL
1
fr =
2π LC
1
1
1
G= ; B= ; Y =
R
X
Z
XC =
f 2 = 1, 6 MHz
fr =
L=
1
2π LC
1
( 2π f r 2 )
Cmax =
2
C
=
1
( 2π f r 2 )
2
L
1
( 2 ⋅ 3,14 ⋅1, 6 ⋅10 )
6 2
=
⋅ 30 ⋅10−12
=
1
1
=
= 0,33 mH
12
−12
100,96 ⋅10 ⋅ 30 ⋅10
3029
1
( 2 ⋅ 3,14 ⋅ 0,52 ⋅10 )
6 2
⋅ 0,33 ⋅10−3
=
10 −9
10−9
=
= 284 pF
10, 66 ⋅ 0, 33 3, 52
______________________________________________________________________
7.100
Dane:
Szukane:
C=
Wzory:
1
1
=
ωC 2π fC
X L = ω L = 2π fL
XC =
C = 30 pF do Cmax
fr =
1
2π LC
1
1
1
G= ; B= ; Y =
R
X
Z
f1 = 0,52 MHz
f 2 = 1, 6 MHz
λ = 367 m
fr =
L=
λ=
f =
C=
1
2π LC
1
( 2π f r 2 )
2
C
=
1
( 2 ⋅ 3,14 ⋅1, 6 ⋅10 )
6 2
⋅ 30 ⋅10−12
=
1
1
=
= 0,33 mH
−12
12
100,96 ⋅10 ⋅ 30 ⋅10
3029
c
f
λ
c
1
1
=
=
1

3 ⋅108 
5,134 ⋅106 ) ⋅ 0,33 ⋅10−3
(
−3
2
3,14
0,
33
10
⋅
⋅
⋅
⋅


367 

1
10 −9
=
=
≈ 0,115 ⋅10−9 F = 115 pF
−3
12
26,35 ⋅10 ⋅ 0, 33 ⋅10
8, 69
( 2πλ c )
2
L
2
2
=
______________________________________________________________________
7.101
Dane:
Szukane:
Wzory:
Gr = 0, 4 S
Rs =
BCr = 0,8 S
X Cs =
BLr = 0, 5 S
X Ls =
1
1
=
ωC 2π fC
X L = ω L = 2π fL
1
fr =
2π LC
1
1
1
G= ; B= ; Y =
R
X
Z
XC =
Yr = Gr 2 + ( BC − BL ) = 0, 42 + ( 0,8 − 0, 5 ) = 0,16 + 0, 09 = 0, 5 S
2
BC − BL 0,8 − 0, 5 0,3
=
=
= 0, 75
Gr
0, 4
0, 4
Ponieważ BC > BL
Czyli prąd wyprzedza napięcie
tgϕ =
2
1
1
=
=2Ω
Yr 0,5
Żeby w obwodzie szeregowym prąd wyprzedzał napięcie to układ musi mieć charakter
pojemnościowy czyli X Cs > X Ls
∆X s
tgϕ =
Rs
Zs =
∆X s = tgϕ Rs
Z s = Rs 2 + ∆X s 2 = Rs 2 + ( tgϕ Rs )
2
Z s = Rs 1 + tg 2ϕ
Rs =
Zs
1 + tg 2ϕ
=
2
1 + 0, 752
=
2
= 1, 6 Ω
1, 25
X Cs − X Ls = tgϕ Rs = 0, 75 ⋅1, 6 = 1, 2
Przyjmuję że cewka jest taka sama jak w obwodzie szeregowym
1
1
X Ls = X Lr =
=
=2Ω
BLr 0,5
X Cs − X Ls = tgϕ Rs
X Cs = tgϕ Rs + X Ls = 0, 75 ⋅1, 6 + 2 = 3, 2 Ω
______________________________________________________________________
7.102
Dane:
Szukane:
Wzory:
R = 100 Ω
Grs =
X C = 300 Ω
BCr =
1
1
=
ωC 2π fC
X L = ω L = 2π fL
1
fr =
2π LC
1
1
1
G= ; B= ; Y =
R
X
Z
Z = R 2 + X C 2 = 100 2 + 3002 = 1 ⋅10 4 + 9 ⋅10 4 = 100 10 Ω
tgϕ =
X C 300
=
=3
R 100
XC =
1
1
10
=
=
S
Z 100 10 1000
B
tgϕ = C r
Gr
Y=
3=
BC r
Gr
BC r = Gr 3
Y 2 = Gr 2 + BCr 2
2
 10 
2
2

