# Complex power - Voltage and current graph



## supra33202 (Feb 16, 2016)

Please see





I followed the whole example except the graph for the voltage and current vector.

Could someone help me out?

@5:50 - How do we draw the graph of voltage and current? How do you know the theta is negative? Can I draw the current vector on the x-axis and the voltage vector at an angle with respect with the current vector? I guess another way to ask the question is - how come the current and voltage vectors are in graph quadrant IV not in quadrant I?﻿

@12:43, how come we have the angle -33.9 for I? Why negative? Is it we just arbitrarily draw the current and voltage vectors at quadrant IV?

I guess the general question is: how do we determine which quadrant to be used when trying to draw the current and voltage vectors on the graph?

Thanks!


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## TWJ PE (Feb 17, 2016)

The load is lagging; therefore it's inductive. So, ELI the ICE man... ELI applies. So current lags voltage.


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## supra33202 (Feb 17, 2016)

I know the power factor is lagging, so current lags voltage.

And I know the graph is correct about showing that relationship (CCW rotation).

But why the current and voltage vectors are in graph quadrant IV not in quadrant I?﻿


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## jgharris P.E. (Feb 18, 2016)

Supra,

the voltage angle is 0 degrees (since this is a single phase circuit and the problem never specified a voltage angle), which is why the voltage is drawn on the real axis.  Therefore the voltage is not in quadrant I or IV.

We know from the power factor that current lags the voltage because the problem says so.  If the pf was .83 LEADING then the current would be in the first quadrant.  In simplistic terms it is a race between the current and the voltage.  The voltage gets to the "finish line" first (angle 0).  At that point, the current is 33 degrees BEHIND the voltage hence the lagging angle of 33 degrees.  Since the Voltage angle=0, abs(33.9) degrees lagging that would be -33.9.  As you rotate around the unit circle (0-360) the current will always be 33 degrees behind/lagging the voltage.     If we had unity power factor at every point in time the current angle would match the voltage angle, ie. pf= cos (theta V-theta i)    pf=cos(0-0), pf=1.     This is critical to understand.

I'm no expert in this, I'm studying for the PE like you are.  Hope this helps...

JGH


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## supra33202 (Feb 18, 2016)

jgharris said:


> the voltage angle is 0 degrees (since this is a single phase circuit and the problem never specified a voltage angle), which is why the voltage is drawn on the real axis.  Therefore the voltage is not in quadrant I or IV.


Thanks for the explanation. I have other questions.

Since the problem never specified a current angle, can I assume the current angle is 0 degree and draw the current vector on the x-axis?

Then, the voltage vector will be in quadrant I and the theta (angle between voltage and current vectors) is positive 33 degree.

Technically, the graph that I illustrated above is also "current lags voltage" or "voltage leads current". Is that correct?

I guess another way to ask the question - Instead of using voltage vector as the reference, can I use current vector as the reference?

Thanks!


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## jgharris P.E. (Feb 18, 2016)

First let me apologize. The statement about the voltage was incorrect.  Review the attached PDF.   I worked the problem in two ways.  First I worked it how I the video did and how I would have if I was doing this problem for the first time.  The second time was with the positive 33.9 degrees instead of the current angle of -33.9 just to show you that it doesn't matter.  Hope this helps.

View attachment pf relation problem.pdf


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## jgharris P.E. (Feb 18, 2016)

> [COLOR= rgb(39, 42, 52)]The second time was with the positive 33.9 degrees instead of the current angle of -33.9 just to show you that it doesn't matter. [/COLOR]


This specific problem it doesn't matter because we have no reference voltage to begin with (since we're solving for V given S and I).  If the problem gave us voltage (such as given V solve for i) and then and you chose a positive current angle that would be incorrect if the problem stated a lagging pf...  

This is a good problem because it really works on the fundamentals.  

Keep studying!  V:


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## supra33202 (Feb 18, 2016)

I watched the video below.



@5:25,  it illustrated the "relative" and "absolute" situation for the voltage and current vectors.

