7.89
Dane:
Szukane:
Wzory:
110
0, 35
220
50
R
L
H
U
V
f
Hz
=
Ω
=
=
=
R
L
Y
I
I
I
ϕ
=
=
=
=
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
2
2
2
Z
R
X
=
+
1
2
r
f
LC
π
=
1
G
R
=
(
)
2
2
2
2
1
1
0, 00909
110
1
1
1
1
1
0, 00909
2
2 3,14 50 0, 35
110
(0, 00909
0, 00909)
0, 00909
0, 00909
0, 00909 2
0, 01285
220 0, 00909
2
(
)
220 (
0, 00909)
2
2
2
L
L
L
L
R
L
L
R
L
G
S
R
B
S
X
L
fL
Y
G
jB
j
S
Y
G
B
I
UG
A
I
U
jB
j
j A
I
I
I
j
A
I
ω
π
= =
=
=
=
=
=
=
=
⋅
⋅ ⋅
= −
=
−
=
+
=
+
=
=
=
=
⋅
=
=
−
=
⋅ −
= −
=
+
= −
=
2
2
2
2
0
2
2
2 2
2,83
0, 00909
tg
1
0, 00909
45
R
L
L
I
I
A
B
G
ϕ
ϕ
+
=
+
=
=
−
−
=
=
= −
= −
______________________________________________________________________
7.90
Dane:
Szukane:
Wzory:
0
70
1, 5
60
R
R
I
A
ϕ
=
Ω
=
=
C
C
Y
B
I
I
=
=
=
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
2
2
2
Z
R
X
=
+
1
2
r
f
LC
π
=
1
G
R
=
(
)
2
2
2
2
2
2
2
2
2
2
2
2
1
1
0, 01429
70
tg
tg
tg
0, 01429
0, 01429
3
0, 01429 1 3
0, 02858
tg
0, 01429 3
0, 02475
1, 5 70 105
105 0, 02475
2, 6
105 0, 02858
3
C
C
C
C
C
G
S
R
B
G
Y
G
B
Y
G
G
Y
G
G
S
B
G
S
U
IR
V
I
UB
A
I
UY
A
ϕ
ϕ
ϕ
ϕ
= =
≈
=
=
+
=
+
=
+
=
+
⋅
=
+ ≈
=
=
≈
=
=
⋅
=
=
=
⋅
≈
=
=
⋅
≈
______________________________________________________________________
7.91
Dane:
Szukane:
Wzory:
2, 5
1, 2
C
I
A
I
A
=
=
R
I
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
2
2
2
Z
R
X
=
+
1
2
r
f
LC
π
=
1
G
R
=
2
2
2
2
2
2
2
2, 5
1, 2
2,193
R
C
R
C
I
I
I
I
I
I
A
=
+
=
−
=
−
≈
U
I
R
I
C
I
______________________________________________________________________
7.92
Dane:
Szukane:
Wzory:
2, 5
2
L
L
X
I
I
=
Ω
=
L
B
R
G
Y
=
=
=
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
2
2
2
Z
R
X
=
+
1
2
r
f
LC
π
=
1
1
1
;
;
G
B
Y
R
X
Z
=
=
=
U
I
R
I
C
I
2
2
2
2
2
2
2
2
2
1
1
0, 4
2, 5
2
2
1
1
2
2
1
2
2
2 0, 4
0,8
0,8
0, 4
0, 48
0, 692
1
1
1, 44
0, 692
L
L
L
L
L
L
L
L
L
L
L
B
S
X
I
I
I
I
U
U
I Z
I
Y
X
Y
U
UB
Y
Y
B
S
Y
G
B
Y
G
B
G
Y
B
S
R
G
=
=
=
=
=
= ⋅ =
⋅ =
⋅
=
⋅
= ⋅
= ⋅
=
=
+
=
+
=
−
=
−
=
=
=
=
≈
Ω
______________________________________________________________________
7.93
Dane:
Szukane:
Wzory:
6
6
500
100
100 10
0,1
0,1 10
50
U
V
I
A
A
C
F
F
f
Hz
µ
µ
=
−
=
−
=
=
=
⋅
=
=
⋅
=
tg
δ
ϕ
=
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
2
2
2
Z
R
X
=
+
1
2
r
f
LC
π
=
1
1
1
;
;
G
B
Y
R
X
Z
=
=
=
tg
R
X
I
I
δ
=
6
6
6
6
6
6
6
6
6
6
0
500
5 10
100 10
1
1
10
2
2 3,14 50 0,1 10
31, 4
1
1
0, 2 10
5 10
1
1
31, 4 10
10
31, 4
31, 4 10
157
0, 2 10
89 38 '
C
C
C
C
U
R
I
X
fC
G
S
R
B
S
X
B
tg
G
π
ϕ
ϕ
=
−
=
−
−
−
−
−
=
=
= ⋅
Ω
⋅
=
=
=
Ω
⋅
⋅ ⋅
⋅
= =
=
⋅
⋅
=
=
=
⋅
⋅
=
=
=
⋅
=
6
6
1
tg
1
10
31, 4
tg
0, 006369
5 10
C
R
X
C
C
X
I
UG
R
I
UB
R
X
δ
δ
=
=
=
=
=
≈
⋅
______________________________________________________________________
7.94
Dane:
Szukane:
Wzory:
1
1
2
380
5, 2
2
cos
0, 3
50
U
V
I
A
I
I
f
Hz
ϕ
=
=
=
=
=
Odbiornik o charakterze
indukcyjnym
C
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
2
2
2
Z
R
X
=
+
1
2
r
f
LC
π
=
1
1
1
;
;
G
B
Y
R
X
Z
=
=
=
1
2
1
2
2
2
2
1
1
2
2
2
2
2
2
1
2
6
5, 2
2, 6
2
2
cos
