Power waste (conceptual doubt)

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I'm currently in a undergrad level course of eletromagnetism and my professor found it useful to remember quickly of some relationships that we see in high scool, among them power waste.
I got the formulas, since they're simple and are direct derivated from the definitions of electric current and ohms law.
wasted power = R i^2 or wasted power = V^2 / R
The thing is that the examples in my textbook always give the total power, the tension and the resistence. For some reason, when I plug V^2/R (which are already given), my answer is incorrect, instead I must first find the current then plug it into R i^2. I feel I'm missing something obvius here, I'll let an exercise from the book.

A transmission line carries a power of 1 GW of a plant, in the voltage of 500kV, to a distance of 400 km. If the cables are made of aluminum and have a diameter of 3 cm, what energy is lost in the cables by the joule effect?

Solution:
1- For simplicity, we will consider the situation where two cables transfer power by direct current. In the round trip, we have 800 km of cable. its resistance is R = pL / (pi (D ^ 2/4)). Pluging the values, we find 30 ohms.

2- The current that travels along the transmission line is given by P = Vi : i = P/V = 2000 A

3- Therefore, the wasted power in heat in the cables is P = R i ^2 = 30 * (2 * 10^3) ^2 = 0.12 GW


In the step 2, why is it wrong to do P = V^2 / R, which would give (5*10^5)^2 / 30, more or less 8.3 GW
 

Answers and Replies

  • #2
CWatters
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WHERE you measure the voltage matters. There are three different voltages in this problem.

The voltage you need to use is the voltage lost in the resistance of the cable not the voltage at the ends...

The voltage at the power station can be used to calculate the power generated (500kV)
The voltage at the destination city can be used to calculate the power that actually arrives.

The voltage difference between the ends is the voltage lost in the cable and is used to calculate the power lost as heat in the cable.
 
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  • #3
CWatters
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As an exercise calculate:

a) the voltage at the destination.
b) the voltage drop down the cable.
c) add up the answer to a and b. Should be 500kV.
 
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  • #4
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Okay, so the current will remain constant in any of these situations
Since the energy drop was 0.12 GW and the current is 2000 A, the voltage drop is 0.12 * 10 ^9 / 2 * 10 ^3 = 0.6 MV
The voltage at the destination will be found throught the total energy that made its way to the city (1GW - 0.12GW = 0.88GW), V = 0.88 * 10^9 / 2*10^3 = 4.4 MV

(4.4 + 0.6)MV = 5MV, as expected, thanks :)
 
  • #5
CWatters
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All looks good. Kirchoff's voltage law proven again :-)
 
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