1.65
Dane:
Szukane:
Wzory:
60
0, 5
4
26
a
w
E
V
I
A
R
R
=
=
= Ω
=
Ω
z
o
I
I
U
=
=
=
U
R
I
=
w
E
U
I R
= + ⋅
60
2
30
60
15
4
52
w
z
w
E
I
A
R
R
E
I
A
R
U
I R
=
=
=
+
=
=
=
= ⋅ =
Ω
___________________________________________________________________________
1.66
Dane:
Szukane:
Wzory:
2, 05
2
10
E
V
U
V
R
=
=
=
Ω
w
R
=
U
R
I
=
w
E
U
I R
= + ⋅
2
0, 2
10
0, 05
0, 25
0, 2
w
U
I
A
R
E U
R
I
=
=
=
−
=
=
=
Ω
___________________________________________________________________________
1.67
Dane:
Szukane:
Wzory:
6
10
w
E
V
R
=
=
Ω
ź
w
I
G
=
=
U
R
I
=
1
G
R
=
1
1
0, 25
4
6
1, 5
4
w
G
S
R
E
I
A
R
= = =
=
= =
___________________________________________________________________________
1.68
Dane:
Szukane:
Wzory:
2
6
2
6
225
0,821
50
24
16
16 10
35 10
/
w
o
Al
E
V
R
l
m
R
S
mm
m
S m
γ
−
=
=
Ω
=
=
Ω
=
= ⋅
= ⋅
o
l
I
U
U
=
=
∆
=
U
R
I
=
l
R
S
γ
=
w
U
E
I R
= − ⋅
6
6
2
2 50
0,179
35 10 16 10
l
l
R
S
γ
−
⋅
=
=
=
Ω
⋅
⋅ ⋅
225
225
9
24 0,821 0,179
25
o
w
l
E
I
A
R
R
R
=
=
=
=
+
+
+
+
9 24
216
225 9 0,821
217, 6
o
o
o
w
U
I R
V
U
E
I R
V
= ⋅
= ⋅
=
= − ⋅
=
− ⋅
≈
9 0,179 1, 6
l
l
U
I R
V
∆
= ⋅ = ⋅
≈
___________________________________________________________________________
1.69
Dane:
Szukane:
Wzory:
240
4
0 12
w
E
V
R
I
A
=
= Ω
= ÷
I
R
=
=
U
R
I
=
w
U
E
I R
= − ⋅
przy U=220 V natężenie prądu wynosi 5 A
220
44
5
U
R
I
=
=
=
Ω
-20
20
60
100
140
180
220
260
0
1
2
3
4
5
6
7
8
9
10
11
12
I [ A ]
[ V ]
Uo=f(I)
U_Rw=f(I)
___________________________________________________________________________
1.70
Dane:
Szukane:
Wzory:
2
6
2
6
240
10
0, 5
6
6 10
248
220
55 10
/
o
w
Cu
E
V
R
R
S
mm
m
l
m
U
V
S m
γ
−
=
=
Ω
=
Ω
=
= ⋅
=
=
= ⋅
ź
o
l
x
I
U
U
U
l
=
=
=
=
=
U
R
I
=
2
4
w
U
E
I R
l
R
S
d
S
γ
π
= − ⋅
=
=
6
6
248
0, 75
55 10 6 10
2
1, 5
p
l
p
l
R
S
R
R
γ
−
=
=
≈
Ω
⋅
⋅ ⋅
= ⋅
=
Ω
11, 5
o
l
R
R
R
=
+
=
Ω
240
20
11, 5 0, 5
w
E
I
A
R
R
=
=
=
+
+
240 20 0, 5
230
20 10
200
230 200
30
ź
w
o
o
l
ź
o
U
E
I R
V
U
I R
V
U
U
U
V
= − ⋅
=
− ⋅
=
= ⋅
=
⋅ =
=
−
=
−
=
230 220 10
x
ź
U
U
U
V
=
− =
−
=
ponieważ są dwa przewody to na jednym przewodzie spadek napięcia do miejsca
wyznaczonego to:
5
2
l
px
U
U
V
=
=
rezystancja przewodu do tego punktu:
5
0, 25
20
px
x
U
R
I
=
=
=
Ω
ponieważ rezystancja przewodu jest proporcjonalna do długości robimy prostą proporcję:
0, 25 248
82, 67
0, 75
x
x
l
x
x
l
R
l
R
l
R
R
l
m
R
=
⋅
=
⋅ =
=
___________________________________________________________________________
1.71
Dane:
Szukane:
Wzory:
1
2
3
8
4
7
5
1
w
U
V
R
R
R
R
=
= Ω
= Ω
= Ω
= Ω
v
U
=
U
R
I
=
w
U
E
I R
= − ⋅
1
2
3
16
R
R
R
R
= +
+
=
Ω
1
8
2
4
U
I
A
R
=
= =
2 16
32
32 2 1 34
ź
ź
w
U
I R
V
E
U
I R
V
= ⋅ = ⋅ =
=
+ ⋅
=
+ ⋅ =
po zwarciu
2
R natężenie prądu zmieni się na:
1
3
1
34
3, 4
1 4 5
3, 4 4 13, 6
z
w
v
z
E
I
A
R
R
R
U
I
R
V
=
=
=
+ +
+ +
= ⋅ =
⋅ =
___________________________________________________________________________
1.72
Dane:
Szukane:
Wzory:
1
2
9
20
5
0, 005
30
0, 03
w
E
V
R
I
mA
A
I
mA
A
=
=
Ω
=
=
=
=
1
2
U
U
=
=
U
R
I
=
w
U
E
I R
= − ⋅
1
2
9 0, 005 20
9 0,1 8, 9
9 0, 030 20
9 0, 6
8, 4
w
w
U
E
I R
V
U
E
I R
V
= + ⋅
= −
⋅
= −
=
= + ⋅
= −
⋅
= −
=
___________________________________________________________________________
1.73
Dane:
Szukane:
Wzory:
1, 25
5
1, 2
z
E
V
I
A
U
V
=
=
=
o
R
=
U
R
I
=
w
U
E
I R
= − ⋅
1, 25
0, 25
5
w
z
E
R
I
=
=
=
Ω
1, 25 1, 2
0, 2
0, 25
1, 2
6
0, 2
w
w
U
E
I R
E U
I
A
R
R
= − ⋅
−
−
=
=
=
=
= Ω
___________________________________________________________________________
1.74
Dane:
Szukane:
Wzory:
5
50
z
I
A
I
A
=
=
w
R
R
=
U
R
I
=
w
U
E
I R
= − ⋅
50
9
5
50
50
w
z
w
w
E
E
R
I
E
I
R
R
E
E
E
E
R
R
I
=
=
=
+
⋅
= −
= −
=
9
9
50
50
9
50
50
w
E
R
E
E
R
E
⋅
⋅ ⋅
=
=
=
⋅
___________________________________________________________________________
1.75
Dane:
Szukane:
Wzory:
6
3
1
E
V
I
A
R
=
=
= Ω
z
I
=
U
R
I
=
w
U
E
I R
= − ⋅
6
1 1
3
6
6
1
w
w
z
w
E
I
R
R
E
R
R
I
E
I
A
R
=
+
= − = − = Ω
=
= =
___________________________________________________________________________
1.76
Dane:
Szukane:
Wzory:
1, 5
1500
1, 2
v
w
R
k
R
=
Ω =
Ω
=
Ω
%
100%
E U
E
−
∆ =
=
U
R
I
=
w
U
E
I R
= − ⋅
w
w
v
v
U
E
I R
I R
E U
E
R
I
E
I R
= − ⋅
⋅
= −
=
= ⋅
1, 2
%
100%
100%
100%
0, 08 %
1500
w
v
I R
E U
E
I R
⋅
−
∆ =
=
=
=
⋅
___________________________________________________________________________
1.77
Dane:
Szukane:
Wzory:
1
2
3
0,1
6
12
4
0, 6
w
R
R
R
R
I
A
=
Ω
= Ω
=
Ω
= Ω
=
1
2
3
E
I
I
I
=
=
=
=
U
R
I
=
1
2
3
1
1
1
1
R
R
R
R
=
+
+
w
U
E
I R
= − ⋅
1
2
3
1
1
1
1
1
1
1
6
6
12
4
12
2
R
R
R
R
R
=
+
+
= +
+ =
= Ω
0, 6 2 1, 2
U
I R
V
= ⋅ =
⋅ =
1, 2 0, 6 0,1 1, 26
w
E
U
I R
V
= + ⋅
=
+
⋅
=
1
1
2
2
3
3
1, 2
0, 2
6
1, 2
0,1
12
1, 2
0, 3
4
U
I
A
R
U
I
A
R
U
I
A
R
=
=
=
=
=
=
=
=
=
___________________________________________________________________________
1.78
Dane:
Szukane:
Wzory:
1
2
2
1,8
2
U
V
U
V
R
=
=
= Ω
w
E
R
=
=
U
R
I
=
1
2
1
1
1
R
R
R
=
+
w
U
E
I R
= − ⋅
1
1
2
2
2
1
2
2
1
2
2
1,8
1,8
1
z
z
U
I
A
R
R
R
U
I
A
R
=
= =
= = = Ω
=
=
=
1
1
2
2
1
1
2
2
1
2
2
1
2
1
1
2
1
2
2
1
(
)
0, 2
0, 25
0,8
w
w
w
w
w
w
w
w
E
U
I R
E
U
I
R
U
I R
U
I
R
U
U
I
R
I R
R I
I
U
U
U
U
R
I
I
=
+ ⋅
=
+ ⋅
+ ⋅
=
+ ⋅
−
= ⋅
− ⋅
−
=
−
−
=
=
=
Ω
−
1
1
2 1 0, 25
2, 25
w
E
U
I R
V
=
+ ⋅
= + ⋅
=
___________________________________________________________________________
1.79
Dane:
Szukane:
Wzory:
2
6, 3
6
0, 2
12
w
E
V
U
V
R
R
=
=
=
Ω
=
Ω
1
R
=
U
R
I
=
1
2
1
1
1
R
R
R
=
+
w
U
E
I R
= − ⋅
6, 3 6
0,3
1,5
0, 2
0, 2
w
w
w
U
E
I R
I R
E U
E U
I
A
R
= − ⋅
⋅
= −
−
−
=
=
=
=
1
2
1
2
2
1
2
6
4
1, 5
1
1
1
1
1
1
4 12
6
12 4
U
R
I
R
R
R
R
R
R
R R
R
R
R
=
=
= Ω
=
+
= −
⋅
⋅
=
=
= Ω
−
−
___________________________________________________________________________
1.80
Dane:
Szukane:
Wzory:
'
120
4
20
20
w
E
V
R
R
R
=
= Ω
=
Ω
=
Ω
'
U
U
=
=
U
R
I
=
1
2
1
1
1
R
R
R
=
+
w
U
E
I R
= − ⋅
'
1
1
1
10
2
z
z
R
R
R
R
R
= +
= =
Ω
'
'
120
120
8, 57
10 4
14
120
120
5
20 4
24
w
z
w
z
w
w
U
E
I R
I R
E
I R
E
I
A
R
R
E
I
A
R
R
= − ⋅
⋅
= − ⋅
=
=
=
≈
+
+
=
=
=
=
+
+
'
5 20 100
8, 57 10
85, 7
z
U
I R
V
U
I R
V
= ⋅ = ⋅
=
= ⋅
=
⋅ =
___________________________________________________________________________
1.81
Dane:
Szukane:
Wzory:
9
10
0,15
w
E
V
R
I
A
=
=
Ω
=
'
''
U
U
U
=
=
=
U
R
I
=
1
2
1
1
1
R
R
R
=
+
w
U
E
I R
= − ⋅
'
'
'
''
''
''
9 0,15 10
7, 5
7, 5
50
0,15
9
9
0, 257
25 10
35
2
9 0, 257 10
6, 43
9
9
0, 3375
16, 67 10
26, 67
3
9 0, 3375 10
5, 63
w
ż
ż
w
w
ż
w
w
U
E
I R
V
U
R
I
E
I
A
R
R
U
E
I R
V
E
I
A
R
R
U
E
I
R
V
= − ⋅
= −
⋅ =
=
=
=
Ω
=
=
=
≈
+
+
= − ⋅
= −
⋅ ≈
=
=
=
≈
+
+
= − ⋅
= −
⋅ ≈
___________________________________________________________________________
1.82
Dane:
Szukane:
Wzory:
120
10
160
3, 2
R
R
S
S
U
V
I
A
U
V
I
A
=
=
=
=
1
2
E
R
R
=
=
=
U
R
I
=
1
2
1
1
1
R
R
R
=
+
w
U
E
I R
= − ⋅
(
)
120 160
40
5,88
3, 2 10
6,8
S
S
w
R
R
w
S
S
w
R
R
w
R
w
S
w
R
S
w
S
R
R
S
R
S
w
R
S
E
U
I
R
E
U
I
R
U
I
R
U
I
R
I
R
I
R
U
U
R I
I
U
U
U
U
R
I
I
=
+ ⋅
=
+ ⋅
+ ⋅
=
+ ⋅
− ⋅
+ ⋅
=
−
−
=
−
−
−
−
=
=
=
≈
Ω
−
−
−
160 3, 2 5,88 178,82
S
S
w
E
U
I
R
V
=
+ ⋅
=
+
⋅
=
160
50
3, 2
120
12
10
S
S
S
R
R
R
U
R
I
U
R
I
=
=
=
Ω
=
=
=
Ω
1
2
1
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
1
1
1
1
1
1
(
)
(
)
0
50
600
0
S
S
R
R
S
S
R
S
R
S
S
R
S
S
S
R
S
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R R
R R
R
R R
R R
R R
R R
R
R R
R R
R
R
= +
=
−
=
+
=
+
−
+
−
=
−
=
−
=
−
−
+
=
−
+
=
2
'
2
2
4
2500 4 600 100
50 10
20
2
2
50 10
30
2
2
b
ac
b
R
a
b
R
a
∆ = −
=
− ⋅
=
− − ∆
−
=
=
=
− + ∆
+
=
=
=
1
2
50 30
20
S
R
R
R
=
−
=
−
=
Ω
___________________________________________________________________________
1.83
Dane:
Szukane:
Wzory:
1, 5
0, 6
16, 2
o
w
E
V
R
R
=
=
Ω
=
Ω
1
2
3
o
U
I
U
U
U
=
=
=
=
=
U
R
I
=
w
U
E
I R
= − ⋅
3
4, 5
4, 5
0, 25
3
1,8 16, 2
18
o
w
E
I
A
R
R
⋅
=
=
=
=
⋅
+
+
1
1
1
2
3
1, 5 0, 25 0, 6
1, 35
w
U
E
I R
V
U
U
U
=
− ⋅
=
−
⋅
=
=
=
0, 25 16, 2
4, 05
o
U
I R
V
= ⋅ =
⋅
=
___________________________________________________________________________
1.84
Dane:
Szukane:
Wzory:
'
o
o
U
U
=
=
U
R
I
=
'
1, 5
1
5
500
o
w
w
E
V
R
R
R
=
= Ω
= Ω
=
Ω
w
U
E
I R
= − ⋅
'
'
'
'
8
12
12
0, 02362
8
8 500
508
11,81
8
12
12
0, 02222
8
40 500
540
11,11
o
w
o
o
w
o
E
I
A
R
R
U
I R
V
E
I
A
R
R
U
I R
V
⋅
=
=
=
=
⋅
+
+
= ⋅ ≈
⋅
=
=
=
=
⋅
+
+
= ⋅ ≈
___________________________________________________________________________
1.85
Czy wartość napięcia źródła może być większa od jego siły elektromotorycznej?
