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Asociacion de resistencias
- 1. ASOCIACION DE RESISTENCIAS
EJERCICOS
Encontrar la Resistencia equivalente de los siguientes circuitos
1.
Req = 푅1 + 푅2 + 푅3 + 푅4 + 푅5
Req = 10 + 5 + 2 + 8 + 20
Req = 45Ω
2.
1
푅푒푞
=
1
푅1
+ 1
푅2
+ 1
푅3
+ 1
푅4
1
푅푒푞
=
1
1,5 푘Ω
+ 1
10 푘Ω
+ 1
4,7 푘Ω
+ 1
100 푘Ω
1
푅푒푞
= 0,98 푘Ω
- 2. 1 = 0,98 푘Ω ∗ 푅퐸
1
= 푅퐸
0,98
1,02 푘Ω = 푅퐸
3.
R1=1000Ω R2=3000Ω R3=8, 2Ω
R4=16Ω R5=75Ω R6=82Ω
R7=150Ω R8=160Ω R9=51Ω
R10=130Ω R11=60Ω R12=50Ω
R13=78, 2Ω R14=15Ω RA=52, 91Ω
RB=36, 63Ω RC=27, 24Ω RD=120,44Ω
Re=127, 91Ω RF=62, 11Ω RG=78, 11Ω
RH=168, 2Ω RI=53, 47Ω RE=4090, 1Ω
- 3. PROCESO
1.
1
푅퐴
=
1
푅6
+
1
푅7
1
푅퐴
=
1
82 Ω
+
1
150 Ω
1
푅퐴
= 0,0189 Ω
1
0,0189 Ω
= 52, 91 Ω
2,
1
푅퐵
=
1
푅9
+
1
푅10
1
푅퐵
=
1
51 Ω
+
1
130 Ω
1
푅퐵
= 0,0273 Ω
1
0,0273 Ω
= 36, 63 Ω
- 4. 3.
1
푅퐶
=
1
푅11
+
1
푅12
1
푅퐶
=
1
60 Ω
+
1
50 Ω
1
푅퐶
= 0,0367 Ω
1
0,0367 Ω
= 27, 24 Ω
4.
푅퐷 = 푅14 + 푅13 + 푅퐶
푅퐷 = 15 Ω + 78, 2 Ω + 27, 24 Ω
푅퐷 = 120, 44 Ω
5.
푅푒 = 푅5 + 푅퐴
푅푒 = 75 Ω + 52, 91 Ω
푅푒 = 127, 91 Ω
RD
- 5. 6.
1
푅퐹
=
1
푅푒
+
1
푅퐷
1
푅퐹
=
1
127, 91 Ω
+
1
120, 44 Ω
1
푅퐹
= 0,0161 Ω
1
0,0161 Ω
= 62, 11 Ω
7.
푅퐺 = 푅4 + 푅퐹
푅퐺 = 16 Ω + 62, 11 Ω
푅퐺 = 78, 11Ω
8.
푅퐻 = 푅3 + 푅8
푅퐻 = 8, 2 Ω + 160 Ω
푅퐻 = 168, 2Ω
- 6. 9.
1
푅퐼
=
1
푅퐺
+
1
푅퐻
1
푅퐼
=
1
78, 11 Ω
+
1
168, 2 Ω
1
푅퐼
= 0,0187 Ω
1
0,0187 Ω
= 53, 47 Ω
10.
푅퐸 = 푅1 + 푅2 + 푅퐼 + 푅퐵
푅퐸 = 1000 Ω + 3000 Ω + 53, 47 Ω + 36, 63 Ω
푅퐸 = 4090,1 Ω