# Why transmission lines are high voltage?

Discussion in 'General Electronics Chat' started by Vorador, Jan 26, 2015.

Oct 5, 2012
87
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I've read on a number of places that, since P=I^2*R, the lower the current the lower the power losses will be, so transmission lines carry low current, high voltage. But Power can also be written as P=V^2/R, so if we want to avoid losses, voltage also needs to be down, not just current. The way I understand it, it doesn't matter which quantity you step up and which you step down, because power is compensated by the reciprocal effect current and voltage have upon each other.

I've always been confused by this. I'll be really grateful for your illuminating response.

Thank you!

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2. ### wayneh Expert

Sep 9, 2010
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Do yourself some math, plugging in examples. For example 1000V and 1 A versus 1V and 1000A. Solve for the dissipated power in a wire with, say, 1Ω of resistance. Maybe plot power loss versus voltage.

You'll never second guess yourself again on this issue.

3. ### #12 Expert

Nov 30, 2010
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You seem to be having difficulty with the part about, "times R" compared to, "divided by R".

and...wayneh beat me again.
Must go buy more coffee today.

4. ### cmartinez AAC Fanatic!

Jan 17, 2007
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make it an espresso for me...

5. ### wayneh Expert

Sep 9, 2010
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On my way to Starbucks now.

Jul 18, 2013
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Some more numbers to digest!.
In the province where I live there are several Hydro dams which supply around 7400Km of total conductor length, the average dia is 4cm, the voltage is between 900Kv and 1000Kv.
Currently the systems are predominately DC transmission now, Less losses and one less conductor.
Max.

7. ### cmartinez AAC Fanatic!

Jan 17, 2007
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You mean one thousand kilo-volts???

Oct 2, 2009
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9. ### cmartinez AAC Fanatic!

Jan 17, 2007
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Jul 18, 2013
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A couple of things you don't hear alot, is Galloping Power lines and a Deer on the line.

Max.

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11. ### cmartinez AAC Fanatic!

Jan 17, 2007
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Resonant movement of power lines due to wind is something I've seen before... can't recall what the solution for that is, though. Does it have to do with the catenaries legth to sag ratio?
I've seen possums walking the tight rope between poles... but a deer atop the line is definitely new to me...

Jul 18, 2013
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Just out of view is a rail line, it is though that a loco struck a herd crossing the tracks, this was a result!.
Galloping lines are often a combo of ice and strong wind.
Max.

13. ### ian field Distinguished Member

Oct 27, 2012
4,447
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You forgot the bit about cost in the equation - sending the power at 230/110V would require conductors several feet thick.

14. ### JMW Member

Nov 21, 2011
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The deer was a hoax. It got tangled when they were pulling the cable.

15. ### JMW Member

Nov 21, 2011
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One other point, look up the "ampacity" of a cable. Say you want 1 KW (1 volt at 1000 amps). I recall a 60's era air search RADAR with a 1 volt magnetron supply, the current was over 100 amps, at any rate the "cables" were copper pipe.

16. ### cmartinez AAC Fanatic!

Jan 17, 2007
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Yeah... we recently discussed that recently in this forum... it's due to the skin-effect

17. ### GopherT AAC Fanatic!

Nov 23, 2012
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I have no doubt that a CN freight train can reach 80+ MPH on the Great Plains of Manitoba and blast a deer 30 feet or more in any direction.

18. ### cabraham Active Member

Oct 29, 2011
82
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The V^2/R you refer to is load power, whereas I^2*R is power lost in line. Big difference. The reason is straight forward as to why high voltage low current is employed.
Insulators are very good compared to conductors with regard to losses. Insulators incur a loss in the form of shunt conductance, denoted by G. G is reciprocal of R, but this R is the insulation resistance, different from load R, or cable R. Power lost in insulation is V^2*G. For any good insulator over a wide span of temperatures in common use, this loss is tiny. But it increases with square of voltage.
In conductors, power lost is I^2*R, where R is the cable resistance and I is load current. Load current and cable current are equal as the cable is in series with the load. As current increases, loss increases with square of current.
Increasing V results in lower I for a constant power load. High V incurs greater insulation loss and lower cable loss. But insulation loss is so many orders of magnitude below cable loss, that it is very advantageous to transmit power at high V values.
If cables could be designed for superconductivity at normal temperatures, then things might be different. The limitation with SC cables is the critical current value where SC is lost. Depending on the values, the transmission voltages could be lowered, but only to a point. I doubt we will ever see 125V with 1.25 million amps. Even SC cables are limited in their current capacity. Instead of 750kV at 200 amp, we may see 75kV at 2,000 amp, depending on the SC cable.
Anyway, as long as insulators are ultra-low loss, and cable losses are much greater, which at this time is the case, higher voltage with lower current will incur much lower losses. So that is why it is done this way. Did I help?

Claude

Last edited: Jan 27, 2015
19. ### WBahn Moderator

Mar 31, 2012
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You can use either expression and you will get the same result. The key is that to use P=(I^2)/R to find the power loss IN THE LINE you need to use the current THROUGH the line. Similarly, to use P=(V^2)/R to find the power loss IN THE LINE you need to use the voltage ACROSS the line. You are trying to use the voltage across B (the load or the generator) to find the power loss in A (the line).

Simple example. You have a line that has 0.1Ω of resistance and you need to deliver 100kW to a load. You can do this either with a 1kV line pulling 100A or you can do it at 100kV pulling 1A. How much power is lost in the line in both cases?

Using P=(I^2)/R:
1kV -> P = (100A)^2 * 0.1Ω = 1000W
100kV -> P = (1A)^2 * 0.1Ω = 0.1W

Using P=(V^2)/R:
1kV -> V = (100A)(0.1Ω) = 10V; P = (10V)^2 / 0.1Ω = 1000W
100kV -> V = (1A)(0.1Ω) = 0.1V; P = (0.1V)^2 / 0.1Ω = 0.1W