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Power Factor

Foreword on power factor.

I've been looking at power factor explanations on the Internet and have been astounded that for something that appears to be straight forward, appears explained in a confusing and laboured way. Even Wikipedia avoids giving a full explanation. In-fact I have only found a single source on Youtube worth listening to.

The best explanation is from a lecture, which clearly explains the concept and calculation from first principles but sadly has some very poor sound. The video is a 57 minute lecture but if you manage to stick with it, you will understand the concept of power factor and how it is derived. Please see my link to references at the end of this post.

So, I will have a go at explaining power factor in this article.  If you are a visual person I would suggest watching the video. This following explanation of power factor is for a single phase domestic installation only.

Warning this article contains some heavy maths including calculus, and I'm not afraid to use it!  I have coloured the more difficult parts in red so if you get "flighty" with maths you can just avoid those sections.  The maths is there more for completeness rather than something that is essential to know. Power factor can be successfully calculated without knowing the heavy mathematics.


Simple explanation of power factor.

Simple explanation of the causes of power factor.

What is Power anyway?

If power is so simple, how do you calculate it for AC mains?

More complex explanation of power factor and its two faces.

What is reactive power factor and how is it calculated?

What is Non-linear power factor and how is it calculated?

Further reading and references.

Simple explanation of power factor.


Power factor is a measure how close to ideal an electrical appliance is to a perfect load. It is expressed as a ratio or percentage between 1 and 0 or 100% and 0% respectively. An ideal load is considered purely resistive and has a power factor equal to 1 or 100%. The worst is PF of 0 or 0% (percent). This is not a statement of efficiency but how effectively it is consuming mains power.

Power factor (PF) = Real power (Pr) / Apparent power (Pa)

OK, it is not clear what "real power" or "apparent power" is, but the above equation is the "fundamental" definition of power factor. Any other explanation is just a short cut and applies in certain scenarios where the above definition applies in "all" scenarios.

Simple explanation of the causes of power factor.

There are two distinct causes of power factor. It is possible to have a mixture of both in any installation. The process to calculate power factor is essentially identical. For this explanation it's simpler to treat them individually. The first is reactive power factor and the second is non-linear power factor.

  1. Reactive power factor is caused by reactive elements in the circuit storing and releasing back energy. The energy is not consumed by the appliance and so represents an extra "unnecessary" burden on the power delivery systems. This means that excess energy appears to be required during one part of the mains cycle and less energy the next. Reactive Power factor is a "representation" of the relative size of this additional burden.

    • The power is being stored and release from reactive elements which may have a net inductance or capacitance.  The additional energy is stored as magnetic flux or electrostatic charge respectively. When one is more dominant than the other the energy is stored temporarily and returned twice per mains cycle.

  2. Non-linear power factor is caused by a distorted, or non-linear loading, by the appliance. This means that excessive power is demanded from the mains delivery systems but only over a small period of the mains cycle. Unlike a purely resistive or reactive load which has a simple and predictable loading shape (linear), the non-linear load will have very low demand for part of the cycle and a much higher demand for a only a small part of the cycle. Non-linear Power Factor is a "representation" of the relative size of this distortion.

    • The move away from linear power supplies, which used hefty power transformers, to switch mode supplies has compounded this problem. Rectified supplies, which include switch mode power supplies (SMPS), all tend to have peak power demand at the corresponding peak voltage of the mains supply. In very bad cases it can cause a flattening of the voltage effectively reducing the peak mains voltage level.

    • European rules IEC EN 61000-3-2 state that any power supply over 75watts must have power factor correction built in at manufacture.

What is Power anyway?

To understand real and apparent powers, you first need to understand what electrical power is, or more accurately how it's calculated. Electrical power is simply voltage in Volts, multiplied by current in Amps, which gives the power in Watts.

Power (P) = Voltage (V) * Current (I)

Yes it's that simple!

If power is so simple, how do you calculate it for a AC mains?

