How does one measure the power factor?
$$\text{power factor}\equiv\frac{\text{power}}{|V||I|}=\frac{R}{\sqrt{R^2+(1/\omega^2C^2)}}$$
for an RC circuit driven with \$V(t)=V_0\cos{(\omega t)}\$.
\$\begingroup\$
\$\endgroup\$
10
-
\$\begingroup\$ Watt meters measure power (as in watts) not power factor. Names have clues! \$\endgroup\$– Andy akaCommented Apr 14, 2015 at 16:50
-
\$\begingroup\$ @Andyaka: Sure, but they're often found in the same device (e.g., in the Kill-a-Watt meters). \$\endgroup\$– GeremiaCommented Apr 14, 2015 at 16:55
-
\$\begingroup\$ Expanding out the part named "power" in the equation will get you the answer. \$\endgroup\$– Ignacio Vazquez-AbramsCommented Apr 14, 2015 at 17:21
-
1\$\begingroup\$ Here's a quick Google search: electrical4u.com/… It looks like they compare the phase of the voltage and current at a point in the circuit. \$\endgroup\$– Greg d'EonCommented Apr 14, 2015 at 18:41
-
1\$\begingroup\$ Try @Gregd'Eon's link for the electromechanical type. The electronic type samples voltage and current much faster than the mains frequency so it can calculate real and apparent power from the current samples, the voltage samples, or the product of each current sample with each voltage sample (discretized instantaneous power). \$\endgroup\$– Spehro 'speff' PefhanyCommented Apr 14, 2015 at 18:58
|
Show 5 more comments
1 Answer
\$\begingroup\$
\$\endgroup\$
6
A circuit's Power Factor is the ratio of the "Real Power" to the "Apparent Power", Pr/Pa. It is also equal to the cos(Voltage Phase Angle - Current Phase Angle). It can be measured in an AC circuit by comparing the Voltage wave form to the Current wave form. Any time the voltage and current wave forms are not exactly in phase there is a power factor < 1.
So if you were to use a circuit to measure the zero crossing points of each wave form (voltage & current), you could then calculate the phase angle difference and the power factor.
For example, with a 50 Hz sine wave there is about 55.6uS per degree, (1/50/360). So if the measured wave forms had a difference of 1000uS this would calculate to a phase difference of 18 degrees, and the power factor would be cos(18), or 0.95
-
2\$\begingroup\$ not quite true... DisplacementPowerFactor = cos(Voltage Phase Angle - Current Phase Angle). PowerFactor is the summation of harmonics WHICH for a linear load (R,L,C) equals teh displacementPowerFactor \$\endgroup\$– user16222Commented Apr 15, 2015 at 12:33
-
\$\begingroup\$ I did say "sine wave" in the example, which would imply no harmonics... If you had other wave forms the harmonics would also need to be considered. \$\endgroup\$– NeddCommented Apr 15, 2015 at 14:19
-
2\$\begingroup\$ doesn't change the fact that the phase difference is the displacementPowerFactor. \$\endgroup\$– user16222Commented Apr 15, 2015 at 15:33
-
\$\begingroup\$ @JonRB What you are telling is that not the phase difference should be takin into account, but the difference of the voltage phase and the current phase. How can one measure this phases by measuring the voltage and current? \$\endgroup\$– otmezgerCommented Feb 10, 2016 at 15:24
-
\$\begingroup\$ @otmezger it depends on what you are after. Are you after PowerFactor or DisplacementPowerFactor. \$\endgroup\$– user16222Commented Feb 10, 2016 at 22:49