# How to Calculate POWER FACTOR

Discussion in 'Homework Help' started by JDR04, May 31, 2012.

1. ### JDR04 Thread Starter Active Member

May 5, 2011
339
4
Hi folks, I'm given the question as;

A circuit has 15V and a current of 2A. Phase angle is 45 degrees, what is the true power?

I know the formula for true power to be; Ptrue =E X I X pf

so; ptrue = 15 x 2 x pf. Am I correct so far?

So, how does one go about working out what the pf is in this question in order to derive the true power. I'm struggling to undersdtand this one as I cannot figure out the difference between true power and power factor

Thanks again guys - JDR04

2. ### #12 Expert

Nov 30, 2010
16,278
6,790
P = EI cos theta
(That's cosine 45 in degrees.)

3. ### WBahn Moderator

Mar 31, 2012
17,737
4,789
True power is the power (joules per second) that is converted from electrical to some other form (mechanical, heat, RF, etc.) and removed from the circuit.

Power factor is the fraction of apparent power (what you get by just blindly multiplying the RMS voltage by the RMS current and ignoring the phase relationship) that becomes true power. For sinusoids, the power factor is the cosine of the angle between voltage and current.

One thing to watch out for is thinking that reactive power (the portion of the apparent power that is not converted to another form but, instead, remains in the circuit and is converted back and forth between electrical and either an electric field or a magnetic field) is simply the "rest" of the apparent power after real power has been accounted for. In other words, thinking that apparent power is simply the sum of real and reactive. Instead, it is the pythagorean sum because the two are 90 degrees out of phase from each other.

4. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
507
Draw the V phasor and the I phasor which are 45 degrees apart: "project" the magnitude of the I phasor onto the V phasor by multiplying by the cosine of the phase angle. Now multiply the V and resulting I magnitude for true power since they are now "in phase" (no more angle between them).

To get PF: divide the true power (from above result) by the apparent power which is simply V times I.

HINT: it's the same as the cosine of the phase angle in this case.

For more complex waveforms it is not.

5. ### JDR04 Thread Starter Active Member

May 5, 2011
339
4
So to see if I'm getting all this;

I got 15V in a circuit with 2A and the phase angle is 45 degrees. What is the true power?

Ptrue = V x I x pf

Therefore, Ptrue = 15V x 2A x pf

pf = cosine 45 degrees = 0.707 so it's 15 x 2 x 0.707 = 30 x 0.707=21.21

I thought a power factor was always below 1? Where did I go wrong?
Thanks JDR04

6. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
507
PF is the ratio of true power (21.21Watts) divided by apparent power (30 VA). The answer is less than one.

7. ### WBahn Moderator

Mar 31, 2012
17,737
4,789

pf = cos(45deg) = 0.707; Ptrue = 15V x 2A x 0.707 = 30W x 0.707 = 21.21W:

NOTE: For a purist, the 15V x 2A has units of VA (volt-amps) because, by convention, watts are expressed as VA for apparent power. In that case, pf has units of W/VA (watts per volt-amp) and everything still works out nicely.

You got a pf of 0.707. That is less than 1.

The 21.21 is not a power factor. Had you carried the units you would have seen 21.21W, which is a big flag that it is a power, and not a power factor (which is dimensionless).

8. ### JDR04 Thread Starter Active Member

May 5, 2011
339
4
Thanks for your time BountyHunter, I'll go through it all again and check it out. Thanks for sharing your knowledge - JDR04

9. ### JDR04 Thread Starter Active Member

May 5, 2011
339
4
Thanks to WBahn, youre so right about how I did'nt track the un its through. Thanks again, I'll go through it all again and try get it right this time. Thanks again JDR04