Hi, I'm sort of new to electronics, but following along the AC ebook quite nicely so far. However, at this page: http://www.allaboutcircuits.com/vol_2/chpt_5/4.html ...can someone explain how one arrives at 131 degrees for IL and IC2 instead of -49 degrees ?? After all, atan(32.516/-28.49) equals -49 degrees. Do I miss something? Thanks, Hens
Hi Bill, thanks for the answer. I do however wonder why someone, with the intent of teaching a difficult subject, would opt for the 131 degrees if -49 degrees is more consistent with what's been taught. What would be the computation from the complex number to 131 degrees if not via the arctan route? And maybe I just fail to see something, but how are 131 degrees and -49 degrees both correct? Say you take a sine wave of amplitude 1.0. At 0 degrees Y equals 0. At 131 degrees, Y would be .755. At -49 degrees (i.e. 311 degres) Y would be -.755. That's MINUS .755. So obviously these are not the same. Anything that would explain the 131 degrees in Khupaldt's excellent AC book would be very welcome, especially because I can follow along anything else up to that point with regular atan() and to_polar()/to_rect() routines. Thanks so much, best regards, Hens
How much trigonometry do you know? Do you know about which trig function returns a positive (or negative value) in which quadrant? You clearly understand the idea of positive y over negative x however you are not correct in the belief that then makes arctan (or arcsin or arcos) angle negative. this may be your difficulty. A negative angle occurs because the rotation arm is rotating in the opposite direction, not because one of the components is negative. Please let us know where your trig is at as this is easy to show with a diagram, but I need to know what sort of diagram to draw.
If you are familiar with phasor representations of complex quantities this may assist in visualizing the result. Consider where the phasor representing the said current would lie on the complex plane. It would be interpreted as a phasor pointing from the origin of the plane into the 3rd quadrant. The usual convention is to consider angles in the 3rd quadrant as lying between +90° and +180°. One is also quite entitled to consider an angle in the third quadrant as a negative value in relation to some reference - such as zero degrees along the real positive axis. So rather than +131.225°, one could rightly consider the angle also to be -228.775°.
Thanks all for the help. My trig is okay. I just fail to see how Khupaldt arrives at 131 degrees for said complex value. The difference between t_n_k's example of +131 and -229 is obvious, because it's just plus or minus 2 pi. But -49 degrees and +131 degrees is another story. Let me rephrase my question. I've been able to follow the AC analyses up to http://www.allaboutcircuits.com/vol_2/chpt_5/4.html with ease. I'm going from polar to rect and vice versa as described. So I just want to know how Khupaldt (or you) would compute a polar representation from the complex number 32.516 - 28.49i. As the book shows, you get the magnitude from the root of the squares of the real and imaginary parts. No problem there; Khupaldt and I both arrive at 43. But then for the angle, all the computations in the AC book take the arctan of the imaginary part divided by the real part. And there, on that above page, is the first time I get -41 degrees and Khupaldt gets 131 degrees. What's the computation that spits out 131 degrees from a complex number 32.516 - 28.49i ? Thanks, Hens
If your calculator handles complex numbers it will most likely show the rectangular form entry for the current will convert to that polar form value shown on the tables. The proviso is whether your calculator actually has that facility.
In building his table Kuphaldt used some negative angles. Whilst it can be a shortcut, I would not recommend it as it is so easy to get one's signs mixed up. tnk has correctly observed that you seem to have done this with the current entries for L and C2. SideNote : The standard convention will always lead to a positive angle, whatever the sign of the tangent, sine or cosine. Each trig function will be negative in two of the four quadrants and positive in two. Have you heard of CAST? With the correct signs, the tangent is negative so a conventional direction would make the tangent second or fourth quadrant. you are led to the second quadrant by the fact that it is the real part that is negative for a positive angle.
Many thanks for the answers! What a good forum this is. The >POLAR function of my TI83+ indeed gives the right angle. Thanks again, Hens