Tak. W przypadku gdy pracuje jako odbiornik.
Na przykład, jest to w przypadku ładowania akumulatora z prostownika.
___________________________________________________________________________
1.86
Dane:
Szukane:
Wzory:
1
1
2
8
0, 35
6,15
0, 02
500
w
w
E
V
R
E
V
R
R
=
=
Ω
=
=
Ω
=
Ω
o
I
U
=
=
U
R
I
=
w
U
E
I R
= − ⋅
2
1
1
2
1
1
8 6,15
1,85
5
0, 35 0, 02
0, 37
8 1, 75
6, 25
w
w
o
w
E
E
I
A
R
R
U
E
I R
V
−
−
=
=
=
=
+
+
=
− ⋅
= −
=
___________________________________________________________________________
1.87
Dane:
Szukane:
Wzory:
0, 06
0,1
0, 04
220
400
g
m
p
R
R
R
U
V
I
A
=
Ω
=
Ω
=
Ω
=
=
g
m
g
E
E
U
=
=
=
U
R
I
=
w
U
E
I R
= − ⋅
220 400 0,1 180
400 0, 04 16
236
236 400 0, 06
260
m
m
p
p
g
p
g
g
m
E
U
I R
V
U
I R
V
U
U
U
V
E
U
I R
V
= − ⋅
=
−
⋅
=
= ⋅
=
⋅
=
= +
=
=
+ ⋅
=
+
⋅
=
___________________________________________________________________________
1.88
Dane:
Szukane:
Wzory:
1
2
110
1, 5
0,5
1, 5
10
w
o
w
R
E
V
R
E
V
R
=
Ω
=
=
Ω
=
=
Ω
1
2
U
U
=
=
U
R
I
=
w
U
E
I R
= − ⋅
1
1
1
1
2
1
2
2
2
20
30
30
0, 25
20
10 110
120
0, 25 110
27, 5
20
3
30 4, 5
34, 5
0, 23
20
3
10 30 110
150
0, 23 110
25, 3
w
o
w
w
E
I
A
R
R
U
I R
V
E
E
I
A
R
R
R
U
I
R
V
⋅
=
=
=
=
⋅
+
+
= ⋅ =
⋅
=
⋅ + ⋅
+
=
=
=
=
⋅
+ ⋅
+
+
+
= ⋅ =
⋅
=
___________________________________________________________________________
1.89
Jest rozwiązane w Zbiorze zadań i nie chciało mi się przepisywać
___________________________________________________________________________
1.90
Dane:
Szukane:
Wzory:
1
2
3
4
5
1
4
2
3
5
1 5
10
8
4
6
15
4
7
1
W
E
E
V
E
V
E
V
E
V
R
R
R
R
R
R
−
=
=
=
=
=
=
=
Ω
=
= Ω
= Ω
= Ω
1 5
1 5
R
E
A I
U
U
V
−
−
−
=
=
=
U
R
I
=
w
U
E
I R
= − ⋅
ponieważ
1
2
4
(
)
E
E
E
+
+
>
3
5
(
)
E
E
+
to
3
5
E i E są odbiornikami dlatego U > E
prąd w obwodzie płynie zgodnie z ruchem wskazówek zegara.
Dlatego przyjąłem ten sam kierunek rozpatrywania potencjałów.
1
2
3
4
5
1
2
3
4
5
1
1
1
1
2
2
2
2
3
3
3
3
4
10 10 8 4 6
10
0, 2
5
5 15 4 4 15 7
50
0, 2 15
3
10 0, 2
9,8
0, 2 4
0,8
10 0, 2
9,8
0, 2 4
0,8
8 0, 2
8, 2
w
R
E
w
R
E
w
R
E
w
R
E
E
E
E
E
I
A
R
R
R
R
R
R
U
I R
V
U
E
I R
V
U
I R
V
U
E
I R
V
U
I R
V
U
E
I R
V
U
I
+
−
+
−
+ − + −
=
=
=
=
⋅
+ +
+
+
+
+ + + + +
= ⋅ =
⋅ =
=
− ⋅
= −
=
= ⋅
=
⋅ =
=
− ⋅
= −
=
= ⋅
=
⋅ =
=
+ ⋅
= +
=
= ⋅
4
4
4
5
5
5
5
0, 2 15
3
4 0, 2
3,8
0, 2 7
1, 4
6 0, 2
6, 2
E
w
R
E
w
R
V
U
E
I R
V
U
I R
V
U
E
I R
V
=
⋅ =
=
− ⋅
= −
=
= ⋅
=
⋅ =
=
+ ⋅
= +
=
___________________________________________________________________________
1.91
Dane:
Szukane:
Wzory:
1
2
3
1
2
3
1 3
2
10
2
0
W
A
E
E
E
V
R
R
R
R
V
−
=
=
=
=
=
=
Ω
= Ω
=
AB
BC
CA
U
U
U
=
=
=
U
R
I
=
w
U
E
I R
= − ⋅
kierunek rozpatrywania przeciwnie do ruchu wskazówek zegara.
1
2
3
1
2
3
1
1
1
1
2
2
2
2
3
3
2 2 2
6
0,1667
3
6 10 10 10
36
0,1667 10 1, 667
2 0,1667 2
2 0, 333 1, 667
0,1667 10 1, 667
2 0,1667 2
2 0, 333 1, 667
0,1667 10 1, 667
w
R
E
w
R
E
w
R
E
E
E
I
A
R
R
R
R
U
I R
V
U
E
I R
V
U
I R
V
U
E
I R
V
U
I R
V
+
+
+ +
=
=
=
≈
⋅
+ +
+
+ + +
= ⋅ =
⋅ =
=
− ⋅
= −
⋅ = −
=
= ⋅
=
⋅ =
=
− ⋅
= −
⋅ = −
=
= ⋅
=
⋅ =
3
3
2 0,1667 2
2 0, 333 1, 667
E
w
U
E
I R
V
=
− ⋅
= −
⋅ = −
=
3
3
3
1
2
2
2
2
1
3
1
1
0
1, 667
1, 667 1, 667
0
0 1, 667
1, 667
1, 667 1, 667
0
0 1, 667
1, 667
1, 667 1, 667
0
CR
R
C
CR
E
BR
C
R
B
BR
E
AR
B
R
A
AR
E
V
U
V
V
V
U
V
V
V
U
V
V
V
U
V
V
V
U
V
V
V
U
V
= −
= −
=
+
= −
+
=
=
−
= −
= −
=
+
= −
+
=
=
−
= −
= −
=
+
= −
+
=
Wykres potencjałów
-2
-1,5
-1
-0,5
0
0
5
10
15
20
25
30
35
40
R [
Ω
]
V [V]
___________________________________________________________________________
1.92
Dane:
Szukane:
Wzory:
1
2
1
2
4, 5
26
0, 4
0, 6
0
p
W
W
O
E
E
V
R
R
R
V
=
=
=
Ω
=
Ω
=
Ω
=
A
B
CO
V
V
U
=
=
=
U
R
I
=
w
U
E
I R
= − ⋅
1
2
1
2
1
1
1
2
2
2
9
9
0, 3333
1 26
27
0, 3333 26
8, 667
4, 5 0, 3333 0, 4
4, 5 0,1333
4, 3667
4, 5 0, 3333 0, 6
4, 5 0, 2
4, 3
w
w
P
P
P
E
w
E
w
E
E
I
A
R
R
R
U
I R
V
U
E
I R
V
U
E
I R
V
+
=
=
=
≈
+
+
+
= ⋅
=
⋅
=
=
− ⋅
=
−
⋅
=
−
=
=
− ⋅
=
−
⋅
=
−
=
1
2
0
4, 3667
4, 3667 8, 667
4, 3
4, 3 4, 3
0
A
E
B
A
p
O
B
E
V
U
V
V
V
U
V
V
V
U
V
= +
=
=
−
=
−
= −
=
−
= −
+
=
1
2
4, 3667
4, 3
CO
A
O
CO
B
O
U
V
V
V
U
V
V
V
=
−
=
=
−
= −
___________________________________________________________________________
1.93
Dane:
Szukane:
Wzory:
1
2
1
2
2
0, 5
0, 7
0
W
W
E
E
E
E
R
R
U
=
=
=
Ω
=
Ω
=
R
=
U
R
I
=
w
U
E
I R
= − ⋅
1
2
1
2
2
2
2
2
1, 2
(1, 2
)
2
(1, 2
)
2
0
(1, 2
)
0, 7
0
2
(1, 2
)
1, 4
0
2
[1, 2
1, 4]
0
2
( 0, 2
)
0
2
0, 2
0
0, 2
w
W
E
w
w
E
E
E
I
R
R
R
R
I
R
E
I
R
E
U
E
I R
E
I R
V
I
R
I
I
R
I
I
R
I
R
R
R
+
=
=
+
+
+
+
=
+
=
= − ⋅
− ⋅
=
+
− ⋅
=
+
− ⋅
=
+ −
=
−
+
=
−
+ =
=
Ω
lub troszkę innym podejściem
1
2
1
1
1
0, 5 0
(1, 2
)
0, 5 0
2
(1, 2
)
2
1, 2
1
2
0, 2
E
E
R
E
W
R
U
U
U
U
E
I R
U
I R
E
I
I R
I
R
I
I R
I
R
I
I
R
R
R
R
+
=
=
− ⋅
= ⋅
− ⋅
+ = ⋅
+
− ⋅
+ = ⋅
+
− = ⋅ ⋅
+ − = ⋅
=
Ω
___________________________________________________________________________
1.94
Dane:
Szukane:
Wzory:
R
=
U
R
I
=
w
U
E
I R
= − ⋅
1, 2
0, 02
2, 5
115
O
W
E
V
R
I
A
U
V
=
=
Ω
=
=
ponieważ bateria jako odbiornik to:
60
60
60 1, 2 2, 5 60 0, 02
72 3
75
2, 5
115
75 2, 5
16
E
w
E
R
R
E
R
U
E
I
R
U
V
U
I R
U
R
U
U
U
R
R
=
+ ⋅
=
⋅
+
⋅ ⋅
=
+ =
= ⋅
=
⋅
=
+
=
+
⋅
=
Ω
___________________________________________________________________________
1.95
Dane:
Szukane:
Wzory:
1
2
3
1
1
9
24
2
E
E
V
E
V
E
E
I
I
−
=
=
= −
⇒
=
1
E
=
U
R
I
=
w
U
E
I R
= − ⋅
1
3
2
1
1
2
3
2
1
1
2
3
2
1
1
2
3
2
1
1
2
3
2
1
3
2
1
1
2
1
2
1
1
1
2
1
2
1
1
1
1
1
1
2
(
) 2
(
) 2
24 9
24 9
(
) 2
15
(15
) 2
15
30 2
3
15
5
E
E
E
E
I
R
R
E
E
E
I
R
R
E
E
E
I
R
R
E
E
E
I
R
R
E
E
E
E
E
E
R
R
R
R
E
E
R
R
R
R
E
E
E
E
E
E
V
−
−
−
=
+
−
+
=
+
−
+
⋅ =
+
−
+
=
+
⋅
−
+
−
−
=
+
⋅
+
− +
− −
=
+
⋅
+
+