 = Gr + ( Gr 3)
 1000 
10 ⋅10−6 = Gr 210
10 ⋅10−6
= 10 −3 S = 0, 001 S
10
= Gr 3 = 0, 001 ⋅ 3 = 0, 003 S
Gr =
BC r
______________________________________________________________________
7.103
Dane:
Szukane:
Wzory:
L=
P=
Psr =
1
1
=
ωC 2π fC
X L = ω L = 2π fL
1
fr =
2π LC
1
1
1
G= ; B= ; Y =
R
X
Z
i = I m sin ωt
R=0
XC =
Na elemencie indukcyjnym napięcie wyprzedza prąd o 900 czyli o
π

p = i ⋅ u = I m sin ωt ⋅ U m sin  ωt +  = I mU m sin ωt ⋅ cos ωt
2

Z wzorów trygonometrycznych wynika:
1
sin α ⋅ cos α = sin 2α
2
Więc
1
p = i ⋅ u = I mU m sin 2ωt
2
Podstawiając wartości skuteczne I m = 2 I i U m = 2U
1
1
p=
2I ⋅
2U ⋅ sin 2ωt = I ⋅ U ⋅ sin 2ωt
2
2
π
2
Moc czynna
P = i 2 ⋅ R = ( I m sin ωt ) ⋅ 0 = 0
2
______________________________________________________________________
7.104
Dane:
Szukane:
Wzory:
C=
i = I m sin ωt
p = f (t )
1
1
=
ωC 2π fC
X L = ω L = 2π fL
1
fr =
2π LC
1
1
1
G= ; B= ; Y =
R
X
Z
Psr =
XC =
Na elemencie pojemnościowym napięcie jest opóźnione względem prądu o - 900 czyli o -
π
2
π

p = i ⋅ u = I m sin ωt ⋅ U m sin  ωt −  = I mU m sin ωt ⋅ ( − cos ωt ) = − I mU m sin ωt cos ωt
2

Z wzorów trygonometrycznych wynika:
1
sin α ⋅ cos α = sin 2α
2
Więc
1
p = i ⋅ u = − I mU m sin 2ωt
2
Podstawiając wartości skuteczne I m = 2 I i U m = 2U
1
1
p=−
2I ⋅
2U ⋅ sin 2ωt = − I ⋅ U ⋅ sin 2ωt
2
2
Moc średnia:
W kondensatorze idealnym moc chwilowa oscyluje względem osi czasu z dwa razy większą
częstotliwością od częstotliwości napięcia i prądu. Przebieg mocy nad osią jest dokładnie tej
samej wielkości co pod osią dlatego Psr = 0
______________________________________________________________________
7.104a
Dane:
Szukane:
C=
i = I m sin ωt
p = f (t )
Psr =
Wzory:
1
1
=
ωC 2π fC
X L = ω L = 2π fL
XC =
fr =
1
2π LC
1
1
1
G= ; B= ; Y =
R
X
Z
Na elemencie pojemnościowym napięcie jest opóźnione względem prądu o - 900 czyli o -
π

p = i ⋅ u = I m sin ωt ⋅ U m sin  ωt −  = I mU m sin ωt ⋅ ( − cos ωt ) = − I mU m sin ωt cos ωt
2

