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My computer isn't using anywhere near 2.5 amps (It's using less than half that) however my kill a watt reports 2.5 amps with a power factor of 0.4. This computer needs to run on a 2 amp circuit, would there be any issues with this?

Or does this mean that because my powersupply is missing a capacitor to bring my power factor closer to one. It will only work on a circuit that can support 2.5 amps. Also in that case is there any kind of plug I could plug into a wall socket that my plug would go into to raise the power factor?

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    \$\begingroup\$ What makes you so sure your kill-a-watt is wrong? \$\endgroup\$
    – brhans
    Commented Jul 7, 2017 at 23:01
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    \$\begingroup\$ @Siedge A power factor less than 1 just means that the volts and the amps are out of phase with each other. 2.5A at a PF of 0.4 is still 2.5A. \$\endgroup\$
    – Simon B
    Commented Jul 7, 2017 at 23:11
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    \$\begingroup\$ It just means that VA and W are not the same. \$\endgroup\$ Commented Jul 7, 2017 at 23:15
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    \$\begingroup\$ @Siedge It may require more than just a capacitor to fix a poor PF on a switched-mode power supply (assuming that's what your computer has). There are off-the-shelf power factor correctors. They are often sold as money saving devices, but rarely make any difference to electricity bills. \$\endgroup\$
    – Simon B
    Commented Jul 7, 2017 at 23:28
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    \$\begingroup\$ A PF as low as 0.4 in an impressive feat. What's the brand and model of your power supply inside it? \$\endgroup\$
    – winny
    Commented Jul 8, 2017 at 6:08

2 Answers 2

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A power factor of 0.4 means that only 40% of the power being drawn is usable (real) power. That would support your observation of drawing more than twice the current expected.

What PC power supply are you currently using? Depending on what you're currently using, it might be worth upgrading to a higher quality unit with better PFC and efficiency.

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A capacitor can be used to correct the power-factor on an inductive load. This is typically done on industrial motor and fluorescent loads which are famously inductive. This works because both the inductive and capacitive currents are sinusoidal and 180° out of phase. If the \$ I_C = I_L \$ then they cancel out. See my answers to Why do we want higher power factor in ac motors? and How do capacitors improve power factor without destroying the capacitor? for more on this.

Your question reads as though you think that adding a capacitor to an unspecified DC power supply will improve the power factor. It may make it worse.

enter image description here

Figure 1. A simple full-wave rectified power supply. Note that current (red) is only drawn from the supply when the rectified AC voltage exceeds the voltage on the capacitor. Increasing the value of \$ C_O \$ only makes this worse. Source: PowerFactor.US.

Adding the capacitor on the AC side can't correct the narrowness of the current pulses. See the linked article for more on active power-factor correction for these situations.

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