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NEON- Re: Power factor & Load Current
Tom,
>>>Telford- A few more questions...
1) Could you clarify what you mean by "natural inductive characteristics" and
why that's a bad thing needing correction?
2) Could you say more about the "transformer's uncompensated inductance" or
define it?
3) How does the added capacitor act to "correct" the power factor?
4) If that loss occurs as a result of the functioning of the transformer, how
does it return through the line and occcur on "their side of the meter?" Tom
U.<<<
What we need is a short course in AC circuit theory. Notice emphesis on the
word "short". Rather than generate a long post on this, I'll break it up into a
couple of posts that build on one-another (besides, I don't have to type it up
all at once this way...). We'll leave out the worst of the math, and go for a
more practical approach.
Part 1 - DC theory.
Note: this is just for review and comparison to AC, later.
a) Resistive Loads. If we take a resistive load, like an automotive tail lamp,
and connect it up to 12 volts, DC, it will draw a current. If it is a 12 watt
lamp, the current will be 12 watts divided by 12 volts = 1 amp. As long as the
battery doesn't go dead or the lamp burn out, it will draw 1 amp. When we
disconnect the battery from the lamp, the voltage across it goes to zero, as
does the current through it. Simple, right?
b) Capacitive Loads. Now suppose we go down to Radio Shack and buy a capacitor,
rated at, say, 1000 uf (or micro-farads - the 'u' is actually a greek letter)
and 16 volts, DC. If we put a voltmeter across it, we'll measure zero volts,
right? Now, we connect it to our 12 volt car battery, with an ammeter in series
so we can measure it's current draw. We notice, when we make the connection, we
get a spark, and the ammeter swings up to a high reading and then returns to
zero. The voltmeter swings from zero up to 12 volts. What happened? The
capacitor charged up to 12 volts, and then stopped drawing current. In fact,
for as long as we keep it connected to the battery, it won't draw any more
current (actually a little tiny bit, but only because it's not perfect). If we
now disconnect the battery, but leave the voltmeter connected to the capacitor,
we will see the meter reading ever so slowly return to zero.
What does this mean? Obviously, the capacitor can store energy. We can charge
it up and we can discharge it (the voltmeter is the resistive load.) The
important thing is this:
A capacitor only draws current when we CHANGE the voltage across it.
The faster we change the voltage, the higher the current will be. As long as
the voltage remains steady, the current into or out of it is ZERO.
c) Inductive Loads. This one is a little harder to demonstrate, but we'll try.
Suppose we have a choke, like the choke in series with the bombarder, for
instance. These are refered to as inductors. It is just a steel core (movable,
in this case, but we'll ignore that) with a lot of wire wound around it. Now,
suppose we have a device known as a 'constant current source'. This is a kind
of power supply which regulates it's output current so it's always at a desired
value. It does this by changing it's output voltage as needed. If we set it
for 1 amp and then short the output through our ammeter, we will measure 1 amp,
at zero volts (it's a short, right?). If we connect it to our lightbulb, we'll
measure 1 amp at 12 volts (refer back to part 'a', above). Just assume it works
as described.
Now, let's connect our choke (let us assume it has no resistance, for
simplicity) to the constant current source, again with the ammeter in series and
the voltmeter in parallel with the choke. What do we measure? We see the
voltmeter jump to a very high value, then return to zero. We see the ammeter
move from zero to 1 amp. Now, for as long as we leave it connected, the ammeter
will read 1 amp and the voltmeter will read zero. When we disconnect the
current source, we get a spark as we break the connection, the voltmeter smacks
against the zero pin and then returns to zero, and the ammeter returns back to
zero.
Like the capacitor, the inductor stores energy (as a magnetic field in it's
steel core). The amount of energy stored is a function of the current passing
through it and the number of turns of wire wound around it's core, among other
things. The important thing is this:
An inductor only has a voltage across it when we CHANGE the current
flowing through it. The faster we change the current, the higher the voltage
will be. As long as the current remains steady, the voltage across it is ZERO.
So, capacitors and inductors are kind of like the yin and yang of the electrical
world - very Zen.
End of part 1. Part 2 tomorrow.
Regards,
Telford Dorr