=
−
⋅
+
=
−
=
=
___________________________________________________________________________
1.96
Dane:
Szukane:
Wzory:
1
2
3
1
2
4
3
2
3
1
24
60
12
9
20
0, 5
1
W
W
W
E
V
E
V
E
V
R
R
R
R
R
R
R
=
=
=
=
= Ω
=
=
Ω
=
=
Ω
= Ω
W tym wydaniu tego Zbioru
Zadań jest małe przeoczenie,
zapomnieli podać R2, w
starszym wydaniu było
podane co teraz jest
uwzględnione w powyższych
danych
1
2
3
V
V
V
U
U
U
I
=
=
=
=
U
R
I
=
w
U
E
I R
= − ⋅
Ponieważ
2
(
)
E >
1
3
(
)
E
E
+
to
1
3
E i E są odbiornikami dlatego dla nich U > E
1
2
3
1
2
3
1
2
3
4
1
1
1
2
2
3
3
24 60 12
24
0, 4
1 0, 5 0, 5 9 9 20 20
60
24 0, 4 1
24, 4
60 0, 4 0, 5
59,8
12 0, 4 0, 5 12, 2
w
w
w
V
w
V
w
E
w
E
E
E
I
A
R
R
R
R
R
R
R
U
E
I R
V
U
E
I R
V
U
E
I R
V
− +
−
− +
−
=
=
=
=
+
+
+ +
+
+
+
+
+ + +
+
=
+ ⋅
=
+
⋅ =
=
− ⋅
=
−
⋅
=
=
+ ⋅
= +
⋅
=
___________________________________________________________________________
1.97
Dane:
Szukane:
Wzory:
1
2
3
1
2
4
3
2
3
1
24
60
12
9
20
0, 5
1
W
W
W
E
V
E
V
E
V
R
R
R
R
R
R
R
=
=
=
=
= Ω
=
=
Ω
=
=
Ω
= Ω
W tym wydaniu tego Zbioru
Zadań jest małe przeoczenie,
zapomnieli podać R2, w
starszym wydaniu było
podane co teraz jest
uwzględnione w powyższych
danych
Wykres potencjałów
U
R
I
=
w
U
E
I R
= − ⋅
1
2
3
1
2
3
1
2
3
4
24 60 12
24
0, 4
1 0, 5 0, 5 9 9 20 20
60
w
w
w
E
E
E
I
A
R
R
R
R
R
R
R
− +
−
− +
−
=
=
=
=
+
+
+ +
+
+
+
+
+ + +
+
prąd w obwodzie płynie zgodnie z ruchem wskazówek zegara.
Dlatego przyjąłem ten sam kierunek rozpatrywania potencjałów.
.
4
1
1
1
2
2
2
3
3
3
0 0, 4 20
8
8 24 0, 4 1
32, 4
32, 4 0, 4 9
36
36 60 0, 4 0, 5
23,8
23,8 0, 4 9
20, 2
20, 2 0, 4 20 12, 2
12, 2 12 0, 4
F
O
E
F
w
D
E
C
D
W
B
C
A
B
O
A
w
V
V
I R
V
V
V
E
I R
V
V
V
I R
V
V
V
E
I R
V
V
V
I R
V
V
V
I R
V
V
V
E
I R
=
− ⋅
= −
⋅
= −
=
− − ⋅
= − −
−
⋅ = −
=
− ⋅ = −
−
⋅ = −
=
+
− ⋅
= − +
−
⋅
=
=
− ⋅
=
−
⋅ =
=
− ⋅
=
−
⋅
=
=
−
− ⋅
=
− −
⋅
0, 5
0 V
=
-40
-30
-20
-10
0
10
20
30
0
10
20
30
40
50
60
70
R [
Ω
]
V [V]
Wykres potencjałów
R4
R1
R2
R3
___________________________________________________________________________
1.98
Dane:
Szukane:
Wzory:
Jak zmienią się wskazania
woltomierzy, po otwarciu
wyłącznika ‘w’
Po otwarciu wyłącznika w obwodzie prąd nie będzie płynął i woltomierze będą s.em. ‘E’.
Czyli wskazanie
2
V zwiększy się, a
1
V i
3
V zmniejszy się.
___________________________________________________________________________
1.99
Dane:
Szukane:
Wzory:
1
2
3
12
24
12
16
U
V
R
R
R
=
=
Ω
=
Ω
=
Ω
1 3
1 3
U
I
−
−
=
=
U
R
I
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
1
2
1
2
3
3
3
3
3
3
1
2
1
1
1
2
2
2
24 12
8
24 12
8 16
24
12
0, 5
24
0, 5 16
8
12 8
4
4
0,1667
24
4
0, 3333
12
z
z
z
z
R R
R
R
R
R
R
R
U
I
A
R
U
I
R
V
U
U
U
V
U
U
U
U
I
A
R
U
I
A
R
⋅
⋅
=
=
= Ω
+
+
=
+
= + =
Ω
=
=
=
= ⋅
=
⋅ =
= −
= − =
=
=
=
=
=
=
=
=
___________________________________________________________________________
1.100
Dane:
Szukane:
Wzory:
2
3
z
R
R
= Ω
= Ω
Jak połączyć
U
R
I
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
1
1
1
2
2
1 2
3
z
R R
R R
R
R
R
R
R
R
R
R
⋅
⋅
=
=
= = Ω
+
= + = + = Ω
Odp. Dwa równolegle z jednym szeregowo.
___________________________________________________________________________
1.101
Dane:
Szukane:
Wzory:
U
R
I
=
1
2
3
4, 4
22
15
14,8
V
V
U
V
R
R
R
R
=
=
Ω
=
Ω
=
Ω
= ∞
1
2
3
O
O
U
I
I
I
I
=
=
=
=
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
=
+
1
1
1
2
2
2
2
3
1
2
3
3
3
3
3
1
4, 4
0, 2
22
0, 2 15
3
4, 4 3
7, 4
7, 4
0, 5
14,8
0, 5 0, 2
0, 7
V
O
O
U
I
A
R
I
I
U
I
R
V
U
U
U
V
U
U
U
I
A
R
I
I
I
A
=
=
=
=
= ⋅
=
⋅ =
=
+
=
+ =
=
=
=
=
= + =
+
=
___________________________________________________________________________
1.102
Dane:
Szukane:
Wzory:
44
AB
R
=
Ω
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
=
+
2
1
1
1
1
44
(
)
(
)
44(
)
2
132
66
AB
R
R
R
R
R
R
R
R
R R
R
R R
R
R
R
R
R
R
=
+
+
+ +
=
+
+
=
+ +
=
=
Ω
___________________________________________________________________________
1.103
Dane:
Szukane:
Wzory:
1
2
40
15
60
Z
R
R
R
=
Ω
=
Ω
=
Ω
X
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
=
+
1
2
1
1
1
1
1
1
40
15
60
1
1
1
40
60
15
3 2
1
120
15
1
1
120
15
15 120
120 15 105
Z
X
X
X
X
X
X
X
R
R
R
R
R
R
R
R
R
R
=
+
+
=
+
+
−
=
+
− =
+
=
+
+ =
=
− =
Ω
___________________________________________________________________________
1.104
Dane:
Szukane:
Wzory:
R
X
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
=
+
a)
4
X
R
R
=
b)
0, 25
4
X
R
R
R
= =
c)
1
1
1
X
R
R
R
R
R
=
+
+
+
1
1
1
2
2
1
2
2
X
X
X
R
R
R
R
R
R
R
=
+
=
=
d)
2
2
X
R
R
R
R
= + =
e)
3
4
3
3
3
X
R
R
R
R
R
R
+
= + =
=
f)
1
1
1
3
X
R
R
R
=
+
1
1
3
3
3
1
4
3
3
0, 75
4
X
X
X
R
R
R
R
R
R
R
R
=
+
=
=
=
g)
5
2
2, 5
2
2
2
X
R
R
R
R
R
R
R
R
= + + =
+ =
=
h)
1
1
1
1
2
X
R
R
R
R
=
+ +
1
1
2
2
2
2
2
1
5
2
2
0, 4
5
X
X
X
R
R
R
R
R
R
R
R
R
=
+
+
=
=
=
i)
X
Z
R
R
R
= +
___________________________________________________________________________
1.105
Dane:
Szukane:
Wzory:
1
2
3
3
2
4
3
R
R
R
I
A
= Ω
= Ω
= Ω
=
1
2
U
U
=
=
I
U
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
=
+
1
1
3 3
9
U
I R
V
= ⋅ = ⋅ =
2
3
2
3
2
2
3
2 4
4
6
3
4
3
4
3
X
X
R R
R
R
R
U
I R
V
U
U
⋅
⋅
=
=
= Ω
+
= ⋅
= ⋅ =
=
___________________________________________________________________________
1.106
Dane:
Szukane:
Wzory:
20
b
R
U
=
=
I
U
R
=
1
2
1
1
1
R
R
R
=
+
2
18
82
20
a
a
I
mA
R
R
I
mA
=
=
Ω
=
Ω
=
1
2
R
R
R
= +
20
(
)
2 (18 82)
200
0, 2
20 2 18
0, 018
0, 2
11,11
0, 018
a
a
a
a
a
b
b
a
a
b
b
U
I
R
R
mV
V
U
U
I
I
I
I
I
I
mA
A
U
R
I
= ⋅
+
= ⋅
+
=
=
=
= +
= − =
− =
=
=
=
=
Ω
___________________________________________________________________________
1.107
Dane:
Szukane:
Wzory:
2
1
2
3
4
2
3
18
3
6
I
A
R
R
R
R
=
= Ω
= Ω
= Ω
= Ω
Z
O
