# Complex Power Current Conjugate S=VI*



## ellen3720

All,

I have what I think may be a really basic question but it keeps tripping me up.

When is the conjugate of the current used with the formula S=VI*?

The 3 phase equations S=sqrt(3)VI and S=3VI don't seem to use the conjugate, but I don't know why. I've also run into some single phase problems that don't use the conjugate.

Am I missing something?

Thanks!

Ellen


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## KJS

Yes, you have to do conjugate of current angle for S. It is easy to remember. If the current angle is given, just take conjugate. If it is not given, then not.


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## justin-hawaii

@ellen3720 

The conjugate is used when you are given the power factor or power factor angle.  The power factor angle tells you the phase angle difference between the voltage and current phasors.  For example, if the power factor angle is 30 degrees lagging and you assume voltage has a phasor angle of 0 degrees, then the current phasor angle will be -30 degrees.  This is because in a lagging power factor, the current phasor angle lags the voltage phasor angle.  If the power factor angle is -30 degrees aka 30 degrees leading and you assume voltage has a  phasor angle of 0 degrees, then current will have a phasor angle of +30 degrees, because the current phasor angle must lead the voltage phasor angle.  

If the power factor angle is not given and you are instead given the current phasor angle and the voltage phasor angle, then you can simply multiply these two phasors to get the single phase apparent power or you can multiply those two phasors and the root 3 term to get the 3 phase apparent power.  

In my above discussion, I am assuming line voltage and current for the references to 3-phase power.


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## Szar

The answer of "why" goes to how Power is defined.

Remember your power triangle. 


Inductive Loads are Positive Angles and Apparent Power is defined as Positive on the Power Triangle. 

Capactive Loads are negative angles and Apparent Power is defined as Negative on the Power Triangle. 

Also remember the Impedance of each type of Load


Inductive Loads are _j_wL or *wL &lt; 90°*

Capactive loads are -jwC or *wC &lt; -90°*



Now back to Math and Mr. Ohms:


V = I * R which makes,

I = V /  R

Accordingly... if V is given a reference angle of Zero and the load is inductive the maths results in a negative phase angle for Current.... 

So with P = V * I... without the conjugate how could apparent power be Positive for an inductive load?  You would have P = V &lt; 0° * I &lt; -90° which would mean your apparent Power would be negative... which its not the case for an inductive load.  The Conjugate is used to correct the sign in your Maths. 

I'd go back and review the Power Basics with the above in mind and just work through the derivations of Single Phase Real Power and Apparent Power, then apply that 3Ø terms to see why the 3Ø power equations don't include the conjugate.  You could probably just Youtube that too...   

I also need a lot more Coffee, otherwise I would have done it above.


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## Zach Stone P.E.

> 19 hours ago, ellen3720 said:
> All,
> 
> I have what I think may be a really basic question but it keeps tripping me up.
> 
> When is the conjugate of the current used with the formula S=VI*?
> 
> The 3 phase equations S=sqrt(3)VI and S=3VI don't seem to use the conjugate, but I don't know why. I've also run into some single phase problems that don't use the conjugate.
> 
> Am I missing something?
> 
> Thanks!
> 
> Ellen
> 
> 
> 2


Hi @ellen3720,

The first formula uses complex numbers and solves for *single phase* *complex power *(both magnitude and angle) as the product of the complex phase voltage and the conjugate of the complex phase current:

S1Φ =VpIp*
S1Φ = |S1Φ|&lt;θ

To solve for three phase power, you can multiply the same formula by three, which also uses complex numbers and the current conjugate:

S3Φ = 3•S1Φ
S3Φ = 3•VpIp*
S3Φ = |S3Φ|&lt;θ

The third formula uses magnitudes only, and just solves for *three phase apparent power*, which by definition is the magnitude of three phase complex power without the angle:

|S3Φ| =√3•|VL|•|IL|

Typically when using this formula, you'll obtain the power angle from the inverse cosine of power factor:

θ = cos-1(PF)

S3Φ = √3•|VL|•|IL|&lt;cos-1(PF)

The following cheat sheet gives examples to help to avoid similar mistakes with power formulas. You can print it out and take it with you to the PE exam:

http://www.electricalpereview.com/biggest-mistake-commonly-made-three-phase-power-formulas/


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## ellen3720

Thanks everyone for the detailed answers. After doing a detailed review,  I think my confusion is in the difference between complex power (magnitude and angle) and apparent power (magnitude only). 

The way I now understand it:

- I don't need to take the conjugate when working with S= sqrt(3)VI because the angle doesn't matter. Only the magnitude is important. The angle would be calculated separately.