Where it gets more complex is where you're calculating power for mains alternating current (AC). For AC mains, the voltage follows what can be considered a sinusoidal path reaching peak voltages (Vpeak) of +/- 340v, in the UK and +/- 170v, in the US. To make the calculation easier the average is used. The problem is that the average of a sine wave over one cycle is 0v.
  • So the electrical engineers decided that if you square the values, take their average and take the root of that average value, you have a result you can work with. That is where the, Root Mean Square (rms) comes from. 

The root mean square can either be laboriously calculated by hand or more simply calculated using mathematics. The result is that the (rms) value of any sine wave is the peak value divided by the square root (Sqrt) of 2.

rms of sine wave =  peak / √2

Using this formula gives (Vrms) as 240v for the UK and 120v for the US. Which is how most people tend to refer to the voltage of their mains.

Generally when calculating power for resistive loads such as lights, heaters, the rms values are used.  Just because we say 240v or 120v does not change the peak voltage or the fact that the frequency of the sinusoidal wave is 50Hz or 60Hz respectively. Often when you look at the power rating of devices, they give the rms current, so the power is simply:

Watts = rms Current * rms Voltage

(Important: That calculation assumes that the both current and voltage are sinusoidal and in phase.)
So, for a sine wave:  

Power (P) = Voltage (Vrms) * Current (I rms)

Vrms = Vpeak / √2
I rms = I peak / √2

     We get:

Power (P) = ( Vpeak * I peak ) / 2

Remember this equation, as we will come back to this equation later! It makes one big assumption, and that is, the power being measured is for a purely resistive or ideal load.  It is this equation we use to calculate apparent power (Pa).  You merely measure the highest current and that becomes (I peak) irrespective of the phase or wave form. As we are making an assumption the apparent power is measured in, "voltage amps" (VA) rather than watts (W). VA is not a real value but the virtual baseline used to calculate power factor.

More complex explanation of power factor and its two faces.

To recap: Power factor is a rating denoting how close to ideal an electrical appliance is to a perfect load. It is expressed as a ratio or percentage between 1 and 0 or 100% and 0% respectively. An ideal load is considered purely resistive and has a power factor equal to 1 or 100%. The worst is PF of 0 or 0% percent. This is not a statement of efficiency but how effectively it is consuming mains power, as stated above

The equation is simple:
Power factor (PF) = Real power (Pr) / Apparent power (Pa)

PF = Pr / Pa

Apparent power is the power that the appliance appears to be using when looking at it simplistically.  It is calculated by just measuring the peak current and the peak voltage and assuming the power is the product of the two, as defined earlier.

So, apparent power is easy to calculate and you just take the peak current for any given load irrespective if it is in phase or simple spike.  As you are assuming it to be a perfect load it is often called VA.

Apparent power (Pa) = ( V peak * I peak ) / 2

Real power is slightly more awkward to calculate. As described above, power is the product of voltage and current but this will only give the power at any one instant.

From earlier we know that power is:

 Power (P) = Voltage (V) * Current (I)

To calculate real power you need to add these products together for one mains cycle and divide by the amount of times that you had taken samples. Unless you have the equipment that can perform the calculation for you, then in reality we take a shortcut for this equation depending on the type of power factor you are calculating.

If I were to show this mathematically it would look something like this:

Please buckle your seat belts! 

Real power (Pr) = Mean power  (Paverage) = ( i = n Vi.Ii ) / n
i = 0

            where "i" represents each instance calculated and "n" represents the amount of instances in total.

The mathematical equation looks worse than it really is.  The "Σ" means the addition (or sum) of the product of each equation for a number "n" times, from the start to the finish. The subscript "i" is just that particular iteration of the equation.  It is finally all divided by "n" because it is the average power we are interested in. This is also known as the arithmetic mean.

As mentioned above a shortcut can be taken depending on the type of power factor but the above equation and painful way to calculate real power is "always" correct.

Where most of the other explanations start to fall down is that there are two main types of power factor.

  1. Reactive power factor, caused by reactive elements such as inductors and capacitors.

  2. Non-linear power factor, caused by loads which take their main power from a small part of the sinusoidal mains cycle.
Most explanations make a fair effort for the first type of power factor but then totally neglect the second or just give it a mention. This is because the first type can be boiled down to a simplistic equation which is sufficiently complex to stop most people in their tracks. I will attempt to explain both.

