I don't understand the importance of high voltage-low current in electricity distribution

** You are confusing yourself by * failing to separate * load ohms with the ohms in the current carrying cable going to the load.

High voltage transmission is ALL about reducing the percentage of power lost in the cables that deliver electrical energy across a country. .

...... Phil

OK, step back a second and I'll attempt to explain this...

Here's the deal: You're building a (really tiny) power grid, and you'd like to distribute *up to* that 100 watts you specified. Assume that you've already decided that your "customer" will be getting 10V... this means they're allowed to pull anywhere from 0 to 10A, or, in resistance, anything from an infinite load (open circuit) down to a 1 ohm resistor.

Now, let's figure out how to get that 100W to the customer. First try the "easy" approach -- just ship them 10V directly! OK, but now let's assume there's a long power line between you and the customer that has a resistance of 1 ohm. If your only has a 10V/1W load (100 ohm or 100mA) -- then you now have a series circuit with 1 ohm resistance in the power line and 100 ohms in the customer's load. That's 101 ohms total, so the load actually draws

10V/101 ohm = 99mA and dissipates 99ma^2*100 = 980mW -- close enough that the customer won't complain. But what if they want to plug in a 10V/100W load? That implies the load's resistance is 1 ohm, but... oh oh... we now have a series circuit with 2 ohms total resistance and, as you can calculate, the customer only gets 5A going through their load or an open-circuit voltage of 5V. In many cases, 5V will be far too low to correctly operate what was designed as a 10V/100W load. Additionally, you can calculate that we're dumping 25W of power into the power line, and *the customer doesn't pay for that* (the power meter only monitors their load), so you're just losing a lot of money if you're the power company.

This problem goes away if we add a transformer to boost the power line voltage to, say, 1kV. Now, to transport 100W over the line, we only need 100mA rather than 10A. Worst case, in the power line we dissipate 100mA^2*1 ohm (the power line resistance) = 10mW -- utterly negligible.

Does this make sense?

The basic idea is that customers get to connect a load *up to* a specified wattage to your power line and that you want to be sure that, at the maximum specified load, only a tiny percentage of the total power gets dumped into the power line and the vast bulk of it goes to the customer.

BTW, the losses in the power lines are referred to as "I-squared-R" losses since that's how you calculated the power lost in them (I^2*R). Since doubling the voltage halves the current (for the same load power), the losses are quartered due to the I^2 term. Nice, huh? 10 times the voltage has 100 times less loss, etc...

---Joel

I think you have your thinking of the actual question given to you a little mixed up. To put it simply as one of the biggest factors are, less conductor material to relay the energy to the end point.

HV circuits like 5k, 12k etc. use very small conductors to transfer the energy to the customer. The current is less of the same ratio. At the end point, it then gets down converted into voltage where the current service will increase and voltage drops. At this point the conductors will have to be increased because a change in ohms in the conductors have more lossy effects at lower voltages verses the higher voltages.

etc.. It's really all about the use of material mass in the end.

if you were to attempt to transfer high levels of power at low voltages for great distances, the amount of copper used would be un thinkable like in supplying great cities for their electrical needs.

--
http://webpages.charter.net/jamie_5"

If the resistance in a circuit is constant, applying more voltage will result in more current, in accordance with Ohm's Law.

However, when dealing with transformers, we are not dealing with a constant resistance. Instead, a transformer passes a constant power - ignoring losses, the power into a transformer will equal the power out. The input voltage/output voltage ratio of a transformer is proportional to the turns ratio of the transformer, and the input current/output current ratio is inversely proportional to the turns ratio, to keep the input and output powers equal.

--
Peter Bennett, VE7CEI  
peterbb4 (at) interchange.ubc.ca  
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

They key is to understand that losses go with i^2, so reducing i (by raising V in proportion since P=3DiV) is a good thing.

Cheers

OK you guys have helped me but what I still don't get is: Why does the equation V^2/R produce a different power than I^2*R, and how do you reduce I and increase V in the equation P=IV without changing the ohms of resistance? I think that the *load*, for instance, the appliances in the house, affect the current right? So that would be the ohms changing? So in P=VI, P is the losses of the load you are powering, and in P=I^2*R that is the resistance of the wire? How do you know whether P is losses or what? I think I'm starting to understand.

ohms

There are two resistances involved. One is the resistance of the wire bringing the power from the station to the consumer. The other is the resistance of any load at the consumer's location. Call them R1 and R2.

There are three voltages to be concerned with. The first is the voltage at the station, the second the voltage=20 dropped along the wire, and the third is the voltage that arrives at the consumer end and is applied across the load resistance. Call them V1, V2, and V3. It is evident that V3 =3D V1 - V2.

The wasted power is given by P1 =3D V2*I =3D V2^2/R1 =3D I^2*R1.

The power delivered to the customer is given by P2 =3D V3*I =3D V3^2/R2 =3D I^2/R2.

Without fully following the thread, your original use of ohms law wasn't about the voltage drop, but was your expectation of a certain resistance because you had voltage and current.