Z wzorów trygonometrycznych wynika:
1
sin α ⋅ cos α = sin 2α
2
Więc
1
p = i ⋅ u = − I mU m sin 2ωt
2
Podstawiając wartości skuteczne I m = 2 I i U m = 2U
p=−
π
2
1
1
I
I2
2I ⋅
2U ⋅ sin 2ωt = − I ⋅ U ⋅ sin 2ωt = − I ⋅
⋅ sin 2ωt = −
⋅ sin 2ωt
2
2
ωC
ωC
Moc średnia:
W kondensatorze idealnym moc chwilowa oscyluje względem osi czasu z dwa razy większą
częstotliwością od częstotliwości napięcia i prądu. Przebieg mocy nad osią jest dokładnie tej
samej wielkości co pod osią dlatego Psr = 0
30
25
20
15
10
i
5
u
0
-5 0
p
0,5
1
1,5
2
2,5
-10
-15
-20
______________________________________________________________________
7.105
Dane:
Szukane:
Wzory:
pmax = 600 W
P=
pmin = −100 W
S=
Q=
cos ϕ =
1
1
=
ωC 2π fC
X L = ω L = 2π fL
1
fr =
2π LC
XC =
pmax − pmin
= UI
2
P = UI cos ϕ
Q = UI sin ϕ
pmax − pmin 600 − (−100)
=
= 350 VA
2
2
=S+P
S = UI =
pmax
P = pmax − S = 600 − 350 = 250 W
P 250
=
= 0, 7143
S 350
Q2 + P2 = S 2
cos ϕ =
Q = S 2 − P 2 = 3502 − 2502 = 60000 ≈ 245 var
______________________________________________________________________
7.106
Dane:
Szukane:
I =4A
S=
U = 217,5 V
Q=
cos ϕ =
P = 522 W
f = 50 Hz
Wzory:
1
1
=
ωC 2π fC
X L = ω L = 2π fL
1
fr =
2π LC
XC =
S = UI = 217,5 ⋅ 4 = 870 VA
Q2 + P2 = S 2
Q = S 2 − P 2 = 8702 − 522 2 = 696 var
P = UI cos ϕ
P 522
=
= 0, 6
S 870
______________________________________________________________________
7.107
cos ϕ =
Dane:
Szukane:
Wzory:
W układzie z rysunku watomierz mierzy moc czynną.
Amperomierz mierzy prąd skuteczny.
Zwarcie kondensatora:
Mamy R X C12 =
1
CC
ω 1 2
C1 + C2
Z12 = R 2 + X C12 2
Z2 = R + X C 2
2
Ponieważ
2
Zwarcie rezystora
XC2 =
1
ω C2


1
= R2 + 
 ω C1C2
 C +C
1
2

 1 
= R +

 ω C2 






2
2
2
C1C2
< C2 to Z12 > Z 2
C1 + C2
Z12 = R 2 + X C12 2
1
CC
ω 1 2
C1 + C2


1
= R2 + 
 ω C1C2
 C +C

1
2
1
CC
ω 1 2
C1 + C2
Z tego widać, że Z12 > Z 2
Z2 = X C 2 =
U
U
I2 =
Z12
Z2
Z tego I12 < I 2
I12 =
U
U
I12 =
I2 =
Z12
Z2
Z tego I12 < I 2
P12 = UI12 cos ϕ = U
Mamy R X C12 =
2
U R
U
= 2R
Z12 Z12 Z12
U R U2
P2 = UI 2 cos ϕ = U
=
R
Z 2 Z 2 Z 22
Z tego wynika że P12 < P2
U R
U2
P12 = UI12 cos ϕ = U
=
R
Z12 Z12 Z12 2
U 0 U2
=
⋅0 = 0
Z 2 Z 2 Z 22
Z tego wynika że P12 > P2
P2 = UI 2 cos ϕ = U
______________________________________________________________________
7.108
Dane:
Szukane:
Wzory:
1
1
=
ωC 2π fC
X L = ω L = 2π fL
XC =






2
i = 2,12 sin (ωt − 300 ) A
u = 360sin (ωt − 900 ) V
f = 50 Hz
u = U m sin (ωt + ϕu )
i = I m sin (ωt + ϕi )
S = UI =
R=
fr =
C=
1
2π LC
P=
S=
u = 360sin (ωt − 900 )
i = 2,12 sin (ωt − 300 )
U m = 360 V
I m = 2,12 A
U=
I=
Um
2
Im
2
U m I m 360 ⋅ 2,12
⋅
=
= 381, 6 VA
2
2 2
P = UI cos ϕ = UI cos (ϕu − ϕi ) = UI cos ( −900 − (−300 ) ) =
360 ⋅ 2,12 1
⋅ = 190,8 W
2
2
P
190,8
= 84, 9 Ω
R= 2 =
2
I
 2,12 


