# Power waste (conceptual doubt)

• Celso
In summary, the voltage at the destination will be 4.4 MV, and the voltage drop down the cable will be 0.6 MV.

#### Celso

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

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.

Last edited:
• Celso
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.

• Celso
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 :)

All looks good. Kirchoff's voltage law proven again :-)

• Zaya Bell

## 1. What is power waste?

Power waste refers to the unnecessary consumption or loss of energy in a system or process. This can occur due to inefficiencies, faulty equipment, or improper use of energy.

## 2. How does power waste affect the environment?

Power waste significantly contributes to environmental degradation by increasing greenhouse gas emissions and contributing to climate change. It also depletes natural resources and can harm ecosystems.

## 3. What are some common sources of power waste?

Common sources of power waste include outdated or poorly maintained equipment, inefficient energy use, and human error. In some cases, power waste can also be caused by design flaws in a system.

## 4. How can power waste be reduced?

Power waste can be reduced by implementing energy-efficient practices and technologies, regularly maintaining equipment, and promoting energy-conscious behavior among individuals. It is also important to conduct energy audits and identify areas where power waste is occurring.

## 5. What are the economic impacts of power waste?

Power waste can have significant economic impacts, including increased energy costs, reduced productivity, and higher maintenance and repair expenses. It can also lead to higher prices for consumers and decreased competitiveness for businesses.