Electrical Power Factor
Power Factor Details - SOURCE
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
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.
True, Reactive, and Apparent power
We know that reactive loads such as inductors and capacitors dissipate zero
power, yet the fact that they drop voltage and draw current gives the deceptive
impression that they actually do dissipate power. This "phantom
power" is called reactive power, and it is measured in a unit called
Volt-Amps-Reactive (VAR), rather than watts. The mathematical symbol for
reactive power is (unfortunately) the capital letter Q. The actual amount of
power being used, or dissipated, in a circuit is called true power, and
it is measured in watts (symbolized by the capital letter P, as always). The
combination of reactive power and true power is called apparent power,
and it is the product of a circuit's voltage and current, without reference to
phase angle. Apparent power is measured in the unit of Volt-Amps (VA) and
is symbolized by the capital letter S.
As a rule, true power is a function of a circuit's dissipative elements,
usually resistances (R). Reactive power is a function of a circuit's reactance
(X). Apparent power is a function of a circuit's total impedance (Z). Since
we're dealing with scalar quantities for power calculation, any complex starting
quantities such as voltage, current, and impedance must be represented by their polar
magnitudes, not by real or imaginary rectangular components. For instance,
if I'm calculating true power from current and resistance, I must use the polar
magnitude for current, and not merely the "real" or
"imaginary" portion of the current. If I'm calculating apparent power
from voltage and impedance, both of these formerly complex quantities must be
reduced to their polar magnitudes for the scalar arithmetic.
There are several power equations relating the three types of power to
resistance, reactance, and impedance (all using scalar quantities):
Please note that there are two equations each for the calculation of true
and reactive power. There are three equations available for the calculation of
apparent power, P=IE being useful only for that purpose. Examine the
following circuits and see how these three types of power interrelate:
Resistive load only:
Reactive load only:
These three types of power -- true, reactive, and apparent -- relate to one
another in trigonometric form. We call this the power triangle:
Using the laws of trigonometry, we can solve for the length of any side
(amount of any type of power), given the lengths of the other two sides, or the
length of one side and an angle.
Power dissipated by a load is referred to as true power. True
power is symbolized by the letter P and is measured in the unit of Watts
Power merely absorbed and returned in load due to its reactive
properties is referred to as reactive power. Reactive power is
symbolized by the letter Q and is measured in the unit of Volt-Amps-Reactive
Total power in an AC circuit, both dissipated and absorbed/returned is
referred to as apparent power. Apparent power is symbolized by the
letter S and is measured in the unit of Volt-Amps (VA).
These three types of power are trigonometrically related to one another.
In a right triangle, P = adjacent length, Q = opposite length, and S =
hypotenuse length. The opposite angle is equal to the circuit's impedance
(Z) phase angle.
Calculating power factor
As was mentioned before, the angle of this "power triangle"
graphically indicates the ratio between the amount of dissipated (or consumed)
power and the amount of absorbed/returned power. It also happens to be the same
angle as that of the circuit's impedance in polar form. When expressed as a
fraction, this ratio between true power and apparent power is called the power
factor for this circuit. Because true power and apparent power form the
adjacent and hypotenuse sides of a right triangle, respectively, the power
factor ratio is also equal to the cosine of that phase angle. Using values from
the last example circuit:
It should be noted that power factor, like all ratio measurements, is a unitless
For the purely resistive circuit, the power factor is 1 (perfect), because
the reactive power equals zero. Here, the power triangle would look like a
horizontal line, because the opposite (reactive power) side would have zero
For the purely inductive circuit, the power factor is zero, because true
power equals zero. Here, the power triangle would look like a vertical line,
because the adjacent (true power) side would have zero length.
The same could be said for a purely capacitive circuit. If there are no
dissipative (resistive) components in the circuit, then the true power must be
equal to zero, making any power in the circuit purely reactive. The power
triangle for a purely capacitive circuit would again be a vertical line
(pointing down instead of up as it was for the purely inductive circuit).
Power factor can be an important aspect to consider in an AC circuit,
because any power factor less than 1 means that the circuit's wiring has to
carry more current than what would be necessary with zero reactance in the
circuit to deliver the same amount of (true) power to the resistive load. If our
last example circuit had been purely resistive, we would have been able to
deliver a full 169.256 watts to the load with the same 1.410 amps of current,
rather than the mere 119.365 watts that it is presently dissipating with that
same current quantity. The poor power factor makes for an inefficient power
Poor power factor can be corrected, paradoxically, by adding another load to
the circuit drawing an equal and opposite amount of reactive power, to cancel
out the effects of the load's inductive reactance. Inductive reactance can only
be canceled by capacitive reactance, so we have to add a capacitor in
parallel to our example circuit as the additional load. The effect of these two
opposing reactances in parallel is to bring the circuit's total impedance equal
to its total resistance (to make the impedance phase angle equal, or at least
closer, to zero).
