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NEON- AC Theory Part 3
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end
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Date: 10 Nov 96 19:54:50 EST
From: tmjdorr <103461.3123@compuserve.com>
To: all neon list people <neon-l@lists.io.com>
Subject: NEON- AC Theory Part 3
Message-ID: <961111005449_103461.3123_IHI39-1@CompuServe.COM>
Sender: owner-neon-l@io.com
Precedence: bulk
Reply-To: neon-l@lists.io.com
X-UIDL: 847751831.002
Greetings all,
This is part 3 of a continuing short course on electrical theory, intended to
help give neon workers a basic theoretical electrical knowlege. Continuing
on...
Part 3: Power Trip
We discussed some of the characteristics of basic electrical components and
their responses to both DC and sinusoidal AC excitation. We now turn our
attention to power. Power is the RATE of energy flow into a load. If a load
operates at a higher power level, it consumes energy at a faster rate. For DC
applied to a resistive load, this is fairly easy to calculate - we know the
formula for power:
Power = Volts x Amps
If we increase the voltage to a resistive load, we know that the amperage
flowing through it will increase and the power level will rise. From ohm's law,
we know:
Amps = Volts / Ohms
If we combine these two equations, we find:
Power = Volts * Volts / Ohms
or:
Power = (Volts ^ 2) / Ohms (where '^' means 'raised to the power of')
in words, power is equal to the square of the voltage, divided by the
resistance. In short, if we double the voltage applied to a resistive load,
it's power level raises by a factor of four. In a resistive load, the units for
power is 'watts'.
What happens when we excite our resistive load with an AC waveform? This is
easiest answered by using a graph (see the included uuencoded file: plots2.gif).
The basic DC equation for power still applies, but for AC, we find the
instantaneous power value, at any given point in time, by multiplying the
voltage and current values for that point in time. The first plot shown is the
voltage waveform and the current waveform, and the fourth is the power waveform.
Notice when the voltage is (mathematically) negative, so is the current,
resulting in a positive power value, as we would expect. Also notice that the
power waveform is sinusoidal, with two differences: (1) it is at twice the
frequency of the voltage and current waveforms, and (2) is entirely above the
'zero' line on our graph. We know that neon goes on and off 120 times a second
- no energy can be transfered when either the voltage or the current is at a
zero level. Now we have graphical proof why neon is stroboscopic (can make
rotating objects appear to stop), as light can only be emitted when energy is
being consumed (assuming: neon gas, clear glass tubing). With (single phase)
AC, the power transfer is not steady, as with DC, but varies sinusoidally with
time.
What happens when we apply AC voltage to an inductor? Using the same method as
with the resistor, above, we find a different result. The second plot shows the
current waveform through the inductor, and the fifth shows the power waveform.
Notice that, like the resistive load waveform, the power waveform for the
inductor is sinusoidal and at twice the frequency of the voltage and current.
On the other hand, notice that the power waveform is centered about the 'zero'
axis of our plot, implying that we have both 'positive' and 'negative' power.
How can this be? We understand that 'positive' power means energy is being
transfered from the source (the power company) to the load. 'Negative' power
must imply the opposite: power is being transfered from the load back to the
power source. In other words, all of the energy received by the inductor during
the second quarter voltage cycle is returned during the third quarter. All
energy received during the fourth quarter is returned on the first quarter of
the next cycle. What this means is that an inductor does not consume energy -
it simply saves it for a while, in it's magnetic field, and then returns it to
the source when conditions are proper.
If we generate the plots for a capacitive load, we find a similar result (but
inverted relative to the inductor). See plots 3 and 6. This means a capacitor,
like the inductor, does not consume energy, but simply returns it later in time.
For both capacitive and inductive loads, the units for power is VARS (stands for
'volt-amps, reactive').
So, you may ask, if inductors and capacitors don't consume energy, what's wrong
with having them connected to our power lines? We answer this next time, when
we discuss vectors, and take a look at Dirk's vector plots. Also: answer to
last quiz next time.
Regards,
Telford Dorr
p.s. is everybody (who's interested) successfully displaying the plots? Speak
up if you're having trouble. I understand Kenny graciously put the first set of
graphs up on his website. Thank, Kenny!