R
U
=
=
I
U
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
2
2
2
34
3
4
2
34
34
2
34
1
1
1
2
2 18
36
3 6
9
36
4
9
2 4
6
6 3 18
18 36
54
54
9
6
z
U
I
R
V
R
R
R
U
I
A
R
I
I
I
A
U
I R
V
U
U
U
V
U
R
I
= ⋅
= ⋅ =
=
+
= + = Ω
=
=
=
= +
= + =
= ⋅ = ⋅ =
=
+
= +
=
=
=
= Ω
___________________________________________________________________________
1.108
Dane:
Szukane:
Wzory:
R
AB
AC
R
R
=
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
a)
rezystancja zastępcza na lewo od R podłączonego do pkt. A B
2
2
2
3
2
5
2
3
3
L
R
R
R
R
R
R
R
R
R
R
⋅
+
= +
=
=
+
po prawej taki sam układ
5
3
P
L
R
R
R
=
=
po takim uproszczeniu zostało tylko połączenie równoległe
1
1
1
1
1
1
1
1
5
5
3
3
1
3
5
3
5
5
5
1
11
5
5
11
AB
L
P
AB
AB
AB
AB
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
=
+ +
=
+ +
=
+
+
=
=
b)
do pewnego momentu jest to samo, czyli:
2
2
2
3
2
5
2
3
3
L
R
R
R
R
R
R
R
R
R
R
⋅
+
= +
=
=
+
1
1
2
1
1
1
1
1
5
3
1
3
5
8
5
5
5
5
8
X
L
X
X
R
R
R
R
R
R
R
R
R
R
R
=
+ =
+
=
+
=
=
2
1
5
13
8
8
X
X
R
R
R
R
R
R
=
+ =
+ =
1
1
1
1
13
2
8
1
8
13
1
13
13
2
1
16
26
13
26
26
26
1
55
26
26
55
O
O
O
O
O
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
=
+ +
=
+
+
=
+
+
=
=
___________________________________________________________________________
1.109
Dane:
Szukane:
Wzory:
1
2
3
4
5
5
12
6
3
4
15
R
R
R
R
R
I
A
= Ω
=
Ω
= Ω
= Ω
= Ω
=
1 5
1 5
O
O
R
U
I
U
−
−
=
=
=
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
4
3
1
4
3
2
1
5
2
2
3
2
2
3
1
18
2
9
2 4
6
6 12
72
4
6 12
18
4 5
9
Z
Z
Z
Z
Z
Z
O
Z
R R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
⋅
=
=
= Ω
+
=
+
= + = Ω
⋅
⋅
=
=
=
= Ω
+
+
=
+ = + = Ω
1
1
1
2
1
2
2
2
5
2
34
5
1
34
4
4
34
3
3
15 9 135
15
15 5
75
135 75
60
60
5
12
15 5 10
10 2
20
20
6, 667
3
20
3, 333
6
O
O
O
Z
U
I R
V
I
I
A
U
I R
V
U
U
U
V
U
I
A
R
I
I
I
A
U
I
R
V
U
I
A
R
U
I
A
R
= ⋅
= ⋅ =
= =
= ⋅ = ⋅ =
=
−
=
−
=
=
=
=
= − = − =
= ⋅
= ⋅ =
=
=
=
=
=
=
___________________________________________________________________________
1.110
Dane:
Szukane:
Wzory:
1
4
2
3
6
5
5
2
4
12
120
R
R
R
R
R
R
U
V
=
= Ω
=
=
= Ω
=
Ω
=
1 5
O
I
U
−
=
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
5
5
5
5
6
6
4
5
6
2
1
3
2
4
2
2
1
1
3
4
4
4
5
4
2
120
10
12
120
30
4
10 30
40
1
1
1
1
1
1
5
4
2 4
12
12
2, 4
5
40 2, 4
96
96
24
4
96
16
6
40 2
80
120 80 96
296
O
U
I
A
R
U
I
A
R
I
I
I
A
R
R
R
R
R
R
U
I
R
V
U
I
A
R
U
I
A
R
R
U
I
R
V
U
U
U
U
V
=
=
=
=
=
=
= + = +
=
=
+
+
= +
=
+
=
Ω =
Ω
= ⋅ =
⋅
=
=
=
=
=
=
=
+
= ⋅
=
⋅ =
=
+
+
=
+ +
=
___________________________________________________________________________
1.111
Dane:
Szukane:
Wzory:
1
2
3
4
5
6
6
1
5
12
12
21
4
36
R
R
R
R
R
R
U
V
= Ω
= Ω
=
Ω
=
Ω
=
Ω
= Ω
=
O
O
I
U
=
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
6
6
6
6
3
3
36
3
6
4
1
2
4
36
5
6
4
5
5
5
5
5
36
36
9
4
36
3
12
3 9 12
1
1
1
1
1
1
3
12
1 5
12
12
4
3
12 4
48
36 48
84
84
4
21
4 12 16
O
O
U
I
A
R
U
I
A
R
I
I
I
A
R
R
R
R
R
R
U
I
R
V
U
U
U
V
U
U
U
I
A
R
I
I
I
A
=
=
=
=
=
=
= + = + =
=
+
+
=
+
=
+
=
Ω = Ω
=
⋅ = ⋅ =
=
+
=
+
=
=
=
=
=
= +
= + =
___________________________________________________________________________
1.112
Dane:
Szukane:
Wzory:
1
2
3
15
12
30
150
600
a
a
I
mA
R
I
mA
I
mA
I
mA
=
=
Ω
=
=
=
1
2
3
R
R
R
=
=
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
1
30
30
1
1
2
3
30
2
150
150
2
3
150
3
1
2
3
600
600
3
15 12 180
0,18
30 15 15
0, 015
0,18
12
0, 015
150 15 135
0,135
0,18 0, 015
0,135 (
)
600
a
a
a
a
b
b
a
a
b
a
b
b
a
a
R
b
a
b
b
a
U
I
R
mV
V
I
I
I
I
I
I
mA
A
U
R
R
R
I
I
I
I
I
I
I
mA
A
U
U
U
R
R
R
I
I
I
I
I
I
= ⋅
= ⋅ =
=
= +
= − =
− =
=
+
+
=
=
=
Ω
= +
= − =
− =
=
+
=
+
⋅
=
⋅
+
= +
= − =
3
2
150
3
2
1
15
585
0, 585
0,18 0, 015
0, 015
0, 585
a
R
R
b
mA
A
U
U
U
U
R
R
R
− =
=
+
+
=
+
⋅ +
⋅
=
⋅
1
2
3
3
1
2
3
2
1
3
1
2
1
2
1
2
1
2
2
1
1
2
12
0,18 0, 015
0,135 (
)
0,18 0, 015
0, 015
0, 585
12 (
)
0,18 0, 015 (12 (
))
0,135 (
)
0,18 0, 015 (12 (
)) 0, 015
0, 585
0,18 0,18
0,135 (
) 0, 015 (
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
+
+
=
+
⋅
=
⋅
+
+
⋅ +
⋅
=
⋅
= −
+
+
⋅
−
+
=
⋅
+
+
⋅
−
+
+
⋅
=
⋅
+
=
⋅
+
+
⋅
1
2
2
1
1
2
1
2
1
1
2
2
1
2
1
1
2
1
2
)
0,18 0,18 0, 015
0, 585
0, 015 (
)
0, 36
0,15
0,15
0, 36
0, 585
0, 015
0, 015
0, 015
0, 36 0,15
0,15
0, 36
0, 6
0, 36 0,15
0,15
0, 6
0, 36 0,15 0, 6
0
0,15
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
+
+
+
⋅
=
⋅ +
⋅
+
=
⋅ +
⋅
=
⋅ +
⋅ +
⋅ −
⋅
−
⋅
=
=
⋅
−
⋅
=
=
Ω
−
⋅
=
=
3
1
2
, 27
1,8
0,15
12 (
) 12 (0, 6 1,8)
9, 6
R
R
R
=
Ω
= −
+
= −
+
=
Ω
A może ? innym podejściem ?
1
30
30
1
2
3
1
1
2
3
1
2
3
2
150
3
2
1
2
3
1
2
3
600
3
2
3
1
3
2
1
15 12 180
0,18
0,18
0, 015
0,18 0, 015
0,135
(
)
0,18 0, 015 (
)
0, 585
0, 0
a
a
a
b
a
a
b
a
a
b
a
a
a
a
b
a
a
a
a
U
I
R
mV
V
I
I
I
U
I
R
R
R
U
I
I
R
R
R
R
R
R
I
I
I
U
I
R
I
I
R
R
R
R
R
I
I
I
U
I
R
R
I
I
R
R
R
R
= ⋅
= ⋅ =
=
=
+
=
+
+
− =
+
+
=
+
+
=
+
+ ⋅
− =
+
+
⋅
=
+
=
+
+ ⋅
+
− =
+
⋅
+
=
1
2
3
3
1
2
3
2
1
3
1
2
1
2
1
2
1
2
2
1
1
2
1
0,18
15
0,18 0, 015
0,135
0,18 0, 015 (
)
0, 585
12
0,18 0, 015 (12
)
0,135
0,18 0, 015 (12
) 0, 015
0, 585
0,135
0,135
0,18 0,18 0, 015
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
=
+
+
+
⋅
=
+
+
⋅
+
=
= − −
+
⋅
− −
=
+
+
⋅
− −
+
⋅
=
⋅ +
⋅
=
+
−
⋅
2
1
1
2
2
1
2
1
2
1
1
1
2
1
0, 015
0, 585
0,18 0,18 0, 015
0, 015
0, 015
0,135
0,135
0, 36 0, 015
0, 015
0, 585
0, 36 0, 015
0,15
0,15
0, 36
0, 6
0, 36
R
R
R
R
R
R
R
R
R
R
R
R
R
R
−
⋅
⋅ =
+
−
⋅ −
⋅ +
⋅
⋅ +
⋅
=
−
⋅ −
⋅
⋅ =
−
⋅
⋅ +
⋅
=
⋅ =
1
0, 36
0, 6
0, 6
R
=
=
Ω
2
2
2
0,15 0, 6 0,15
0, 36
0,15
0, 36 0, 09
0, 27
1,8
0,15
R
R
R
⋅
+
⋅
=
⋅
=
−
=
=
Ω
3
12 0, 6 1,8
9, 6
R
= −
−
=
Ω
Eee, nic szybciej nie poszło. Dobrze że chociaż wynik tak sam wyszedł ☺.
