# Lagging Power Factor



## va_gator (Apr 4, 2011)

Given the power factor, we can find the power factor angle (@), with: @ = arccos(PF).

Then if we are given a current (I) and it is said to be lagging the voltage, we can write the current as I&lt;[email protected] ('&lt;' is the angle symbol, for polar representation). In some of the calculations in the sample exam, the '-' is applied to the angles for lagging power factors (I&lt;[email protected]), but not all the time. In other solutions, the '-' for the angles is omitted on lagging power factors. It appears on the motor problems, and I can’t seem to find a trend when to include and when to drop the negative for the angles for lagging PFs

This is one of the last things I’m struggling with, I may be overlooking a very simple concept, and would appreciate it if any of you can help me out. Thanks.


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## Flyer_PE (Apr 4, 2011)

It all depends on if you are dealing with the angle of current with respect to voltage or the angle for apparent power.

Apparent power: S=VI*. A load with a lagging power factor will have a negative angle for current and a positive angle for apparent power.


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## cableguy (Apr 4, 2011)

Lagging means the current lags the voltage. Therefore the angle is negative whenever you see the word lagging.

35 degrees lagging means -35 degrees.


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## va_gator (Apr 4, 2011)

Flyer_PE said:


> It all depends on if you are dealing with the angle of current with respect to voltage or the angle for apparent power.
> Apparent power: S=VI*. A load with a lagging power factor will have a negative angle for current and a positive angle for apparent power.


That is probably the closest "trend" that I can use, when dealing with apparent power (S), even on a system with lagging PF, S is represented with a positive angle:

S&lt;@

But when you are using the current, I to solve for example V=IR, if the PF is lagging, use I= I&lt;[email protected]

But there are still situations when that rule is broken by the sample test solution (Problem 517 of the 2009 NCEES Sample). Where S and the negative angle representation for the PF is used to solve for Reactive Power.


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## sam314159 (Apr 4, 2011)

Y-System:

The power factor angle is the angle by which V_phase LEADS the corresponding I_line

Delta-System:

The power factor angle is the angle by which V_line LEADS the corresponding I_phase

Just in general though, the power factor angle is the angle by which voltage LEADS current.

*Lagging Power Factor (Inductive Load)*

I&lt;[email protected] : Negative Angle (If you take your voltage as the zero reference, then I needs to be LAGGING or [email protected] behind the 0)

S&lt;[email protected]: Positive Angle (P+jQ) (A load taking in S&lt;[email protected], or P+jQ, means it is taking in WATTS and VARS, both are same sign)

Z&lt;[email protected]: Positive Angle (R+jX) (X is inductive, X=jwL which is positive)

*Leading Power Factor (Capacitive Load)*

I&lt;[email protected]: Positive Angle (If you take your voltage as the zero reference, then I needs to be LEADING or [email protected] ahead of the 0)

S&lt;[email protected] : Negative Angle (P-jQ) (A load taking in S&lt;[email protected], or P-jQ, means it is taking in WATTS and putting out VARS, opposite signs)

Z&lt;[email protected] : Negative Angle (R-jX) (X is capacitive, X=-j(1/wC) which is negative)


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## Flyer_PE (Apr 4, 2011)

va_gator said:


> Flyer_PE said:
> 
> 
> > It all depends on if you are dealing with the angle of current with respect to voltage or the angle for apparent power.
> ...


They didn't break the rule in their solution to #517, they just didn't include all the details.

The synchronous motor has a leading power factor which gives a negative angle for apparent power (S).

What they didn't show for the induction motor is that I = 14.43/-53.13o. However, S=sqrt(3)VI* -&gt; SInd Mot = sqrt3*480*14.43/+53.13o


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## va_gator (Apr 4, 2011)

Thanks for the help, I somewhat understand it now, but more importantly, I can use the above as a guideline. Just a few minutes ago I found a thread with a related topic.

http://engineerboards.com/index.php?showtopic=11450


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## Jonjo (Apr 6, 2011)

va_gator said:


> Thanks for the help, I somewhat understand it now, but more importantly, I can use the above as a guideline. Just a few minutes ago I found a thread with a related topic.
> http://engineerboards.com/index.php?showtopic=11450


More over , if your problem is what to use? , find first is (+A°) Cap Circuit , or (- A°) Inductive Circuit , and than no problem because

Cos(+A°) = Cos(-A°) =&gt; , P = 1.7321 V.I Cos(+A°) = 1.7321 V.I Cos(-A°) is the Module of the Power assets (Watts)


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