# Help with complex power and power factor correction problem

I'm having a bit of trouble trying to understand what is being asked of me in a relatively basic circuit theory question involving complex power and power factor correction.

Now, part (b) and (c) seem easy enough to answer given enough information but I cannot figure out whether I'm understanding (a) right. Without knowledge of the power S2 drawn by the corrective device, I cannot see how I can determine the PF (power factor) of the source let alone its total power delivered S = S1 + S2. Is it presumed for (a) that only the motor is drawing power here?

• The corrective device comes into the circuit only in b). So calculate it just without the corrective device. Dec 7 '14 at 17:29

Now, part (b) and (c) seem easy enough to answer given enough information but I cannot figure out whether I'm understanding (a) right.

The question reads to me as if part(a) applies before the correction so....

Try looking up power factor phasor diagrams: -

You have an apparent power (volt*amps or VA) of 150 and, you have an angle of 24 degrees. Just apply pythagoras to calculate real (active) power. As bonus you can even calculate reactive power.

• right, if (a) applies prior to correction, it follows readily that pf = cos(Arg(S1)) = cos(24 deg), yes? but without more knowledge about V, I there appears to be virtually no way to compute the required impedance in (b) for pf correction it seems Dec 7 '14 at 10:39
• Yes it does i.e 137 watts of active power. Dec 7 '14 at 10:40
• but without further knowledge of the problem, such as V or I, it seems impossible to determine the necessary impedance for the corrective device to reach the desired pf. I mean, it's relatively easy to leap to Arg(S_1 + S_2) = -arccos(0.98) but without more knowledge it seems impossible to ascertain S_2 let alone Z_2 (where S_2 = V_eff^2 / Z_2 = I_eff^2 Z_2) Dec 7 '14 at 10:45
• I thought you said (b) and (c) were easy? In the example, if you added pure reactive power in the opposite direction to that shown in my diagram then clearly (c) can be answered and, yes, without knowledge of the motor's impedance or the source voltage you cannot answer (b). Dec 7 '14 at 10:49
• thank you! You confirmed my suspicion that there's something missing from my understanding of the problem preventing me from solving (b). I just wanted to verify there wasn't something I was missing from part (a) as well. Dec 7 '14 at 20:49

Well first off I never done this for a motor. But if you could model the motor as a simple lumped element. (like a resistor is series with an inductor.) Then you can ask what impedance in parallel makes the load look purely resistive to the source.

For the L- R case above I know the answer is a series R-C with equal R's and RC=L/R. (equal time constants.) Now for motor use you probably don't want another resistance on the load... so maybe just the C? at least that's a place to start.

• If all the data was there to be able to calculate impedances, you would only need a capacitor - it's exactly the same as a // tuned circuit - impedance becomes resistive at or about $\dfrac{1}{2\pi\sqrt{LC}}$ Dec 7 '14 at 21:44
• OK, I think the R in the C line makes the Q ~ 1, or less. Dec 7 '14 at 21:58
• It certainly makes the phase angle zero but Q is dependant on the total resistance. Power factor correction is nothing other than designing a tuned circuit!!! Dec 7 '14 at 22:25
• @Andyaka why do you say 'at or about' $\omega_0$? Is it potentially the case that the impedance is undefined at $\omega_0$? Dec 8 '14 at 1:43
• @oldrinb - I was thinking of a RLC parallel circuit with R in series with L. This would have a slight phase error but in PFC this is not quite the same - RLC are all in parallel therefore if Xc = XL then phase will be 0 degrees and impedance will be purely resistive. Dec 8 '14 at 10:09