___________________________________________________________________________
1.113
Dane:
Szukane:
Wzory:
120
30
U
V
R
=
=
Ω
1
2
AC
FB
EB
I
I
R
U
U
=
=
=
=
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
=
+
1
2
1
1
1
2
4
2
4
8
8
40
4
2
6
6
120
2
2
60
120
1
4
120
AC
AC
R
R
R
R
R
R R
R
R
R
R
R
U
I
A
R
U
I
A
R
=
+
⋅
⋅
=
=
=
=
Ω
+
=
=
=
=
=
=
1
2
2
2 30
60
3
1 3 30
90
30
2
1 2 30
60
60 60
0
CB
CF
FB
CF
CB
CE
EB
CE
CB
U
I R
V
U
I
R
V
U
U
U
V
U
I
R
V
U
U
U
V
= ⋅ = ⋅
=
= ⋅
= ⋅ ⋅
=
=
−
=
= ⋅
= ⋅ ⋅
=
=
−
=
−
=
___________________________________________________________________________
1.114
Dane:
Szukane:
Wzory:
1
2
3
4
5
I
I
I
I
I
=
=
=
=
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
=
+
1
6
2
3
4
5
120
45
220
76
240
360
U
V
R
R
R
R
R
R
=
=
=
Ω
=
Ω
=
Ω
=
Ω
=
Ω
4
5
1
4
5
2
1
3
1
2
3
1
2
1
3
6
1
6
1
2
6
2
1
6
1
1
2
2
240 360
86400
144
600
600
144 76
220
220 220
110
220 220
45 110 45
200
120
0, 6
200
120 0, 6 45 0, 6 45
66
Z
Z
Z
Z
Z
Z
Z
R R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
U
I
I
A
R
U
U
U
U
U
U
U
U
U
I R
I
R
V
I
⋅
⋅
=
=
=
=
Ω
+
=
+
=
+
=
Ω
⋅
⋅
=
=
=
Ω
+
+
= +
+
=
+
+
=
Ω
= =
=
=
=
+
+
= −
−
= − ⋅ − ⋅
=
−
⋅ −
⋅
=
2
2
2
1
2
3
3
1
2
3
3
3
2
45
3
45
2
3
45
4
4
45
5
5
66
0, 3
220
0, 6 0, 3
0, 3
0, 3 76
22,8
66 22,8
43, 2
43, 2
0,18
240
43, 2
0,12
360
U
A
R
I
I
I
I
I
I
A
U
I
R
V
U
U
U
U
U
U
V
U
I
A
R
U
I
A
R
=
=
=
= +
= + =
−
=
= ⋅
=
⋅
=
=
+
=
−
=
−
=
=
=
=
=
=
=
___________________________________________________________________________
1.115
Dane:
Szukane:
Wzory:
1
2
4
3
5
6
44
40
120
20
35
45
U
V
R
R
R
R
R
R
=
=
Ω
=
=
Ω
=
Ω
=
Ω
=
Ω
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
1
2
3
4
5
6
I
I
I
I
I
I
I
=
=
=
=
=
=
=
1
1
3
1
2
2
1
2
3
5
6
3
4
4
3
4
5
4
2
5
4
4
4
4
4
4
5
6
3
40 20
60
60 120
40
60 120
35 45
80
80 120
48
80 120
40 48
88
44
0, 5
88
0, 5 48
24
24
0, 2
120
24
0,
80
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
U
I
A
R
U
I R
V
U
I
A
R
U
I
I
R
= +
=
+
=
Ω
⋅
⋅
=
=
=
Ω
+
+
=
+
=
+
=
Ω
⋅
⋅
=
=
=
Ω
+
+
=
+
=
+
=
Ω
=
=
=
= ⋅
=
⋅
=
=
=
=
= =
=
=
2
2
2
2
2
2
1
3
1
3
0, 5 40
20
20
0,1667
120
20
0, 3333
60
Z
Z
Z
Z
Z
A
U
I R
V
U
I
A
R
U
I
I
A
R
= ⋅
=
⋅
=
=
=
=
= =
=
=
___________________________________________________________________________
1.116
Dane:
Szukane:
Wzory:
1
2
5
7
8
3
4
6
3
4
12
6
12
R
R
R
R
R
R
R
R
= Ω
=
=
=
= Ω
=
Ω
= Ω
=
Ω
AE
CF
R
R
=
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
1
4
1
4
2
5
2
5
6
6
3
8
3
8
7
7
3 6
2
3 6
4 4
2
4 4
2 2
4
4 12
3
4 12
12 4
3
4 12
3 3
6
6 4
2, 4
4 6
AB
BC
AC
AB
BC
AC
AD
AC
DE
FE
AD
DE
FE
AE
FE
R R
R
R
R
R R
R
R
R
R
R
R
R
R
R
R
R
R R
R
R
R
R
R
R
R
R
R
R
R
⋅
⋅
=
=
= Ω
+
+
⋅
⋅
=
=
= Ω
+
+
=
+
= + = Ω
⋅
⋅
=
=
= Ω
+
+
⋅
⋅
=
=
= Ω
+
+
=
+
= + = Ω
⋅
⋅
=
=
=
Ω
+
+
1
4
1
4
2
5
2
5
6
6
3
8
3
8
7
3 6
2
3 6
4 4
2
4 4
2 2
4
4 12
3
4 12
12 4
3
4 12
4 3
7
7 3
2,1
7 3
AB
BC
AC
AB
BC
AC
AD
AC
DE
DF
DE
DF
AD
CF
DF
AD
R R
R
R
R
R R
R
R
R
R
R
R
R
R
R
R
R
R R
R
R
R
R
R
R
R
R
R
R
R
⋅
⋅
=
=
= Ω
+
+
⋅
⋅
=
=
= Ω
+
+
=
+
= + = Ω
⋅
⋅
=
=
= Ω
+
+
⋅
⋅
=
=
= Ω
+
+
=
+
= + = Ω
⋅
⋅
=
=
=
Ω
+
+
___________________________________________________________________________
1.117
Dane:
Szukane:
Wzory:
1
2
5
7
8
3
4
6
24
3
4
12
6
12
AE
U
V
R
R
R
R
R
R
R
R
=
= Ω
=
=
=
= Ω
=
Ω
= Ω
=
Ω
5
8
I
I
=
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
1
4
1
4
2
5
2
5
6
6
3
8
3
8
7
7
3 6
2
3 6
4 4
2
4 4
2 2
4
4 12
3
4 12
12 4
3
4 12
3 3
6
6 4
2, 4
4 6
AB
BC
AC
AB
BC
AC
AD
AC
DE
FE
AD
DE
FE
AE
FE
R R
R
R
R
R R
R
R
R
R
R
R
R
R
R
R
R
R R
R
R
R
R
R
R
R
R
R
R
R
⋅
⋅
=
=
= Ω
+
+
⋅
⋅
=
=
= Ω
+
+
=
+
= + = Ω
⋅
⋅
=
=
= Ω
+
+
⋅
⋅
=
=
= Ω
+
+
=
+
= + = Ω
⋅
⋅
=
=
=
Ω
+
+
7
7
7
7
8
8
5
5
24
10
2, 4
24
6
4
10 6
4
4 3 12
12
3
4
24 12 12
12
3
4
3 2
6
6
1, 5
4
FE
FE
FE
AC
FE
AE
R
R
DE
R
DE
DE
AD
DE
AD
DE
DE
R
AC
BC
R
BC
BC
U
I
A
R
U
I
A
R
I
I
I
I
I
I
A
U
I
R
V
U
I
A
R
U
U
U
U
U
U
V
U
I
A
R
U
I
R
V
U
I
A
R
=
=
=
=
=
=
= +
= − = − =
=
⋅
= ⋅ =
=
=
=
=
+
= −
=
− =
=
=
=
=
⋅
= ⋅ =
=
= =
___________________________________________________________________________
1.118
Dane:
Szukane:
Wzory:
5
1
2
5
7
8
3
4
6
1
3
4
12
6
12
I
A
R
R
R
R
R
R
R
R
=
= Ω
=
=
=
= Ω
=
Ω
= Ω
=
Ω
CE
U
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
1
4
1
4
2
5
2
5
6
6
3
8
3
8
7
7
3 6
2
3 6
4 4
2
4 4
2 2
4
4 12
3
4 12
12 4
3
4 12
3 3
6
6 4
2, 4
4 6
AB
BC
AC
AB
BC
AC
AD
AC
DE
FE
AD
DE
FE
AE
FE
R R
R
R
R
R R
R
R
R
R
R
R
R
R
R
R
R
R R
R
R
R
R
R
R
R
R
R
R
R
⋅
⋅
=
=
= Ω
+
+
⋅
⋅
=
=
= Ω
+
+
=
+
= + = Ω
⋅
⋅
=
=
= Ω
+
+
⋅
⋅
=
=
= Ω
+
+
=
+
= + = Ω
⋅
⋅
=
=
=
Ω
+
+
5
5
5
5
2
2
5
2
5
6
6
6
7
7
7
1 4
4
4
1
4
1 1
2
2 2
4
4 4
8
8
2
0, 667
12
3
2
8
2
2, 667
3
3
8
2
4 10
3
3
2
2
8 10
18
3
3
AB
AB
AB
AB
AC
AB
AC
AD
AB
AD
EC
AC
U
I
R
V
U
I
A
R
I
I
I
A
U
I
R
V
U
U
U
V
U
I
A
A
R
I
I
I
A
A
U
I
R
V
U
U
U
V
= ⋅
= ⋅ =
=
= =
= + = + =
=
⋅
= ⋅ =
=
+
= + =
=
=
=
=
=
+ = + =
=
=
⋅
= ⋅ =
=
+
= +
=
___________________________________________________________________________
1.119
Dane:
Szukane:
Wzory:
5
1
2
3
4
5
2
2
3
14
30
25
50
1
o
w
I
A
R
R
R
R
R
R
R
=
= Ω
= Ω
=
Ω
=
Ω
=
Ω
=
Ω
= Ω
z
U
E
I
R
=
=
=
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
45
5
5
45
4
4
3
4
5
3
3
3
2
3
45
2
2
2
2
3
1
1
2
1
2 50 100
100
4
25
4 2
6
6 30 180
180 100
280
280
20
14
20 6
26
26 2
52
26 3
78
52 280 78
410
410 26 1
436
o
o
o
w
z
U
I R
V
U
I
A
R
I
I
I
A
U
I R
V
U
U
U
V
U
I
A
R
I
I
I
A
U
IR
V
U
IR
V
U
U
U
U
V
E
U
IR
V
R
=
= ⋅
=
=
=
=
= + = + =
=
= ⋅
=
=
+
=
+
=
=
=
=
= + =
+ =
=
=
⋅ =
=
=
⋅ =
=
+
+
=
+
+
=
= +
=
+ ⋅ =
410
15, 77
26
U
I
=
=
=
Ω
___________________________________________________________________________
1.120
Dane:
Szukane:
Wzory:
1
2
3
200
100
300
U
V
R
R
R
=
=
Ω
=
=
Ω
2
2
Z
O
U
U
=
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
1
2
2
2
200
200
0, 5
100 300
400
0, 5 300 150
O
O
O
U
I
A
R
R
U
I
R
V
=
=
=
=
+
+
= ⋅
=
⋅
=
2
3
2
3
1
2
300 300
150
300 300
200
200
0,8
100 150
250
0,8 150 120
Z
ZW
Z
W
ZW
Z
R R
R
R
R
U
I
A
R
R
U
I
R
V
⋅
⋅
=
=
=
Ω
+
+
=
=
=
=
+
+
=
⋅
=
⋅
=
___________________________________________________________________________
1.121
Dane:
Szukane:
Wzory:
1
2
3
3
2
100
900
1
9, 5
AB
U
V
R
R
R
k
R
k
=
=
Ω
=
Ω
= Ω
=
Ω
U
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
=
+
1
1
2
1
1
2
3
1
2
3
1
2
1
2
1
9,5
9,5
9,5
2
9,5
2
9,5
9,5
9,5
1
9,5
2
2
0, 02
100
(
)
0, 02 (100 900)
20
(
)
1 1
0, 5
(
)
1 1
20 9, 5
380
0, 5
380 20
400
AB
Z
k
k
Z
k
k
k
Z
k
k
k
Z
k
k
k
U
I
A
R
U
I
R
R
V
R
R
R
R
k
R
R
R
U
U
U
I
R
U
I
R
U
U
R
R
U
R
U
V
R
U
U
U
U
U
V
=
=
=
= ⋅
+
=
⋅
+
=
⋅
+
⋅
=
=
=
Ω
+
+
+
=
=
=
=
⋅
⋅
=
=
=
=
+
=
+
=
+
=
___________________________________________________________________________
1.122
Dane:
Szukane:
Wzory:
120
40
80
O
P
d
g
U
V
R
R
R
R
=
=
Ω
=
Ω
=
O
O
U
I
=
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
=
+
2
40
2
40 40
20
40
40
40 20
60
120
2
60
2 20
40
40
1
40
P
d
g
d
d
d
P
d
d
O
Z
d
O
g
Z
O
d
Z
O
O
O
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
U
I
A
R
U
U
I R
V
U
I
A
R
=
+
=
+
=
=
=
Ω
⋅
⋅
=
=
=
Ω
+
+
=
+
=
+
=
Ω
=
=
=
=
= ⋅
= ⋅
=
=
=
=
___________________________________________________________________________
1.123
Dane:
Szukane:
Wzory:
120
40
80
O
P
d
U
V
R
R
R
const
x
=
=
Ω
=
Ω
=
wykresy
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
=
+
x =liniowe położenie suwaka potencjometru, zakres < 0;1 >, tzn. 1 oznacza położenie suwaka
w maksymalnym położeniu czyli 100% zakresu.
a)
Odbiornik podłączony.
