On Sun, 12 Jun 2005 16:35:48 -0800, (Floyd L.
Davidson) wrote:

John Fields wrote:
On Sun, 12 Jun 2005 14:00:23 -0800, (Floyd L.
Davidson) wrote:

My point still stands, that if the current is changing, it is by
definition AC, and current not changing is DC. Trying to look
at it as DC is all in one direction and anything else is AC,
doesn't work.

---
Your point is flawed. Alternating Current, by definition, causes
electrons to move in one direction for a time, and then to reverse
direction for a time.

That isn't true.

---
Yes, it is. If you have proof, instead of just a statement to the
effect that it isn't, I'd love to see it.
---

The sinusoidally varying unipolar voltage under consideration _always_
forces electrons to move in one direction only.

A non-sequitor.

---
"Non sequitur." No. A non-sequitur is an inference or a conclusion
that does not follow from the premises, or a comment that is unrelated
to a preceding one. My error was the omission of a reference, Mr.
Lancaster's: "1 volt peak sinewave with a 0.6 volt dc term"
---

Since the voltage varies, the current will also, but the _direction_
in which the electrons are travelling will never change.

If it varies, it's AC.

---
No, it isn't. What's necessary is the polarity reversal before it can
be considered AC.
---

That means that the signal is DC. A varying DC, but DC nonetheless.

If there is such a think as "varying DC", connect a load to
it... through a capacitor. Now, how do you describe the effect
that load has on your "varying DC". The load see's *only* AC,
even according to your definition. That AC came from somewhere,
and it certainly was not generated by the capacitor.

---
It most certainly was!

Consider a DC coupled audio amplifier running from a single 12V
supply, with its output set to Vcc/2 and feeding an 8 ohm load with a
4VPP sinusoidal signal. Like this:

+12
|
+--o--+
| AMP |---+--Vout
+--o--+ |
| [4R]
| |
GND GND

Now, with phi being equal to the phase angle of the signal and zero
degrees corresponding the voltage halfway between the most positive
and least positive output voltage, the output voltage excursions will
look like this:

phi Vout
-----+------
0° 6V
90° 8V
180° 6V
270° 4V
360° 6V

Now, connect that magical capacitor between the amp and the load, as
shown below, and watch what happens:

+12
|
+--o--+
| AMP |---[cap]--+--Vout
+--o--+ |
| [4R]
| |
GND GND


phi Vout
-----+------
0° 0V
90° +2V
180° 0V
270° -2V
360° 0V

Why?

Well,for starters, consider that under quiescent conditions the
left-hand side of the cap will be charged to 6V and the right hand
side will be at zero volts since there is no galvanic path to the
output of the amp through the cap.

Now, imagine that the voltage at the amp's outout starts to go
positive. What will happen is that the amp will start sucking
electrons out of the cap, generating a potential difference across the
cap's plates which causes electrons to flow through the resistor,
making the top of the resistor more positive than the bottom.

Continuing in time, a point will be reached where the output of the
amp will start forcing electrons _into_ the resistor, at which point
the direction of travel of the electrons will be reversed. This
periodic reversal will cause the polarity of the signal into the
resistor to alternate. This alternating voltage will then give rise
to an _alternating current_ in the resistor.
---

That's because AC is *not* defined by any change in direction,
but only by a rate of movement change.

---
Poppycock. It's precisely the alternations in the direction of charge
flow which cause it to be called "Alternating Current".

Your way would have it be called AC by assigning some arbitrary rate
of change, irrespective of direction as the delineation point, which
makes no sense at all. That is, what would you specify as the rate of
change which would delineate between between AC and DC? 0.5A/s?
0.001A/s? 0.1V/s?

--
John Fields
Professional Circuit Designer