I 8 4 v r i 8 4 60 8 4 i 8 4 7 143 a i 8 60 8 i 8 7 5

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I 8 , 4 = V R
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I 8 , 4 = 60 8 , 4 I 8 , 4 = 7 , 143 [ A ] I 8 = 60 8 I 8 = 7 , 5 [ A ] Ahora:
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Hallamos la corriente sobre la resistencia de 6Ω, como esta en serie con la de 2,4 la corriente es la misma. I 6 = 7 , 143 [ A ] Entonces la corriente que pasa sobre la resistencia de 6Ω es de 7,143 [A]. Hallamos el voltaje sobre la resistencia de 2,4Ω. V = I × R
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V = 7 , 143 × 2 , 4 V = 17 , 1432 [ V ] Ahora tenemos: Como la resistencia de 4Ω y la de 6Ω están en paralelo su voltaje es el mismo y es de ( V = 17 , 1432 [ V ]) Ahora pasamos a calcular la corriente sobre la resistencia de 6Ω.
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I 6 = V R I 6 = 17 , 1432 6 I 6 = 2 , 8572 [ A ] Y el circuito final es: Como la resistencia de 2Ω y la de 4Ω están en serie la corriente es la misma ye de 2,8572 [A] , por lo tanto, la corriente sobre la resistencia de 2Ω es de 2,8572 [A].
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C. En relación con el enunciado b, cuánta potencia se disipa en el resistor 4 Ω que está en la parte superior del circuito. Tenemos que la potencia eléctrica es: P = V × I El voltaje que consume la resistencia es de 17,1432 [A], y la corriente será: I 4 = 17 , 1432 4 I 4 = 4 , 2858 [ A ] Y la potencia será: P = 17 , 1432 × 4 , 2858
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  • Fall '19
  • Punto, Recta, Pendiente, Función lineal, Potencia eléctrica

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