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Sci.Electronics.Basics -> I don't understand the importance of high voltage-low current in electricity distribution

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Date: 19:55 14-04-08

My problem is this: You are distributing 100 watts. You start out with
10 volts, 10 amps. You increase the volts by a factor of 2 through a
transformer. Now you must have 20 volts, 5 amps right? Using Ohm's
Law, 10 volts = 10 amps * 1 ohms, in the first case there is 1 ohm.
But, in the next case you have 20 volts = 5 amps * 4 ohms. So why did
the ohms change? When you use a transformer to increase voltage,
doesn't that mean ohms must increase if amps decrease? I am thoroughly
confused. I know I am thinking about it the wrong way, can someone
explain it to me? If you want me to clarify, let me know. Basically my
question is this: How do you create more voltage without creating more
current?

Author: Phil Allison
Date: 20:24 14-04-08

<chesemonkyloma@gmail.com>
>
> My problem is this: You are distributing 100 watts. You start out with
> 10 volts, 10 amps. You increase the volts by a factor of 2 through a
> transformer. Now you must have 20 volts, 5 amps right? Using Ohm's
> Law, 10 volts = 10 amps * 1 ohms, in the first case there is 1 ohm.
> But, in the next case you have 20 volts = 5 amps * 4 ohms. So why did
> the ohms change? When you use a transformer to increase voltage,
> doesn't that mean ohms must increase if amps decrease? I am thoroughly
> confused. I know I am thinking about it the wrong way, can someone
> explain it to me? If you want me to clarify, let me know. Basically my
> question is this: How do you create more voltage without creating more
> current?

** 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. .

http://en.wikipedia.org/wiki/Electric_power_transmission#Losses

...... Phil

Author: Joel Koltner
Date: 20:26 14-04-08

<chesemonkyloma@gmail.com> wrote in message
news:e89f08c4-b769-4600-8fb1-b2bc0adf4658@24g2000hsh.googlegroups.com...
> How do you create more voltage without creating more
> current?

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

Author: Jamie
Date: 20:30 14-04-08

chesemonkyloma@gmail.com wrote:

> My problem is this: You are distributing 100 watts. You start out with
> 10 volts, 10 amps. You increase the volts by a factor of 2 through a
> transformer. Now you must have 20 volts, 5 amps right? Using Ohm's
> Law, 10 volts = 10 amps * 1 ohms, in the first case there is 1 ohm.
> But, in the next case you have 20 volts = 5 amps * 4 ohms. So why did
> the ohms change? When you use a transformer to increase voltage,
> doesn't that mean ohms must increase if amps decrease? I am thoroughly
> confused. I know I am thinking about it the wrong way, can someone
> explain it to me? If you want me to clarify, let me know. Basically my
> question is this: How do you create more voltage without creating more
> current?
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";

Author: Peter Bennett
Date: 22:06 14-04-08

Author: Varactor
Date: 06:32 15-04-08

On Apr 15, 11:55=A0am, chesemonkyl...@gmail.com wrote:
> My problem is this: You are distributing 100 watts. You start out with
> 10 volts, 10 amps. You increase the volts by a factor of 2 through a
> transformer. Now you must have 20 volts, 5 amps right? Using Ohm's
> Law, 10 volts =3D 10 amps * 1 ohms, in the first case there is 1 ohm.
> But, in the next case you have 20 volts =3D 5 amps * 4 ohms. So why did
> the ohms change? When you use a transformer to increase voltage,
> doesn't that mean ohms must increase if amps decrease? I am thoroughly
> confused. I know I am thinking about it the wrong way, can someone
> explain it to me? If you want me to clarify, let me know. Basically my
> question is this: How do you create more voltage without creating more
> current?

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

Date: 15:16 15-04-08

On Apr 15, 6:32 am, Varactor <Morefl...@gmail.com> wrote:
> On Apr 15, 11:55 am, chesemonkyl...@gmail.com wrote:
>
> > My problem is this: You are distributing 100 watts. You start out with
> > 10 volts, 10 amps. You increase the volts by a factor of 2 through a
> > transformer. Now you must have 20 volts, 5 amps right? Using Ohm's
> > Law, 10 volts = 10 amps * 1 ohms, in the first case there is 1 ohm.
> > But, in the next case you have 20 volts = 5 amps * 4 ohms. So why did
> > the ohms change? When you use a transformer to increase voltage,
> > doesn't that mean ohms must increase if amps decrease? I am thoroughly
> > confused. I know I am thinking about it the wrong way, can someone
> > explain it to me? If you want me to clarify, let me know. Basically my
> > question is this: How do you create more voltage without creating more
> > current?
>
> They key is to understand that losses go with i^2, so reducing i (by
> raising V in proportion since P=iV) 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.