 2 
Q2 + P2 = S 2
= UI cos 600 =
Q = S 2 − P 2 = 381, 6 2 − 190,82 ≈ 330, 5 var
Ponieważ napięcie jest opóźnione w stosunku do prądu obwód ma charakter pojemnościowy
Q
330, 5
330, 5
=
= 147 Ω
XC = 2 =
2
I
2, 247
 2,12 


 2 
1
XC =
2π fC
1
1
C=
=
≈ 21, 66 µ F
2π fX C 2 ⋅ 3,14 ⋅ 50 ⋅147
______________________________________________________________________
7.109
Dane:
Szukane:
R = XC
S=
P = 440 W
Q=
U = 220 V
cos ϕ =
f = 50 Hz
R=
XC =
C=
Wzory:
1
1
=
ωC 2π fC
X L = ω L = 2π fL
1
fr =
2π LC
XC =
Z = R2 + X C 2 = R2 + R2 = R 2
I=
U
U
=
Z R 2
2
U2
 U 
P=I R=
 R = 2R
R 2
U2
2202
48400
R=
=
==
= 55 Ω
2 P 2 ⋅ 440
880
X C = R = 55 Ω
2
1
2π fC
1
1
C=
=
= 57,9 µ F
2π fX C 2 ⋅ 3,14 ⋅ 50 ⋅ 55
XC =
Z = R 2 = 55 ⋅1, 41 = 77, 55 Ω
cos ϕ =
R
R
1
2
=
=
=
= 0, 7071
Z R 2
2
2
2
2
U 2 2202
 U 
 U 
Q = I XC = 
X
=
R
=
=
= 440 var
 C 

