Diferencia entre revisiones de «1soin/ECA/UD2/01 Ariketa. Zirkuitu elektrikoak»
Línea 14: | Línea 14: | ||
<math>P_1 = \frac {V^2}{R} = \frac {1}{500} = 0,002W</math> | <math>P_1 = \frac {V^2}{R} = \frac {1}{500} = 0,002W</math> | ||
R2: | |||
I2 =I | |||
<math>V_2=I \cdot R=0.002 \cdot 1000 = 2V</math> | |||
<math>P_2 = \frac {V^2}{R} = \frac {1}{1000} = 0,004W</math> | |||
R3: | |||
I3 =I | |||
<math>V_3=I \cdot R=0.002 \cdot 1500 = 3V</math> | |||
<math>P_3 = \frac {V^2}{R} = \frac {1}{1500} = 0,006W</math> | |||
R4: | |||
I4 =I | |||
<math>V_4=I \cdot R=0.002 \cdot 2000= 4V</math> | |||
<math>P_4 = \frac {V^2}{R} = \frac {1}{2000} = 0,008W</math> |
Revisión del 08:39 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]
R2:
I2 =I
[math]\displaystyle{ V_2=I \cdot R=0.002 \cdot 1000 = 2V }[/math]
[math]\displaystyle{ P_2 = \frac {V^2}{R} = \frac {1}{1000} = 0,004W }[/math]
R3:
I3 =I
[math]\displaystyle{ V_3=I \cdot R=0.002 \cdot 1500 = 3V }[/math]
[math]\displaystyle{ P_3 = \frac {V^2}{R} = \frac {1}{1500} = 0,006W }[/math]
R4:
I4 =I
[math]\displaystyle{ V_4=I \cdot R=0.002 \cdot 2000= 4V }[/math]
[math]\displaystyle{ P_4 = \frac {V^2}{R} = \frac {1}{2000} = 0,008W }[/math]