2
2
2
2
2
2
2
(1
)
(1
)
(
)
(1
)
(1
) (
)
C
g
d
g
d
d
O
Z
d
O
O
Z
O
O
O
O
C
g
Z
O
O
O
O
O
O
O
C
O
O
O
C
R
R
R
R
x R
R
x R
R
R
R
R
R
x R R
R
x R
R
x R R
x R
x R
R
x R R
R
R
R
x R
x R
R
x R
R
x
x R
R R
x R R
x R
R R
x
R
x R R
x R R
R
x R
R
x R
R
x
R
x R
R R
R
x
=
+
= − ⋅
= ⋅
⋅
=
+
⋅ ⋅
=
⋅ +
⋅ ⋅
− ⋅ ⋅ ⋅ +
+ ⋅ ⋅
=
+
= − ⋅ +
=
⋅ +
⋅ +
− ⋅ ⋅
+ ⋅
+ ⋅ ⋅
⋅
+ ⋅
− ⋅
− ⋅ ⋅
+ ⋅ ⋅
=
=
⋅ +
⋅ +
− ⋅
+ ⋅
+ ⋅
=
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
(
)
(1
)
(
)
(
) (1
)
(1
)
(
) (1
)
(
O
C
O
O
O
g
g
O
O
O
O
O
O
O
O
O
R
R
U
I
R
U
x R
R
I
x
R
x R
R R
U
U
U
U
I R
U
I
x R
U
x R
R
U
x R
R
x R
U
U
x R
U
x
R
x R
R R
x
R
x R
R R
U
x R
R R
x
U
x R
R R
x
R
U
U
U
x
R
x R
R R
⋅ +
=
⋅ ⋅ +
=
− ⋅
+ ⋅
+ ⋅
= −
= − ⋅
= − ⋅ − ⋅
⋅ ⋅ +
⋅ ⋅ +
⋅ − ⋅
= −
⋅ − ⋅ = −
− ⋅
+ ⋅
+ ⋅
− ⋅
+ ⋅
+ ⋅
⋅ ⋅
+ ⋅
⋅ −
⋅ ⋅
+ ⋅
− ⋅
= −
= −
− ⋅
+ ⋅
+ ⋅
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
)
(
)
(
)
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
x R R
x
R
x R
R R
U
x
R
x R
R R
U
x R
R R
x
R
x R R
U
x
R
x R
R R
x
R
x R
R R
x R
R R
x
R
x R R
U
U
x
R
x R
R R
x R R
x R
R
U
U
U
x
R
x R
R R
R
x
R
x R
R
− ⋅ ⋅
− ⋅
+ ⋅
+ ⋅
⋅ − ⋅
+ ⋅
+ ⋅
− ⋅ ⋅
+ ⋅
− ⋅
− ⋅ ⋅
=
− ⋅
+ ⋅
+ ⋅
− ⋅
+ ⋅
+ ⋅
− ⋅
− ⋅
+ ⋅
+ ⋅ ⋅
= ⋅
− ⋅
+ ⋅
+ ⋅
⋅ ⋅
⋅
= ⋅
= ⋅ ⋅
− ⋅
+ ⋅
+ ⋅
− ⋅ + ⋅ +
2
2
2
2
200
40
80
80 40
8000
200
80
80
40
2
2
1
O
O
O
O
x R
x
U
U
x
R
x R
R
x
x
x
x
U
x
x
x
x
⋅
⋅ ⋅
= ⋅
=
− ⋅ + ⋅ +
− ⋅ + ⋅ +
⋅
=
=
−
+
+
−
+
+
b)
Przerwa na odbiorniku, czyli niepodłączony.
(1
)
(1
)
(1
)
(1
)
(1
)
200
g
d
g
d
d
O
g
g
O
O
O
R
R
R
R
x R
R
x R
R
R
x
U
I
R
U
U
U
U
I R
U
I
x R
U
U
x R
U
U
x R
U
U
U
x
U
U
xU
R
R
U
xU
U
x
=
+
= − ⋅
= ⋅
=
=
= −
= − ⋅
= − ⋅ − ⋅
⋅ − ⋅
= − ⋅ − ⋅ = −
= − ⋅ − = − +
=
=
⋅
Uo=f(x)
0
50
100
150
200
0
0,2
0,4
0,6
0,8
1
X
odb.jest
odb. odł
ą
czony
___________________________________________________________________________
1.124
Dane:
Szukane:
Wzory:
300
120
200
100
2
O
P
P
d
g
d
U
V
U
V
R
R
R
R
R
const
x
=
=
=
Ω
=
=
=
Ω
=
O
R
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
(
)
(
)
300 100
120
10000 100
100
30000
120
10000
d
O
Z
d
O
g
Z
O
Z
O
Z
g
Z
d
O
O
d
O
d
O
g
d
O
d
O
O
g
d
O
d
O
d
O
d
O
d
O
d
O
O
g
d
O
d
O
g
d
g
O
d
O
O
O
O
O
R
R
R
R
R
U
I
R
R
U
I R
U
U
R
R
R
R
R
U
U
R
R
R
R
R
R
R
R
R
U
U
R R
R
R
R
R
R
R
R
U R
R
U R
R
U
R R
R
R
R
R R
R R
R
R
R
R
R
R
⋅
=
+
=
+
= ⋅
=
⋅
+
⋅
=
⋅
⋅
+
+
+
⋅
=
⋅
+
+
⋅
+
+
⋅
⋅
⋅ ⋅
=
=
+
+
⋅
+
+
⋅
⋅
⋅
=
+
+
⋅
⋅
=
200
120 (100 2
)
300
12000 240
300
12000
300
240
12000
200
60
O
O
O
O
O
O
O
O
R
R
R
R
R
R
R
R
+
⋅
+
=
⋅
+
=
⋅
=
⋅
−
=
=
Ω
___________________________________________________________________________
1.125
Dane:
Szukane:
Wzory:
100
24
P
AB
R
R
=
Ω
=
Ω
d
g
R
R
=
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
2
2
2
2
1
2
(
)
0
100
2400
0
4
10000 4 2400
400
100 20
40
2
2
100 20
60
2
2
10
P
d
g
g
P
d
d
g
AB
d
g
d
P
d
AB
d
P
d
d
P
d
d
AB
P
AB
P
d
P
d
d
d
P
AB
P
d
d
d
d
g
P
d
g
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
b
ac
b
R
a
b
R
a
R
R
R
R
=
+
=
−
⋅
=
+
⋅
−
=
+
−
⋅
−
⋅
=
⋅
=
⋅
−
−
⋅
+
⋅
=
−
+
=
∆ = −
=
− ⋅
=
− − ∆
−
=
=
=
− + ∆
+
=
=
=
Ω
=
−
=
0 60
40
−
=
Ω
___________________________________________________________________________
1.126
Dane:
Szukane:
Wzory:
P
R
l
=
=
( )
AB
R
f y
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
2
2
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
P
P
P
d
P
P
d
g
g
P
d
g
P
d
g
AB
d
g
P
P
AB
P
P
P
P
AB
P
P
AB
P
AB
P
AB
R
l
l
R
R
l
y
R
R
l
R
R
R
R
R
R
l
y
R
R
l
R
R
R
R
R
y
l
y
R
R
l
l
R
y
l
y
R
R
l
l
R
R
y
l
y
l
l
R
R
y l
y
l
R
y
l
y
l
R
y l
y
R
y l
y
R
l
l
R
y
R
y
l
l
=
⋅
=
⋅
=
+
=
−
−
=
⋅
⋅
=
+
−
⋅ ⋅
⋅
=
−
⋅ +
⋅
⋅ ⋅
⋅ −
=
⋅ + −
⋅
⋅ −
=
+ −
⋅ −
=
⋅
=
⋅ −
∼
( )
AB
R
f y
=
AB
R to wartość funkcji , y to argument
jest to parabola, taka odwrócona z maksimum przy
2
l
y
=
AB
R > 0 to
0 ;
y
l
∈
〈
〉
___________________________________________________________________________
1.127
Dane:
Szukane:
Wzory:
1
2
4
3
6
1
6
3
AC
U
V
R
R
R
R
=
= Ω
=
= Ω
= Ω
BD
U
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
1
1
3
2
2
4
1
3
2
4
6
6
1, 5
1 3
4
6
6
0, 5
6 6
12
1, 5 3
4, 5
0, 5 6
3
4, 5 3 1, 5
AC
AC
CD
CB
BD
CD
CB
U
I
A
R
R
U
I
A
R
R
U
I R
V
U
I
R
V
U
U
U
V
=
=
= =
+
+
=
=
=
=
+
+
= ⋅
=
⋅ =
= ⋅
=
⋅ =
=
−
=
− =
___________________________________________________________________________
1.128
Dane:
Szukane:
Wzory:
1
2
4
3
3
1
6
3
BD
B
D
U
V
V
V
R
R
R
R
=
−
=
= Ω
=
= Ω
= Ω
AB
U
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
Uziemiamy pkt D,
zaznaczamy spadki napięcia na rezystorach (UWAGA na kierunki prądów),
przesuwamy się w jedną stronę np.: D_C_B
1
3
1
2
4
2
1
3
2
4
1
2
3
3
6
3
3
6
BD
CD
BC
CD
BC
CD
BC
BD
U
U
U
U
U
U
I R
I
U
I
R
I
U
I R
I
R
I
I
= −
+
= −
+
= ⋅
= ⋅
= ⋅
= ⋅
= ⋅ − ⋅
= − ⋅ + ⋅
potem w drugą: D_A_B
1
1
1
2
2
2
1
1
2
2
1
2
3
1
6
3
3
1
6
BD
AD
AB
AD
AB
AD
AB
U
U
U
U
U
U
I R
I
U
I
R
I
I R
I
R
I
I
=
−
=
−
= ⋅ = ⋅
= ⋅
= ⋅
= ⋅ − ⋅
= ⋅ − ⋅
1
2
1
2
1
2
2
2
2
2
2
2
1
3
3
6
3
1
6
3
6
3
(3
6) 3
6
3
9
18
6
12
12
1
3 ( 1) 6
3 6
3
I
I
I
I
I
I
I
I
I
I
I
I
A
I
A
= − ⋅ + ⋅
= ⋅ − ⋅
= + ⋅
= − + ⋅ ⋅ + ⋅
= − − ⋅ + ⋅
⋅ = −
= −
= + − ⋅ = − = −
1
1
3
(
)
3 (1 3)
12
AC
U
I R
R
V
=
+
= − ⋅ + = −
napięcie wyszło ujemne dlatego że jest odwrotna biegunowość w stosunku do założenia.
Polega to na przyjęciu odpowiedniej biegunowości
BD
U
, bo w treści nie ma o tym mowy.
gdyby w tym rozwiązaniu przyjąć że
BD
U
na potencjał odwrotny tzn.
3
BD
B
D
U
V
V
V
=
−
= −
,
lub przyjąć kierunek prądów na odwrotny to wynik byłby dodatni . Nie zmienia to faktu, że i
ten wyliczony jest dobry (patrz rozwiązany w Zbiorze Zadań przykład 1.127).
Lub troszkę innym spojrzeniem.
Dla niektórych może prostszy sposób , a dla innych może być bardziej nieczytelnym.
Rysujemy koło rezystorów wektory spadków napięć. Jeżeli przy lewym dolnym jest dłuższy
to prawy dolny krótszy. A u góry na odwrót ( lewy górny krótszy , prawy dłuższy) . Teraz
równania: od dolnego dłuższego/większego odejmujemy dolny krótszy i to równa się
naszemu danemu ‘U’.
U góry tak samo od dłuższego/większego odejmujemy krótszy i to równa się też podanemu
‘U’. Oczywiście wektor to odpowiedni prąd razy właściwy rezystor.
