- #1
Celso
- 33
- 1
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 resistance. 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 GWIn 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
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 resistance. 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 GWIn 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