Based on the video, since the original problem didn't specified the phase angle of the input voltage, we can use the "relative" situation. We can assume that the voltage is at 0 degree. And we can just use P=VI. No need to use S=VI*.

No need to worry about the conjugate.

And the theta (angle between voltage and current vectors) is negative 33 degree.

The voltage at the load is 438.6V at 67.8 degree.

Is that correct? I am still confused.

Are you saying that since the original problem didn't specified the phase angle of the input voltage or input current, the answer (voltage at the load) can be at 0 degree and 67.8 degree? (The problem says the generator generate 100kw only; no mention about phase angel of the input voltage or current)

I know the 2 calculations that you did the theta (angle between voltage and current vectors) is always 33 degree in magnitude. Whether it is positive or negative, it depends on how we draw the voltage and current vectors on the graph.

So at the end of the day, we don't need to worry about phase angle of the voltage of the load?

Thanks!


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## DK PE (Feb 19, 2016)

Maybe I'll help or make worse but I'll give it a whirl.  If you prefer to think of 1st quadrant, then simply choose the current as the reference phasor.  Therefore, I is I at 0 degrees. Then the voltage at the load has to be at POSITIVE 33 degrees (in 1st quad) (due to lagging PF)

Now just apply S = V I* ,   since I is all real the conjugate is the same and the VI* is the product of VI at an angle of 33 degrees.  This is the same answer as the video got choosing V as the reference phasor so just proceed further the same.

Hope I helped and didn't make things worse?


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## supra33202 (Feb 22, 2016)

Hi DK PE, it helps. Thanks!

I guess I am still not 100% clear.

This is my understanding.

1) If the problem didn't specified voltage angle, we can assume the voltage angle is at 0 degree. And draw the current vector based on if the problem specified if current lagging or leading voltage or if power factor is lagging or leading.

2) Based on Complex Imaginary's conjugate videos, the angle between voltage and current vectors is what we are interested in. That is why we need to do conjugate of current if we have "absolute" situation".

3) If the problem didn't specified voltage angle or current angle, you can pick either the voltage or current as your reference (either voltage vector or current vector can be at 0 degree).

4) You can draw the current and voltage vectors in either quadrant I or IV.

Can I say the followings?

If the current and voltage vectors are in quadrant I, no need to do conjugate of current?

If the current and voltage vectors are in quadrant IV, do conjugate of current?

Thanks!


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## DK PE (Feb 22, 2016)

supra33202 said:


> Hi DK PE, it helps. Thanks!





> supra33202 said:
> 
> 
> > Hi DK PE, it helps. Thanks!
> ...



1) True

2) False... the relationship S = VI* MUST be used because positive complex power was defined a long time back for a lagging or inductive load and they wanted S in 1st quadrant.  As you could see by the video you attached and my explanation...  Whether you choose voltage reference at 0 degrees or current phasor at 0 degrees, you get S at ~ 33 degrees in each case.  It doesn't have anything to do with relative anything or quadrants, you must use S = VI* (unless the numbers just happen to work... by luck)

3) true

4) true

One thing to keep in mind is when you are choosing your reference phasor, it is just a matter of a snapshot in time, if you choose V @ 0 degrees and then current is lagging... just wait a moment for phasors to rotate and I is now at 0 and V is leading by the same amount.  What quadrant you are drawing your phasors in doesn't impact the S = VI* relationship, you must use it....


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## supra33202 (Feb 22, 2016)

I know the magnitude of the angle between voltage and current is ~33 degrees.

How do we know if the angle between voltage and current vectors is positive or negative?

Do we always do voltage angle minus current angle or current angle minus voltage angle? And then do conjugate of current?

Thanks!


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## DK PE (Feb 22, 2016)

Here's the way I think about it... The phasor vectors ALWAYS rotate CCW.  Imagine you are sitting stationary on the arrow at end of +x (real) axis... So the voltage phasor comes by and then a small fraction of time later the current vector sweeps by the same place.   The current is lagging (that is arriving at the x=axis) at a later point in time (lagging).  So both of these phasors are rotating at the operating frequency with a fixed ~33 degree space between them. And S is always = VI*


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## knight1fox3 (Feb 23, 2016)




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