5, 2 0, 3 1, 56
5, 2
1, 56
4, 96
2, 6
1, 56
2, 08
4, 96 2, 08
2,88
2,88
0, 00758
380
0, 00758
0, 00758
24,1 10
24,1
2
2 3,14 50
314
R
L
R
L
R
C
L
L
C
C
C
C
I
I
A
I
I
A
I
I
I
A
I
I
I
A
I
I
I
A
I
B
S
U
B
B
C
F
f
ϕ
µ
ω
π
−
=
=
=
=
=
⋅
=
=
−
=
−
=
=
−
=
−
=
=
−
=
−
=
=
=
=
=
=
=
=
=
⋅
=
⋅
⋅
F
______________________________________________________________________
7.95
Dane:
Szukane:
Wzory:
1
2
380
5, 2
minimum
cos
0, 3
50
U
V
I
A
I
f
Hz
ϕ
=
=
=
=
=
Odbiornik o charakterze
indukcyjnym
C
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
2
2
2
Z
R
X
=
+
1
2
r
f
LC
π
=
1
1
1
;
;
G
B
Y
R
X
Z
=
=
=
(
)
(
)
1
2
2
2
1
1
2
2
2
2
2
2
2
2
2
2
1
1
1
1
1
1
1
cos
cos
cos
1 cos
R
R
L
L
R
I
I
I
I
I
I
I
I
I
I
I
I
I
ϕ
ϕ
ϕ
ϕ
=
=
+
=
−
=
−
=
−
=
−
Jeżeli składowa bierna indukcyjna zostanie zrównoważona równolegle dołączoną składową
bierną pojemnościową to natężenie prądu pobierane przez układ będzie najmniejsze (co
wynika z trójkąta prądów)
(
)
(
)
(
)
(
)
(
)
(
)
2
2
2
2
2
2
2
2
2
2
1
1
1
1
1
1
1
1
2
2
1
2
1
2
2
1
6
cos
cos
1 cos
1 cos
1
1
2
1
2
2
1 cos
2
1 cos
5, 2
1 0, 3
5, 2 0, 9539
41, 6 10
41, 6
2
2 3,14 50 380
119320
L
R
L
C
C
C
C
C
C
I
I
I
I
I
I
I
I
I
I
I
I
B
fC
X
fC
I
B U
fCU
I
fCU
I
C
F
F
fU
ϕ
ϕ
ϕ
ϕ
π
π
π
ϕ
π
ϕ
µ
π
=
−
=
−
=
−
=
−
=
−
=
=
=
=
=
=
−
=
−
−
⋅
=
=
=
≈
⋅
=
⋅
⋅ ⋅
______________________________________________________________________
7.96
Dane:
Szukane:
Wzory:
50
0, 03
0,16
0,12
L
C
U
V
G
S
B
S
B
S
=
=
=
=
; ;
R C L
Y
I
I
ϕ
=
=
=
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
2
2
2
Z
R
X
=
+
1
2
r
f
LC
π
=
1
1
1
;
;
G
B
Y
R
X
Z
=
=
=
(
)
(
)
(
)
2
2
2
2
2
2
2
2
2
2
4
4
0, 03
0,16 0,12
9 10
16 10
0, 05
L
C
L
C
Y
G
B
Y
G
B
B
Y
G
B
B
S
−
−
=
+
=
+
−
=
+
−
=
+
−
=
⋅
+ ⋅
=
0
50 0, 03 1, 5
50 0,16
8
50 0,12
6
0, 04
tg
1, 3333
0, 03
53 07 '
R
L
L
C
C
L
C
R
I
UG
A
I
UB
A
I
UB
A
I
I
I
ϕ
ϕ
=
=
⋅
=
=
=
⋅
=
=
=
⋅
=
−
=
=
=
=
______________________________________________________________________
7.97
Dane:
Szukane:
Wzory:
50
0, 03
0,16
0,12
L
C
U
V
G
S
B
S
B
S
=
=
=
=
; ;
R C L
Y
I
I
ϕ
=
=
=
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
1
1
1
;
;
G
B
Y
R
X
Z
=
=
=
(
)
(
)
(
)
2
2
2
2
2
2
2
2
2
2
4
4
0, 03
0,16 0,12
9 10
16 10
0, 05
L
C
L
C
Y
G
B
Y
G
B
B
Y
G
B
B
S
−
−
=
+
=
+
−
=
+
−
=
+
−
=
⋅
+ ⋅
=
0
50 0, 03 1, 5
50 0,16
8
50 0,12
6
0, 04
tg
1, 3333
0, 03
53 07 '
R
L
L
C
C
L
C
R
I
UG
A
I
UB
A
I
UB
A
I
I
I
ϕ
ϕ
=
=
⋅
=
=
=
⋅
=
=
=
⋅
=
−
=
=
=
=
U
I
R
I
C
I
I
L
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:
Wzory:
3
6
20
20 10
0,12
0,12 10
C
L
mH
H
C
F
F
R
µ
−
−
=
=
⋅
=
=
⋅
= ∞
r
f
Z
Y
=
=
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
1
2
r
f
LC
π
=
1
1
1
;
;
G
B
Y
R
X
Z
=
=
=
3
6
6
6
6
6
6
1
1
1
2
2 3,14 20 10
0,12 10
6, 28 20000 10
0,12 10
1
10
3250
3, 25
6, 28 48, 99
6, 28 20000 10
0,12 10
r
f
LC
Hz
kHz
π
−
−
−
−
−
−
=
=
=
=
⋅
⋅
⋅
⋅
⋅
⋅
⋅
=
=
≈
=
⋅
⋅
⋅
⋅
U
I
R
I
C
I
I
L
3
3
3
6
3
2
2
2
3
3 2
1
1
0
1
1
1
10
2, 449 10
2
2 3,14 3250 20 10
408200
1
2
2 3,14 3250 0,12 10
2, 449 10
(
)
0
(2, 449 10
2, 449 10 )
0
0
1
1
0
L
L
C