___________________________________________________________________________
1.129
Dane:
Szukane:
Wzory:
1
2
4
3
6
1
6
3
AC
U
V
R
R
R
R
=
= Ω
=
= Ω
= Ω
A
B
I
I
=
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
A) otwarty
1
3
2
4
1
1
1
1
1
1
1 3
6 6
1
1
1
4
12
1
3
1
12
12
12
3
4
6
2
3
R
R
R
R
R
R
R
R
R
U
I
A
R
=
+
+
+
=
+
+
+
= +
=
+
=
= Ω
=
= =
B) zamknięty
3
4
1
2
1
2
3
4
1 6
3 6
6
18
2,857
1 6
3 6
7
9
6
2,1
2,857
R R
R R
R
R
R
R
R
U
I
A
R
⋅
⋅
⋅
⋅
=
+
=
+
= +
=
Ω
+
+
+
+
=
=
=
___________________________________________________________________________
1.130
Dane:
Szukane:
Wzory:
1
2
3
4
36
2
3
6
4
AC
U
V
R
R
R
R
=
= Ω
= Ω
= Ω
= Ω
BD
I
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
3
4
1
2
1
2
3
4
1
1
1
1
2
2
2
2
1
1
1
2
2
1
1
2
1
2
2
2
2
2
2
2
1
2 3
6 4
6
24
3, 6
2 3
6 4
5
10
36
10
3, 6
3
2
10
10
3
(10
) 2
3
20 2
5
20
4
10 4
6
R R
R R
R
R
R
R
R
U
I
A
R
U
I R
U
I
R
I
R
I R
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
A
I
A
⋅
⋅
⋅
⋅
=
+
=
+
= +
=
Ω
+
+
+
+
=
=
=
= ⋅
= ⋅
⋅
= ⋅
= +
⋅ = ⋅
= +
= −
⋅ =
−
⋅
⋅ =
− ⋅
⋅ =
=
= − =
2
3
3
2
4
4
3
3
4
4
3
4
3
4
3
4
3
4
4
4
4
4
4
3
6
4
10
10
(10
) 6
4
60 6
4
6
10 6
4
U
I
R
U
I
R
I
R
I
R
I
I
I
I
I
I
I
I
I
I
I
I
I
I
A
I
A
= ⋅
= ⋅
⋅
= ⋅
= +
⋅ = ⋅
= +
= −
−
⋅ = ⋅
− ⋅ = ⋅
=
= − =
1
3
1
3
6 4
2
BD
BD
I
I
I
I
I
I
A
= +
= − = − =
___________________________________________________________________________
1.131
Dane:
Szukane:
Wzory:
U
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
3
4
3
2
8
8
2
BD
I
A
R
R
R
R
=
= Ω
= Ω
= Ω
= Ω
1
2
R
R
R
= +
1
3
4
2
1
1
1
1
2
2
2
2
1
1
2
1
2
3
3
2
4
4
3
3
4
4
3
4
8
2
8
2
BD
BD
I
I
I
I
I
I
U
I R
U
I
R
I
R
I R
I
I
U
I
R
U
I
R
I
R
I
R
I
I
= +
= +
= ⋅
= ⋅
⋅
= ⋅
⋅ = ⋅
= ⋅
= ⋅
⋅
= ⋅
⋅ = ⋅
2
1
3
4
1
3
4
2
1
3
4
2
2
3
3
2
2
3
3
2
3
2
3
2
8
2
8
2
3
3
3
3
8
(
3) 2
8
(
3) 2
8
2
6
8
2
6
2
6
8
8
2
6
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
⋅ = ⋅
⋅ = ⋅
= +
= +
= +
= +
⋅ =
+ ⋅
⋅ =
+ ⋅
⋅ = ⋅ +
⋅ = ⋅ +
⋅ +
=
⋅ = ⋅ +
3
3
3
3
3
3
3
3
2
6
8
2
6
8
4
12
8
6
8
64
4
12 48
60
60
1
I
I
I
I
I
I
I
I
A
⋅ +
⋅ = ⋅
+
⋅ +
⋅ =
+
⋅
= ⋅ + +
⋅
=
=
3
2
1
3
4
2
1
1
1
2
3
3
1
2
2
6
2 1 6
1
8
8
3 1 3
4
3 1 3
4
4 2
8
1 8
8
8 8 16
I
I
A
I
I
A
I
I
A
U
I R
V
U
I
R
V
U
U
U
V
⋅ +
⋅ +
=
=
=
= + = + =
= + = + =
= ⋅ = ⋅ =
= ⋅
= ⋅ =
=
+
= + =
___________________________________________________________________________
1.132
Dane:
Szukane:
Wzory:
1
2
3
1
2
15
30
20
20
20
30
V
V
U
V
R
k
R
k
R
k
R
k
R
k
=
=
Ω
=
Ω
=
Ω
=
Ω
=
Ω
1
2
1
2
OV
OV
ZV
ZV
U
U
U
U
=
=
=
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
a)
1
2
1
1
2
2
15
15
0, 3
30
20
50
0, 3
30
9
0, 3
20
6
O
V
V
OV
O
V
OV
O
V
U
V
V
I
mA
R
R
k
k
k
U
I
R
mA
k
V
U
I
R
mA
k
V
=
=
=
=
+
Ω +
Ω
Ω
= ⋅
=
⋅
Ω =
= ⋅
=
⋅
Ω =
b)
1
1
1
1
1
2
2
2
2
2
1
2
1
1
2
2
20 30
600
12
20 30
50
30 20
600
12
30 20
50
15
15
0, 625
12
12
24
0, 625
12
7, 5
0, 625
12
7, 5
V
V
V
V
Z
ZV
Z
ZV
Z
R R
R
k
R
R
R R
R
k
R
R
U
V
V
I
mA
R
R
k
k
k
U
I
R
mA
k
V
U
I
R
mA
k
V
⋅
⋅
=
=
=
=
Ω
+
+
⋅
⋅
=
=
=
=
Ω
+
+
=
=
=
=
+
Ω +
Ω
Ω
= ⋅ =
⋅
Ω =
= ⋅
=
⋅
Ω =
___________________________________________________________________________
1.133
Dane:
Szukane:
Wzory:
1
2
OV
OV
W
U
U
I
=
=
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
3
1
2
30
30
20
20
20
30
V
V
U
V
R
k
R
k
R
k
R
k
R
k
=
=
Ω
=
Ω
=
Ω
=
Ω
=
Ω
1
2
R
R
R
= +
a)
1
2
2
2
1
2
2
2
2
2
30
30
0, 6
30
20
50
0, 6
20
12
30
30
0, 6
20
30
50
0, 6
30
18
18 12
6
O
V
V
OV
O
V
R
OR
R
W
OR
OV
U
V
V
I
mA
R
R
k
k
k
U
I
R
mA
k
V
U
V
V
I
mA
R
R
k
k
k
U
I
R
mA
k
V
U
U
U
V
=
=
=
=
+
Ω +
Ω
Ω
= ⋅
=
⋅
Ω =
=
=
=
=
+
Ω +
Ω
Ω
= ⋅
=
⋅
Ω =
=
−
= − =
b)
1
1
1
1
1
2
2
2
2
2
1
2
1
1
2
2
1
1
1
2
2
2
20 30
600
12
20 30
50
30 20
600
12
30 20
50
30
30
1, 25
12
12
24
1, 25
12
15
1, 25
12
15
15
0, 5
30
15
2
V
V
V
V
Z
Z
Z
Z
Z
Z
V
V
Z
V
V
R R
R
k
R
R
R R
R
k
R
R
U
V
V
I
mA
R
R
k
k
k
U
I
R
mA
k
V
U
I
R
mA
k
V
U
V
I
mA
R
k
U
V
I
R
⋅
⋅
=
=
=
=
Ω
+
+
⋅
⋅
=
=
=
=
Ω
+
+
=
=
=
=
+
Ω +
Ω
Ω
= ⋅ =
⋅
Ω =
= ⋅
=
⋅
Ω =
=
=
=
Ω
=
=
1
2
2
1
0, 75
0
0, 75
0, 5
0, 25
V
W
V
W
V
V
mA
k
I
I
I
I
I
I
mA
mA
mA
=
Ω
+
=
=
+
=
−
=
___________________________________________________________________________
1.134
Dane:
Szukane:
Wzory:
16
25
21
AB
BC
CA
R
R
R
=
Ω
=
Ω
=
Ω
1
2
3
R
R
R
=
=
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
Zadanie niby proste , ale w rzeczywistości wielu padło próbując to rozwiązać!
spirala_grzejna/rezystor pomiędzy zaciskami
1
AB
R
=
,
2
BC
R
=
,
3
AC
R
=
standardowo to wchodzi się w wielomiany wysokiego stopnia i tam się grzęźnie.
Dlatego trzeba inaczej, zauważ powtarzające się
1
2
3
R
R
R
+
+
i tu trzeba szukać ratunku.
Tak jak na powyższym wyliczeniu, takie grupowe podstawienie (można powiedzieć troszkę
na intuicję, ale jest to metoda zwykłego podstawiania) dało efekt znacznego uproszczenia i
układ stał się wyliczalny.
Rozwiązanie 2 metodą nie dało korzystniejszych rezultatów.
2
3
1
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
3
2
1
2
3
2
1
3
2
1
3
2
1
3
2
1
3
1
2
3
2
1
3
1
2
3
3
1
2
1
1
1
(
)
(
)
(
)
(
)
16
1
1
1
(
)
(
)
(
)
(
)
25
1
1
1
AB
AB
BC
BC
CA
R
R
R
R
R
R
R
R
R
R
R R
R
R R
R
R R
R
R
R
R
R
R R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R R
R
R R
R
R R
R
R
R
R
R
R R
R
R
R
R
R
R
R
R
+
+
+
+
=
+
=
=
+
+
+
+
=
+
+
+
=
+
+
+
+
+
+
=
+
=
=
+
+
+
+
=
+
+
+
=
+
+
=
+
=
+
1
2
3
3
1
2
3
1
2
1
2
3
3
1
2
1
2
3
(
)
(
)
(
)
21
CA
R
R
R
R R
R
R R
R
R
R
R
R
R R
R
R
R
R
+
+
+
+
=
+
+
+
=
+
+
1
2
3
1
2
3
2
1
3
1
2
3
3
1
2
1
2
3
(
)
16
(
)
25
(
)
21
R R
R
R
R
R
R R
R
R
R
R
R R
R
R
R
R
+
=
+
+
+
=
+
+
+
=
+
+
1
2
3
1
2
3
(
)
16
R R
R
R
R
R
+
+
+
=
2
1
3
1
2
3
3
1
2
1
2
3
(
)
25
(
)
16
(
)
21
(
)
16
R R
R
R R
R
R R
R
R R
R
+
=
+
+
=
+
1
2
3
2
1
3
1
2
3
3
1
2
1
2
1
3
1
2
2
3
1
2
1
3
1
3
2
3
1
2
1
3
2
3
1
2
1
3
2
3
1
2
1
3
1
2
1
3
1
3
1
2
3
2
25
(
) 16
(
)
21
(
) 16
(
)
25
25
16
16
21
21
16
16
9
25
16
21
5
16
9
25
21
5
20
12
12
3
20
R R
R
R R
R
R R
R
R R
R
R R
R R
R R
R R
R R
R R
R R
R R
R R
R R
R R
R R
R R
R R
R R
R R
R R
R R
R R
R R
R
R
⋅
+
= ⋅
+
⋅
+
= ⋅
+
+
=
+
+
=
+
+
=
+
=
+
=
+
=
=
=
2
1
2
1
2
2
2
1
2
1
2
2
2
1
2
2
2
1
2
1
2
2
5
3
3
21
5
16
5
5
3
21
3
16
5
3
24
16
5
48
24
5
48
2
120
5
R
R R
R
R
R
R
R R
R R
R
R
R R
R
R
R
R
R
R
R
+
=
+
=
=
=
=
=
1
3
1
2
3
1
2
3
:
(
)
16
teraz
R
i
R
podstawiamy
do
R R
R
R
R
R
+
+
+
=
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
3
(
)
2
3
5
5
5
5
16
2
3
(
)
5
5
2
16
16
32
25
16
32
25
50
R R
R
R
R
R
R R
R
R
R
R R
R
R
+
+
+
=
+
=
=
=
=
Ω
3
2
1
2
3
3
50
30
5
5
2
2
50
20
5
5
R
R
R
R
=
= ⋅
=
Ω
=
= ⋅
=
Ω
2) Sposób. Początek podobnie a potem troszkę inaczej , tak jakby bardziej w kierunku
matematycznych przekształceń ,metodą przeciwnych współczynników.
