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In phasor or vector diagram, a capacitor that is parallel to the supply can improve power factor. I know this is practically true but I don't understand the mathematical equation:

The total impedance (Z) of the following circuit has imaginary part i=root(-1). That means it has a reactants and it will consume reactive power.

If XL = Xc , the reacance should be infinity or has a very large value so it will consume large reactive power.

I feel like algebra does not support phasor diagram... Would you tell me what I'm missing?

Thank you, enter image description here

Edit:

Here is the opposite case:

If a very high reactance is good for power factor, Here is a circuit with a very low reactance. Does it also improve power factor? If yes, which one is better? enter image description here

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    \$\begingroup\$ Another way to look at this -- put just a plain resistor supplied by the voltage source, what would happen if the resistance goes to infinity (which would be a open circuit). That is, power goes down when impedance goes up. \$\endgroup\$
    – rioraxe
    Commented Apr 9, 2016 at 19:09
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    \$\begingroup\$ You need a load resistance in the circuit to appreciate the implications of power factor fully \$\endgroup\$
    – Chu
    Commented Apr 9, 2016 at 22:00

1 Answer 1

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If the impedance of the capacitor and that of the inductor were perfectly equal, the reactive power would be exactly zero.

The formula for the reactive power has two components, current squared and impedance. And while impedance goes to infinite, as you correctly pointed out, current goes twice as fast to zero, so reactive power goes to zero.

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  • \$\begingroup\$ Thank you very much for your answer, I really apprecieate that. Would you look at the edit of the question, Please? \$\endgroup\$
    – Michael George
    Commented Apr 9, 2016 at 20:46
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    \$\begingroup\$ If you are asking wether the series or parallel solution is preferred? The parallel. Why? Because a series impedance would lower the voltage presented to the load. And you don't want that, you want the load to receive the voltage it was rated for. And, BTW, you are welcome! \$\endgroup\$
    – Claudio Avi Chami
    Commented Apr 9, 2016 at 20:53
  • \$\begingroup\$ In an industrial environment connecting a capacitor in series with the load is totally impractical. Also, the load is largely resistive (the bit that does the work) with a series inductance that arises from, for example, motor windings. \$\endgroup\$
    – Chu
    Commented Apr 9, 2016 at 22:35

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