C
L
C
G
R
B
S
X
fL
B
fC
S
X
Y
G
B
B
Z
Y
π
π
−
−
−
−
−
−
= =
=
∞
=
=
=
=
=
⋅
⋅
⋅
⋅ ⋅
=
=
= ⋅
⋅
⋅
⋅
≈
⋅
=
+
−
=
+
⋅
−
⋅
≈
=
= =
≈ ∞
ց
______________________________________________________________________
7.99
Dane:
Szukane:
Wzory:
max
1
2
30
0, 52
1, 6
C
pF do C
f
MHz
f
MHz
=
=
=
max
L
C
=
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
1
2
r
f
LC
π
=
1
1
1
;
;
G
B
Y
R
X
Z
=
=
=
(
)
(
)
(
)
(
)
2
2
12
12
6
12
2
9
9
max
2
2
6
3
2
1
2
1
1
1
1
0, 33
100, 96 10
30 10
3029
2
2 3,14 1, 6 10
30 10
1
1
10
10
284
10, 66 0, 33
3, 52
2
2 3,14 0, 52 10
0, 33 10
r
r
r
f
LC
L
mH
f
C
C
pF
f
L
π
π
π
−
−
−
−
−
=
=
=
=
=
=
⋅
⋅ ⋅
⋅
⋅
⋅
⋅ ⋅
=
=
=
=
=
⋅
⋅
⋅
⋅
⋅
⋅
______________________________________________________________________
7.100
Dane:
Szukane:
Wzory:
C
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
max
1
2
30
0, 52
1, 6
367
C
pF do C
f
MHz
f
MHz
m
λ
=
=
=
=
1
2
r
f
LC
π
=
1
1
1
;
;
G
B
Y
R
X
Z
=
=
=
(
)
(
)
(
)
(
)
2
2
12
12
6
12
2
2
2
2
8
6
3
3
9
9
12
3
1
2
1
1
1
1
0, 33
100, 96 10
30 10
3029
2
2 3,14 1, 6 10
30 10
1
1
1
2
3 10
5,134 10
0, 33 10
2 3,14
0, 33 10
367
1
10
0,115 10
115
26, 35 10
0, 33 10
8, 69
r
r
f
LC
L
mH
f
C
c
f
f
c
C
c
L
F
π
π
λ
λ
πλ
−
−
−
−
−
−
−
=
=
=
=
=
=
⋅
⋅ ⋅
⋅
⋅
⋅
⋅ ⋅
=
=
=
=
=
=
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
=
=
≈
⋅
=
⋅
⋅
⋅
pF
______________________________________________________________________
7.101
Dane:
Szukane:
Wzory:
0, 4
0,8
0, 5
r
Cr
Lr
G
S
B
S
B
S
=
=
=
s
Cs
Ls
R
X
X
=
=
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
1
2
r
f
LC
π
=
1
1
1
;
;
G
B
Y
R
X
Z
=
=
=
(
)
(
)
2
2
2
2
0, 4
0,8 0, 5
0,16 0, 09
0, 5
0,8 0, 5
0, 3
0, 75
0, 4
0, 4
r
r
C
L
C
L
r
Y
G
B
B
S
B
B
tg
G
ϕ
=
+
−
=
+
−
=
+
=
−
−
=
=
=
=
Ponieważ
C
L
B
B
>
Czyli prąd wyprzedza napięcie
1
1
2
0, 5
s
r
Z
Y
=
=
= Ω
Ż
eby w obwodzie szeregowym prąd wyprzedzał napięcie to układ musi mieć charakter
pojemnościowy czyli
Cs
Ls
X
X
>
(
)
2
2
2
2
2
2
2
1
2
2
1, 6
1, 25
1
1 0, 75
0, 75 1, 6 1, 2
s
s
s
s
s
s
s
s
s
s
s
s
s
Cs
Ls
s
X
tg
R
X
tg R
Z
R
X
R
tg R
Z
R
tg
Z
R
tg
X
X
tg R
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
∆
=
∆ =
=
+ ∆
=
+
=
+
=
=
=
=
Ω
+
+
−
=
=
⋅
=
Przyjmuję że cewka jest taka sama jak w obwodzie szeregowym
1
1
2
0, 5
Ls
Lr
Lr
X
X
B
=
=
=
= Ω
0, 75 1, 6 2
3, 2
Cs
Ls
s
Cs
s
Ls
X
X
tg R
X
tg R
X
ϕ
ϕ
−
=
=
+
=
⋅
+ =
Ω
______________________________________________________________________
7.102
Dane:
Szukane:
Wzory:
100
300
C
R
X
=
Ω
=
Ω
rs
Cr
G
B
=
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
1
2
r
f
LC
π
=
1
1
1
;
;
G
B
Y
R
X
Z
=
=
=
2
2
2
2
4
4
100
300
1 10
9 10
100 10
300
3
100
C
C
Z
R
X
X
tg
R
ϕ
=
+
=
+
=
⋅
+ ⋅
=
Ω
=
=
=
( )
2
2
2
2
2
2
6
2
6
3
1
1
10
1000
100 10
tg
3
3
10
3
1000
10 10
10
10 10
10
0, 001
10
3
0, 001 3
0, 003
C r
r
C r
r
C r
r
r
Cr
r
r
r
r
C r
r
Y
S
Z
B
G
B
G
B
G
Y
G
B
G
G
G
G
S
S
B
G
S
ϕ
−
−
−
= =
=
=
=
=
=
+
=
+
⋅
=
⋅
=
=
=
=
=
⋅ =
______________________________________________________________________
7.103
Dane:
Szukane:
Wzory:
sin
m
L
i
I
t
ω
=
=
0
sr
P
P
R
=