2 R 2 ⋅ 55
R 2
R 2
2
2
U 2 2 220 2 ⋅ 2
 U 
S =I Z =
 R 2 = 2 R = 2 ⋅ 55 ≈ 622 VA
R 2
2
______________________________________________________________________
7.110
Dane:
Szukane:
Wzory:
Turbina na moc czynną, generator i transformator na moc pozorną.
______________________________________________________________________
7.111
Dane:
Szukane:
Wzory:
S=
Q=
I=
IR =
P = 2 kW
U = 220 V
f = 50 Hz
cos ϕ = 0,8
1
1
=
ωC 2π fC
X L = ω L = 2π fL
XC =
IL =
cos ϕ =
S=
P
S
P
2000
=
= 2500 VA
cos ϕ
0,8
QL = S 2 − P 2 = 2500 2 − 20002 = 1500 var
S 2500
=
= 11,36 A
220
U
I
cos ϕ = R
I
I R = I cos ϕ = 11,36 ⋅ 0,8 = 9, 09
I=
I L = I 2 − I R 2 = 11, 362 − 9, 092 = 46, 42 = 6,8 A
______________________________________________________________________
7.112
Dane:
Szukane:
P = 440 W
Ss =
Sr =
U = 220 V
Qs =
Qr =
cos ϕ s =
cos ϕr =
Rs =
X Ls =
G=
BLr =
Ls =
Lr =
I = 2,5 A
f = 50 Hz
a)
Wzory:
1
1
=
ωC 2π fC
X L = ω L = 2π fL
XC =
S = UI = 220 ⋅ 2,5 = 550 VA
Q2 + P2 = S 2
Q = S 2 − P 2 = 5502 − 4402 = 330 var
P 440
=
= 0,8
S 550
U 220
= 88 Ω
Z= =
I
2,5
R
cos ϕ =
R = Z cos ϕ = 88 ⋅ 0,8 = 70, 4 Ω
Z
cos ϕ =
X L = Z 2 − R 2 = 882 − 70, 42 = 52,8 Ω
X L = 2π fL
L=
XL
52,8
=
≈ 0,168 H
2π f 2 ⋅ 3,14 ⋅ 50
b)
S = UI = 220 ⋅ 2,5 = 550 VA
Q2 + P2 = S 2
Q = S 2 − P 2 = 5502 − 4402 = 330 var
P 440
=
= 0,8
S 550
2, 5
I
= 11, 36 mS
Y= =
U 220
G
cos ϕ =
G = Y cos ϕ = 11,36 ⋅ 0,8 = 9,1 mS
Y
cos ϕ =
BL = Y 2 − G 2 = 11, 362 − 9,12 = 6,8 mS
XL =
1
1
=
= 147 Ω
BL 6,8 ⋅10−3
X L = 2π fL
L=
XL
147
=
≈ 0, 468 H
2π f 2 ⋅ 3,14 ⋅ 50
______________________________________________________________________
7.113
Dane:
Szukane:
P = 250 W
U = 220 V
IN = 2 A
Ul =
f = 50 Hz
Rd = 4 Ω
Ls =
Ud =
cos ϕ s =
Połączenie szeregowe dławika i lampy
Wzory:
1
1
=
ωC 2π fC
X L = ω L = 2π fL
XC =
Lampa to rezystancja
Dławik to połączenie szeregowo rezystancji i reaktancji indukcyjnej.
P
250
Rl = 2 = 2 = 62, 5 Ω
IN
2
U l = I N Rl = 2 ⋅ 62,5 = 125 V
S = UI = 220 ⋅ 2 = 440 VA
Pd = Rd I N 2 = 4 ⋅ 2 2 = 16 W
P = Pd + Pl = 16 + 250 = 266 W
P 266
=
= 0, 6045
S 440
U 220
Z= =
= 110 Ω
I
2
cos ϕ =
X L = Z 2 − ( Rl + Rd ) = 110 2 − ( 62, 5 + 4 ) = 12100 − 4422, 25 = 87, 6 Ω
2
X L = 2π fL
L=
2
XL
87, 6
=
≈ 0, 279 H
2π f 2 ⋅ 3,14 ⋅ 50
Z d = Rd 2 + X l 2 = 4 2 + 87, 6 2 = 87, 69 Ω
U d = I N Z d = 2 ⋅ 87, 69 ≈ 175, 4 Ω
______________________________________________________________________
7.114
Dane:
Szukane:
P = 40 W
Us =
U = 220 V
Ud =
I = 0, 41 A
f = 50 Hz
Rd =
cos ϕ = 0, 6
cos ϕ d =
Wzory:
1
1
=
ωC 2π fC
X L = ω L = 2π fL
XC =
Xd =
Połączenie szeregowe dławika i lampy
