rectangular to polar conversions

Thread Starter

lemon

Joined Jan 28, 2010
125
Hi:
I am reading/working through the following page on ac Series-parallel R, L, and C circuits. On this page they make a conversion from rectangular to polar forms but I am having difficulty understanding how they came to the eventual polar angles. I am ok with doing basic conversions. Could someone explain that for me please, or direct me to the page that does. I read it somewhere on these pages earlier but can't recall it and can't seem to find it now. Thank you.

http://www.allaboutcircuits.com/vol_2/chpt_5/4.html
 

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Thread Starter

lemon

Joined Jan 28, 2010
125
yeah I have studied that page before but am still confused. The thing that is confusing me is that in the conversion from rectangular form to polar form, we use arctan(b/a), which for the first example in the attached image, Zc1=0-j564.38, would be arctan(564.38/0), which is of course undefined. Yet their solution is 564.38<-90.
 

Papabravo

Joined Feb 24, 2006
12,501
The tan of 90 degrees is not undefined it is infinite; so the arctan of infinity is 90 degrees and the arctan of minus infinity is -90 degrees.

Undefined is something 0/0

Plot the points on the complex plane to see what is going on.
 

Thread Starter

lemon

Joined Jan 28, 2010
125
The tan of 90 degrees is not undefined it is infinite; so the arctan of infinity is 90 degrees and the arctan of minus infinity is -90 degrees.

Undefined is something 0/0

Understood.
But - 564.38 on the complex plane puts it in the lower left quadrant. While that is minus, it isn't -90. Isn't is -155.62?
 

Papabravo

Joined Feb 24, 2006
12,501
The tan of 90 degrees is not undefined it is infinite; so the arctan of infinity is 90 degrees and the arctan of minus infinity is -90 degrees.

Undefined is something 0/0

Understood.
But - 564.38 on the complex plane puts it in the lower left quadrant. While that is minus, it isn't -90. Isn't is -155.62?
No that is not correct. The complex number has two(2) components, just like a vector. 0 - j564.38 has a Real part of 0 so you stay on the vertical axis and you go 564.38 units in the negative j (aka y) direction and you arrive at the point (0, -564.38) in Cartesian coordinates which is equivalent to the the complex number 0 -j564.38 on the complex plane. Got it now?
 

Thread Starter

lemon

Joined Jan 28, 2010
125
yep! I think so. I don't go round I go up and down and left and right :)

But that is only when there is a zero involved though, right?
 
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