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

Discussion in 'General Electronics Chat' started by Robert Smith_1437948150, Jul 26, 2015.

1. ### Robert Smith_1437948150 Thread Starter Member

Jul 26, 2015
38
1
Hello,

First of all, electronics is not my area so I have very little knowledge in the subject, but I was hoping someone could help clear something up for me in the simplest terms possible.

I understand power lines use a high voltage and low current to improve efficiency, and the formula for this is 'P = VI'. For a fixed amount of power if you increase the voltage then the current is reduced. To deliver 100W you can either have 50V and 2I or 25V and 4I.

...but looking at Ohms law, V =IR, if we want to have a higher voltage and lower current the resistance has to go up, e.g 50v = 2I x 25R or 25V = 4I x 6.25R.

Do power lines then need high resistance to send at a high voltage whilst maintaining a low current? Aren't power lines meant to be low resistance, where does the resistance come from?

If anyone can help clear up my confusion in the simplest terms possible it would be greatly appreciated.

Thanks,
Robert

2. ### dl324 Distinguished Member

Mar 30, 2015
3,384
653
Wires are rated by current carrying capability; higher capacity means a larger diameter wire and more weight. Power companies use high voltage transmission lines and step down the voltage at the pole just before connecting to building service so they can use smaller gauge wire for transmission; reducing both wire cost and weight.

As you mention, P=IV; to deliver 1000W, you can have V=1000V and I = 1A, V=100V and I = 10A, etc. A 1A current can be carried in a much smaller diameter wire than 10A.

Last edited: Jul 26, 2015
3. ### Robert Smith_1437948150 Thread Starter Member

Jul 26, 2015
38
1
Hi dl324, thanks for the response.

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?

Regards.

4. ### ISB123 Well-Known Member

May 21, 2014
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Voltage comes from the transformers.Wire resistance is to blame for power loss it doesn't dictate output voltage.

5. ### Brownout Well-Known Member

Jan 10, 2012
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First, the efficiency comes from P = V*I, or more specifically, P = I^2*R, where R is the resistance of the transmission wire. Now, the 'resistance' you're asking about comes at the end user, either a business or residence. Before the power is connected to the end user, it is transformed to a lower level, so that it really doesn't matter what the voltage was at the high voltage segment, the voltage at the user is the conventional low residence or business level.

6. ### Robert Smith_1437948150 Thread Starter Member

Jul 26, 2015
38
1
Thanks ISB123 and Brownout.

To be honest I still don't understand it.

Can anybody provide a simple example using some arbitrary values.

If not I'll just leave it, as I've done quite a lot of looking online and can't seem to grasp the concept, so I'm obviously out of my depth with this stuff.

Jul 18, 2013
10,859
2,525
One other issue or side effect of HV transmissions, is if it is over considerable distances there are also radiation losses to be considered.
To offset this now long distance power distribution is to convert to DC and convert back to AC on the consumer end.
Also one less conductor.
Max.

8. ### jpanhalt AAC Fanatic!

Jan 18, 2008
5,699
909
How very sad. Please be more specific as to what you want.

John

9. ### dl324 Distinguished Member

Mar 30, 2015
3,384
653
They want wire resistance to be as low as possible to minimize power transmission losses. Ideally there would be no IR drop; that's why they use high voltages for transmission lines and step it down with transformers before entering the building.

In an ideal transformer, power is conserved; 1000V @ 1A will give you 100V @ 10A.

10. ### cmartinez AAC Fanatic!

Jan 17, 2007
3,698
2,792
It's not as hard as you think... discard the equation for a moment and consider this:

• Power is created by the generator
• It is then transported through the lines, that is, copper wires.
• Copper wire has resistance. A very, very small resistance but it has resistance nevertheless.
• There are two things that limit power arriving on the receiving end:
1. The first one, is the power being input in the generator itself.
2. The second one, is the power that lost in the copper wire, due to its resistance. This means that the wire will heat up, and that heat is the power lost that won't be able to arrive to its destination.
• So even if the wire had zero resistance (a super conductor) and even if the generator's windings did too, you'd still have a limit on the amount of power you'd be able to tap into the receiving end, since the power received can never be greater than the power that was input in the generator in the first place.

11. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998
If a residence runs on 120V RMS and is using 100Amps, then the effective resistance ( though we never really call it that ) is 120V/100A = 1.2Ohms. And as I said, it doen't matter what the high voltage segment is because a transformer is used to transform the voltage to 120V.

12. ### Robert Smith_1437948150 Thread Starter Member

Jul 26, 2015
38
1
Thanks, so to reduce the power loss they will send at a high voltage and low current.

Going back to the equation V = IR, how does this work out when you also have the low resistance from the copper wire if you see what I mean? How can the power line have a really high voltage when both the current and resistance is low?

Thanks.