=
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
1
2
r
f
LC
π
=
1
1
1
;
;
G
B
Y
R
X
Z
=
=
=
Na elemencie indukcyjnym napięcie wyprzedza prąd o
0
90
czyli o
2
π
sin
sin
sin
cos
2
m
m
m
m
p
i u
I
t U
t
I U
t
t
π
ω
ω
ω
ω
= ⋅ =
⋅
+
=
⋅
Z wzorów trygonometrycznych wynika:
1
sin
cos
sin 2
2
α
α
α
⋅
=
Więc
1
sin 2
2
m
m
p
i u
I U
t
ω
= ⋅ =
Podstawiając wartości skuteczne
2
m
I
I
=
i
2
m
U
U
=
1
1
2
2
sin 2
sin 2
2
2
p
I
U
t
I U
t
ω
ω
=
⋅
⋅
= ⋅ ⋅
Moc czynna
(
)
2
2
sin
0
0
m
P
i
R
I
t
ω
= ⋅ =
⋅ =
______________________________________________________________________
7.104
Dane:
Szukane:
Wzory:
sin
m
C
i
I
t
ω
=
=
( )
sr
p
f t
P
=
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
1
2
r
f
LC
π
=
1
1
1
;
;
G
B
Y
R
X
Z
=
=
=
Na elemencie pojemnościowym napięcie jest opóźnione względem prądu o -
0
90 czyli o -
2
π
(
)
sin
sin
sin
cos
sin
cos
2
m
m
m
m
m
m
p
i u
I
t U
t
I U
t
t
I U
t
t
π
ω
ω
ω
ω
ω
ω
= ⋅ =
⋅
−
=
⋅ −
= −
Z wzorów trygonometrycznych wynika:
1
sin
cos
sin 2
2
α
α
α
⋅
=
Więc
1
sin 2
2
m
m
p
i u
I U
t
ω
= ⋅ = −
Podstawiając wartości skuteczne
2
m
I
I
=
i
2
m
U
U
=
1
1
2
2
sin 2
sin 2
2
2
p
I
U
t
I U
t
ω
ω
= −
⋅
⋅
= − ⋅ ⋅
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
0
sr
P
=
______________________________________________________________________
7.104a
Dane:
Szukane:
Wzory:
sin
m
C
i
I
t
ω
=
=
( )
sr
p
f t
P
=
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
1
2
r
f
LC
π
=
1
1
1
;
;
G
B
Y
R
X
Z
=
=
=
Na elemencie pojemnościowym napięcie jest opóźnione względem prądu o -
0
90 czyli o -
2
π
(
)
sin
sin
sin
cos
sin
cos
2
m
m
m
m
m
m
p
i u
I
t U
t
I U
t
t
I U
t
t
π
ω
ω
ω
ω
ω
ω
= ⋅ =
⋅
−
=
⋅ −
= −
Z wzorów trygonometrycznych wynika:
1
sin
cos
sin 2
2
α
α
α
⋅
=
Więc
1
sin 2
2
m
m
p
i u
I U
t
ω
= ⋅ = −
Podstawiając wartości skuteczne
2
m
I
I
=
i
2
m
U
U
=
2
1
1
2
2
sin 2
sin 2
sin 2
sin 2
2
2
I
I
p
I
U
t
I U
t
I
t
t
C
C
ω
ω
ω
ω
ω
ω
= −
⋅
⋅
= − ⋅ ⋅
= − ⋅
⋅
= −
⋅
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
0
sr
P
=
-20
-15
-10
-5
0
5
10
15
20
25
30
0
0,5
1
1,5
2
2,5
i
u
p
______________________________________________________________________
7.105
Dane:
Szukane:
Wzory:
max
min
600
100
p
W
p
W
=
= −
cos
P
S
Q
ϕ
=
=
=
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
1
2
r
f
LC
π
=
max
min
max
min
max
max
2
2
2
2
2
2
2
2
cos
sin
600 ( 100)
350
2
2
600 350
250
250
cos
0, 7143
350
350
250
60000
245 var
p
p
UI
P
UI
Q
UI
p
p
S
UI
VA
p
S
P
P
p
S
W
P
S
Q
P
S
Q
S
P
ϕ
ϕ
ϕ
−
=
=
=
−
− −
=
=
=
=
= +
=
− =
−
=
= =
=
+
=
=
−
=
−
=
≈
______________________________________________________________________
7.106
Dane:
Szukane:
Wzory:
4
217, 5
522
50
I
A
U
V
P
W
f
Hz
=
=
=
=
cos
S
Q
ϕ
=
=
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
1
2
r
f
LC
π
=
2
2
2
2
2
2
2
217, 5 4
870
870
522
696 var
cos
522
cos
0, 6
870
S
UI
VA
Q
P
S
Q
S
P
P
UI
P
S
ϕ
ϕ
=
=
⋅ =
+
=
=
−
=
−
=
=
= =
=
______________________________________________________________________
7.107
Dane:
Szukane:
Wzory:
W układzie z rysunku watomierz mierzy moc czynną.
