# About power factor in circuits

We have an AC circuit with rms voltage V, and resistor R. Assuming that the circuit does noy contain any reactive components, Z=R therefore the rms current flowing in the circuit can be obtained by I = V/Z = V/R. If we add reactive component(s) to the circle, we expect that Z will increase, therefore less current will flow in the circuit after the change.

I have a problem with this statement: "In an electric power system, a load with a low power factor draws more current than a load with a high power factor for the same amount of useful power transferred."

How is this possible when the current in the second circuit was lower? Also, how can a circuit 'draw' more current, if the value of the current is fixed and can be found after calculating impedence using ohm's law? Would appreciate to have a sample circuit to demonstrate this.

You are making false assumptions.

If you have a resistive circuit, and then add reactive components, it is now a higher impedance circuit to begin with and does not use same amount of useful power.

But yet you are comparing it with a circuit that has a reactive components and does use the same amount of useful power.

And you also assumed that adding reactive components increase the impedance, which means putting it im series with load.

If you add a capacitor in parallel to the resistance, it decreases the impedance. The resistor will now dissipate equal power than before, but charging and discharging the capacitor takes extra current.

• Can you elaborate further on the last paragraph? Why not add the capacitor in series. Also, if added in parallel why the resistor dissipates the same power as before? Commented May 22 at 13:36
• @SleepingOwl132 If you add capacitor in series with the original resistor, your current drops and original resistor gets less power, so you would have to do something to get same real power dissipation. Also if you add capacitor in parallel with the original resistor, the resistor dissipates same power because it has same AC sine wave voltage over it, the capacitor just draws extra current as reactive load. Commented May 22 at 16:13

How is this possible when the current in the second circuit was lower?

It isn't. You are assuming that the reactive impedance always is in series with the load resistance.

Also, there is nothing in the statement you quoted that says the low and high power factor loads have the same resistance in series with different amounts of reactance. It says that the real powers in the loads are equal. That is a very different thing.

The hypothetical is with the same amount of useful (that is, real) power transferred.

The "second" circuit has lower current, so it is obvious that the real power in the load is reduced.

We have an AC circuit with rms voltage V, and resistor R. Assuming that the circuit does not contain any reactive components, Z=R therefore the rms current flowing in the circuit can be obtained by I = V/Z = V/R. If we add reactive component(s) to the circle, we expect that Z will increase

No, it may change, depending on what reactive component you add where, it might even stay the same

therefore less current will flow in the circuit after the change.

No, it may change, depending on what reactive component you add where, it might even stay the same

I have a problem with this statement: "In an electric power system, a load with a low power factor draws more current than a load with a high power factor for the same amount of useful power transferred."

This is almost completely unrelated to your first paragraph. It's essentially a definition of power factor. One way to define it is to relate the VAr that the load draws to the useful power. If they are equal, the current and voltage are in phase, the power factor is 1, and the load appears pure resistive to the supply. If it draws more VAr from the supply than the useful load, the power factor is less than 1, and the voltage and current are not in phase.