Correcting power factor without affecting active power.......

[Voltageisntreal]

When correcting power factor without affecting active power, wouldn't the current and voltage of the system have to change due to

Active power =VrmsIrmscos(theta), where cos(theta) is the power factor?

 

[Rive]

Most cases the voltage can be considered as given, so what changes is the current.

 

[anorlunda]

Most cases the voltage can be considered as given, so what changes is the current.

@Rive is right considering voltage magnitude, but the phase angle might change.

@Voltageisntreal , welcome to PF.

You might understand better after reading these two PF Insights articles.
https://www.physicsforums.com/insights/ac-power-analysis-part-1-basics/
https://www.physicsforums.com/insights/ac-power-analysis-part-2-network-analysis/

 

[Voltageisntreal]

@Rive is right considering voltage magnitude, but the phase angle might change.

@Voltageisntreal , welcome to PF.

You might understand better after reading these two PF Insights articles.
https://www.physicsforums.com/insights/ac-power-analysis-part-1-basics/
https://www.physicsforums.com/insights/ac-power-analysis-part-2-network-analysis/

Hello to you too and thanks :D.

I'm not sure if I understand better now or not :p.

Steinmetz said, if I percieve correctly from the article, that z=A+jB, where jB is the phase shift. I've always just seen imaginary power as a phase shift between voltage and current in AC, where j is 90 degrees. Essentially, voltage is applied and then electromagnetic energy lags or leads it depending on whether the imaginary power is positive or negative which comes down to how capacitors and inductors act in AC.

My issue was that with a transformer in which I want to correct the power factor , but I want to keep active power the same, then you change reactive/imaginary power to get the result- yes. You do that by changing the phase angle indirectly by adding a capacitance or inductance.

I thought then, however, would that not result in a change in the current (as Rive said) in terms of it's magnitude to balance the formula: VIcos(theta)= Active Power?

I now feel, from the article, that because of this shift due to the change in reactive power-- V and I do change at instantaneous points because of the shift--- however, the rms remains the same and the average active power is therefore the same?

If so that does make more sense and thank you for you help :).

 

[Rive]

Let's take a simple example. If the PSU in your PC draws 460W, that would mean 2A at 230V.
Of course, PC power supplies has power factor correction these days. Without correction that PSU would draw 575VA (not W, but VA!), in case the power factor // cos(theta) is 0.8.
Since the voltage is the same 230V, it would mean 2.5A current.

So, we can say that in this case the PFC reduced the current to 2A from 2.5A, with the effective power remaining 460W. The very meaning of the PFC is that you modify the cos(theta) from 0.8 to 1 (so theta will change).

PS.: In PC PSUs the correction is done with electronics these days, not with transformers or simple additional inductive/capacitive elements.

 

[Voltageisntreal]

Let's take a simple example. If the PSU in your PC draws 460W, that would mean 2A at 230V.
Of course, PC power supplies has power factor correction these days. Without correction that PSU would draw 575VA (not W, but VA!), in case the power factor // cos(theta) is 0.8.
Since the voltage is the same 230V, it would mean 2.5A current.

So, we can say that in this case the PFC reduced the current to 2A from 2.5A, with the effective power remaining 460W. The very meaning of the PFC is that you modify the cos(theta) from 0.8 to 1 (so theta will change).

PS.: In PC PSUs the correction is done with electronics these days, not with transformers or simple additional inductive/capacitive elements.

I see, I had gotten confused ! Tyvm :3 :D, I understand now. It does reduce the current, which it would have to because rms is average anyway so the shift wouldn't have affected it.<3.