Diferencia entre revisiones de «1soin/ECA/UD2/01 Ariketa. Zirkuitu elektrikoak»
| Línea 39: | Línea 39: | ||
<math>P_4 = \frac {V^2}{R} = \frac {1}{2000} = 0,008W</math> | <math>P_4 = \frac {V^2}{R} = \frac {1}{2000} = 0,008W</math> | ||
== 2) == | == 2) == | ||
| Línea 49: | Línea 48: | ||
<math>P = \frac{V^2}{R} \Rightarrow P = \frac{9^2}{600} =0,135W</math> | <math>P = \frac{V^2}{R} \Rightarrow P = \frac{9^2}{600} =0,135W</math> | ||
<math>I = \frac{V}{R} \Rightarrow I = \frac{9}{600} =0, | <math>I = \frac{V}{R} \Rightarrow I = \frac{9}{600} =0,015A</math> | ||
==3)== | ==3)== | ||
R<sub>2</sub> eta R<sub>3</sub>, haien artean paraleloan daudenez, hauen erresistentzia baliokidea kalkulatuko dugu, gero R<sub>1</sub>ekin egin behar diren kalkuluak errazteko. | R<sub>2</sub> eta R<sub>3</sub>, haien artean paraleloan daudenez, hauen erresistentzia baliokidea kalkulatuko dugu, gero R<sub>1</sub>ekin egin behar diren kalkuluak errazteko. | ||
<math>R_{23} \Rightarrow \frac{1}{R_{23}} = \frac{1}{R_2} + \frac{1}{R_3} \Rightarrow \frac{1}{R_{23}} = \frac{1}{1500} + \frac{1}{1000} = \frac{5}{3000} \Rightarrow R_{23} = 600 \Omega</math> | |||
<math>R_T = 600 + 400 = 1000\Omega</math> | |||
<math>I = \frac{V}{R} \Rightarrow I = \frac{10}{1000} = 0,01A</math> | |||
Revisión del 07:29 9 nov 2016
1)
[math]\displaystyle{ R_T = R_1+ R_2 + R_3 + R_4 \Rightarrow 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]
2)
Paraleloan jarritako zirkuituetan tentsioa konstantea jarraitzen da.
[math]\displaystyle{ R_T \Rightarrow \frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} \Rightarrow \frac{1}{R_T} = \frac{1}{2000} + \frac{1}{2000} + \frac{1}{3000} + \frac{1}{3000} = \frac{10}{6000} \Rightarrow R_T = 600 \Omega }[/math]
[math]\displaystyle{ P = \frac{V^2}{R} \Rightarrow P = \frac{9^2}{600} =0,135W }[/math]
[math]\displaystyle{ I = \frac{V}{R} \Rightarrow I = \frac{9}{600} =0,015A }[/math]
3)
R2 eta R3, haien artean paraleloan daudenez, hauen erresistentzia baliokidea kalkulatuko dugu, gero R1ekin egin behar diren kalkuluak errazteko.
[math]\displaystyle{ R_{23} \Rightarrow \frac{1}{R_{23}} = \frac{1}{R_2} + \frac{1}{R_3} \Rightarrow \frac{1}{R_{23}} = \frac{1}{1500} + \frac{1}{1000} = \frac{5}{3000} \Rightarrow R_{23} = 600 \Omega }[/math]
[math]\displaystyle{ R_T = 600 + 400 = 1000\Omega }[/math]
[math]\displaystyle{ I = \frac{V}{R} \Rightarrow I = \frac{10}{1000} = 0,01A }[/math]
