Dr. Kelly,

Oversight? TIA

Dr. Kelly,

Oversight? TIA

--------------

OK - real power/apparent power = power factor. If you look at the apparent power S(VA) real power P(W) and reactive "power" Q(VAR) and draw the power triangle.

S /| PF=cos (a) =P/S=P/root{P^2 +Q^2] or generally but not always R/root[R^2+X^2] / |Q a = angle =arccos(PF) = arctan Q/P = arcsin Q/S (1) /a_| P You have to use trig functions as available on most calculators to get the angle as the nice relationship is actually between the cosine and the power factor - not the angle and the pf.

PF angle Pf lead, angle is - PF lag, angle is + 1 0 0.8 36.87 degrees 0.6 53.12 0 90 Hence the relationship between pf and angle is as above angle = arccos(pf) etc That is it in the simplest form.

--

Don Kelly snipped-for-privacy@shawcross.ca

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Don Kelly snipped-for-privacy@shawcross.ca

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Thanks. I keep the following for reference:

AC Electrical Circuits:

voltage E = 120 volts

resistance R = 10.8 ohms

current I = 10 amps

From which:

impedance Z = E / I = 12 ohms reactance X = Z - jR = 5.23 ohms

power P = I^2 * R = 1080 W

reactive power Q = I^2 * X = 523 VAR

apparent power S = E * I = 1200 VA

Phase Angle Between Voltage and Current:

angle = arc tan X

angle = arc cos R

angle = arc sin X

Power Factor

power factor PF = cos angle = P

angle = arc cos PF = 25.84 degrees

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