1soin/ECA/UD2/01 Ariketa. Zirkuitu elektrikoak

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]