Amperomierz mierzy prąd skuteczny.
Zwarcie kondensatora:
Zwarcie rezystora
Mamy R
12
1
2
1
2
1
C
X
C C
C
C
ω
=
+
2
2
1
C
X
C
ω
=
2
2
2
2
12
12
1
2
1
2
2
2
2
2
2
2
2
1
1
C
C
Z
R
X
R
C C
C
C
Z
R
X
R
C
ω
ω
=
+
=
+
+
=
+
=
+
Ponieważ
1
2
2
1
2
C C
C
C
C
<
+
to
12
2
Z
Z
>
12
2
12
2
U
U
I
I
Z
Z
=
=
Z tego
12
2
I
I
<
2
12
12
2
12
12
12
2
2
2
2
2
2
2
cos
cos
U
R
U
P
UI
U
R
Z
Z
Z
U R
U
P
UI
U
R
Z Z
Z
ϕ
ϕ
=
=
=
=
=
=
Z tego wynika że
12
2
P
P
<
Mamy R
12
1
2
1
2
1
C
X
C C
C
C
ω
=
+
2
2
2
2
12
12
1
2
1
2
2
2
1
2
1
2
1
1
C
C
Z
R
X
R
C C
C
C
Z
X
C C
C
C
ω
ω
=
+
=
+
+
=
=
+
Z tego widać, że
12
2
Z
Z
>
12
2
12
2
U
U
I
I
Z
Z
=
=
Z tego
12
2
I
I
<
2
12
12
2
12
12
12
2
2
2
2
2
2
2
cos
0
cos
0
0
U
R
U
P
UI
U
R
Z
Z
Z
U
U
P
UI
U
Z Z
Z
ϕ
ϕ
=
=
=
=
=
=
⋅ =
Z tego wynika że
12
2
P
P
>
______________________________________________________________________
7.108
Dane:
Szukane:
Wzory:
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
(
)
(
)
0
0
2,12 sin
30
360 sin
90
50
i
t
A
u
t
V
f
Hz
ω
ω
=
−
=
−
=
R
C
P
S
=
=
=
=
1
2
r
f
LC
π
=
(
)
sin
m
u
u
U
t
ω ϕ
=
+
(
)
0
360 sin
90
u
t
ω
=
−
360
m
U
V
=
2
m
U
U
=
(
)
sin
m
i
i
I
t
ω ϕ
=
+
(
)
0
2,12 sin
30
i
t
ω
=
−
2,12
m
I
A
=
2
m
I
I
=
360 2,12
381, 6
2
2
2
m
m
U
I
S
UI
VA
⋅
=
=
⋅
=
=
(
)
(
)
0
0
0
cos
cos
cos
90
( 30 )
360 2,12 1
cos 60
190,8
2
2
u
i
P
UI
UI
UI
UI
W
ϕ
ϕ ϕ
=
=
−
=
−
− −
=
⋅
=
=
⋅ =
2
2
190,8
84, 9
2,12
2
P
R
I
=
=
=
Ω
2
2
2
2
2
2
2
381, 6
190,8
330, 5 var
Q
P
S
Q
S
P
+
=
=
−
=
−
≈
Ponieważ napięcie jest opóźnione w stosunku do prądu obwód ma charakter pojemnościowy
2
2
330, 5
330, 5
147
2, 247
2,12
2
1
2
1
1
21, 66
2
2 3,14 50 147
C
C
C
Q
X
I
X
fC
C
F
fX
π
µ
π
=
=
=
=
Ω
=
=
=
≈
⋅
⋅ ⋅
______________________________________________________________________
7.109
Dane:
Szukane:
Wzory:
440
220
50
C
R
X
P
W
U
V
f
Hz
=
=
=
=
cos
C
S
Q
R
X
C
ϕ
=
=
=
=
=
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
1
2
r
f
LC
π
=
2
2
2
2
2
2
2
2
2
2
2
2
2
220
48400
55
2
2 440
880
55
1
2
1
1
57, 9
2
2 3,14 50 55
2
55 1, 41
77, 55
1
2
cos
0, 7071
2
2
2
C
C
C
C
Z
R
X
R
R
R
U
U
I
Z
R
U
U
P
I R
R
R
R
U
R
P
X
R
X
fC
C
F
fX
Z
R
R
R
Z
R
π
µ
π
ϕ
=
+
=
+
=
=
=
=
=
=
=
=
==
=
Ω
⋅
= =
Ω
=
=
=
=
⋅
⋅ ⋅
=
= ⋅
=
Ω
= =
=
=
=
2
2
2
2
2
2
2
2
2
220
440 var
2
2 55
2
2
2
220
2
2
622
2
2 55
2
C
C
U
U
U
Q
I X
X
R
R
R
R
U
U
S
I Z
R
VA
R
R
=
=
=
=
=
=
⋅
⋅
=
=
=
=
≈
⋅
______________________________________________________________________
7.110
Dane:
Szukane:
Wzory:
Turbina na moc czynną, generator i transformator na moc pozorną.
