Gemini AI providing wrong info related to basic power factor calculations

While performing some research, I noticed that Gemini claimed that: power factor = 0.60 = active power / apparent power and this means 60% of the apparent power is made out of active power and remaining 40% is reactive power (see figure below).

Although seems convincing and can misguide some of the younger folks in the industry, this is not true and we can validate this by just looking at the reactive and active power relationships and working out the example:

Assume P = 1MW ?and our power factor is 0.60pf.

Then knowing, Pf = active power / apparent power (as also indicated by Gemini), the apparent power S = 1.66MVA (which is true).

?However, in order to calculate Q, we need to use the power triangle relationship.

So solving for Q will give you, 1.33MVAr.

So if you calculate 1.33/1.66, this is equal to 0.83 (or 83%, not 40% as Gemini indicated!).

This discrepancy occurs because Active power and Reactive power are added vectorially (and they perform work in different planes per say). They are two different types of power, and we can only add two same entities (1 apples + 1 apples = 2 apples) and not two different items such as 1 apple + 1 orange = perhaps a cocktail but surely not 2 apples.

So as per the power triangle:

Well this makes me question, why on earth the P and Q get added vectorially in first place and how did we get the above equations:

To answer the above questions, I would like to re-write the above equations as there are key assumptions that are often not explicitly called out:

• P here is the average power over the full cycle. Power flows from generation to load, so there has to be an average amount of power which is delivered.
• Q here is NOT the average power over the full cycle, but rather just the PEAK of its waveform. Q average in the system over its waveform is ZERO because in ? of cycle, it is delivered to the reactive loads (such as inductor and motor assuming lagging power for this example ) to form their magnetic field and the other ? cycle, those fields are collapsed and reactive power is returned to the reactive suppliers such as capacitors to form electrical fields. Keep in mind that I mentioned ? because Q oscillates twice the frequency of the fundamental frequency.
• V and I in the above equations are RMS voltages (and not peak or instantaneous voltages).

So lets re-write power triangle equations:

?

Back to real question, how did we come up with the above equations. Here is an example:

We know that Power = Voltage x Current or it becomes easier to remember if you realize that:

-????????? Power = Watts = Energy or joules per second

-????????? V = joules per coulomb (coulomb can be thought of a way to easily write 6.25 x 10^18 electrons in a simpler way as this is how many electrons it takes to make up on coulomb).

-????????? I = coulomb per second.

So now just multiply instantaneous voltage and current together to get instantaneous power, which we call p(t):

Use some trig identifies or perhaps ask wolfram to do it for you, you will get (notice the red font terms):

You basically get three terms and the ones that are cos waveforms are related to active power and the ones that are related to sin waveform are for reactive power.

So Active Instantaneous Power? = lets say capital P with respect to time

And Reactive Instantaneous Power = CAPITAL Q(t):

So now take the average of active power (the black font term will become zero after you take the anti derivative of a sin(2wt) ):

Now take the magnitude or peak of the Q(t):

Now if you draw the phasor diagram of these two equations, they will be orthogonal to each other (one is cos and the other one sin) and this will give you a power triangle!

BONUS CONTENT:

You may have also seen another equation:

You may think that it’s the same power triangle equation with V^2 showing up because Irms is replaced with V/X?

Well no, this is a completely different equation which represents the Power output of a synchronous generator (or

Let me add additional context

-????????? P here is also average Power

-????????? I and V are also in RMS

-????????? But θ is not θv - θi but rather the angle difference between generator rotor angle and the grid voltage angle (so this equation is subtracting angles of two different voltage quantities whereas the power triangle was subtracting voltage angle from the current angle).