Efficiency of Power Line Transmission

In summary: Please go back to this hint on c). Do you know what a voltage divider is? If not, when you looked it up on wikipedia.org (or elsewhere), what about it did you not understand?A voltage divider is a simple circuit that divides a source voltage into smaller voltages using resistors. In this case, we are given the resistance of the supply line and we need to calculate the source voltage that would give us the target voltage of 240V at the consumer. This can be done using the formula for a voltage divider: Vout = Vin * (R2 / (R1 + R2)), where Vin is the source voltage, R1 is the resistance of the supply line, R2 is the
  • #1
yeah:)
25
0

Homework Statement



4561507462_0ebfea103e_b.jpg


Homework Equations



N/A



The Attempt at a Solution



I have been struggling with these questions for the past week, with not much result at all. I have to study at home, through the use of various textbooks, and with no teacher to guide me through the process. I would be extremely grateful to the person who could explain each question to me step-by-step, eventually reaching the answer. Thank you in advance.
 
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  • #2


yeah:) said:

Homework Statement



4561507462_0ebfea103e_b.jpg


Homework Equations



N/A



The Attempt at a Solution



I have been struggling with these questions for the past week, with not much result at all. I have to study at home, through the use of various textbooks, and with no teacher to guide me through the process. I would be extremely grateful to the person who could explain each question to me step-by-step, eventually reaching the answer. Thank you in advance.

The Relevant Equations are not "N/A". For a), use the relationship between power, current and voltage. For part b), use Ohm's Law.

Please show us your work on at least those two parts...
 
  • #3


OK, here is my attempt at the first two parts (I seriously cannot do the others - PLEASE help me!):

a) watts = volts*amps

48,000=240*amps

a = 200 amps

b) V=IR

R=V/I=48000/200=240 Ohms



PLEASE, PLEASE, PLEASE help me on the other parts!
 
  • #4


Hint on c) -- you calculated the effective resistance of the consumer. You are given the resistance of the supply line. Use a voltage divider equation to tell you what the source voltage must be to get the target 240V at the consumer.
 
  • #5


berkeman said:
Hint on c) -- you calculated the effective resistance of the consumer. You are given the resistance of the supply line. Use a voltage divider equation to tell you what the source voltage must be to get the target 240V at the consumer.

Sorry, I cannot understand that - could you explain it to me in more detail please?
 
  • #6


yeah:) said:
Sorry, I cannot understand that - could you explain it to me in more detail please?

What do you not understand? If you are not familiar with the term "voltage divider", you could use wikipedia.org to look it up.
 
  • #7


Could you or someone else please just do the last three parts, explaining them as you go? That way, I would stand a chance of understanding how to do them...
 
  • #8


yeah:) said:
Could you or someone else please just do the last three parts, explaining them as you go? That way, I would stand a chance of understanding how to do them...

Nope. We don't help you cheat here. You'll need to try another website for that feature.

If you are willing to do some work on your own, this is a great website. If you want to be spoonfed now, and fail later, then you need to try some other websites.

BTW, I'm going to change the title of your thread. Momentum Conservation seems a bit off-topic.
 
  • #9
Look, however you treat me on this forum, the last thing I want you to say about me is that I 'cheat'. As I have already mentioned, I have to study MYSELF, AT HOME, with only textbooks to help me - I wrote these questions out myself, because I was struggling with them and needed full explanation on them - could you please do me that favour and explain the last three parts fully? I myself should know best the way in which I learn, and that way has always been by example. So please, please show me how to do those - they should probably take around 30 seconds for a physicist of your standard.
 
  • #10
yeah:) said:
Look, however you treat me on this forum, the last thing I want you to say about me is that I 'cheat'. As I have already mentioned, I have to study MYSELF, AT HOME, with only textbooks to help me - I wrote these questions out myself, because I was struggling with them and needed full explanation on them - could you please do me that favour and explain the last three parts fully? I myself should know best the way in which I learn, and that way has always been by example. So please, please show me how to do those - they should probably take around 30 seconds for a physicist of your standard.

When you supply power down a trasmission line (or just over a wire), there is some power loss in the wire due to I^2R loss from the current having to pass through some (hopefully low) resistance. One of the ways you minimize those losses in AC Mains power distribution is to step up the voltage for the actual transmission line, and then step it back down again at the consumer location. Why would stepping up the voltage cut down power losses in the transmission line, assuming a constant power load requirement at the consumer location?
 
  • #11
Could you please explain this concept to me, using the questions as the example? I would understand this much better if you showed me how to do it.
 
  • #12
yeah:) said:
Could you please explain this concept to me, using the questions as the example? I would understand this much better if you showed me how to do it.

This wikipedia article should help you to understand:

http://en.wikipedia.org/wiki/Power_transmission_line

.
 
  • #14
yeah:) said:
I have read the article, and I still cannot figure out how to do the questions - please just show me!

No can do -- it's against the rules here.

berkeman said:
Hint on c) -- you calculated the effective resistance of the consumer. You are given the resistance of the supply line. Use a voltage divider equation to tell you what the source voltage must be to get the target 240V at the consumer.

Please go back to this hint on c). Do you know what a voltage divider is? If not, when you looked it up on wikipedia.org (or elsewhere), what about the explanation did you not understand?
 

1. What is the efficiency of power line transmission?

The efficiency of power line transmission refers to the percentage of energy that is successfully transmitted from the power plant to the end user. It takes into account factors such as resistance, voltage drop, and other losses that occur during the transfer of electrical energy.

2. How is the efficiency of power line transmission calculated?

The efficiency of power line transmission is calculated by dividing the amount of energy delivered to the end user by the amount of energy generated by the power plant. This value is then multiplied by 100 to obtain a percentage.

3. What factors affect the efficiency of power line transmission?

There are several factors that can affect the efficiency of power line transmission, including the distance between the power plant and the end user, the quality of the power lines and equipment used, and weather conditions. In addition, the type of energy being transmitted (e.g. AC or DC) can also impact efficiency.

4. How does the efficiency of power line transmission impact energy costs?

The efficiency of power line transmission can have a significant impact on energy costs. A lower efficiency means that more energy is lost during transmission, resulting in higher costs for the end user. In order to maintain a high level of efficiency, regular maintenance and upgrades to the power grid are necessary.

5. Can the efficiency of power line transmission be improved?

Yes, the efficiency of power line transmission can be improved through various methods such as using higher voltage transmission lines, implementing smart grid technology, and reducing the distance between the power plant and end user. Additionally, investing in renewable energy sources can also help to improve the overall efficiency of the power grid.

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