What is reactive power factor and how is it calculated?

The first type, reactive power factor is caused by reactive loads predominantly inductors such as inductive motors in air conditioners, fridges, washing machines, transformers and inductive ballasts from "old" fluorescent light installations. Capacitors can be found in LED lights and dimmer switches but tend to be small values compared to the inductive components of other appliances.

  • The fluorescent light tube itself is not inductive and neither are CFL lights. CFL do not feature here as CFLs tend to use the same switching circuitry as electronic transformers.

The physical reason for reactive power factor is that capacitors and inductors do not consume power, per say, but store it as electrostatic charge and magnetic flux respectively. As they are connected to AC mains they are constantly building, losing and rebuilding this energy reserve. This storing and releasing of energy means that at points of the mains cycle the current flow is higher (storing energy reserve) than would be for a normal resistive or ideal load. Equally at other points the cycle current flow is less and in some cases can be negative, effectively returning power back into the mains.

  • This is especially true for unloaded induction motors and unloaded transformers.

It is frequently argued that this additional current is bad because energy infrastructure has to deal with the phantom load, putting additional stress on it. This is especially true of large companies with inductive heaters or heavy machinery.

It is not so clear cut for typical domestic installations, as the energy would be consumed by a multitude of other resistive appliances which have resistive loads, such as electric heaters or tungsten filament based lights. If not in your house, then it would be consumed by your neighbours before it ever got as far as the local substation. This is why commercial firms are penalised for poor power factor and domestic meters do not even bill reactive power (yet).

so what is the calculation for this type of power factor?

as above:

power factor = real power / apparent power

PF = Pr / Pa

The real power has to be measured as either work done or energy used averaged. Alternatively calculated manually from the waveform using average power.

from earlier:
Real power (Pr) = ( i = n Vi.Ii ) / n
i = 0

Assuming that the load is purely reactive there will be no distortion, so it would be easy to calculate this mathematically... If you were a mathematician.

If you were to look at the current waveform with an oscilloscope it would look like the sinusoidal trace has shifted with relation to the voltage trace. This is known as a phase shift and is the telltale sign of a reactive element to the load.

This sounds way too simple! What about the complicated trigonometric explanations that scare most people off.

Apparent power is actually a combination of real and reactive power because they are related. Sometimes the real power to apparent power is shown as a vector diagram and that's where you get the unhelpful explanation of the power factor being the cosine of the phase angle. It makes one crucial assumption, that the power factor is, "only" caused by a purely reactive element.

The explanation goes something like this:

With the real power being one side of the right-angle triangle (on x-axis) and the reactive power being perpendicular and forming the other side (on y-axis). Using "vector" addition, the apparent power is the hypotenuse linking the two. With this diagram and some basic school trigonometry; cosine(θ) is adjacent/ hypotenuse (A/H) which is the calculation of power factor.

A = Real power (Pr)
H = Apparent power (Pa)

cosine(θ) = A / H =  Pr / Pa = PF

Therefore:                  cosine(θ) = PF

Where (θ) is the phase angle between the real and apparent power.

So if you hear someone saying power factor is the cosine of the phase angle, run for the hills because they're about to try and sell you something. At this moment in time, external power factor correction is not needed in a typical domestic installations. EU law dictates that bad PF or removal of excessive harmonic loads must be designed out of the appliance by the manufacturer for loads over 75w.  This rule has been voluntarily adopted by both the US and China in the main part.

Mathematical derivation of PF = cosine(θ) from first principles.

Warning, the following information contains a derivation of the formula for "Real power" (Pr) and "Power factor" (PF) based on a purely reactive load such as an inductor or capacitor. If you do not or cannot understand this information, it makes little difference as it does not really matter for the understanding of power factor. This information is here mainly because I can't find it in other descriptions of power factor on the Internet.  Please note that this derivation contains weapons grade calculus.

