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NEON- AC Theory Part 4 - Vectors
Greetings all,
This is the fourth part in a short course on AC circuit theory. In this part,
hopefully, we will finally get to the point. So far, we have characterized the
three standard electrical components (resistance, capacitance, and inductance)
and their responses to sinusoidal AC voltage excitation. Up to this point, we
have worked with each component by itself. Unfortunately, in the real world,
this doesn't happen. Most circuits have at least two, or sometimes all three of
the basic component types. Also, we have been working with plots of the
responses plotted versus time. While these look nice, and help one to
understand what's going on, they are really not all that useful in everyday
work. As is usually the case, a mathematician, somewhere and some time ago,
came up with a method of expressing these plotted responses in such a way that
they're easy to deal with. The method is refered to as vector arithmetic.
In order to use this method, we have to have one thing constant, and that is the
frequency of the sinusoudal excitation voltage. Since we all use either 60 or
50 hertz (cycles per second) power, and since it's pretty constant in frequency,
this isn't a problem. (On the other hand, if you're working deep in the back
woods somewhere, running off of a Honda generator, then maybe it is a factor.)
Refer to the first plot in the uuencoded file 'plots3.gif' (included as a
separate file). What we have here is a plot of the current flowing through a
resistor, drawn as a vector. The line segment with the arrow head is the
vector. Mathematically, a vector is a line segment which has length AND
direction. The length of the vector represents the magnitude of the current
flowing through the resistor, in amps. The direction of the vector represents
the phase angle of that current relative to the voltage. As we know, the phase
angle of resistive current is zero - the current and voltage are 'in phase'.
Refer to the waveform plots back in session #2. I have labeled the angle values
on the first plot - they also apply to the other plots, and have been omitted
for clarity.
The second plot is for a current flowing through an inductor, again relative to
the applied voltage. We know that an inductor's current LAGS it's voltage by 90
degrees, so we plot the vector pointing to -90. It's length represents the
magnitude of the current, again in amps.
Now suppose we have a resistor AND an inductor, wired in parallel, connected to
an AC voltage source. What does the current look like? Let us assume that the
previous two plots represent the individual currents for the resistor and
inductor by themselves. If we plot both vectors on the same chart, we get the
third chart. Now here's the good part - we can add vectors graphically by
moving them so the head of one vector connects to the tail of another vector,
for as many vectors as we have. As long as we don't accidentally rotate a
vector, this method works. But since all of our vectors will point towards 0
(resistive), -90 (inductive) or +90 (capacitive), degrees, it shouldn't be a
problem keeping them straight.
In plot #4, we have moved the inductive current vector to connect to the end of
the resistive current vector. Now, we draw a final vector, from the tail of the
first vector, to the head of the last vector. The length of this final vector
represents the magnitude of the expected total current flowing into our parallel
circuit (in amps). The angle of the final vector will be the phase angle of the
total current relative to the voltage. Notice that it may be at any angle, from
-90 to +90 degrees, depending on the components in our circuit.
The plot shown is representitive of a typical uncompensated neon transformer.
The vector labeled 'B' represents the neon load connected to the transformer and
thermal transformer losses. The vector labeled 'A' represents the (equivalent)
inductive current which flows because of the magnetic shunt, built into the
transformer, which is responsible for the ballasting action of the transformer.
The final vector 'C' represents the actual line current draw of the transformer.
Notice that it is larger than just the resistive vector by itself. In other
words, our transformer is drawing more current than is needed to light the neon.
Since it is inductive in nature, the power is returned back to the line, but it
still requires us to use larger primary wiring and fuses than would be required
it it wasn't there.
If we divide the magnitude of the resistive vector 'B' by the magnitude of the
total current vector 'C', we get a number known as the 'power factor' of our
transformer. This is usually printed on the transformer rating plate, and
labeled "PF". The angle between the vectors can be found from trigonometry:
cos(Z) = B/C where Z in the angle.
or:
cos(Z) = PF where PF = power factor
If we know the power factor value, we can find the angle:
Z = arc-cos(PF)
Note: we normally use the greek letter 'phi' for the angle, but because using it
as an ASCII character would probably cause email on three continents to crash
and disappear, for safety, we'll use the letter 'Z' instead.
--------------------
Answer to last quiz:
Quiz question: Suppose we have an inductor of such a size that, when 120 volts
AC is applied to it, it draws a current of 1 amp. Suppose we also have a
capacitor of such a size that, when 120 volts AC is applied to it, it also draws
a current of 1 amp. What would we measure on an ammeter if we connected both
of them, in parallel, to the 120 VAC source at once? (The ammeter is on one of
the wires going from the wall socket to the paralleled inductor and capacitor.)
ANS: ZERO! Looking at our current plots for inductive and capacitive current,
we see that they are mirror images of each other. When one of them is positive
in magnitude, the other is negative. Since they are the same absolute magnitude
in all places, when added, they cancel each other out. Does this mean the
components draw no current? No, what current each component draws is supplied
in turn by the other. They just sit there and pass energy back and fourth. If
there was no circuit resistance or radiation losses, once started, they would do
this forever. This is similar to the swinging pendulum on a clock - the
mainspring only has to supply just enough energy to compensate for friction and
wind losses to keep it swinging forever.
-------------------------------
Exercise for the reader: Take the resultant current vector plot and add a third
vector. Connect this vector to the end of vector 'A'. Let this vector be the
same length as the inductive current vector 'A', but let it point in the
opposite direction of the inductive vector. What kind of component does this
vector represent? A capacitor, of course! If you draw this correctly, the
final vector 'C' will now overlay the resistor vector 'B' and be the same length
as 'B'. The inductive and capacitive vectors have cancelled each other! This
is what adding a compensating capacitor to our neon transformer does!
-------------------------------
One of my favorite sayings is:
"In theory, theory works in practice; in practice, it doesn't."
What this means is there is a fly in the ointment here. The problem occurs
because of a characteristic of steel-cored transformers: the inductive current
through them isn't quite sinusoidal. There is a slight distortion, due to the
fact that steel is magnetically slightly non-linear. This effect is refered to
as hysteresis. The distortion shows up as a third (and higher odd) harmonic
current flow. In practice this means you can't correct a transformer's power
factor all the way to 1.0 - about 0.9 is as close as you can get. This is why
power-factor corrected neon transformers still have a power factor rating of 0.9