Lampa to rezystancja
Dławik to połączenie szeregowo rezystancji i reaktancji indukcyjnej.
P
40
=
≈ 238 Ω
2
I
0, 412
U s = IRs = 0, 41 ⋅ 238 ≈ 97, 6 V
Rs =
S = UI = 220 ⋅ 0, 41 = 90, 2 VA
Pd + P
S
Pd = S cos ϕ − P = 90, 2 ⋅ 0, 6 − 40 = 14,14 W
cos ϕ =
Pd 14,14
=
= 84,1 Ω
I 2 0, 412
U 220
= 536, 6 Ω
Z= =
I 0, 41
Rd =
X L = Z 2 − ( Rs + Rd ) = 536, 62 − ( 238 + 84,1) ≈ 429 Ω
2
2
Z d = Rd 2 + X l 2 = 84,142 + 429 2 ≈ 437 Ω
cos ϕd =
Rd 84,1
=
= 0,1924
Z d 437
U d = IZ d = 0, 41 ⋅ 437 ≈ 179 V
______________________________________________________________________
7.115
Dane:
Szukane:
P = 40 W
C=
U = 220 V
I = 0, 41 A
Wzory:
1
1
=
ωC 2π fC
X L = ω L = 2π fL
XC =
f = 50 Hz
cos ϕ sd = 0, 6
cos ϕ = 1
Połączenie szeregowe dławika i lampy
Lampa to rezystancja. Dławik to połączenie szeregowo rezystancji i reaktancji indukcyjnej.
P
40
=
≈ 238 Ω
2
I
0, 412
U s = IRs = 0, 41 ⋅ 238 ≈ 97, 6 V
Rs =
S = UI = 220 ⋅ 0, 41 = 90, 2 VA
Pd + P
S
Pd = S cos ϕ − P = 90, 2 ⋅ 0, 6 − 40 = 14,14 W
cos ϕ =
Pd 14,14
=
= 84,1 Ω
I 2 0, 412
U 220
= 536, 6 Ω
Z= =
I 0, 41
Rd =
X L = Z 2 − ( Rs + Rd ) = 536, 62 − ( 238 + 84,1) ≈ 429 Ω
2
2
Zamiana na równoważny równoległy
1
1
Y= =
= 1,863 mS
Z 536, 6
G
cos ϕ sd =
G = cos ϕ sd Y = 0, 6 ⋅1,863 = 1,118 mS
Y
BL = Y 2 − G 2 = 1,8632 − 1,1182 = 1, 49 mS
Po dodaniu kondensatora
______________________________________________________________________
7.116
Dane:
Szukane:
P = 200 W
C=
U = 220 V
f = 50 Hz
cos ϕ s = 0, 6
cos ϕ = 0, 9
Połączenie szeregowe rezystancji i reaktancji indukcyjnej.
P
cos ϕ s =
S
P
200
S=
=
= 333,33 VA
cos ϕ s 0, 6
Z=
U2
2202
=
= 145, 2 Ω
S
333,33
Wzory:
1
1
=
ωC 2π fC
X L = ω L = 2π fL
XC =
1
1
=
= 6,89 mS
Z 145, 2
G
cos ϕ s =
G = Y cos ϕ s = 6,89 ⋅ 0, 6 = 4,13 mS
Y
Y 2 = G 2 + BL 2
Y=
BL = Y 2 − G 2 = 6,89 2 − 4,132 = 5, 51 mS
G
cos ϕ =
Y
G
cos ϕ =
2
G 2 + ( BL − BC )
(
cos 2 ϕ G 2 + ( BL − BC )
( BL − BC ) =
2
2
)=G
2
G2
− G2
2
cos ϕ
BL − BC =
G2
− G2
cos 2 ϕ
BC = BL −
4,132
G2
2
−
G
=
5,
51
−
− 4,132 = 5,51 − 2 = 3,51 mS
cos 2 ϕ
0, 92
BC =
C=
1
= 2π fC
XC
BC
3,51 ⋅10−3
=
≈ 11,18 µ F
2π f 2 ⋅ 3,14 ⋅ 50
______________________________________________________________________
7.117
Dane:
Szukane:
P1 = 3 kW
I=
U1 = 220 V
P=
cos ϕ1 = 0, 6
Q=
S=
P2 = 1, 6 kW
U 2 = 220 V
cos ϕ =
cos ϕ2 = 0,8
U = 220 V
P
cos ϕ1 = 1
S1
Z1 =
S1 =
P1
3000
=
= 5000 VA
cos ϕ1
0, 6
U12 2202
=
= 9, 68 Ω
S1 5000
Wzory:
1
1
=
ωC 2π fC
X L = ω L = 2π fL
XC =
1
1
=
= 103,3 mS
Z1 9, 68
Y1 =
G1
Y1
cos ϕ1 =
G1 = Y1 cos ϕ1 = 103, 3 ⋅ 0, 6 = 61,98 mS
Y 2 = G 2 + BL 2
