# Re: Basic Question

On Thu, 4 Sep 2008 16:41:33 +0000 (UTC) EskWIRED@xxxxxxxxxxxxxxxxxxx wrote:

| I'm wondering how one can squeeze more current through a given diameter
| wire by means of increasing the voltage.
|
| I'm wondering a bunch of other stuff too, but the answer to this question
| seems basic to my eventual understanding of other questions.
|
| "More current" means that more coulombs of charge pass by a given point
| per unit of time. I take this to mean that more electrons flow past the
| point.

More current does not mean more work done (energy) nor does it mean a
greater rate of work done (power). Current is a consequence of power
transmission.

Consider using a hydraulic power system, where the flow of water is used
to perform work. Voltage would be the pressure of the system. Current
would be the rate of flow. If the load is geared up to require a lot of
turns of a turbine, it will take a lot of water flow, but not a lot of
pressure, to do the work. But if the gearing requires fewer turns to do
the same work, you need more pressure, but less water will flow.

For a given load, of course increasing pressure increases the flow. That
is an overall increase in work done (even if it is work destroying the
load, such as running a 120 volt light bulb on 480 volts).

| If more electrons pass by the point per unit of time, does this mean that
| the velocity of the electrons is greater at higher voltages, all other
| things being equal? Is the means of getting more electrons through a
| given wire to increase the velocity of the electrons by means of
| increasing the voltage?

For a given current, the electron flow is the same. For a thinner wire
it will be a smaller cross section, and thus a faster veloity. For a
thicker wire, it will be a greater cross section, and thus a slower
velocity.

Think of the hydraulic water pipe driving a given mechanical load. If
the pipe leading to the load is smaller, water flows at the same rate in
terms of volume (e.g. number of liters or gallons per minute), but the
water is flowing at a faster velocity (each molecule traveling a greater
number of meters or feet per second).

| If so, then how fast do electrons travel?

I recall seeing this number somewhere, but that it was not really very large.
Keep in mind that for AC, the electrons will be going back and forth.

| I (think I) understand that the electrons do not necessarily pass through
| the entire length of the wire, but instead, they travel between adjacent
| copper atoms. When one atom near the source gets extra electrons, it
| passes some electrons to its neighboring atom. This propogates down the
| wire from one end to the other, until eventually, the current reaches the
| load at the other end.

And with AC they turn around and come back.

And with either AC or DC, when they are going to the load on one wire,
they are coming from the load on the other.

| And am I correct that at higher voltages, this process happens more
| quickly? If one had a really long wire, could the time difference be
| noticible and significant?

The distance isn't really significant. The voltage does not affect the
rate of electrons flow (directly). It is the current that does, and the
thickness of the wire affects the velocity.

| So how fast does this effect propogate down the length of the wire? And
| is velocity of travel the mechanism that allows more electrons to "pass
| through" a thin wire at higher voltage per unit of time?

The effect propogates at a speed approaching the speed of light. A given
conductor has a "velocity factor" that is expressed as a percentage of the
speed of light. This is the velocity of propogation.

| OK, so that is my basic question, and I guess it assumes DC voltages. Is
| that true?

AC causes the electrons to "slosh" back and forth.

| Taking it one step further, ISTM that with AC, the electons starting at
| the source might not get too far down the wire before polarity changes.
| Here in North America, we switch polarity 60 times a second.

Right. They will slow down, stop, then start going back the other way.

At radio frequencies, which is basically AC changing at rates of millions
or billions of times a second, the PROPOGATION can still be on the way down
the wire when the polarity of the source reverses. In this case the distance
the electrons move is microscopic.

| So are the same electrons going back and forth, back and forth, along the
| same physical chunk of wire at AC voltages?

There is some variation over the cross section of the wire. AC favors the
outer portion of the wire. At radio frequencies, and at high current levels,
it's significant and needs to be considered.

| AND - what happens when we use low voltage and high frequency?

The same things happen, on a different scale.

| Specifically, what would happen if the switching frequency is
| significantly quicker than the time required for the voltage change to
| propogate all the way down the wire from the generator to the load? Does
| the transmission system suffer weird phase problems?

It's not really a problem. More problems happen as a result of some of the
propogation reflecting back. This reflection exists for DC switching on and
for all AC frequencies. It's insignificant except at high frequencies where
the length of conductor is a significant portion of the wavelength at that
frequency.

| And am I starting to figure out the term "Power Factor" using this line of
| reasoning, or am I going in weird circles, and beginning my reasoning
| process with defective premises?

No. Power factor is when the phase angle of the sine waveform of the current
is offset from the phase angle of the sine waveform of the voltage. The
effect of this issue is basically that the load acts as a generator for part
of the AC cycle. It takes power for part of the cycle and gives it back for
part of the cycle. The power factor value is the proportion of power that it
takes that it keeps. The difference is what it gives back. A power factor of
1 (one) is a case where the load keeps all the power it takes. A power factor
of 0.5 means it takes twice as much as it keeps.

Note that keeping can be a misnomer above. Most loads dissipate the power in
the form of work done, heat, or radiated energy, or stores in batteries, etc.

The problem with a low power factor is that it means the current is higher
for the same amount of work being done. That means there is greater waste
of power because current on conductors does "work" in the form of heating
the wire.

Current through a resistance has a voltage drop that is proportional to the
resistance times the current. Wires have rather low resistance in general,
but there is some, so over a long distance this can be a significant waste
at high current levels. One thing to consider is that as the voltage drop
increases, this means the POWER dissipated in the wire increases and this
is proportional to the voltage (drop) times the current. Since the voltage
(drop) is already proportional to the current, we see current involved in
this TWICE. So ultimately, the POWER lost in the wire is proportional to
the SQUARE of the current. If you double the current in a resistance or
wire, the power dissipated is four times as much. This sounds bad and it
can be. But the inverse can be very good. If you cut the current in half
(by making the system voltage twice as much), you dissipate only 1/4 the
power in the wire if they stay the same size. This can save money by using
a thinner wire. Power engineers designing transmission lines figure out
the best economic balance for things like wire thickness vs. power lost at
various load levels, and figure that against costs of system voltage (more
insulation, for example), and costs of supporting the wires in the form of
stronger supporting towers. Wind load is also a factor. They have computer
programs to model this to help them figure out what is best for each new
transmission line being built.

| OK. Lots more than one question. But I seem to have some basic
| misunderstandings or ignorance that is not letting me figure this stuff
| out in a manner I am confident of.
|
| Any correction of basic misuderstandings will be greatly appreciated.

Let me know what questions pop up, or go unanswered, from the above.

--
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| Phil Howard KA9WGN (email for humans: first name in lower case at ipal.net) |
.

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