______________________________________________________________________
7.111
Dane:
Szukane:
Wzory:
2
220
50
cos
0,8
P
kW
U
V
f
Hz
ϕ
=
=
=
=
R
L
S
Q
I
I
I
=
=
=
=
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
2
2
2
2
2
2
2
2
cos
2000
2500
cos
0,8
2500
2000
1500 var
2500
11, 36
220
cos
cos
11, 36 0,8
9, 09
11, 36
9, 09
46, 42
6,8
L
R
R
L
R
P
S
P
S
VA
Q
S
P
S
I
A
U
I
I
I
I
I
I
I
A
ϕ
ϕ
ϕ
ϕ
=
=
=
=
=
−
=
−
=
=
=
=
=
=
=
⋅
=
=
−
=
−
=
=
______________________________________________________________________
7.112
Dane:
Szukane:
Wzory:
440
220
2, 5
50
P
W
U
V
I
A
f
Hz
=
=
=
=
cos
s
s
s
s
s
L
s
S
Q
R
X
L
ϕ
=
=
=
=
=
=
cos
r
r
r
r
L
r
S
Q
G
B
L
ϕ
=
=
=
=
=
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
a)
2
2
2
2
2
2
2
2
2
2
2
220 2, 5
550
550
440
330 var
440
cos
0,8
550
220
88
2, 5
cos
cos
88 0,8
70, 4
88
70, 4
52,8
52,8
2
0,168
2
2 3,14 50
L
L
L
S
UI
VA
Q
P
S
Q
S
P
P
S
U
Z
I
R
R
Z
Z
X
Z
R
X
X
fL
L
H
f
ϕ
ϕ
ϕ
π
π
=
=
⋅
=
+
=
=
−
=
−
=
= =
=
=
=
=
Ω
=
=
= ⋅
=
Ω
=
−
=
−
=
Ω
=
=
=
≈
⋅
⋅
b)
2
2
2
2
2
2
2
2
2
2
2
3
220 2, 5
550
550
440
330 var
440
cos
0,8
550
2, 5
11, 36
220
cos
cos
11, 36 0,8
9,1
11, 36
9,1
6,8
1
1
147
6,8 10
147
2
0, 468
2
2 3,14 50
L
L
L
L
L
S
UI
VA
Q
P
S
Q
S
P
P
S
I
Y
mS
U
G
G
Y
mS
Y
B
Y
G
mS
X
B
X
X
fL
L
H
f
ϕ
ϕ
ϕ
π
π
−
=
=
⋅
=
+
=
=
−
=
−
=
= =
=
=
=
=
=
=
=
⋅
=
=
−
=
−
=
=
=
=
Ω
⋅
=
=
=
≈
⋅
⋅
______________________________________________________________________
7.113
Dane:
Szukane:
Wzory:
250
220
2
50
4
N
d
P
W
U
V
I
A
f
Hz
R
=
=
=
=
= Ω
cos
l
d
s
s
U
U
L
ϕ
=
=
=
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
Połączenie szeregowe dławika i lampy
Lampa to rezystancja
Dławik to połączenie szeregowo rezystancji i reaktancji indukcyjnej.
(
)
(
)
2
2
2
2
2
2
2
2
250
62, 5
2
2 62, 5 125
220 2
440
4 2
16
16 250
266
266
cos
0, 6045
440
220
110
2
110
62, 5 4
12100 4422, 25
87, 6
87, 6
2
0
2
2 3,14 50
l
N
l
N
l
d
d
N
d
l
L
l
d
L
L
P
R
I
U
I R
V
S
UI
VA
P
R I
W
P
P
P
W
P
S
U
Z
I
X
Z
R
R
X
X
fL
L
f
ϕ
π
π
=
=
=
Ω
=
= ⋅
=
=
=
⋅ =
=
= ⋅ =
=
+ = +
=
= =
=
=
=
=
Ω
=
−
+
=
−
+
=
−
=
Ω
=
=
=
≈
⋅
⋅
2
2
2
2
, 279
4
87, 6
87, 69
2 87, 69 175, 4
d
d
l
d
N
d
H
Z
R
X
U
I Z
=
+
=
+
=
Ω
=
= ⋅
≈
Ω
______________________________________________________________________
7.114
Dane:
Szukane:
Wzory:
40
220
0, 41
50
cos
0, 6
P
W
U
V
I
A
f
Hz
ϕ
=
=
=
=
=
cos
s
d
d
d
d
U
U
R
X
ϕ
=
=
=
=
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
Połączenie szeregowe dławika i lampy
Lampa to rezystancja
Dławik to połączenie szeregowo rezystancji i reaktancji indukcyjnej.
