
#1
Apr2907, 12:53 AM

P: 23

how do you minimize resistance on a power line if there is a large amount voltage and low amperage being sent through it to a house and the house ends up with a lot of power?




#2
Apr2907, 01:44 AM

Mentor
P: 21,998

Transformers step up then down the voltage to give you 120/240 V in your house, while the high tension lines carry much larger (500 kV, I think) voltages.
Resistance kills amperage, not voltage, so you raise the voltage to minimize the power loss. Resistance is just a physical a property of the wire. 



#3
Apr2907, 01:56 AM

P: 388

The best will be to use superconducting wires. If not, you could try very thick silver conductors.
Of course this is just a joke. It is not necessary to minimize the resistance in a power line. It is a tradeoff between the cost of the energy lost and the cost of the line itself. This is the reason why no power lines work at millions of volts. They would be too expensive and a nuisance. Most power lines work at much lower voltages. Moreover, you cannot feed a house with mains power supply at thousands of volts. Soon you wouldn't have a single customer left! The tradeoff consists to use hundreds of thousands volts for long distances, tens of thousands for local distribution, and mains voltage at block level. You use transformers to lower the voltage at each transition. 



#4
Apr2907, 02:17 AM

P: 388

high voltage, low amperage, large amount of powerI ignore the relative small capacitive current on the line. 



#5
Apr2907, 09:48 PM

Mentor
P: 21,998

My first sentence there is perhaps badly worded, but I think we are talking about two different things. I think you are talking about voltage drop along the wire in a single case, but I'm talking about the engineering decision to select a higher voltage to transmit the required power. By selecting a higher voltage and lower amperage, you minimize the power loss. What I meant by "resistance kills amperage" is just that in most circuits the voltage is a fixed property of the voltage source, so if you throw in more resistance, the voltage stays the same and current drops.




#6
Apr2907, 11:02 PM

P: 308

So what was the question? 



#7
Apr3007, 01:35 PM

P: 6

Maybe I can add my question then
Like the previous posters people always talk about how you can transport a large amount of power with minimal losses because you use a low current, and a high voltage in your wires. Then they say that the heat losses are the following: I^2 x R x t = Q This is find very weird, because no matter what combination of numbers you plug in the Q/t is always the same as the total power (U x I = P). So every last bit of energy you put in is turned into heat. (Short circuit?) Now I assume this means that this equation isnt the whole story, because this all seems rather weird... I guess since in a normal situation there are actually multiple power consumers on the end of the wire, you should really use the equation like you use it in a serial circuit, so the wire (object/machine/etc) with the most resistance gets the most power delivered to it, and thus the least power delivered is in the object with the least resistance (powerline, where we would call the power delivered "loss" ) Am I missing something completely obvious? Or I get it right? Or am I just misapplying the equations? Anyways, thanks for your patience, somehow I always found this very confusing. Especially when people simply repeat Joule's Law without clarification. 



#8
Apr3007, 02:07 PM

P: 388

You are making confusion between the resistance of the power line and the equivalent resistance of consumers. If the resistance of the line is R and the resistance of consumers is Rc the total power is:
P=(R+Rc) I^2 = RI^2 + RcI^2 the first term represent the loses in the line and the second the power in the consumer's home. It is important that R (line) be very small compared to Rc. But that is the problem of the electricity provider. 



#9
Apr3007, 02:18 PM

P: 6

Right, so it is the difference between the resistance of the powerline and the resistance of the end user that makes the real difference.
Thanks for the quick reply 



#10
Apr3007, 02:29 PM

P: 388

Yes. Just, as I said, it is the equivalent resistance of all end users. And it is not the difference that counts but the actual value of both




#11
Apr3007, 02:36 PM

P: 6

D^oh
Sorry, I should use my terms more accurately. I got it now 



#12
May107, 10:38 PM

P: 31

What makes it possible is that it's DC! Only resistive losses. As you enter the LA area, you pass by the power station that converts it back to AC. The local impulse EM radiation is intense. 1800 MW power. Blanks out your car radio for a short while. 



#13
May107, 10:50 PM

P: 308





#14
May207, 12:39 AM

P: 388

I searched Google, but the only thing I found is a 1.15 MV power line in Russia. I would also know what you do with 1MV DC. As you cannot use transformers to lower the voltage, you need something like a motorgenerator that works at 1MV. I'm not electrical engineer and I am very curious to know how such a thing works. 



#15
May207, 07:42 PM

P: 308

You definitely don't need a motor/generator. Solid state is regarded as preferable. You just use a "valve hall" or something (a building full of giant transistors or other switching devices) to invert the signal 60 times a second. 



#16
May207, 10:29 PM

P: 308

OK, I checked, and couldn't find any MV AC lines other than the russian one. I thought I had seen some others somewhere, but I guess maybe not.
As for DC, the abovelinked Wikipedia page on HVDC lists several lines (in the tables near the bottom) greater than or equal to +500kV, which is a megavolt all the way across. 



#17
May307, 01:53 AM

P: 388




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