Diferencia entre revisiones de «1soin/ECA/UD2/01 Ariketa. Zirkuitu elektrikoak»

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< 1soin‎ | ECA‎ | UD2
(Página creada con «1) Rt = R1+ R2 + R3 + R4 -> Rt = 500+1000+1500+2000 = 5000 Ohm I = V/R = 10/5000 = 0.002A P = V2/R = 0.02w R1: V=I*R=0.002*500»)
 
 
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1)


Rt = R1+ R2 + R3 + R4 -> Rt = 500+1000+1500+2000 = 5000 Ohm
== 1) ==


I = V/R = 10/5000 = 0.002A
<math>R_T = R_1+ R_2 + R_3 + R_4 \Rightarrow R_T = 500+1000+1500+2000 = 5000 \Omega</math>


P = V2/R = 0.02w
<math>I = \frac{V}{R} = \frac{10}{5000} = 0,002A</math>
 
<math>P = \frac {V^2}{R} = \frac {10}{500} = 0,02W</math>


R1:
R1:


V=I*R=0.002*500
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>
 
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>
 
== 2) ==
 
Paraleloan jarritako zirkuituetan tentsioa konstantea jarraitzen da.
 
<math>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>P = \frac{V^2}{R} \Rightarrow P = \frac{9^2}{600} =0,135W</math>
 
<math>I = \frac{V}{R} \Rightarrow I = \frac{9}{600} =0,015A</math>
 
==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.
 
<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 = 400 + 600 = 1000\Omega</math>
 
<math>I = \frac{V}{R} \Rightarrow I = \frac{10}{1000} = 0,01A</math>
 
R<sub>1</sub>:
 
I1=I
 
<math>V_1 = I_1 \cdot R_1 \Rightarrow V_1 = 0,01 \cdot 400 = 4V</math>
 
R<sub>23</sub>:
 
I<sub>23</sub>=I
 
<math>V_{23} = I_{23} \cdot R_{23} \Rightarrow V_{23} = 0,01 \cdot 600 = 6V</math>
 
R<sub>2</sub>:
 
<math>I_2 = \frac{V_2}{R_2} \Rightarrow I = \frac{6}{1500} = 0,004A</math>
 
R<sub>3</sub>:
 
<math>I_3 = \frac{V_3}{R_3} \Rightarrow I = \frac{6}{1000} = 0,006A</math>

Revisión actual del 07:38 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 = 400 + 600 = 1000\Omega }[/math]

[math]\displaystyle{ I = \frac{V}{R} \Rightarrow I = \frac{10}{1000} = 0,01A }[/math]

R1:

I1=I

[math]\displaystyle{ V_1 = I_1 \cdot R_1 \Rightarrow V_1 = 0,01 \cdot 400 = 4V }[/math]

R23:

I23=I

[math]\displaystyle{ V_{23} = I_{23} \cdot R_{23} \Rightarrow V_{23} = 0,01 \cdot 600 = 6V }[/math]

R2:

[math]\displaystyle{ I_2 = \frac{V_2}{R_2} \Rightarrow I = \frac{6}{1500} = 0,004A }[/math]

R3:

[math]\displaystyle{ I_3 = \frac{V_3}{R_3} \Rightarrow I = \frac{6}{1000} = 0,006A }[/math]

Obtenido de «https://inigo.sendino.org/academico/index.php?title=1soin/ECA/UD2/01_Ariketa._Zirkuitu_elektrikoak&oldid=5065»