13. ### ISB123 Well-Known Member

May 21, 2014
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For example lets say you have a small heater that draws 1000W at 220V(That voltage comes from a transformer box which is hooked up to high power lines).
High Power Line-->Trafo. Box-->Home-->Device

Ohms law says that for heater to draw 1000W you would need 50Ω at 220V which would result in current drawn of 4.5A. You would be drawing
4.5A through your Breaker Panel which can draw 200A max.(Fuse rated) from the Trafo. Box.

Basically high power line doesn't care how much resistance the carrier wire has since it provides high current at low voltage from the trafo. box(Power Distribution Box in your street).

220V*4.5A=1000W

7000V*4.5A=31500W

Without transformer you wouldn't have to worry about power line resistance since there would be no wire left.

Last edited: Jul 26, 2015
14. ### OBW0549 Well-Known Member

Mar 2, 2015
1,395
941
Actually, you're missing several things, not just one. You've got V=IR, the relationship between voltage, resistance and current right; but you've managed to get yourself thoroughly confused as to which voltage, which current, and which resistance you should be looking at.

Here's the real deal: the amount of current flowing through the transmission wires, times the resistance of those wires, will determine how much voltage is lost along the transmission wires from the generator end to the load end. The higher the resistance, the more voltage will be lost; and likewise, the higher the current, the more voltage will be lost. Now here's the key: the voltage applied by the generator to the transmission line-- that is, the voltage between one conductor of the transmission line and the other conductor-- has absolutely no bearing whatsoever on that calculation. It is completely irrelevant. V=IR determines how much voltage is lost due to that current going through the transmission line's resistance, NOT the voltage applied to the line by whatever is driving it.

For example, suppose I have a hydroelectric generator that delivers 500,000 volts to a transmission line stretching a hundred miles, and that transmission line has a resistance of 10 ohms from one end to the other; and at the far end of the transmission line, the load (a city, for example) has a nominal power demand of 500 megawatts and therefore draws a thousand amperes. 1000 amperes times 10 ohms equals 10,000 volts, which is the voltage lost going down the transmission line from the generator to the load. The generator is delivering 500,000 volts to the line, but only 490,000 volts appears a the other end where the load is.

The problem, of course, is that since power equals voltage times current, that lost voltage represents 10 million watts of wasted energy-- 10,000 volts times 1000 amperes equals 10,000,000 watts. Still, at this high voltage, the transmission system is losing only 2% of the power being pumped into it. Not bad.

Now lets try to send that same 500 megawatts down that exact same transmission line, but at a much lower voltage-- say, 100,000 volts. This time, 5,000 amperes are needed to transmit the same amount of power. What happens to the voltage drop along the transmission line? The line drop is equal to 5000 amps times 10 ohms, or 50,000 volts; and 50,000 volts times 5,000 amps is 250 megawatts. Now, we're losing fully 50% of our power in the transmission line!

And there you have the basic reason why we try to transmit electrical power at the highest voltage practical: it minimizes power loss in the transmission wire.

15. ### cmartinez AAC Fanatic!

Jan 17, 2007
3,698
2,792
Another simple concept, really. The equation of the relation between resistance, current and voltage (ohm's law) is V=IR
But the equation of the relation between power (Watts) voltage and current is: P=VI
So you can always trade between voltage and current to get the same power... for example, if you have 5 amps of current at 10 volts, the power being produced is 50 watts... but if you were to double the amps, and halve the volts, and had instead 10 amps of current at 5 volts... you'd still end up with 50 watts of power!

16. ### Robert Smith_1437948150 Thread Starter Member

Jul 26, 2015
38
1
Perfect, this is exactly what I was looking for.
Thanks for taking the time to answer and give the examples, made it much easier to understand.

Thanks to everyone else as well, sorry if I didn't explain what I was trying to figure out clear enough!

Regards,
Robert

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17. ### LDC3 Active Member

Apr 27, 2013
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There is one thing that OBW0549 didn't say, but dl324 did say. At large currents, the wire heats up, causing more resistance, which causes a higher voltage drop. A larger wire can carry more current with less heating and less voltage drop. But a larger wire increases the cost of running HV lines.

So, there is a trade-off between how much current to transfer and the voltage on the lines. Too much voltage and you have electric arcs shorting out the transmission.

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18. ### Robert Smith_1437948150 Thread Starter Member

Jul 26, 2015
38
1
Thanks for pointing that out

Regards,
Robert

19. ### #12 Expert

Nov 30, 2010
16,704
7,354
Without reading every post: I think the missing part might be the high resistance of the load in a residence. If I use 10 amps from a 120 volt transformer on a power pole, that looks like 12 ohms. The transformer on the pole reflects that resistance as much as it transforms the voltage. If the feed line to the local transformer is 7200 volts, the resistance in my house is interpreted as 43,200 ohms by the transformer.

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20. ### crutschow Expert

Mar 14, 2008
13,508
3,385
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.

Last edited: Jul 26, 2015
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