I have included an emergency bypass button which will allow you to neatly jump this section

To see the full horror of this mathematically derived from first principals, then buckle in for this white knuckle ride....

note that from earlier:

Real power (Pr) = Mean power  (Paverage) = ( i = n Vi.Ii ) / n
i = 0

This calculation is going to be performed in radians so we use the letter "t" rather than "i" to denote time.

Real power (Pr) = ( t = Vt.It ) / 2π
t = 0

The steps t are T/2π where T is the time for one cycle of the mains.

The "2π" is the mathematical length of a single mains cycle in "radians," which is correct at any frequency.
For a pure sinusoid the voltage "V"and current "I", they can be described with the following equations:

V = Vpeak.sin(ωt)
I = I peak.sin(ωt - θ)

Where θ is the phase angle between the voltage and the current when looking at the trace on an oscilloscope.

Real power (Pr)  = (1/2π ). t = Vpeak.sin(ωt).I peak.sin(ωt - θ)
t = 0

=  (Vpeak.I peak/2π). t= sin(ωt).sin(ωt - θ)
t= 0

Expand using the trigonometric identity:   sin(x - y) = sin(x).cos(y) - cos(x).sin(y)  gives:

=  (Vpeak.I peak/2π). t= cos(θ).sin2(ωt) - sin(θ).sin(ωt).cos(ωt)
t= 0

Simplify Using the double angle identity:     sin(2x) = 2sin(x).cos(x)   gives:

=  (Vpeak.I peak/2π). t= cos(θ).sin2(ωt) - sin(θ).sin(2ωt)/2
t= 0


=  (Vpeak .Ipeak/2π).|cos(θ)((ωt/2) - sin(2ωt)/4) - (sin(θ)/2).(-(cos(2ωt)/2))  t=

t= 0

Applying limits:
                             =  (Vpeak . Ipeak/2π).|cos(θ)((2π/2) - sin(2.(2π))/4) + (sin(θ)/4).(cos(2.(2π))) - cos(θ)(0/2) - sin(2.(0))/4 - (sin(θ)/4).(cos(2.(0))) |  

Resolving terms:
                             =  (Vpeak . Ipeak/2π).|cos(θ)(2π/2 - (0) + (sin(θ)/4) (1) - (0) - (0) - (sin(θ)/4) (1) |

We get:

                             =  (Vpeak . Ipeak/).cos(θ)(/2)


Real power (Pr) = ( V peak * Ipeak ).cos(θ) / 2

From in an earlier section in this article we know that:

Apparent power (Pa) = ( Vpeak * I peak ) / 2

Since power factor:               PF = Pr / Pa

So combining the derivations of Real and Apparent power gives:

                        PF = Pr / Pa= ((V peak* I peak).cos(θ) / 2) / (( V peak * I peak ) / 2)

                                           = cos(θ)

Therefore in the case of a "pure" sinusoidal wave the power factor

PF = cos(θ)

Where "θ" is the phase angle between the current and voltage waveforms

Hit me with some extra equations, I can take it!

  • When looking at the vector diagram it is possible to understand how the capacitance cancels out the reactive load of the inductance. If you are adding capacitance, it either needs to be matched to the reactive inductance, usually on the same side of the power button or actively managed. As the vector diagram is a right angle triangle we can use Pythagoras's rule:

      Apparent power (Pa)2Real power (Pr)2Reactive power (Pvr)2
    Pvr2 = Pa2 - Pr2
    Pvr = √ ( Pa2 - Pr2 )

    Of course this assumes that you have successfully calculated both the apparent and real powers.

    Also if we accept that in the special case that:
      cos(θ) = Pr / Pa

    Then we know that:
    Pr = Pa.cos(θ)

    Remembering some old school trigonometry, as Pvr  is the opposite side to θ means that:

    Pvr = Pa.sin(θ)

    so :
    Reactive power (Pvr) = ( V peak * Ipeak ).sin(θ) / 2

  • These are handy equations if you need to know the reactive component of the apparent power.  Once you have established the reactive power, you can calculate the capacitance or inductance to balance the reactive element.

    • But be warned, if you match the capacitance so it exactly cancels out the inductive element, you are likely to get unwanted resonance because you have just built a tuned circuit by accident. So it is advised that you should not go to unity power factor but to a power factor of around 0.9

What is Non-linear power factor and how is it calculated?

