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Suppose

The power network 1 has a resistive load and therefore the current supplied by the generator 1 to the load is purely resistive.

The power network 2 has a resistive + inductive load and therefore the current supplied by the generator 2 is resistive and reactive.

My question is what does it mean for the generator 2 to supply both resistive and reactive current to its load? Is it a disadvantage for that generator to do so? Are there more losses?

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Yes, there are more losses in the second case. This is because there is higher current flowing thru the system but with the same amount of real power transferred.

The extra current doesn't cause a higher load on the generator itself. This is because it pushes the generator forward during part of the cycle and backward during another part of the cycle. These balance out to no net load over whole cycles.

However, there are inevitable series resistances in the system. These are in the generator windings, transformers, transmission lines, and other places. More current thru these resistances means more real loss.

So reactive power flowing thru a system doesn't by itself cause more load. However, it causes more current, which causes more losses because the system components aren't perfect.

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  • \$\begingroup\$ So in the second case there would be both (I_resistive^2)* R losses and (I_reactive)^2*R losses? \$\endgroup\$
    – Taven
    Commented May 18, 2017 at 12:17
  • \$\begingroup\$ @Taven The losses are more \$ I_{Total}^2\times R\$ and \$ I_{Total}^2\times X\$. But overall, Yes. \$\endgroup\$ Commented May 18, 2017 at 13:29
  • \$\begingroup\$ @StainlessSteelRat Even though the reactive power does not contribute to the real power losses, in this case, the reactive currents and impedances contribute to losses? How is that possible? \$\endgroup\$
    – Taven
    Commented May 18, 2017 at 13:41
  • \$\begingroup\$ There are inductive and resistive losses to transmission lines, motor windings, transformer windings, etc. The power you get out is less than the power in. Hence losses. \$\endgroup\$ Commented May 18, 2017 at 14:23
  • \$\begingroup\$ @Taven Reactive current corresponds to reactive power, which is related to a lossless transfer of energy between the source and the load. However, it indeed causes losses in the wires, because they have resistance, and whenever current passes (be it active current or reactive current) it will cause heat as per Joule's law. \$\endgroup\$
    – alejnavab
    Commented Nov 4, 2020 at 15:38

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