(
)
(
)
2
2
2
2
2
2
2
2
2
2
2
2
40
238
0, 41
0, 41 238
97, 6
220 0, 41 90, 2
cos
cos
90, 2 0, 6 40 14,14
14,14
84,1
0, 41
220
536, 6
0, 41
536, 6
238 84,1
429
84,14
429
4
s
s
s
d
d
d
d
L
s
d
d
d
l
P
R
I
U
IR
V
S
UI
VA
P
P
S
P
S
P
W
P
R
I
U
Z
I
X
Z
R
R
Z
R
X
ϕ
ϕ
=
=
≈
Ω
=
=
⋅
≈
=
=
⋅
=
+
=
=
− =
⋅
−
=
=
=
=
Ω
=
=
=
Ω
=
−
+
=
−
+
≈
Ω
=
+
=
+
≈
37
84,1
cos
0,1924
437
0, 41 437
179
d
d
d
d
d
R
Z
U
IZ
V
ϕ
Ω
=
=
=
=
=
⋅
≈
______________________________________________________________________
7.115
Dane:
Szukane:
Wzory:
40
220
0, 41
50
cos
0, 6
cos
1
sd
P
W
U
V
I
A
f
Hz
ϕ
ϕ
=
=
=
=
=
=
C
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
Połączenie szeregowe dławika i lampy
Lampa to rezystancja. Dławik to połączenie szeregowo rezystancji i reaktancji indukcyjnej.
(
)
(
)
2
2
2
2
2
2
2
2
40
238
0, 41
0, 41 238
97, 6
220 0, 41 90, 2
cos
cos
90, 2 0, 6 40 14,14
14,14
84,1
0, 41
220
536, 6
0, 41
536, 6
238 84,1
429
s
s
s
d
d
d
d
L
s
d
P
R
I
U
IR
V
S
UI
VA
P
P
S
P
S
P
W
P
R
I
U
Z
I
X
Z
R
R
ϕ
ϕ
=
=
≈
Ω
=
=
⋅
≈
=
=
⋅
=
+
=
=
− =
⋅
−
=
=
=
=
Ω
=
=
=
Ω
=
−
+
=
−
+
≈
Ω
Zamiana na równoważny równoległy
2
2
2
2
1
1
1,863
536, 6
cos
cos
0, 6 1,863 1,118
1,863
1,118
1, 49
sd
sd
L
Y
mS
Z
G
G
Y
mS
Y
B
Y
G
mS
ϕ
ϕ
= =
=
=
=
=
⋅
=
=
−
=
−
=
Po dodaniu kondensatora
______________________________________________________________________
7.116
Dane:
Szukane:
Wzory:
200
220
50
cos
0, 6
cos
0, 9
s
P
W
U
V
f
Hz
ϕ
ϕ
=
=
=
=
=
C
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
Połączenie szeregowe rezystancji i reaktancji indukcyjnej.
2
2
cos
200
333, 33
cos
0, 6
220
145, 2
333, 33
s
s
P
S
P
S
VA
U
Z
S
ϕ
ϕ
=
=
=
=
=
=
=
Ω
2
2
2
2
2
2
2
1
1
6,89
145, 2
cos
cos
6,89 0, 6
4,13
6,89
4,13
5, 51
s
s
L
L
Y
mS
Z
G
G
Y
mS
Y
Y
G
B
B
Y
G
mS
ϕ
ϕ
= =
=
=
=
=
⋅
=
=
+
=
−
=
−
=
(
)
(
)
(
)
(
)
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
3
cos
cos
cos
cos
cos
4,13
5, 51
4,13
5, 51 2
3, 51
cos
0, 9
1
2
3, 51 10
11,18
2
2 3,14 50
L
C
L
C
L
C
L
C
C
L
C
C
C
G
Y
G
G
B
B
G
B
B
G
G
B
B
G
G
B
B
G
G
B
B
G
mS
B
fC
X
B
C
F
f
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
π
µ
π
−
=
=
+
−
+
−
=
−
=
−
−
=
−
=
−
−
=
−
−
=
− =
=
=
⋅
=
=
≈
⋅
⋅
______________________________________________________________________
7.117
Dane:
Szukane:
Wzory:
1
1
1
2
2
2
3
220
cos
0, 6
1, 6
220
cos
0,8
220
P
kW
U
V
P
kW
U
V
U
V
ϕ
ϕ
=
=
=
=
=
=
=
cos
I
P
Q
S
ϕ
=
=
=
=
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
1
1
1
1
1
1
2
2
1
1
1
3000
cos
5000
cos
0, 6
220
9, 68
5000
P
P
S
VA
S
U
Z
S
ϕ
ϕ
=
=
=
=
=
=
=
Ω
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
1
1
1
1
1
103, 3
9, 68
cos
cos
103, 3 0, 6
61, 98
103, 3
61, 98
82, 64
L
L
Y
mS
Z
G
G
Y
mS
Y
Y
G
B
B
Y
G
mS
ϕ
ϕ
=
=
=
=
=
=
⋅
=
=
+
