How do power lines have a high voltage and a low current?

AnalogKid

Joined Aug 1, 2013
10,987
I think I'm missing something here, how can they transmit at a high voltage and low current when there needs to be a high resistance according to 'V = IR', where does the high resistance come from?
Your basic premise is incorrect. You are assuming that the purpose of the wire is to limit the current to hold the voltage constant. In fact, it is the wire resistance that is (relatively) constant, and the voltage and current vary around it according to a permutation of Ohm's Law: R = E / I For a constant resistance, like a fixed length of wire, if you increase the voltage, the current increases also.

More important is Watt's Law, P = E x I This is what drives power company's designs. If you want to deliver 2000 W, this can be 1000 V at 2 A or 100 V at 20 A. With just this equation there is no implicit advantage to either high or low voltage. But next comes Joule's Law, P = I^2 x R This falls out from a combination of Ohm's and Watt's Laws. With Joule's Law, it is clear that for any particluar current in a wire (the wire is the resistor in this case), as the resistance increases so does power dissipated in the wire. This is why there is an advantage to high voltage/low current when transporting large amounts of power over long distances. In the previous example, if the 2000 watts is going through a wire with 2 ohm resistance, then for 1000 V at 2 A, the loss in the wire is 8 W, but for 100 V at 20 A, the loss is 800 W.

The important relationship is that the power loss in a resistor or wire increases as the square of the current. That adds up.

ak
 

boatsman

Joined Jan 17, 2008
187
I remember reading in a science magazine from the 1950s that in the Soviet Union long distance electrical power transmission was using DC with the negative wire connected to the earth and only the positive wire being used on the pylons. Incidently is it possible theoretically to route high tension cables underground instead of the overhead pylons being blots on the landscape? Or would there be in addition to the higher cost problems of insulation and radiation?
 

ISB123

Joined May 21, 2014
1,236
High tension cables are not underground because they generally stretch for miles across the country.The cost would simply be too much since it would require digging trenches,cable having thick isolation and people could get hurt much more easier since they are not aware of power lines under the ground unless they have plans.
 

boatsman

Joined Jan 17, 2008
187
High tension cables are not underground because they generally stretch for miles across the country.The cost would be simply too much since it would require digging trenches,cable having thick isolation and people could get hurt much more easier since they are not aware of power lines under the ground unless they have the plans.
Maybe one day every household will be self sustaining electrically. I recall in the UK gas was supplied to each household in pipes from the local gasworks. Nowadays liquified propane gas is either supplied to each house in cylinders or several houses are connected to a large underground storage tank that is periodically filled up by the gas supplier, each house having a meter.
 

ISB123

Joined May 21, 2014
1,236
I doubt about that,government would lose massive amounts of money if everyone produced its own power.In case we manage to get self sustained power country would impose some kind of tax and price probably would remain the same like you used their power.
 

blocco a spirale

Joined Jun 18, 2008
1,546
I recall in the UK gas was supplied to each household in pipes from the local gasworks. Nowadays liquified propane gas is either supplied to each house in cylinders or several houses are connected to a large underground storage tank that is periodically filled up by the gas supplier.
This is completely untrue.
 

AnalogKid

Joined Aug 1, 2013
10,987
Underground high power electrical distribution certainly is possible, and is common in the downtown areas of most large and medium sized cities, and in many more upscale suburbs that want to have a "better" look about them. The higher installation and maintenance costs are offset by the lower costs of not having to shut down or work around the high tension lines when building buildings. But the big, continuing, long term cost is the loss of electricity due to increased capacitance among the three phase legs. The smaller the tunnel, the lower the installation cost, but the higher the power loss.

ak
 

Thread Starter

Robert Smith_1437948150

Joined Jul 26, 2015
38
As #12 implied, when a power line is carrying a certain amount of power as determined by I*V then, for the expression V = I*R, the R is not the line resistance, it is the equivalent resistance of all the loads connected at the far end of the line.
...and now I'm confused.

In OBW0549's example he used V = IR to calculate the voltage lost on the transmission wires, and therefore R was the resistance of the wire. Now your saying R is the resistance of all the loads at the end of the line?

Regards.
 
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Brownout

Joined Jan 10, 2012
2,390
...and now I'm confused.

In OBW0549's example he used V = IR to calculate the voltage lost on the transmission wires, and therefore R was the resistance of the wire. Now your saying R is the resistance of all the loads at the end of the line?