P
cos ϕ2 = 2
S2
BL1 = Y12 − G12 = 103, 32 − 61,982 = 82, 64 mS
P2
1600
S2 =
=
= 2000 VA
cos ϕ 2
0,8
U 2 2 2202
=
= 24, 2 Ω
2000
S2
1
1
Y2 =
=
= 41, 32 mS
Z 2 24, 2
Z2 =
cos ϕ2 =
G2
Y2
G2 = Y2 cos ϕ2 = 41,32 ⋅ 0,8 = 33, 06 mS
Y 2 = G 2 + BL 2
Y=
BL 2 = Y2 2 − G2 2 = 41,322 − 33, 062 = 24, 79 mS
( G1 + G2 ) + ( BL1 + BL 2 )
2
cos ϕ =
2
=
( 61,98 + 33, 06 ) + (82, 64 + 24, 78 )
2
2
= 143, 42 mS
G G1 + G2 61,98 + 33, 06
=
=
= 0, 6626
Y
Y
143, 42
U
= UY = 220 ⋅143, 42 ⋅10−3 = 31,55 A
Z
P = P1 + P2 = 3000 + 1600 = 4600 W = 4, 6 kW
I=
S=
P
4, 6
=
= 6, 94 kVA
cos ϕ 0, 6626
Q = S 2 − P 2 = 5,19 k var
______________________________________________________________________
7.118
Dane:
P1 = 2 kW
cos ϕ1 = 0,8
P2 = 1, 5 kW
cos ϕ2 = 0, 7
P3 = 3 kW
U = 220 V
∆U % = 3%
L = 70 m
λCu = 55 ⋅106 S / m
Szukane:
S=
Wzory:
1
1
=
ωC 2π fC
X L = ω L = 2π fL
XC =
U w = U − 3%U = 97%U = 213, 4 V
P
P
2000
cos ϕ1 = 1
S1 = 1 =
= 2500 VA
S1
cos ϕ1
0,8
U12 2202
Z1 =
=
= 19, 36 Ω
S1 2500
1
1
Y1 =
=
= 51, 65 mS
Z1 19, 36
cos ϕ1 =
G1
Y1
Y 2 = G 2 + BL 2
P
cos ϕ2 = 2
S2
G1 = Y1 cos ϕ1 = 51, 65 ⋅ 0,8 = 41, 32 mS
BL1 = Y12 − G12 = 51, 652 − 41,322 = 31,3 mS
P2
1500
S2 =
=
= 2143 VA
cos ϕ2
0, 7
U 2 2 2202
=
= 22,59 Ω
S2
2143
1
1
Y2 =
=
= 44, 27 mS
Z 2 22, 59
Z2 =
cos ϕ2 =
G2
Y2
Y 2 = G 2 + BL 2
R=
G2 = Y2 cos ϕ2 = 44, 26 ⋅ 0, 7 = 30,99 mS
BL 2 = Y2 2 − G2 2 = 44, 27 2 − 30, 992 = 31, 61 mS
U 2 220 2
=
= 16,13 Ω
P3 3000
G3 =
1
1
=
= 61, 98 mS
R 16,13
______________________________________________________________________
7.119
Dane:
Szukane:
U = 380 V
S=
Z = (24 + j 60) Ω
P=
Q=
Wzory:
S =UI *
I=
380 ( 24 − j 60 )
U
380
=
=
= ( 2,18 − j 5, 46 ) A
Z 24 + j 60
24 2 + 60 2
S = U I * = 380 ( 2,18 + j 5, 46 ) = ( 828, 4 + j 2075 ) VA
P = 828, 4 W
Q = 2075 var
S = P 2 + Q 2 = 828, 42 + 20752 = 4990000 = 2234 VA
______________________________________________________________________
7.120
Dane:
I = 25e
Szukane:
j150
A
Wzory:
S=
U = 220e j 45 V
0
S =UI *
P=
Q=
S = U I * = 220e j 45 ⋅ 25e− j15 = 5500e j 30 VA
0
0
0
S = 5500 VA
S = 5500(cos 300 + j sin 300 ) = 5500(
3
1
+ j ) = ( 4757,5 + j 2750 ) VA
2
2
P = 4757, 5 W = 4, 76 kW
Q = 2750 var = 2, 75 k var
______________________________________________________________________
7.121
Dane:
WP = 300000 kWh
WQ = 100000 k var
Szukane:
Wzory:
cos ϕ =
S =UI *
WP = P ⋅ t
WP
t
WQ = Q ⋅ t
P=
Q=
WQ
t
2
2
W  W 
S = P +Q =  P  + Q  =
 t   t 
2
2
( 3 ⋅10 ) + (1⋅10 )
5 2
t2
5 2
t2
WP
3 ⋅105
P
3 ⋅105
t
t
cos ϕ = =
=
=
= 0,9486
S
10 ⋅105
10 ⋅105
10 ⋅105
t
t
1
10 ⋅105
10
= 10 ⋅10 =
t
t