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Because I know the value for the power factor for the series RL circuit I've tried to find the ratio between the series circuit power factor and the parallel circuit power factor, expressed in terms of resistance and impedance, and I found that:

$$\frac {\cos_s^2\phi}{\cos_p^2\phi} = \frac {R^2}{X_L^2},$$

which doesn't really help me, because I need an actual value.

Is there another way to get the power factor for the parallel circuit?

The values of the components are not known, but are the same in the series and parallel circuit.

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The PF of a series RL circuit is found from $\tan(PF) = R/\omega L$ while the PF of a parallel RL circuit is found from $\tan (PF) = \omega L/R$. Thus the tangents of the power factors of the 2 circuits are reciprocals of each other. For example, if the series RL circuit has a power factor of 0.4 (or 21.8 degrees), the parallel RL circuit has a a power factor of $\arctan (1/0.4)$ or 68.2 degrees.

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  • $\begingroup$ Thank you, this was exactly what I needed, knowing this I was able to continue from my ideea and get the answer. $\endgroup$ – Mihai Feb 19 '14 at 12:25
  • $\begingroup$ I have one thing to add tough, from what I know the power factor is the cosine of the phase angle, so shouldn't this $tan(PF)=\frac{ωL}{R}$ be like this : $tan(arccos(PF))=\frac{wL}{R}$ ? $\endgroup$ – Mihai Feb 19 '14 at 12:36

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