Regards.
I think you had it after OBW549's post. Just stop there and let it sink in before getting confused again.
 

Thread Starter

Robert Smith_1437948150

Joined Jul 26, 2015
38
Haha yes, I should have left it there.

Again, in simple terms can someone just explain why "In OBW0549's example he used V = IR to calculate the voltage lost on the transmission wires, and therefore R was the resistance of the wire. Now your saying R is the resistance of all the loads at the end of the line?"

Then I'm just going to leave it...

Regards.
 

cmartinez

Joined Jan 17, 2007
8,220
Haha yes, I should have left it there.

Again, in simple terms can someone just explain why "In OBW0549's example he used V = IR to calculate the voltage lost on the transmission wires, and therefore R was the resistance of the wire. Now your saying R is the resistance of all the loads at the end of the line?"

Then I'm just going to leave it...

Regards.
I agree... once your understanding "sinks in" then you can investigate further... otherwise it might cause more confusion for you to dig deeper at this point.
 

ISB123

Joined May 21, 2014
1,236
R just means a resistive load.It could be a wire,light bulb,TV,computer,power drill,etc..
Basically you have series of transformers converting high voltage/low current to low voltage/high current the closer it gets to it's final destination(settlements).
 

OBW0549

Joined Mar 2, 2015
3,566
Alright, I'm going to give this one more shot and then I'm done.

Haha yes, I should have left it there.
Yes, you should have, because you're making the same mistake now that you made in the very beginning.

Again, in simple terms can someone just explain why "In OBW0549's example he used V = IR to calculate the voltage lost on the transmission wires, and therefore R was the resistance of the wire. Now your saying R is the resistance of all the loads at the end of the line?"
OK, listen carefully:

If you are trying to relate the voltage drop along the wires and the current running through the wires via the formula V = I*R, the resistance of the wires is what's relevant.

If you are trying to relate the generator voltage and the current running through the circuit via V = I*R, it's the total resistance that's relevant; that is, the resistance of the load plus the resistance of the wires connecting it to the generator (resistances in series add).

I don't know any other way to put it: when using the formula V = I*R, you need to use the right V, the right I, and the right R; otherwise you get nonsense results.
 

Thread Starter

Robert Smith_1437948150

Joined Jul 26, 2015
38
Alright, I'm going to give this one more shot and then I'm done.


Yes, you should have, because you're making the same mistake now that you made in the very beginning.


OK, listen carefully:

If you are trying to relate the voltage drop along the wires and the current running through the wires via the formula V = I*R, the resistance of the wires is what's relevant.

If you are trying to relate the generator voltage and the current running through the circuit via V = I*R, it's the total resistance that's relevant; that is, the resistance of the load plus the resistance of the wires connecting it to the generator (resistances in series add).

I don't know any other way to put it: when using the formula V = I*R, you need to use the right V, the right I, and the right R; otherwise you get nonsense results.
Right, got it now, thanks again.

Regards.
 

dl324

Joined Mar 30, 2015
16,846
High tension cables are not underground because they generally stretch for miles across the country.The cost would simply be too much since it would require digging trenches,cable having thick isolation and people could get hurt much more easier since they are not aware of power lines under the ground unless they have plans.
Regions that experience severe weather sometimes bury transmission lines so they don't have to deal with outages related to trees touching lines and ice causing them to sag and short. My neighborhood should have underground lines but, unfortunately, it's not cost effective for the power company to do it. When we have bad weather, we can be without power for a week at a time...

In the USA, people are supposed to use cable locator companies to mark underground utilities before digging. Several of my neighbors didn't know better and interrupted some of the phone service to our neighborhood when they were installing a culvert.
 
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ISB123

Joined May 21, 2014
1,236
Well I don't know how it works in the USA but where I live all power cables are underground in newly built settlements generally on left or right side of the road.But 500kV lines are suspended on towers.
 

wayneh

Joined Sep 9, 2010
17,496
Right, got it now, thanks again.
I'm late to this thread but the way I think of it is very simple (for me) to remember: The power loss from a conductor goes up with the square of current driven through that conductor.

That's it. A square law for power-loss versus current. It's immediately obvious then, that for a given power to the load though a wire, you want to minimize current in the wire. You accomplish that by increasing the voltage until you reach practical limits.

Power loss = I^2•R where R is the resistance of the conductor and I is the current passing through it. It is NOT the R of your load, which was confusing you. ∆V of the load = I•Rload.
 
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