Mesh Current Method (complex numbers part)

[andymarra]

Basically, part of my uni course is electronics and some of the mesh current method questions involve complex numbers, mainly the use of j and other things. I would get the answer as the solution gets it down to 4-j7 and then it says this equals 8.0623<-60.2551° ......... ...< is supposed to be like/_ but couldn't find the proper symbol. I just can't understand the meaning of j and such things. How does it get from 4-j7 to this other answer? Just wondering if anyone could explain the conversions of these symbols to normal numbers and degrees as it is saying the final answer is? can provide full solution if needed.
Thanks, Andy

I can do the question, and understand most of it, its a solution from a past paper in university, but I am having trouble understanding only the complex numbers part, mainly the way in which it changes from j over to a number and degrees? i understand only that j is the sqrt of -1. Any help appreciated! Thanks again! Andy

 

[xcvxcvvc]

j is the imaginary component. If you've taken mechanical physics, and I assume you have, you can think of anything with a j as being the "y-component" and anything without it being the "x-component"

http://en.wikipedia.org/wiki/Imaginary_number

Look at that graph. Like I said earlier, the imaginary axis is the y-axis, and the real axis is the x-axis. You can represent any rectangular coordinate as a magnitude and an angle.

 

[andymarra]

okay I understand what you're saying, but how does this make 4-j7=8.0623<-60.2551? is there a further calculation that is needed to convert this? Thanks

 

[xcvxcvvc]

okay I understand what you're saying, but how does this make 4-j7=8.0623<-60.2551? is there a further calculation that is needed to convert this? Thanks

the magnitude is the squares of the real and the square of the imaginary added and then square rooted:
[tex]\sqrt{4^2 + (-7)^2} = 8.0623[/tex]

the angle is sort of the arctan of the imaginary over the real. Sometimes, you need to modify the angle slightly, because your calculator doesn't know the difference between
arctan[ (-y)/x] and arctan[ y/(-x)]
[tex]arctan(\frac{-7}{4}) = -60.2551^o[/tex]

so the final answer is
[tex]8.0623 \angle -60.2551^o=8.0623e^{-1.0517j}[/tex]

 

[berkeman]

okay I understand what you're saying, but how does this make 4-j7=8.0623<-60.2551? is there a further calculation that is needed to convert this? Thanks

You convert from rectangular to polar coordinates.

The horizontal (real) component is 4, so that is a vector that points to the right along the x axis. The vertical (imaginary) component is -7, so that is a vector that points down. Adding the two vectors gives you the complex result, which is a vector that starts at the origin, and points down to the right. Its length is the hypotenuse of the right triangle (base = 4, height = 7 down), and the angle it forms with the positive x-axis is -60 degrees. The convention for that angle in the rectangular-to-polar conversion is that the angle is positive starting at the x-axis and going in the counter-clockwise direction. So going in the opposite direction makes the angle negative by convention.

Does that make sense?


EDIT -- Ack, beat out again by xcvxcvvc!