2
3
1
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
3
2
1
2
3
2
1
3
2
1
3
2
1
3
2
1
3
1
2
3
2
1
3
1
1
1
1
(
)
(
)
(
)
(
)
16
(
)
16
1
1
1
(
)
(
)
(
)
(
)
25
AB
AB
BC
BC
R
R
R
R
R
R
R
R
R
R
R R
R
R R
R
R R
R
R
R
R
R
R R
R
R
R
R
R R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R R
R
R R
R
R R
R
R
R
R
R
R R
R
R
R
+
+
+
+
=
+
=
=
+
+
+
+
=
+
+
+
=
+
+
+
+
+
=
+
+
+
+
=
+
=
=
+
+
+
+
=
+
+
+
=
+
2
3
2
1
3
1
2
3
1
2
3
3
1
2
3
1
2
3
1
2
1
2
3
3
1
2
1
2
3
3
1
2
1
2
3
(
)
25
1
1
1
(
)
(
)
(
)
21
(
)
21
CA
CA
R
R R
R
R
R
R
R
R
R
R
R
R
R
R R
R
R R
R
R
R
R
R
R R
R
R
R
R
R R
R
R
R
R
+
+
+
+
=
+
+
=
+
=
+
+
+
=
+
+
+
=
+
+
+
+
+
=
wyznaczam
1
2
3
R
R
R
+
+
i porównuję stronami
1
2
3
1
2
3
2
1
3
1
2
3
3
1
2
1
2
3
1
2
3
2
1
3
1
2
3
3
1
2
2
1
3
3
1
2
1
2
3
2
1
3
1
2
3
3
1
2
2
1
(
)
16
(
)
25
(
)
21
(
)
(
)
16
25
(
)
(
)
16
21
(
)
(
)
25
21
25
(
) 16
(
)
21 (
) 16
(
)
21
(
R R
R
R
R
R
R R
R
R
R
R
R R
R
R
R
R
R R
R
R R
R
R R
R
R R
R
R R
R
R R
R
R R
R
R R
R
R R
R
R R
R
R R
+
+
+
=
+
+
+
=
+
+
+
=
+
+
=
+
+
=
+
+
=
+
=
+
+
=
+
3
3
1
2
1
2
1
3
1
2
2
3
1
2
1
3
1
3
2
3
1
2
2
3
1
3
2
3
)
25
(
)
25
25
16
16
21
21
16
16
21
21
25
25
R
R R
R
R R
R R
R R
R R
R R
R R
R R
R R
R R
R R
R R
R R
+
=
+
+
=
+
+
=
+
+
=
+
1
2
1
3
2
3
1
2
1
3
2
3
1
2
1
3
2
3
9
25
16
21
5
16
21
25
4
R R
R R
R R
R R
R R
R R
R R
R R
R R
+
=
+
=
=
+
1
2
1
3
2
3
1
2
1
3
2
3
1
2
1
3
2
3
1
3
2
3
1
3
2
3
3
1
2
3
1
2
2
2
1
1
2
1
3
1
2
3
9
25
16
21
5
16
21
25
4
2
3
30
12
30
12
0
(30
12
)
0
0
30
12
0
12
2
30
5
1
2
12
20
0
( 12
20
)
0
R R
R R
R R
R R
R R
R R
R R
R R
R R
od
odjąć
R R
R R
R R
R R
R
R
R
poniewaz
R
R
R
R
R
R
od
odjąć
R R
R R
R
R
R
poniewaz
+
=
+
=
−
=
=
−
=
−
=
〉
−
=
=
=
−
+
=
−
+
=
1
2
3
2
2
3
0
12
20
0
12
3
20
5
R
R
R
R
R
R
〉
−
+
=
=
=
po wielu męczarniach układ , skrócił się do takiego prostego:
2
1
2
3
2
5
3
5
R
R
R
R
=
=
2
R wyznacza się z wzoru pierwotnego.
1
3
1
2
3
1
2
3
2
2
2
2
2
2
2
2
2
2
2
2
(
)
16
2
3
(
)
2
3
5
5
5
5
16
3
2
(
)
32
80
48
5
80
80
80
80
wyznaczone R i R podstawiam do jednego z pierwszych wzorów
R R
R
R
R
R
R
R
R
R
R
R
R
R R
R
R
R
+
+
+
=
+
+
+
=
+
+
+
=
2
2
2
2
2
2
2
2
2
1
2
3
3
160
2
(
)
5
5
3
80
(
)
5
5
400
8
50
2
2 50
20
5
5
3
3 50
30
5
5
R
R
R R
R
R
R
R
R
R
R
R
=
+
=
+
=
=
Ω
⋅
=
=
=
Ω
⋅
=
=
=
Ω
3)Sposób , troszkę dziwaczny, początek jak zwykle podobny
2
3
1
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
1
3
2
1
2
3
2
1
3
2
1
3
2
1
3
2
1
3
1
2
3
2
1
3
1
1
1
1
(
)
(
)
(
)
(
)
16
16(
)
(
)
1
1
1
(
)
(
)
(
)
(
)
25
AB
AB
BC
BC
R
R
R
R
R
R
R
R
R
R
R R
R
R R
R
R R
R
R
R
R
R
R R
R
R
R
R
R
R
R
R R
R
R
R
R
R
R
R
R
R
R
R
R R
R
R R
R
R R
R
R
R
R
R
R R
R
R
+
+
+
+
=
+
=
=
+
+
+
+
=
+
+
+
=
+
+
+
+
=
+
+
+
+
+
=
+
=
=
+
+
+
+
=
+
+
+
=
2
3
1
2
3
2
1
3
1
2
3
3
1
2
3
1
2
3
1
2
1
2
3
3
1
2
1
2
3
1
2
3
3
1
2
25(
)
(
)
1
1
1
(
)
(
)
(
)
21
21(
)
(
)
CA
CA
R
R
R
R
R
R R
R
R
R
R
R
R
R
R
R R
R
R R
R
R
R
R
R
R R
R
R
R
R
R
R
R
R R
R
+
+
+
+
=
+
+
+
=
+
=
+
+
+
=
+
+
+
=
+
+
+
+
=
+
1
2
3
1
2
3
1
2
3
2
1
3
1
2
3
3
1
2
16(
)
(
)
25(
)
(
)
21(
)
(
)
R
R
R
R R
R
R
R
R
R R
R
R
R
R
R R
R
+
+
=
+
+
+
=
+
+
+
=
+
1
2
3
R
R
R
S
+
+
=
to taka niewiadoma pomocnicza
1
2
3
2
1
3
3
1
2
16
(
)
25
(
)
21
(
)
S
R R
R
S
R R
R
S
R R
R
=
+
=
+
=
+
1
2
1
3
1
2
2
3
1
3
2
3
16
25
21
S
R R
R R
S
R R
R R
S
R R
R R
=
+
=
+
=
+
teraz przerabiamy stronami, przez sumowanie dwóch i odjęcie trzeciego.
ale dlaczego tak ? może to jakaś wariacja wg metody przeciwnych współczynników .
1
2
1
3
1
2
2
3
1
3
2
3
1
2
1
3
1
2
2
3
1
3
2
3
1
2
1
3
1
2
2
3
1
3
2
3
(1) (2) (3)
(1) (2) (3)
(1) (2) (3)
16
25
21
16
25
21
16
25
21
S
S
S
R R
R R
R R
R R
R R
R R
S
S
S
R R
R R
R R
R R
R R
R R
S
S
S
R R
R R
R R
R R
R R
R R
+
−
−
+
− +
+
+
−
=
+
+
+
−
−
−
+
=
+
−
−
+
+
−
+
+
= −
−
+
+
+
+
1
2
1
3
2
3
20
2
12
2
30
2
S
R R
S
R R
S
R R
=
=
=
teraz dzielimy strony równań parami: 1 przez 2, 1 przez 3, 2 przez 3
1
2
1
3
1
2
2
3
1
3
2
3
3
2
3
1
2
1
1
3
1
2
2
20
12
2
2
20
30
2
2
12
30
2
5
3
2
3
2
5
3
2
5
2
R R
S
S
R R
R R
S
S
R R
R R
S
S
R R
R
R
R
R
R
R
R
R
R
R
=
=
=
=
=
=
=
=
podstawiamy do równania pierwotnego
1
2
3
1
2
3
1
1
1
1
1
1
1
1
1
1
1
3
1
2
16(
)
(
)
5
3
5
3
16(
)
(
)
2
2
2
2
10
8
16
2
2
20
3
3 20
30
2
2
5
5 20
50
2
2
R
R
R
R R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
+
+
=
+
+
+
=
+
=
=
Ω
⋅
=
=
=
Ω
⋅
=
=
=
Ω
___________________________________________________________________________
1.135
Dane:
Szukane:
Wzory:
2
3
4
175
100
20
0
G
R
R
R
I
=
Ω
=
Ω
=
Ω
=
1
R
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
wg wzoru na równowagę mostka
3
2
1
4
3
2
1
4
100 175
17500
875
20
20
R R
R R
R R
R
R
⋅
= ⋅
⋅
⋅
=
=
=
=
Ω
lub troszkę naokoło
1
1
1
2
2
2
3
3
3
4
4
4
1
2
3
4
1
3
1
1
3
3
3
3
1
1
2
4
2
2
4
4
1
2
3
4
3
3
2
3
4
1
3
2
1
4
3
2
1
4
175 100
875
20
U
I R
U
I
R
U
I
R
U
I
R
I
I
I
I
U
U
I R
I
R
I
R
I
R
U
U
I
R
I
R
I R
I
R
I
R
R
I
R
R
R R
R R
R R
R
R
= ⋅
= ⋅
= ⋅
= ⋅
=
=
=
⋅ = ⋅
⋅
=
=
⋅
= ⋅
⋅
= ⋅
⋅ ⋅ = ⋅
⋅
= ⋅
⋅
⋅
=
=
=
Ω
___________________________________________________________________________
1.136
Dane:
Szukane:
Wzory:
1
2
3
4
875
175
100
20
0
6
G
R
R
R
R
I
E
V
=
Ω
=
Ω
=
Ω
=
Ω
=
=
1
2
I
I
=
=
U
I
R
=
1
2
1
1
1
R
R
R
=
+
1
2
R
R
R
= +
1
1
2
2
2
4
6
5, 71
875 175
6
50
100 20
E
I
mA
R
R
E
I
mA
R
R
=
=
=
+
+
=
=
=
+
+
___________________________________________________________________________
1.137
Dane:
Szukane:
Wzory:
1
2
1
2
1 2
100
100
20
50
52
1
0, 004
x
l
o
Cu
o
R
R
rezystor miedziany
R
rezystor drutowy
l
l
l
mm
C
l
l
mm
l
mm
C
ϑ
ϑ
ϑ
ϑ
α
=
Ω
=
=
= + =
=
=
=
=
=
2
ϑ
=
1
2
R
R
R
= +
1
2
(1
(
))
R
R
α ϑ ϑ
= ⋅ +
−
1
1
1
2
l
l
l
l
l
R
R
l
R
R
ϑ
= ⋅
=
2
1
1
1
1
2
2
2
100
100
100 50
100
50
x
l
l
l
l
x
l
l
R
R
R R
l
R
R R
l
l
R
l
R
l
R
l
ϑ
ϑ
ϑ
ϑ
ϑ
ϑ
⋅
= ⋅
⋅ ⋅
⋅
⋅
⋅
=
=
=
=
=
Ω
⋅
w temperaturze pracy:
1 2
2 2
2 2
1 2
2 2
2 2
1 2
1 2
2
2 2
1 2
1 2
1 2
1 2
2
2 2
2 2
2 2
100 52
48
100
100
100 52
108, 33
48
l
l
l
l
x
l
l
l
l
x
l
l
l
l
l
l
l
l
mm
l
R
R
l
l
R
R
l
R
R
R R
l
R
R R
l
l
R
l
R
l
R
l
ϑ
ϑ
ϑ
ϑ
ϑ
ϑ
ϑ
ϑ
ϑ
ϑ
ϑ
ϑ
ϑ
ϑ
ϑ
ϑ
ϑ
ϑ
=
+
= −
=
−
=
= ⋅
= ⋅
⋅
= ⋅
⋅ ⋅
⋅
⋅
⋅
=
=
=
=
=
Ω
⋅
2
2
2
2
2
2
2
2
(1
(
))
1
(
)
1
(
)
108, 33
1
1
100
20
20,83 20
40,83
0, 004
x
x
x
x
x
x
x
o
x
R
R
R
R
R
R
R
R
C
ϑ
ϑ
ϑ
ϑ
ϑ
ϑ
ϑ
ϑ
α ϑ ϑ
α ϑ ϑ
ϑ ϑ
α
ϑ
ϑ
α
=
⋅ +
−
= +
−
−
=
−
−
−
=
+ =
+
=
+
=
___________________________________________________________________________
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