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Circuito mixto

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Cómo resolver un problema de circuito mixto.

Cómo resolver un problema de circuito mixto.


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  • 1. Circuito MixtoResolución de problemas.
  • 2. I1 R1 =40 Ω B AEn el siguiente circuito calcula lassiguientes magnitudes: R2 =10 ΩLa resistencia total.Los voltajes: VA-B, VB-C. V =80v I2Las intensidades: R3 =12ΩIt, I1 y I2 It C
  • 3. I1 R1 =40 Ω1. Calculamos la resistencia A B equivalente de la parte del circuito con conexión e paralelo, Rp R2 =10 Ω a la que llamaremos Rp:1/Rp=1/R1+1/R2; V =80v I2 R3 =12Ω1/Rp=1/40+1/10=1+4/40= 5/40;1/Rp=5/40 ⇒ Rp=40/5=8Ω; It C Rp= 8 Ω
  • 4. I1 R1 =40 Ω2. Calculamos la resistencia equivalente (total), sumando Rp A B con R3 Rp R2 =10 ΩRt= Rp + R3 ; Rt= 8 + 12 = 20 ΩLa resistencia equivalente del V =80v I2 R3 =12Ω circuito es de 20 Ω It C
  • 5. I1 R1 =40 Ω3. Calculamos ahora la intensidad total (It), aplicando la ley de Ohm A B a todo el circuito: Rp R2 =10 ΩV=It × Rt; It= V/Rt = 80/20 = 4 A V =80v I2 R3 =12ΩLa intensidad total del circuito es de 4 amperios. It C
  • 6. I1 R1 =40 Ω4. Calculamos ahora los voltajes parciales. VAB y VBC A B Rp R2 =10 ΩVAB=It × Rp = 4 × 8= 32 v V =80v I2 R3 =12ΩVBC=It × R3 = 4 × 12= 48 vComprobamos: It CV= VAB + VBC = 32 + 48 = 80 v5. Calculamos la intensidades parciales I1 e I2I1= VAB/R1= 32/40 = 0,8 AI2= VAB/R2= 32/10 = 3,2 AComprobamos:It=I1+I2= 0,8+3,2 = 4 A
  • 7. I1 R1 =40 Ω4. Calculamos ahora los voltajes parciales. VAB y VBC A B Rp R2 =10 ΩVAB=It × Rp = 4 × 8= 32 v V =80v I2 R3 =12ΩVBC=It × R3 = 4 × 12= 48 vComprobamos: It CV= VAB + VBC = 32 + 48 = 80 v5. Calculamos la intensidades parciales I1 e I2I1= VAB/R1= 32/40 = 0,8 AI2= VAB/R2= 32/10 = 3,2 AComprobamos:It=I1+I2= 0,8+3,2 = 4 A