# Power Factor Correction

by KhalDirth
Tags: correction, factor, power
P: 2,751
 Quote by KhalDirth This is true. I was stuck in the thought of maintaining impedance. I suppose we all come across mental roadblocks. However, this doesn't change the fact that skeptic's solution and the books solution both happen to be right, but contain different real impedances. Skeptic maintains a .4 ohms real impedance; the solution in the book gives 3.842 ohms as the answer. Both have the same power factor, but different real and imaginary components. Deserving of thought, yes?
What? Skeptic only came up with that alternate solution at your insistence of keeping the real impedance at 0.4 ohms. He did this by adding not only parallel capacitance but by using additional parallel resistance too. What you end up with is a circuit which draws no reactive power but wastes a whooping 90% of it's power in the "power correction" network. Not it's not a practical solution, but you asked for it.
P: 30
 Quote by uart What? Skeptic only came up with that alternate solution at your insistence of keeping the real impedance at 0.4 ohms. He did this by adding not only parallel capacitance but additionally using parallel resistance so as to artificially lower the real impedance and waste a lot of power. What you end up with is a circuit which draws no reactive power but wastes 90% of it's power in the "power correction" network. Not it's not a practical solution , but you asked for it.
Lol, you missed my point. The book insists on keeping the real admittance constant. I assumed that the same method would be true for evaluating the circuit as an impedance. I found that by trying to hold the real impedance constant, I was not getting the same answer (inverse, of course) as the book. This is what was stymying me.
P: 2,751
 Quote by KhalDirth Lol, you missed my point. The book insists on keeping the real admittance constant. I assumed that the same method would be true for evaluating the circuit as an impedance. I found that by trying to hold the real impedance constant, I was not getting the same answer (inverse, of course) as the book. This is what was stymying me.
Ok I see what you mean. The important point to consider is that the addition of a parallel reactance keeps the real part of admittance unchanged while the addition of a series reactance keeps the real part of the impedance unchanged. However only the parallel solution keeps the original load voltage and hence load power unchanged!

Hope that helps. :)
P: 30
 Quote by uart Ok I see what you mean. The important point to consider is that the addition of a parallel reactance keeps the real part of admittance unchanged while the addition of a series reactance keeps the real part of the impedance unchanged. However only the parallel solution keeps the original load voltage and hence load power unchanged! Hope that helps. :)
I see that now, and I totally understand where you're coming from about load power. I guess the part I'm still having trouble with is: why should it matter whether we view an object as an impedance or an admittance? We add a reactance in parallel with a load, but whether or not we think of the load as an admittance or impedance changes whether or not the real part is affected? The math works out, but the intuition doesn't.

Thank you for being so helpful.
 Sci Advisor P: 2,751 Ok, it just comes down to the different way impedances combine in series versus parallel. Series impedances add together in a direct algebraic manner, while parallel impedances dont. Parallel admittances add together in a direct algebraic manner, while series admittances dont. I don't really think there's anything more than that to it.

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