Power Factor Calculations
Power Factor Calculation
The power factor of an alternating current circuit is the cosine of the
phase angle. This topic is useful to any electrician working with transformers
and motors. Inductive loads have various power factors which are very important
in incorporating into calculations for conductor sizing.
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
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
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
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.
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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