- Under any other circumstances (S = VI*, S = 3VI*), the current angle should be reversed.

If the problem statement gives the current as I = 10&lt;*-*15, I would use 10&lt;*+*15 in S = VI* or S = 3VI*.

If the problem statement gives the current as 50A @ 0.80 lagging power factor, the current would be I = 50&lt;*-*36.87. However, in S = VI* or S = 3VI*, I would use 50&lt;*+*36.87.

If the problem statement gives S and V and asked to solve for current, I would reverse the sign of whatever current I calculated using S and V.

What had been screwing me up was the following problem from Spin Up:

A 3 phase, 480V generator has a balanced 600kVA load with a PF of 0.75 lagging. What current does the load draw?

I had calculated I = 600kVA/sqrt(3)*480 = 721.69 A. The current angle is -41.41 degrees, based on the PF provided. 

Originally, I had then taken the conjugate of the angle for an answer of 721.69 &lt;* + *41.41 A. Which is incorrect.

In this case the angle is given by the PF and only the magnitude is given by the apparent power equation.

Is there ever a case when S= sqrt(3)VI* is used for magnitude and angle, though? In Graffeo's study guide, he lists S = 3VphIph* = sqrt(3)VlineIline* as the "apparent or complex power" equations. This is what had led me to use +41.41 degrees as the angle above.


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## Electrical786

@ellen3720: The problem might give you realistic data. Its upto you to decipher between phase and line voltages,currents. MAKE SURE YOU WRITE FULL EQUATIONS in PHASOR FORMAT AND NOT JUST EQUATIONS LIKE S=V*I or S=sqrt(3)*V*I


for example the problem you mentioned,

A 3 phase, 480V generator has a balanced 600kVA load with a PF of 0.75 lagging. What current does the load draw?

1st of all, they havent mentioned if its a DELTA or WYE System.  I am pretty sure NCEES does clarify what type of system before asking to calculate current (SEE NCEES Problem 121)

2. I would take 480 V as the LINE VOLTAGE  (it works for both WYE and DELTA system)

3. they are asking for current Load draws, this could be Phase or Line current. 

4. the complex power S provided is 600 kVA ,

Use the equation 

|S3Φ| = √3•|VL|•|IL|

Plugging in the values, you can easily get the magnitude of LINE CURRENT to be 721.69 A.

the angle for S3Φ would be cos﻿-1﻿(0.75) which is +41.41﻿, and like folks have already clarified to you the Current angle would be Negative because we are assuming Voltage has a ZERO Angle. so IL would be 721.69 &lt;* - *41.41﻿﻿ A

5. I dont always agree with GRAFFEOs formula

S3Φ=sqrt﻿(3)VlineIline﻿﻿* ﻿

sometimes you get right answer, and sometimes not. So best thing is to write the Voltage and Current in phasor format and understand LAG/LEAD Relationship between Voltage and Current, and you will never get the angle wrong.

Like Zach wrote the formula

|S3Φ| =√3•|VL|•|IL| (I use for MAGNITUDE ONLY)

S3Φ = 3•VpIp*
 

I use this for Complex power MAGNITUDE and ANGLE . this will always work!


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## rmsg

ellen3720 said:


> All,
> 
> I have what I think may be a really basic question but it keeps tripping me up.
> 
> When is the conjugate of the current used with the formula S=VI*?
> 
> The 3 phase equations S=sqrt(3)VI and S=3VI don't seem to use the conjugate, but I don't know why. I've also run into some single phase problems that don't use the conjugate.
> 
> Am I missing something?
> 
> Thanks!
> 
> Ellen


Conjugate comes into play because normally current's phase angle is opposite sign of actual power factor angle. Power factor angle is normally same angle as impedance angle

Power is also written as

P = (V x V)/Z  but since the other formula where V and I are taken, we have to consider the conjugate of current angle not the actual current angle

So in essence just remember that the conjugate I angle is same as Power Angle or Impedance angle Z, which should be used for Power


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## Imran

If 3 Phase Complex power is S=50MVA Angle 30degrees lagging and the *LINE voltage in star connection* is 30kV rms then Mod Z per phase  = 30x30/S* . Am I right ? This is a question please since I am confused about sqrt 3 , whether to apply or not


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## Imran

Hi Friends , I got the above resolved by further study .

in 3 Phase star system Vph =VLine/sqrt 3 and S phaser is for a 3Ph system ( given). Then  Mod (*VLine/sqrt3) ^2) /( I S I &lt; -Phi) /3* reduces it to 1 Phase equivalent and square root 3 gets eliminated .


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