Respuesta :
Answer:
a) [tex]P_{L}=199.075W[/tex]
b) [tex]P_{L}=1.991x10^{-4}W[/tex]
Explanation:
1) Notation
Power on the refrigerator: [tex]P=IV=3Ax110V=330W[/tex]
Voltage [tex]V=110V[/tex]
[tex]D=8.252mm[/tex], so then the radius would be [tex]r=\frac{8.252}{2}=4.126mm[/tex]
[tex]L=2x10km=20km=20000m[/tex], representing the length of the two wires.
[tex]\rho=2.65x10^{-8}\Omega m[/tex], that represent the resistivity for the aluminum founded on a book
[tex]P_L[/tex] power lost in the transmission.
2) Part a
We can find the total power adding all the individual values for power:
[tex]P_{tot}=(330+100+60+3)W=493W[/tex]
From the formula of electric power:
[tex]P=IV[/tex]
We can solve for the current like this:
[tex]I=\frac{P}{V}[/tex]
Since we know [tex]P_{tot}[/tex] and the voltage 110 V, we have:
[tex]I=\frac{493W}{110V}=4.482A[/tex]
The next step would be find the cross sectional are for the aluminum cables with the following formula:
[tex]A=\pi r^2 =\pi(0.004126m)^2=5.348x10^{-5}m^2[/tex]
Then with this area we can find the resistance for the material given by:
[tex]R=\rho \frac{L}{A}=2.65x10^{-8}\Omega m\frac{20000m}{5.348x10^{-5}m^2}=9.910\Omega[/tex]
With this resistance then we can find the power dissipated with the following formula:
[tex]P_{L}=I^2 R=(4.482A)^2 9.910\Omega=199.075W[/tex]
And if we want to find the percentage of power loss we can use this formula
[tex]\% P_{L}=\frac{P_L}{P}x100[/tex]
3) Part b
Similar to part a we just need to change the value for V on this case to 110KV.
We can solve for the current like this:
[tex]I=\frac{P}{V}[/tex]
Since we know [tex]P_{tot}[/tex] and the voltage 110 KV=110000V, we have:
[tex]I=\frac{493W}{110000V}=4.482x10^{-3}A[/tex]
The cross sectional area is the same
The resistance for the material not changes.
With this resistance then we can find the power dissipated with the following formula:
[tex]P_{L}=I^2 R=(4.482x10^{-3}A)^2 9.910\Omega=1.991x10^{-4}W[/tex]
