A copper transmission cable 100 km long and 10.0 cm in diameter carries a current of 125 A. (b) How much electrical energy is dissipated as thermal energy every hour?

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Hi everyone today, we are going to determine the amount of energy lost as heat each day. Of an aluminum cable used in a low voltage power line that has 74.8 diameter strands. So the D. Is going to be 4.8 millimeters, which is going to be 4.8 times 10 to the power of minus three m. And it will transmit an I. Or an electric current of 4:30 p.m. And it will have the length of 1.3 km, which is essentially 1.3 times 10 to the power of three m in S. I. So now recall that to calculate the amount of energy lost. We can do that by multiplying power but it's time and the time here is actually going to be one day because it is asking for the amount of energy loss each day. So I'm just gonna confer that into seconds which is the S. I. Units 24 hours in a day, 60 minute in an hour and seconds in a minute. Make sure that the units cancel out. And this will correspond to 400 seconds. Like. So. Okay, so now we know what the T. Is but we do not know what the P. Is. And recall that power can actually be calculated by multiplying ay and fee together. But we have to notice that in this case we do not know what the voltage is or the potential difference. But what we can calculate is what we're given is the diameter and the length. So that will correspond to the assistance. So we want to connect this to the resistance using home slaw recall that slaw have fi two B equals two. I am multiplied by R. And in this case we want to actually substitute defeat so that we can only we can actually express power in terms of I. And our only so P equals two pi square multiplied by our when we substitute defeat with this equation here, I am going to indicate that by just making a line there just like so Okay so now we want to try to determine what R. S using receptivity and area. And using this equation right here we can actually do that. So recall that the R. Is going to equal to row multiplied by L. Over A. And row is going to be the restive city of the aluminum cable. L. Is going to be the length and A. Is going to be the cables cross sectional area. Okay, so we can actually start by calculating the cables cross sectional area. So in this case the A. Is going to be remember we will have seven different stands. Right? So A. Is going to be seven multiplied by the area of the cross sectional of the cable which is just by our square where the R. Is the radius of the cable. So because the D. S. 4.8 times 10 to the power of minus three m. The R. Which is only half of it are is then going to be 2.4 times then to the power of minus three m just like so so A is then going to be seven multiplied by p multiplied by 2.4 times 10 to the part of minus three m squared. Just like so okay, I'm just going to leave it like that for now. So another thing that we need to know is the row or the receptivity of the aluminum, which in this case is a known fact that the row is going to be 2.75 times then to the power of -8, multiplied by meter. And this is a constant that we know. And now we also know what the L. Is. So we all we know all the variables needed to calculate the resistance. Let's start doing that. So the Are here is row L divided by eight. So this is going to be 2.75 times 10 to the power of -8. Uh multiple perimeter Will supply by L. Which is going to be 1.3 times 10 to the power of three m. Like so divided by the area which is seven times pi times 2.4 times 10 to the part of minus three m square. Don't forget the square. And this will actually come out to be 0.2822. Oh, just like so okay, now that we get what the resistance is, we also know what the eye is. We can start calculating what the power is. Okay so the power is then going to be I squared multiplied by R. And therefore the I. S. 4 30 AM pier Squared multiplied by 0.28 - two. And that will cause the power and we can just directly and put that into the E. Which is going to be P multiplied by T. So this will be 4:30 AM Squared multiplied by 0.822 multiplied by the time which is 86, seconds just like so and we can plug this into our calculator to get the E. 4.51 times 10 to the power of nine jewels. And that will be the final answer for this particular problem which is 4.51 times 10 to the power of nine jules. That will be answer a. Like. So So if we notice here the resistance of 0.2822 is actually a very low value but the large large current of 430 amp is actually what is responsible for the really large energy loss that we see here which is in the 10th report of nine magnitude which is pretty extreme. Okay so if you guys have any more problems or any question on this, make sure to watch our other videos. We have similar videos on similar topics. That will definitely help you guys to better your understanding of this topic. Awesome. Thank you.

A copper transmission cable 100 km long and 10.0 cm in diameter carries a current of 125 A. (b) How much electrical energy is dissipated as thermal energy every hour?

Verified Solution

Key Concepts

Ohm's Law

Resistance of a Conductor

Power Dissipation