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Classical Physics
Mechanics
Power waste (conceptual doubt)
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[QUOTE="Celso, post: 6078107, member: 652839"] 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 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 [/QUOTE]
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Power waste (conceptual doubt)
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