=
−
=
−
=
2
2
2
2
2
2
2
2
2
2
2
1600
cos
2000
cos
0,8
220
24, 2
2000
P
P
S
VA
S
U
Z
S
ϕ
ϕ
=
=
=
=
=
=
=
Ω
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
1
41, 32
24, 2
cos
cos
41, 32 0,8
33, 06
41, 32
33, 06
24, 79
L
L
Y
mS
Z
G
G
Y
mS
Y
Y
G
B
B
Y
G
mS
ϕ
ϕ
=
=
=
=
=
=
⋅
=
=
+
=
−
=
−
=
(
) (
)
(
) (
)
2
2
2
2
1
2
1
2
1
2
3
1
2
2
2
61, 98 33, 06
82, 64 24, 78
143, 42
61, 98 33, 06
cos
0, 6626
143, 42
220 143, 42 10
31, 55
3000 1600
4600
4, 6
4, 6
6, 94
cos
0, 6626
5,19 var
L
L
Y
G
G
B
B
mS
G
G
G
Y
Y
U
I
UY
A
Z
P
P
P
W
kW
P
S
kVA
Q
S
P
k
ϕ
ϕ
−
=
+
+
+
=
+
+
+
=
+
+
=
=
=
=
=
=
=
⋅
⋅
=
= + =
+
=
=
=
=
=
=
−
=
______________________________________________________________________
7.118
Dane:
Szukane:
Wzory:
1
1
2
2
3
6
2
cos
0,8
1, 5
cos
0, 7
3
220
%
3%
70
55 10
/
Cu
P
kW
P
kW
P
kW
U
V
U
L
m
S m
ϕ
ϕ
λ
=
=
=
=
=
=
∆
=
=
= ⋅
S
=
1
1
2
2
C
L
X
C
fC
X
L
fL
ω
π
ω
π
=
=
=
=
3%
97%
213, 4
w
U
U
U
U
V
= −
=
=
1
1
1
1
1
1
2
2
1
1
1
2000
cos
2500
cos
0,8
220
19, 36
2500
P
P
S
VA
S
U
Z
S
ϕ
ϕ
=
=
=
=
=
=
=
Ω
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
1
1
1
1
1
51, 65
19, 36
cos
cos
51, 65 0,8
41, 32
51, 65
41, 32
31, 3
L
L
Y
mS
Z
G
G
Y
mS
Y
Y
G
B
B
Y
G
mS
ϕ
ϕ
=
=
=
=
=
=
⋅
=
=
+
=
−
=
−
=
2
2
2
2
2
2
2
2
2
2
2
1500
cos
2143
cos
0, 7
220
22, 59
2143
P
P
S
VA
S
U
Z
S
ϕ
ϕ
=
=
=
=
=
=
=
Ω
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
1
44, 27
22, 59
cos
cos
44, 26 0, 7
30, 99
44, 27
30, 99
31, 61
L
L
Y
mS
Z
G
G
Y
mS
Y
Y
G
B
B
Y
G
mS
ϕ
ϕ
=
=
=
=
=
=
⋅
=
=
+
=
−
=
−
=
2
2
3
3
220
16,13
3000
1
1
61, 98
16,13
U
R
P
G
mS
R
=
=
=
Ω
= =
=
______________________________________________________________________
7.119
Dane:
Szukane:
Wzory:
(24
60)
380
Z
j
U
V
=
+
Ω
=
S
P
Q
=
=
=
*
S
U I
=
(
) (
)
(
) (
)
2
2
2
2
2
2
380 24
60
380
2,18
5, 46
24
60
24
60
*
380 2,18
5, 46
828, 4
2075
828, 4
2075 var
828, 4
2075
4990000
2234
j
U
I
j
A
Z
j
S
U I
j
j
VA
P
W
Q
S
P
Q
VA
−
=
=
=
=
−
+
+
=
=
+
=
+
=
=
=
+
=
+
=
=
______________________________________________________________________
7.120
Dane:
Szukane:
Wzory:
0
0
15
45
25
220
j
j
I
e
A
U
e
V
=
=
S
P
Q
=
=
=
*
S
U I
=
(
)
0
0
0
45
15
30
0
0
*
220
25
5500
5500
3
1
5500(cos 30
sin 30 )
5500(
)
4757, 5
2750
2
2
4757, 5
4, 76
2750 var
2, 75 var
j
j
j
S
U I
e
e
e
VA
S
VA
S
j
j
j
VA
P
W
kW
Q
k
−
=
=
⋅
=
=
=
+
=
+
=
+
=
=
=
=
______________________________________________________________________
7.121
Dane:
Szukane:
Wzory:
300000
100000 var
P
Q
W
kWh
W
k
=
=
cos
ϕ
=
*
S
U I
=
(
) (
)
2
2
2
5
5
2
5
2
2
10
2
2
5
5
5
5
5
3 10
1 10
1
10 10
10 10
3 10
3 10
cos
0, 9486
10 10
10 10
10 10
P
P
Q
Q
Q
P
P
W
P t
W
P
t
W
Q t
W
Q
t
W
W
S
P
Q
t
t
t
t
t
t
W
P
t
t
S
t
t
ϕ
= ⋅
=
= ⋅
=
⋅
⋅
⋅
=
+
=
+
=
+
=
⋅
=
⋅
⋅
= =
=
=
=
⋅
⋅
⋅