Recent content by KhalDirth

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...

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...

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...

I never liked it much myself. My professor swore by it. Said he looked at all the other books around. He was a very sharp guy, so I just took his word for it. Do you teach, as well?

Ahh, I see. Imaginary numbers violate the properties of the square root.

Also, sqrt(-1*-1) = sqrt(-1) *sqrt(-1) ? Is this some sort of mathematical weirdness? My calculator certainly doesn't like the idea if I subtract one from the other, I get -2 or 2 (depending on which way I subtract). I must also be stumped.

Electric Circuit Analysis, Third Edition David E. Johnson, Johnny R. Johnson, John L. Hilburn, Peter D. Scott ISBN 0-471-36571-8 I emailed Peter Scott (no reply, yet). I guess, reading the problem now, the authors never did say that they meant to maintain the real impedance. I just think...

Ah. This is interesting. You see, I found this problem in a text on power engineering (again, not HW, as I'm not in school). The authors did their calculations just as I did mine for admittance. You converted what you found should be your final impedance into an admittance, and then solved for...

I'm saying, via parallel impedances method Yt = Yc + YL. I'm asking this, because it seems to me that the solution to this problem is determined by the way you solve it. In other words, can you do the math (Yt = Yc + Yl, as seen in my attachment) in such a way that the result yields a real...

Okay. So how would you run the calculation for parallel admittances, such that the final result would give you an impedance with a real part of .4?

So, you're saying that if someone asked me to correct the power factor of a known load without changing the real impedance, I would not be able to do so by running the calculations via parallel admittance values (as I tried to do)?

Alright, I understand your math. I agree that you can place 0.349284 - j0.200955 in parallel with the load to get the desired impedance. What I don't understand is why you can also add an admittance of j0.801 in parallel and get the same circuit. You see in my right-hand calculations, the YC is...

Perhaps this touches on some concept of duality that I'm not getting?

Alright. So you say that the second way is to add an impedance of 0.349284 - j0.200955 in parallel to my load impedance. It is safe to say that this would be the same as adding an admittance of 1/(0.349284 - j0.200955) in parallel to my load admittance (.25 - j0.75)? The total admittance at...

I had my file size too large. I've attached it as a .jpg. However, the quality might be too poor. So I've included a URL as well. http://i993.photobucket.com/albums/af55/khaldirth/PFProblem2.jpg