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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
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
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
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
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
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

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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