Engineering Sinusoids- Fundamentals of Electric Circuits

[kostantina]

Find the phase angle between i_1= - 4 sin(377t+55) and i_2= 5 cos(377t-65)

The Attempt at a Solution

Ok, so I converted i_1= 4 cos(377t+145) and i_2=5cos(377t-65)

According to lectures in class the 'most' negative leads. So in this case i_2 leads by -65-145=210. However, if you look at the publishers solution it says that i_1 leads by 210. Why? Any feedback would be greatly appreciated. Thank you.

Here is a link to the publishers solution posted online on his student edition help/tools website:

Its problem 9.2

http://highered.mcgraw-hill.com/sites/dl/free/0073380571/938372/Chapt09PP_120121.pdf

 

[Simon Bridge]

Draw the two as phasors and you'll see.

 

[kostantina]

I did and according to my lectures i_2 should lead. +x axis is the +cos(wt) and the -y axis is the +sin(wt)
So i_1 = cos(377t+145) is in the II quadrant, and i_2=5cos(377-65 is in the IV quadrant). we start from the cos(wt) axis which is the + x-axis and go counterclockwise for polar angle so first we wee i_1 and then i_2. So i_1 is lagging hence i_2 leading.
Can anyone clarify this for me please?

 

[Simon Bridge]

i_2 leads by -65-145=210

... you missed out a minus sign.
if i₂ leads by a negative amount...

 

[kostantina]

Yes, we were told in class that result will be negative. And the most negative leads. So everything just contradicts here. This approach solved all the previous examples I tried. Now it all gets negated? Graphically how can you tell which one leads and which one follows? Its an angle 210 >180 so why not the other one lead?

 

[Simon Bridge]

If a phasor leads by angle A degrees then it also follows by angle 360-A degrees right?
But if it leads by -A then doesn't that mean it follows by A.

-65 degrees is the same as 295 degrees isn't it?
210 is the same as -150?

If the phasors were separated by 180 degrees, which would lead?
Which is the "most negative"?

i.e. is there a physical significance to "leading"?