Since we know that the (uncorrected) reactive power is 119.998 VAR
(inductive), we need to calculate the correct capacitor size to produce the same
quantity of (capacitive) reactive power. Since this capacitor will be directly
in parallel with the source (of known voltage), we'll use the power formula
which starts from voltage and reactance:
Let's use a rounded capacitor value of 22 µF and see what happens to our
The power factor for the circuit, overall, has been substantially improved.
The main current has been decreased from 1.41 amps to 994.7 milliamps, while the
power dissipated at the load resistor remains unchanged at 119.365 watts. The
power factor is much closer to being 1:
Since the impedance angle is still a positive number, we know that the
circuit, overall, is still more inductive than it is capacitive. If our power
factor correction efforts had been perfectly on-target, we would have arrived at
an impedance angle of exactly zero, or purely resistive. If we had added too
large of a capacitor in parallel, we would have ended up with an impedance angle
that was negative, indicating that the circuit was more capacitive than
It should be noted that too much capacitance in an AC circuit will result in
a low power factor just as well as too much inductance. You must be careful not
to over-correct when adding capacitance to an AC circuit. You must also be very
careful to use the proper capacitors for the job (rated adequately for power
system voltages and the occasional voltage spike from lightning strikes, for
continuous AC service, and capable of handling the expected levels of current).
If a circuit is predominantly inductive, we say that its power factor is lagging
(because the current wave for the circuit lags behind the applied voltage wave).
Conversely, if a circuit is predominantly capacitive, we say that its power
factor is leading. Thus, our example circuit started out with a power
factor of 0.705 lagging, and was corrected to a power factor of 0.999 lagging.
Poor power factor in an AC circuit may be ``corrected,'' or
re-established at a value close to 1, by adding a parallel reactance
opposite the effect of the load's reactance. If the load's reactance is
inductive in nature (which is almost always will be), parallel capacitance
is what is needed to correct poor power factor.
Practical power factor correction
When the need arises to correct for poor power factor in an AC power system,
you probably won't have the luxury of knowing the load's exact inductance in
henrys to use for your calculations. You may be fortunate enough to have an
instrument called a power factor meter to tell you what the power factor
is (a number between 0 and 1), and the apparent power (which can be figured by
taking a voltmeter reading in volts and multiplying by an ammeter reading in
amps). In less favorable circumstances you may have to use an oscilloscope to
compare voltage and current waveforms, measuring phase shift in degrees
and calculating power factor by the cosine of that phase shift.
Most likely, you will have access to a wattmeter for measuring true power,
whose reading you can compare against a calculation of apparent power (from
multiplying total voltage and total current measurements). From the values of
true and apparent power, you can determine reactive power and power factor.
Let's do an example problem to see how this works:
First, we need to calculate the apparent power in kVA. We can do this by
multiplying load voltage by load current:
As we can see, 2.308 kVA is a much larger figure than 1.5 kW, which tells us
that the power factor in this circuit is rather poor (substantially less than
1). Now, we figure the power factor of this load by dividing the true power by
the apparent power:
Using this value for power factor, we can draw a power triangle, and from
that determine the reactive power of this load:
To determine the unknown (reactive power) triangle quantity, we use the
Pythagorean Theorem "backwards," given the length of the hypotenuse
(apparent power) and the length of the adjacent side (true power):
If this load is an electric motor, or most any other industrial AC load, it
will have a lagging (inductive) power factor, which means that we'll have to
correct for it with a capacitor of appropriate size, wired in parallel.
Now that we know the amount of reactive power (1.754 kVAR), we can calculate the
size of capacitor needed to counteract its effects:
Rounding this answer off to 80 µF, we can place that size of capacitor in
the circuit and calculate the results:
An 80 µF capacitor will have a capacitive reactance of 33.157 Ω,
giving a current of 7.238 amps, and a corresponding reactive power of 1.737 kVAR
(for the capacitor only). Since the capacitor's current is 180o
out of phase from the the load's inductive contribution to current draw, the
capacitor's reactive power will directly subtract from the load's reactive
power, resulting in:
This correction, of course, will not change the amount of true power
consumed by the load, but it will result in a substantial reduction of apparent
power, and of the total current drawn from the 240 Volt source:
The new apparent power can be found from the true and new reactive power
values, using the standard form of the Pythagorean Theorem:
This gives a corrected power factor of (1.5kW / 1.5009 kVA), or 0.99994, and
a new total current of (1.50009 kVA / 240 Volts), or 6.25 amps, a substantial
improvement over the uncorrected value of 9.615 amps! This lower total current
will translate to less heat losses in the circuit wiring, meaning greater system
efficiency (less power wasted).
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