# Efficiency of Power Line Transmission

1. Apr 28, 2010

### yeah:)

1. The problem statement, all variables and given/known data

2. Relevant equations

N/A

3. 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.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 28, 2010

### Staff: Mentor

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

3. Apr 28, 2010

### yeah:)

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

4. Apr 28, 2010

### Staff: Mentor

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. Apr 28, 2010

### yeah:)

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

6. Apr 28, 2010

### Staff: Mentor

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. Apr 28, 2010

### yeah:)

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. Apr 28, 2010

### Staff: Mentor

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. Apr 28, 2010

### yeah:)

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. Apr 29, 2010

### Staff: Mentor

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. Apr 29, 2010

### yeah:)

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. Apr 29, 2010

### Staff: Mentor

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

.

13. Apr 30, 2010

### yeah:)

14. Apr 30, 2010

### Staff: Mentor

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

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?