Power in resistive and reactive AC circuits
For
comparison, let's consider a simple AC circuit with a purely reactive load in
Figure below.
AC
circuit with a purely reactive (inductive) load.
Power
is not dissipated in a purely reactive load. Though it is alternately absorbed
from and returned to the source.
Note
that the power alternates equally between cycles of positive and negative.
(Figure above ) This means that power is being alternately absorbed from and
returned to the source. If the source were a mechanical generator, it would
take (practically) no net mechanical energy to turn the shaft, because no power
would be used by the load. The generator shaft would be easy to spin, and the
inductor would not become warm as a resistor would.
Now,
let's consider an AC circuit with a load consisting of both inductance and
resistance in Figure blow.
AC
circuit with both reactance and resistance.
At a
frequency of 60 Hz, the 160 millihenrys of inductance gives us 60.319 Ω of
inductive reactance. This reactance combines with the 60 Ω of resistance to
form a total load impedance of 60 + j60.319 Ω, or 85.078 Ω ∠ 45.152o.
If we're not concerned with phase angles (which we're not at this point), we
may calculate current in the circuit by taking the polar magnitude of the
voltage source (120 volts) and dividing it by the polar magnitude of the
impedance (85.078 Ω). With a power supply voltage of 120 volts RMS, our load
current is 1.410 amps. This is the figure an RMS ammeter would indicate if
connected in series with the resistor and inductor.
We
already know that reactive components dissipate zero power, as they equally
absorb power from, and return power to, the rest of the circuit. Therefore, any
inductive reactance in this load will likewise dissipate zero power. The only
thing left to dissipate power here is the resistive portion of the load
impedance. If we look at the waveform plot of voltage, current, and total power
for this circuit, we see how this combination works in Figure below.
A
combined resistive/reactive circuit dissipates more power than it returns to
the source. The reactance dissipates no power; though, the resistor does.
As with
any reactive circuit, the power alternates between positive and negative
instantaneous values over time. In a purely reactive circuit that alternation
between positive and negative power is equally divided, resulting in a net
power dissipation of zero. However, in circuits with mixed resistance and
reactance like this one, the power waveform will still alternate between
positive and negative, but the amount of positive power will exceed the amount
of negative power. In other words, the combined inductive/resistive load will
consume more power than it returns back to the source.
Looking
at the waveform plot for power, it should be evident that the wave spends more
time on the positive side of the center line than on the negative, indicating
that there is more power absorbed by the load than there is returned to the
circuit. What little returning of power that occurs is due to the reactance;
the imbalance of positive versus negative power is due to the resistance as it
dissipates energy outside of the circuit (usually in the form of heat). If the
source were a mechanical generator, the amount of mechanical energy needed to
turn the shaft would be the amount of power averaged between the positive and
negative power cycles.
Mathematically
representing power in an AC circuit is a challenge, because the power wave
isn't at the same frequency as voltage or current. Furthermore, the phase angle
for power means something quite different from the phase angle for either
voltage or current. Whereas the angle for voltage or current represents a relative
shift in timing between two waves, the phase angle for power represents a ratio
between power dissipated and power returned. Because of this way in which AC
power differs from AC voltage or current, it is actually easier to arrive at
figures for power by calculating with scalar quantities of voltage, current,
resistance, and reactance than it is to try to derive it from vector, or
complex quantities of voltage, current, and impedance that we've worked with so
far.
0 comments:
Post a Comment