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.
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REVIEW:
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In a purely resistive circuit, all circuit
power is dissipated by the resistor(s). Voltage and
current are in phase with each other.
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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.
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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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