You are actually dealing with the voltage drop along the cable, or the resistance of that cable which will cause a voltage drop.

The higher the current through a circuit, the more affect that resistance has. Lower the current, and the resistance of the cable becomes less of a factor.

But of course, lower the current and you can't supply as much power to the load. Which is why they raise the voltage to compensate for the lower current, the power in wattage being passed along the cable being the same if both are changed by the same factor.

Try a different angle. In the days of tubes, the voltages were all high voltage, while the current levels were really quite low. Except for circuits where really high power was used, like transmitters, you rarely saw large diameter wire in the wiring, since it didn't need to pass much current, and the resistance of that narrow diameter wire was not a factor. For a lot of equipment, the power supply would offer up 350v, if that much, but the current drain would never be more than a few hundred milliamps.

Then along came solid state devices. They all ran at quite low voltage, but the current drain was pretty high. So 12 volts or even 5volts, but it was often common to see an amp or so needed. Suddenly, you had to be careful of the wiring, because the resistance of the narrower gauge wire would become a factor. Bad connectors too, if they didnt' make good contact their resistance would be more significant. The resistance of the #20 wire or whatever was used did not change from when it was used in tube circuits, but if it had a resistance of 1ohm in the tube equipment, the 1ohm in solid state equipment might start being a problem because of the needed current passed through it.

Michael

--
If you have a 10 volt source which can deliver 100 watts of power to a
load, then the circuit looks like this:


         I--->    
    +--------------+

--- It doesn't.

Let's look at your earlier circuit again:

10V>--------+ | [1R] | GND>--------+

The current in the resistor will be:

E 10V I = --- = ----- = 10 amperes R 1R

and the power disspated by the resistor will be either:

E² 10²V P = ---- = ----- = 100 watts, or R 1R

P = I²R = 10²V * 1R = 100 watts

--- You can't.

---

--- Right.

---

--- Yes

---

--- The notation is largely immaterial. Watts is watts, so it just depends on what you're talking about or what you're looking at.

Using your 10V source we can look at it like this:

10V>--[WIRING RESISTANCE]---+ | [LOAD RESISTANCE] | GND>------------------------+

and if we assign an arbitrary resistance of, say, 0.1 ohm to the wiring resistance and leave the load at 1 ohm, then we'll have:

R1

10V>--[0.1R]--+ | [1R]R2 | GND>----------+

Now the current in the circuit will be:

E 10V I = --------- = ------ = 9.09 amperes. R1 + R2 1.1R

The voltage dropped across R1 will be:

E = IR = 9.09A * 0.1R = 0.91 volts

and the power the wiring will dissipate will be:

P = IE = 9.09A * 0.91V ~ 8.3 watts

or P = I²R = 9.09A² * 0.1R ~ 8.3 watts

E² 0.91V² or P = ---- = --------- ~ 8.3 watts R 0.1R

Since the supply puts out 10 volts and the wiring drops 0.91V of it, that means that the load has 9.09 volts across it, so it will dissipate:

P = IE = 9.09A * 9.09V ~ 82.6 watts

Just for fun, work it out using P = I²R and P = E²/R and you'll see that it all comes out the same.

The point is that one [general] notation doesn't stand for line losses and another one stand for load dissipation.

JF

and to extend this a bit more:

If we want to deliver 10A to a one ohm load, then the circuit looks like this:

I---> +----------------------------------------------------+

--- ^^^^ Oops... 10²A

JF

And it's easy to prove from V=IR

V^2/R = V * V / R = IR * IR / R = IR*I = I^2*R

Tim.

--
God said, "div D = rho, div B = 0, curl E = - @B/@t, curl H = J + @D/@t," 
and there was light.

     http://tjw.hn.org/      http://www.locofungus.btinternet.co.uk/

Thank you so much! I haven't actually read all the posts yet, but understanding that it is the voltage DROP clears things up for me. You guys are so helpful! I'm pretty new to electronics/electricity, but I'm starting to understand it much better. I'll probably have more questions in the future.

Also, my problem was that I KNEW the equations work, and I KNEW you couldn't change those without resistance, I did that with the equations, but now I get it!

Ohms don't enter into it for this calculation, except for the resistance of the transmission wires themselves.

Say you've got 1000 feet of AWG 10, resistance .9989 ohms/1000 ft. which is close enough to 1 ohm for this discussion (source:

}.

So, with a 10 volt, 10 A supply, the one ohm of wire will drop the whole

10 volts, leaving nothing for the load (actually, you'd get a voltage divider consisting of the wire resistance and the actual resistance of the load, but let's set that aside for now.)

With a 20 volt, 5A supply, the line drop is only 5 volts, leaving 15 volts for the load. With a 100V, 1A supply, the line drop is 1V, leaving 99V for the load, and so on.

The load itself has a resistance, which is, indeed, R = E/I; if you apply Ohm's law to the supply, it's referred to as "impedance", but that's a whole nother topic as well.

The point is, the higher voltage/lower current supply incurs less IR (E = IR) losses in the lines themselves.

Hope This Helps! Rich