The second form of power factor is caused by non-linear loads, often called rectified loads. Actually this is incorrect. Most switch-mode power supplies initially rectify the mains and use a capacitor to store energy as a reservoir. It's this capacitor where the problem lies. If there is no capacitor (or small capacitance) such as in an electronic transformer, then there isn't a problem. Dimmers are technically responsible for causing non-linear loads but they are used to reduce power consumption of mainly resistive loads, so tend not compound a problem but rather alleviate it.

The affect of the capacitor in a rectified load is to act as an energy reservoir for the switch mode regulation part of the circuit. As the energy is depleted the voltage across the capacitor drops but is recharged by the next rectified mains cycle, but only when the mains voltage is greater than the voltage of the capacitor, thus allowing it to recharge. This results in a large current draw for only a tiny portion of the mains sine-wave where the capacitor is recharging. This can be seen as a current spike when looked at the trace on an oscilloscope. The larger the capacitor the smaller recharge period resulting in a larger current spike.
  • The load is still using the same energy but the period that the circuit recovers the energy from each mains cycle is smaller as the capacitance is made larger. So choice of the appropriate capacitor for the load is important.

  • Another thing to note is that, there is no phase shift between the voltage and current, so it can be considered in phase. Using cosine(θ) would lead you to an incorrect conclusion that the power factor was unity.  There is no reactive component so no phase angles, thus cosine(θ) does not apply for non-linear loads
Because the current is a spike, when the voltage is at its peak, the whole of the energy needed to power the supply is absorbed in this small time period. This sudden draw of power puts an excessive strain on the power delivery systems and is arguably worse than a reactive load of the same power capability for a linear supply. But how do we calculate the power factor of this type of load? 

So what is the calculation for this type of power factor?

as before:

power factor = real power / apparent power

PF = Pr / Pa

We effectively assume that the peak current is the peak current of a sine wave and calculate the apparent power as RMS voltage multiplied by the peak current /root 2. So we use the equation we saw earlier.

Apparent power (Pa) = ( V peak * I peak ) / 2

The real power is more difficult to calculate as it is the sum of the current spike multiplied by that portion of the voltage when the current is being drawn. This gives us the real power but we must use the average power over the full cycle.

The real power has to be measured as either work done or energy used averaged. Alternatively calculated manually from the waveform using average power.

from earlier:
Real power (Pr) = ( i = n Vi.Ii ) / n
i = 0

Assuming that the load is purely nonlinear, there will be no phase shift and so, there is no simple calculation such as  "cos(θ)".  It would not be as easy to  calculate this mathematically as with the reactive power factor above.


                                rms of square wave = Peak

                            rms of triangular wave = Peak / √3

  • To calculate the power using mathematics is more involved as both the voltage and portion of the sine wave have asymmetric duty cycles. If you look at my references you will see a link to this information on how these may be calculated mathematically. A simpler option is to measure the DC output at its full rated load.  This will always be a value slightly smaller than the real power at the input as there will be losses in the regulation. Using this value would have the effect of giving a slightly worse PF than if it the real power had been accurately measured/calculated at the input.

Power factor is always:

Power Factor = Real power / Apparent power

European rules IEC EN 61000-3-2 state that any manufactured power supply over 75watts must have power factor correction built in. This has the affect of spreading the current draw over the whole mains cycle. With modern switch mode regulation, the output voltage can be varied precisely (and independently) as the input voltage cycles allowing for power factor correction which is inherent in the design.

For further reading and references:  Please see my resource page

Excellent video but with dodgy sound. Here are some salient points:

00:00 musical introduction.

01:15 verbal introduction.

10:00 explanation of calculation of power.

19:30 power in mixed reactive and resistive circuit.

24:40 description of power factor (reactive component).

27:22 How to measure power factor (reactive component).

34:50 Examples of calculations.

38:40 description non-linear load.

40:50 how to calculate power factor (non-linear loads).

45:20 Power factor for arbitrary load.

1 comment:

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