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If I know the Watts, the Power factor, the VA, or anything the standard Kill-A-Watt meter can measure, how can I determine the best capacitor to use for a heavily inductive load?

I have some Microwave Oven transformers I am using and I am hoping to make the power factor better. Once I have a good power factor, what changes would one expect? Would the MOTs have a higher power ability? Or will it just cause the power bill to increase because I am actually pulling a power that the company measures?

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Most residential customers do not have to pay for apparent power, but only active power.

Things which may change the active power drawn (off the top of my head):

  1. Real wires have resistance. Increasing the power factor will reduce the currents, meaning for the same system real impedance you will have lower losses. This may or may not be a significant saving (usually for residential applications it isn't).
  2. Inductors have series resistance, and capacitors have leakage (modeled as a high value parallel resistance) as well as ESR. Adding either of these could increase the active power drawn, but picking an appropriate inductor/capacitor can make this negligible.

Note: In the following calcuations I am assuming a linear AC system (all sources are sinusoidal, only have resistors, capacitors, and inductors). As soon as you introduce some non-linearity (say, add a bridge rectifier) most of this analysis no longer applies.

Looking at the specs, the Kill-a-Watt is capable of measuring the active power, apparent power, frequency, Vrms, and Irms (among other things, but these are the useful components). This is technically not enough information by itself to determine whether you need to add an inductor or capacitor to increase the power factor, but this can be mitigated if you know your system and/or can do a bit of experimentation.

Knowing the rms voltages/currents and the active power P (or we can use the apparent power |S|), we can find the phase angle (to within a sign): \begin{align} \phi &= \pm \cos^{-1}\left(\frac{P}{V_{rms} I_{rms}}\right) = \pm \cos^{-1}\left(\frac{P}{|S|}\right) \end{align}

Now calculating the equivalent system complex impedance Z=R+Xi: \begin{align} R &= \frac{V_{rms}}{I_{rms}} \cos(\phi)\\ X &= \frac{V_{rms}}{I_{rms}} \sin(\phi) \end{align}

We're only interested in the imaginary component X. We want to add another reactive element in such a way that the corrected system impedance has X=0. This is where not knowing the sign of \$\phi\$ is somewhat detrimental. If we know that all we have is a large inductive load, we can make the assumption that \$\phi\$ really is positive, and at the end after we've added the appropriate capacitor, check to see if that improved the power factor. Likewise, if we find that the system is capacitive, then \$\phi\$ must be negative.

To fix an inductive load you need to add in parallel a capacitor who's value is: \begin{align} C_{pfc} = \frac{1}{2\pi f X} \end{align}

Likewise, to fix a capacitive load you need to add in series an inductor who's value is: \begin{align} L_{pfc} = \frac{-X}{2\pi f} \end{align}

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    \$\begingroup\$ Do circuit breakers pop with only active power, not large amounts of apparent power? \$\endgroup\$ – skyler Feb 16 '15 at 1:47
  • \$\begingroup\$ circuit breakers are current limit triggered so the peak current level is the figure of merit (i.e. current over a some short-ish integration time). This means even large inrush current might trigger the circuit breaker. \$\endgroup\$ – helloworld922 Feb 16 '15 at 5:17

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