Author: Greg Neill
Date: 15:47 15-04-08

<chesemonkyloma@gmail.com> wrote in message
news:808aa209-d4f8-4b27-a83b-dac3b61c8f2a@t54g2000hsg.googlegroups.com=20

> 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=3DIV 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=3DVI, P is the losses of the load you are powering,
> and in P=3DI^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.

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.

Date: 16:49 15-04-08

On Tue, 15 Apr 2008, chesemonkyloma@gmail.com wrote:

> On Apr 15, 6:32 am, Varactor <Morefl...@gmail.com> wrote:
>> On Apr 15, 11:55 am, chesemonkyl...@gmail.com wrote:
>>
>>> My problem is this: You are distributing 100 watts. You start out with
>>> 10 volts, 10 amps. You increase the volts by a factor of 2 through a
>>> transformer. Now you must have 20 volts, 5 amps right? Using Ohm's
>>> Law, 10 volts = 10 amps * 1 ohms, in the first case there is 1 ohm.
>>> But, in the next case you have 20 volts = 5 amps * 4 ohms. So why did
>>> the ohms change? When you use a transformer to increase voltage,
>>> doesn't that mean ohms must increase if amps decrease? I am thoroughly
>>> confused. I know I am thinking about it the wrong way, can someone
>>> explain it to me? If you want me to clarify, let me know. Basically my
>>> question is this: How do you create more voltage without creating more
>>> current?
>>
>> They key is to understand that losses go with i^2, so reducing i (by
>> raising V in proportion since P=iV) 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.
>
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

Author: John Fields
Date: 17:34 15-04-08

On Mon, 14 Apr 2008 16:55:17 -0700 (PDT), chesemonkyloma@gmail.com
wrote:

>My problem is this: You are distributing 100 watts. You start out with
>10 volts, 10 amps. You increase the volts by a factor of 2 through a
>transformer. Now you must have 20 volts, 5 amps right? Using Ohm's
>Law, 10 volts = 10 amps * 1 ohms, in the first case there is 1 ohm.
>But, in the next case you have 20 volts = 5 amps * 4 ohms. So why did
>the ohms change? When you use a transformer to increase voltage,
>doesn't that mean ohms must increase if amps decrease? I am thoroughly
>confused. I know I am thinking about it the wrong way, can someone
>explain it to me? If you want me to clarify, let me know. Basically my
>question is this: How do you create more voltage without creating more
>current?

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

I--->
+--------------+ <--------+
| | |
[SOURCE] [LOAD]R [E]
| | |
+--------------+ <--------+

and, if the load _is_ dissipating 100 watts, then the current in it
must be:

P 100W
I = --- = ------ = 10 amperes
E 10V

and its resistance must be:

E 10V
R = --- = ----- = 1 ohm
I 10A

Then, we can write, for convenience:

10V
10A--> /
+--------------+
| |
[SOURCE] [1R]
| |
+--------------+

Now, if we add a transformer in order to double the voltage into the
load, we'll have:

20V
20-->A /
+----+ +----------+
| P||S |
[SOURCE] R||E [1R]
| I||C |
+----+ +----------+

Notice that, since we've increased the voltage into the load to 20V,
the current through the load will be:

E 20V
I = --- = ----- = 20 amperes
R 1R

and the power it'll want to dissipate is:

P = IE = 20V * 20A = 400 watts.

Now we have a problem, since the source can only supply 100 watts.

So what do we do?

Change the load resistance so that it'll be dissipating 100 watts.

Like this:

E² 20²
R = ---- = ------ = 4 ohms
P 100W

So you can see that you were on the right track.

Matter of fact, in the real world the problem becomes one of changing
the resistance of the load as power supply voltages change. For
example, in countries where 120VAC mains are standard, a 100 watt
incandescent lamp filament has a resistance of about 80 ohms when the
lamp is hot while in countries with 240VAC mains a 100 watt
incandescent lamp will have a filament resistance of about 320 ohms.

So, the answer to your question:

"How do you create more voltage without creating more current?"

is: "Increase the load resistance."

JF

Author: John Fields
Date: 18:25 15-04-08

On Tue, 15 Apr 2008 12:16:17 -0700 (PDT), chesemonkyloma@gmail.com
wrote:

>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,

---
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

>and how do you
>reduce I and increase V in the equation P=IV without changing the ohms
>of resistance?

---
You can't.
---

>I think that the *load*, for instance, the appliances
>in the house, affect the current right?

