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.
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:
Resistive/reactive load:
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.
- REVIEW:
- 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 (W).
- 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 (VAR).
- 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 quantity.
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 length.
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
delivery system.
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 circuit:
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
inductive.
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.
- REVIEW:
- 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).
Contributors
Contributors to this chapter are listed
in chronological order of their contributions, from most recent to
first. See Appendix 2 (Contributor List) for dates and contact
information.
Jason Starck
(June 2000): HTML document formatting, which led to a much
better-looking second edition.
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