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I made a DIY solar system including a batteries and a pure sine-wave inverter, and I wanted to increase the power factor (which is typically 0.6 ~ 0.8) with a laptop power adapter load.

After calculating the right capacitor value, I tried to connect a 1.5µF in parallel to increase the power factor; unfortunately the power factor was actually decreased.

On the other hand, when I connect a fan containing an induction motor, the power factor goes above 0.9!

I don't get why this happens.

By the way, I don't have an oscilloscope.

My meter reads: enter image description here

The calculations I made: enter image description here

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  • \$\begingroup\$ "...I tried to connect a 1.5µF in parallel..." in parallel with what? The nature of the load determines the power factor. The load that you failed to describe. Is the key to the answer. It sounds like it must have been another capacitor or something that draws harmonic currents. \$\endgroup\$
    – user80875
    Commented May 23, 2020 at 19:47
  • \$\begingroup\$ There was only a laptop power adapter that was connected to the inverter. Just making sure I understand, could it be that the adapter is a capacitive load, and hence my capacitor didn't increased the power factor? @CharlesCowie \$\endgroup\$
    – WoeIsMe
    Commented May 23, 2020 at 20:14
  • \$\begingroup\$ You should revise the question to describe the load," a power adaptor for a laptop." I posted an answer that may explain. \$\endgroup\$
    – user80875
    Commented May 23, 2020 at 20:44

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Electronic loads, such as voltage converters and electronic equipment that contain rectifiers and voltage converters generally draw current from the AC source that is distorted, not a pure sine wave. That current consists of a sine wave mixed with sine waves that are integral multiples of the source sine wave frequency. Those are harmonic currents. They are similar to phase shifted currents in thatchy reduce the power factor. However they can not be reduced by adding a capacitor in parallel. Unless capacitors are added as part of a tuned circuit that filters the harmonic currents, they would only reduce the power factor.

Adding any other load that has a higher power factor would tend to increase the power factor.

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    \$\begingroup\$ This describes my problem very well. Thank you so much! \$\endgroup\$
    – WoeIsMe
    Commented May 23, 2020 at 21:36

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