Diferencia entre revisiones de «1soin/ECA/UD2/01 Ariketa. Zirkuitu elektrikoak»
De Portfolio Academico
Línea 5: | Línea 5: | ||
<math>I = \frac{V}{R} = \frac{10}{5000} = 0,002A</math> | <math>I = \frac{V}{R} = \frac{10}{5000} = 0,002A</math> | ||
<math>P = \frac {V^2}{R} = 0,02W</math> | <math>P = \frac {V^2}{R} = \frac {10}{500} = 0,02W</math> | ||
R1: | R1: | ||
I1 =I | |||
<math>V_1=I \cdot R=0.002 \cdot 500 = 1V</math> | |||
<math>P_1 = \frac {V^2}{R} = \frac {1}{500} = 0,002W</math> |
Revisión del 08:33 4 nov 2016
1)
[math]\displaystyle{ R_T = R_1+ R_2 + R_3 + R_4 -\gt R_T = 500+1000+1500+2000 = 5000 \Omega }[/math]
[math]\displaystyle{ I = \frac{V}{R} = \frac{10}{5000} = 0,002A }[/math]
[math]\displaystyle{ P = \frac {V^2}{R} = \frac {10}{500} = 0,02W }[/math]
R1:
I1 =I
[math]\displaystyle{ V_1=I \cdot R=0.002 \cdot 500 = 1V }[/math]
[math]\displaystyle{ P_1 = \frac {V^2}{R} = \frac {1}{500} = 0,002W }[/math]