# What You Need to Know About Power Factor Basics In my last blog, I have discussed the 3 Things You Need to Know About Electrical Power in AC Circuit, where I talked about the three different types of power in the AC electrical system, which are the Real power, Reactive power and Apparent power. All these powers combine to give us a very significant value called the Power Factor. In this blog, I will tell you what you need to know about power factor basics

## A Quick Recap About the Three Powers that Affect the Power Factor

• Real power is the power that gets converted into another form of useful power such as mechanical power, heat, and light.
• Reactive power is the power that does not perform any useful work; however, it is required by some equipment such as motors and transformers to create a magnetic field to operate.
• Apparent power is the power you supply to the load. ## Importance of Knowing the Power Factor

Understanding Power Factor (PF) is important as it gives us a clear understanding of the efficiency of the electrical system. It’s the ratio of true power and the apparent power, which represents how much power supplied by the source will perform useful work.
The cosine of the angle between the Apparent power and the True power will reveal the power factor of a system. The bigger the reactive power in the system, the bigger the phase angle, and the lower the power factor. It’s represented by a number between 0 and 1 and doesn’t have a unit.

PF = cos θ

When the PF is 1, it means that the resistive load will consume all the power supplied by the source and converts into another form of useful energy. If the PF is 0, it means that all the power from the source is entirely reactive and will be stored in the reactive load and return to the source. A perfect power factor would be 1.0 or unity PF, but except for a purely resistive load, this would be counterproductive because that will create resonance. A good PF would be 1.0 to 0.95, an OK PF is between 0.95 and 0.8, and anything below 0.8 needs improvement or power factor correction (PFC).

## Let's See the Power Factor in Action

The best way to explain and understand the effects in ac electrical system is to connect and observe it, so that’s what I have done. Let me show you the effects of various loads and understand the PF based on them.

### Power Factor in Resistive Loads

Purely resistive loads convert all the supplied power into useful power, that is, Apparent power is equal to the True power. This makes the PF of 1. Calculations are as follows

V = 5.1V, I = 1A and PF = 1
Apparent power = S = V.I = 5.1 x 1 = 5.1 VA
True power = P = V.I.PF = 5.1 x 1 x 1 = 5.1 W
Phase angle θ = cos-1(PF) = cos-1(1) = 0 deg
Reactive power = Q = V.I.sin θ = 5.1 x 1 x 0 = 0 VAR

### Power Factor in Inductive Loads

Purely inductive loads use all the supplied power as reactive power. I’m using a real inductive load, which have some resistance due to the windings so some power will convert to useful and some used as reactive.

In the video, I’m using a 6Ω 29mH inductor first, then 18Ω 92mH and lastly 18Ω 680mH inductor from our AC pracbox. Using the same formula as before, here are the measured and calculated values

Component Voltage Current Power Factor Phase Angle True Power Apparent Power Reactive Power
29mH Inductor (6Ω)
8.5V
0.756A
0.599
53.2 deg
3.8W
6.43VA
5.15VAR
92mH Inductor (18Ω)
11.9
0.352A
0.544
57.04 deg
2.2W
4.18VA
3.51VAR
680mH Inductor (18Ω)
13.8V
0.101A
0.299
72.6 deg
0.4W
1.39VA
1.33VAR

As you can see, as the inductance increased, the PF decreased, the phase angle between apparent and true powers increased, and the reactive power also increased.

### Power factor in capacitive loads

Capacitive loads that are pure convert all the apparent power into reactive power. You can see in the video, that the 80uF capacitor from the AC pracbox was drawing current, but there was no True power, which means all the power reactive and the PF was 0.
Calculations are as follows

V = 14.6V, I = 0.382A and PF = 0
Apparent power = S = V.I = 14.6 x 0.382 = 5.57 VA
True power = P = V.I.PF = 14.6 x 0.382 x 0 = 0 W
Phase angle θ = cos-1(PF) = cos-1(0) = 90 deg
Reactive power = Q = V.I.sin θ = 14.6 x 0.382 x 90 = 5.57 VAR

### Power factor in a 3 phase induction motor

Motors are known to have the best efficiency when they are fully loaded and the worst when not loaded. And discussed earlier, the power factor and efficiency are quite closely related. To prove this point, I’m connecting a 3 phase induction motor in this video.

Let’s see the calculated and measured values

Component Voltage Current Power Factor Phase Angle True Power Apparent Power Reactive Power
41.1V
0.616A
0
90 deg
OW
25.32VA
25.32VAR
40.8V
0.720A
0.230
76.70 deg
6.6W
29.34VA
28.56VAR

The larger motors will have a small PF even when it not loaded.

## Effects of Low Power Factor

When the power factor is low, this means that the electrical system has low efficiency. The major effects of low power factor are:

• The system would draw a much higher current for real and reactive power, which increase line losses and hence the heat
• Larger conductor sizes and switch gears are required to compensate for the line losses, which make this a costly exercise

Low power factor not only causes an issue on the system but will also cost a lot of money as the power company will charge an extra cost on the top of the demand charge if the power factor falls below a specific limit.

So how do we fix this problem? Power factor correction, which is the topic for another post.

## Conclusion

Understanding the power factor is essential as it gives us insight into the efficiency and the balance of the reactive load in the system. It also helps us to avoid any issues caused by the poor power factor like extra installation cost and low power factor penalty or charges.

A low power factor can be easily fixed using a method called “Power Factor Correction (PFC)” which I will discuss in my next blog post.

Thank you for reading, and I hope you found the post helpful. 