Power in resistive and reactive AC circuits
Consider a circuit for a single-phase AC
power system, where a 120 volt, 60 Hz AC voltage source is delivering
power to a resistive load:
In this example, the current to the load
would be 2 amps, RMS. The power dissipated at the load would be 240
watts. Because this load is purely resistive (no reactance), the current
is in phase with the voltage, and calculations look similar to that in
an equivalent DC circuit. If we were to plot the voltage, current, and
power waveforms for this circuit, it would look like this:
Note that the waveform for power is
always positive, never negative for this resistive circuit. This means
that power is always being dissipated by the resistive load, and never
returned to the source as it is with reactive loads. If the source were
a mechanical generator, it would take 240 watts worth of mechanical
energy (about 1/3 horsepower) to turn the shaft.
Also note that the waveform for power is
not at the same frequency as the voltage or current! Rather, its
frequency is double that of either the voltage or current
waveforms. This different frequency prohibits our expression of power in
an AC circuit using the same complex (rectangular or polar) notation as
used for voltage, current, and impedance, because this form of
mathematical symbolism implies unchanging phase relationships. When
frequencies are not the same, phase relationships constantly change.
As strange as it may seem, the best way
to proceed with AC power calculations is to use scalar notation,
and to handle any relevant phase relationships with trigonometry.
For comparison, let's consider a simple
AC circuit with a purely reactive load:
Note that the power alternates equally
between cycles of positive and negative. 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:
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 my 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:
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.
- REVIEW:
- In a purely resistive circuit, all
circuit power is dissipated by the resistor(s). Voltage and current
are in phase with each other.
- In a purely reactive circuit, no
circuit power is dissipated by the load(s). Rather, power is
alternately absorbed from and returned to the AC source. Voltage and
current are 90o out of phase with each other.
- In a circuit consisting of resistance
and reactance mixed, there will be more power dissipated by the load(s)
than returned, but some power will definitely be dissipated and some
will merely be absorbed and returned. Voltage and current in such a
circuit will be out of phase by a value somewhere between 0o
and 90o.
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