---
Right.
---

>So that would be the ohms changing?

---
Yes
---

>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.

---
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

Author: Tim Woodall
Date: 18:33 15-04-08

On Tue, 15 Apr 2008 16:34:45 -0500,
John Fields <jfields@austininstruments.com> wrote:
<snip>

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--->
+----------------------------------------------------+ <--------+
| | |
[SOURCE]10A [wire]r [LOAD]R [E]
| | |
+----------------------------------------------------+ <--------+

Now we've got some very long wires connecting our source to our load
with non-negligible resistance r.

So the source needs to have a voltage of 10*(1+r)

and the power we waste heating the wires is I * Vwire = 10 * 10 * r

We want to minimise this. One option is to make the wire thicker and
reduce r. But this gets expensive buying all that metal for the wires.

So instead we'll do the following with two identical (assumed perfect)
transformers

I--->
+---------+ +-----------------------------+ +------+ <--------+
| P||S S||P | |
[SOURCE]10A R||E [wire]r E||R [LOAD]R [E]
| I||C C||I | |
+---------+ +-----------------------------+ +------+ <--------+

No assume that SEC has 10x as many turns as PRI. So the voltage in the
middle section will be 10x what it is at the source and the load.

Because the voltage is 10x and energy must be conserved, the current in
the middle section must be 1/10 the current from the source.

When we convert back we go back to 1/10 the voltage, so we must have 10x
the current, again to conserve energy.

So now the voltage drop in the wire is 10*1/10*r and the power loss is
10*1/10 * 10*1/10*r = r

So we've reduced the power lost from 100r to r just by adding two
transformers.

The main problem with using higher voltages is that they are harder to
keep in the wires. You're going to have to have better insulators
supporting or surrounding the wire etc. So there comes a point where the
money you save by reducing the current is offset by the extra money you
spend insulating the wires.

I think the grid goes up to 650000V. So (approximately) 3000x the wall
voltage. That means that the current in those wires is about 1/3000 of
what the end user is using and so the power wasted heating the wires is
about 1/9000000 of what it would be if it was 240V all the way from
powerstation to home.

One final point. It's important to realise that transformers conserve
energy. In the ideal case, if the secondary is open circuit and no
current is flowing then no power is being drawn from the transformer so
no energy is going into the primary either.

So here:

+---------+ +----------
| P||S
[SOURCE] R||E
| I||C
+---------+ +----------

There is no power being drawn from the source. (It's not true to say
there is no current flowing in the primary, just that there is no power
being drawn from the source)

In practice there are losses in the transformer which is why a wall wart
still gets warm if it is left plugged in with no load attached. But you
may have noticed that it gets warmer when there is a load attached - of
if you plug one into a power meter, you'll be able to see the power
being drawn from the mains increase as the power drawn from the load
increases.

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/

Author: John Fields
Date: 18:35 15-04-08

On Tue, 15 Apr 2008 17:25:37 -0500, John Fields
<jfields@austininstruments.com> wrote:

>On Tue, 15 Apr 2008 12:16:17 -0700 (PDT), chesemonkyloma@gmail.com
>wrote:
>
>
>>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,
>
>---
>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
--- ^^^^
Oops... 10²A

JF

Author: Tim Woodall
Date: 18:37 15-04-08

On Tue, 15 Apr 2008 17:25:37 -0500,
John Fields <jfields@austininstruments.com> wrote:
> On Tue, 15 Apr 2008 12:16:17 -0700 (PDT), chesemonkyloma@gmail.com
> wrote:
>
>
>>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,
>
> ---
> It doesn't.

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/

Date: 19:03 15-04-08

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.

Date: 22:02 15-04-08

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!

Author: Rich Grise
Date: 15:15 17-04-08

On Mon, 14 Apr 2008 16:55:17 -0700, chesemonkyloma wrote:

> My problem is this: You are distributing 100 watts. You start out with
> 10 volts, 10 amps. You increase the volts by a factor of 2 through a
> transformer. Now you must have 20 volts, 5 amps right? Using Ohm's
> Law, 10 volts = 10 amps * 1 ohms, in the first case there is 1 ohm.
> But, in the next case you have 20 volts = 5 amps * 4 ohms. So why did
> the ohms change? When you use a transformer to increase voltage,
> doesn't that mean ohms must increase if amps decrease? I am thoroughly
> confused. I know I am thinking about it the wrong way, can someone
> explain it to me? If you want me to clarify, let me know. Basically my
> question is this: How do you create more voltage without creating more
> current?

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: http://www.thelearningpit.com/elec/tools/tables/Wire_table.htm }.

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

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